Plot scaled and rotated bivariate distribution using matplotlib












2















I am trying to plot a bivariate gaussian distribution using matplotlib. I want to do this using the xy coordinates of two scatter points (Group A), (Group B).



I want to adjust the distribution by adjusting the COV matrix to account for each Groups velocity and their distance to an additional xy coordinate used as a reference point.



I've calculated the distance of each groups xy coordinate to that of the reference point. The distance is expressed as a radius, labelled [GrA_Rad],[GrB_Rad].



So the further they are away from the reference point the greater the radius. I've also calculated velocity labelled [GrA_Vel],[GrB_Vel]. The direction of each group is expressed as the orientation. This is labelled [GrA_Rotation],[GrB_Rotation]



Question on how I want the distribution to be adjusted for velocity and distance (radius):



I'm hoping to use SVD. Specifically, if I have the rotation angle of each scatter, this provides the direction. The velocity can be used to describe a scaling matrix [GrA_Scaling],[GrB_Scaling]. So this scaling matrix can be used to expand the radius in the x-direction and contract the radius in the y-direction. This expresses the COV matrix.



Finally, the distribution mean value is found by translating the groups location (x,y) by half the velocity.



Put simply: the radius is applied to each group's scatter point. The COV matrix is adjusted by the radius and velocity. So using the scaling matrix to expand the radius in x-direction and contract in y-direction. The direction is measured from the rotation angle. Then determine the distribution mean value by translating the groups location (x,y) by half the velocity.



Below is the df of these variables



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib.animation as animation

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data = d)


I've made an animated plot of each xy coordinate.



GrA_X = [10,12,17,16,16,14,12,8]
GrA_Y = [10,12,13,7,6,7,8,8]

GrB_X = [5,8,13,16,19,15,13,5]
GrB_Y = [6,15,12,10,8,9,10,8]

Item_X = [6,8,14,18,13,11,16,15]
Item_Y = [10,12,8,12,15,12,10,8]

scatter_GrA = ax.scatter(GrA_X, GrA_Y)
scatter_GrB = ax.scatter(GrB_X, GrB_Y)
scatter_Item = ax.scatter(Item_X, Item_Y)

def animate(i) :
scatter_GrA.set_offsets([[GrA_X[0+i], GrA_Y[0+i]]])
scatter_GrB.set_offsets([[GrB_X[0+i], GrB_Y[0+i]]])
scatter_Item.set_offsets([[Item_X[0+i], Item_Y[0+i]]])

ani = animation.FuncAnimation(fig, animate, np.arange(0,9),
interval = 1000, blit = False)









share|improve this question




















  • 2





    I've re-read the question several times and experimented with the code, but I'm afraid I still don't grasp the essence of the question. For example, what is Item? The animation shows that Item moves over time, along with GrA and GrB, but there's no discernible relationship between the three. Could you rephrase and/or simplify the question, and maybe provide specific output that would result from specific example input?

    – Peter Leimbigler
    Nov 23 '18 at 1:59











  • The put simply part is what I'm after. I've changed item to the reference point. I basically want to 1) apply the radius to each group. 2) Use the orientation to provide direction. 3) using the scaling factor to expand the radius in the x-direction and contract in the y-direction.

    – user9410826
    Nov 23 '18 at 11:14











  • Thanks for looking at this. The relationship between the 3 is distance between the two groups and that of the reference. That's what determines the radius. To provide a real world application, the two group scatters are people. The radius represents influence over a certain area. This influence should adjust for their velocity and distance to the reference point. Does this make sense?

    – user9410826
    Nov 23 '18 at 11:18






  • 1





    I also do not understand the question. It seems all the variables you talk about in the text do not even exist in the code?

    – ImportanceOfBeingErnest
    Nov 23 '18 at 18:30











  • @ImportanceOfBeingErnest. I haven't included how I worked them out in the code. I've just added them in the df. I didn't want to confuses readers with code that isn't pertinant to the question (something I've failed to do). But I wanted to provide context on how I calculated the variables. Should I strip the question right back?

    – user9410826
    Nov 24 '18 at 4:15
















2















I am trying to plot a bivariate gaussian distribution using matplotlib. I want to do this using the xy coordinates of two scatter points (Group A), (Group B).



I want to adjust the distribution by adjusting the COV matrix to account for each Groups velocity and their distance to an additional xy coordinate used as a reference point.



I've calculated the distance of each groups xy coordinate to that of the reference point. The distance is expressed as a radius, labelled [GrA_Rad],[GrB_Rad].



So the further they are away from the reference point the greater the radius. I've also calculated velocity labelled [GrA_Vel],[GrB_Vel]. The direction of each group is expressed as the orientation. This is labelled [GrA_Rotation],[GrB_Rotation]



Question on how I want the distribution to be adjusted for velocity and distance (radius):



I'm hoping to use SVD. Specifically, if I have the rotation angle of each scatter, this provides the direction. The velocity can be used to describe a scaling matrix [GrA_Scaling],[GrB_Scaling]. So this scaling matrix can be used to expand the radius in the x-direction and contract the radius in the y-direction. This expresses the COV matrix.



Finally, the distribution mean value is found by translating the groups location (x,y) by half the velocity.



Put simply: the radius is applied to each group's scatter point. The COV matrix is adjusted by the radius and velocity. So using the scaling matrix to expand the radius in x-direction and contract in y-direction. The direction is measured from the rotation angle. Then determine the distribution mean value by translating the groups location (x,y) by half the velocity.



Below is the df of these variables



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib.animation as animation

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data = d)


I've made an animated plot of each xy coordinate.



GrA_X = [10,12,17,16,16,14,12,8]
GrA_Y = [10,12,13,7,6,7,8,8]

GrB_X = [5,8,13,16,19,15,13,5]
GrB_Y = [6,15,12,10,8,9,10,8]

Item_X = [6,8,14,18,13,11,16,15]
Item_Y = [10,12,8,12,15,12,10,8]

scatter_GrA = ax.scatter(GrA_X, GrA_Y)
scatter_GrB = ax.scatter(GrB_X, GrB_Y)
scatter_Item = ax.scatter(Item_X, Item_Y)

def animate(i) :
scatter_GrA.set_offsets([[GrA_X[0+i], GrA_Y[0+i]]])
scatter_GrB.set_offsets([[GrB_X[0+i], GrB_Y[0+i]]])
scatter_Item.set_offsets([[Item_X[0+i], Item_Y[0+i]]])

ani = animation.FuncAnimation(fig, animate, np.arange(0,9),
interval = 1000, blit = False)









share|improve this question




















  • 2





    I've re-read the question several times and experimented with the code, but I'm afraid I still don't grasp the essence of the question. For example, what is Item? The animation shows that Item moves over time, along with GrA and GrB, but there's no discernible relationship between the three. Could you rephrase and/or simplify the question, and maybe provide specific output that would result from specific example input?

    – Peter Leimbigler
    Nov 23 '18 at 1:59











  • The put simply part is what I'm after. I've changed item to the reference point. I basically want to 1) apply the radius to each group. 2) Use the orientation to provide direction. 3) using the scaling factor to expand the radius in the x-direction and contract in the y-direction.

    – user9410826
    Nov 23 '18 at 11:14











  • Thanks for looking at this. The relationship between the 3 is distance between the two groups and that of the reference. That's what determines the radius. To provide a real world application, the two group scatters are people. The radius represents influence over a certain area. This influence should adjust for their velocity and distance to the reference point. Does this make sense?

    – user9410826
    Nov 23 '18 at 11:18






  • 1





    I also do not understand the question. It seems all the variables you talk about in the text do not even exist in the code?

    – ImportanceOfBeingErnest
    Nov 23 '18 at 18:30











  • @ImportanceOfBeingErnest. I haven't included how I worked them out in the code. I've just added them in the df. I didn't want to confuses readers with code that isn't pertinant to the question (something I've failed to do). But I wanted to provide context on how I calculated the variables. Should I strip the question right back?

    – user9410826
    Nov 24 '18 at 4:15














2












2








2


1






I am trying to plot a bivariate gaussian distribution using matplotlib. I want to do this using the xy coordinates of two scatter points (Group A), (Group B).



I want to adjust the distribution by adjusting the COV matrix to account for each Groups velocity and their distance to an additional xy coordinate used as a reference point.



I've calculated the distance of each groups xy coordinate to that of the reference point. The distance is expressed as a radius, labelled [GrA_Rad],[GrB_Rad].



So the further they are away from the reference point the greater the radius. I've also calculated velocity labelled [GrA_Vel],[GrB_Vel]. The direction of each group is expressed as the orientation. This is labelled [GrA_Rotation],[GrB_Rotation]



Question on how I want the distribution to be adjusted for velocity and distance (radius):



I'm hoping to use SVD. Specifically, if I have the rotation angle of each scatter, this provides the direction. The velocity can be used to describe a scaling matrix [GrA_Scaling],[GrB_Scaling]. So this scaling matrix can be used to expand the radius in the x-direction and contract the radius in the y-direction. This expresses the COV matrix.



Finally, the distribution mean value is found by translating the groups location (x,y) by half the velocity.



Put simply: the radius is applied to each group's scatter point. The COV matrix is adjusted by the radius and velocity. So using the scaling matrix to expand the radius in x-direction and contract in y-direction. The direction is measured from the rotation angle. Then determine the distribution mean value by translating the groups location (x,y) by half the velocity.



Below is the df of these variables



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib.animation as animation

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data = d)


I've made an animated plot of each xy coordinate.



GrA_X = [10,12,17,16,16,14,12,8]
GrA_Y = [10,12,13,7,6,7,8,8]

GrB_X = [5,8,13,16,19,15,13,5]
GrB_Y = [6,15,12,10,8,9,10,8]

Item_X = [6,8,14,18,13,11,16,15]
Item_Y = [10,12,8,12,15,12,10,8]

scatter_GrA = ax.scatter(GrA_X, GrA_Y)
scatter_GrB = ax.scatter(GrB_X, GrB_Y)
scatter_Item = ax.scatter(Item_X, Item_Y)

def animate(i) :
scatter_GrA.set_offsets([[GrA_X[0+i], GrA_Y[0+i]]])
scatter_GrB.set_offsets([[GrB_X[0+i], GrB_Y[0+i]]])
scatter_Item.set_offsets([[Item_X[0+i], Item_Y[0+i]]])

ani = animation.FuncAnimation(fig, animate, np.arange(0,9),
interval = 1000, blit = False)









share|improve this question
















I am trying to plot a bivariate gaussian distribution using matplotlib. I want to do this using the xy coordinates of two scatter points (Group A), (Group B).



I want to adjust the distribution by adjusting the COV matrix to account for each Groups velocity and their distance to an additional xy coordinate used as a reference point.



I've calculated the distance of each groups xy coordinate to that of the reference point. The distance is expressed as a radius, labelled [GrA_Rad],[GrB_Rad].



So the further they are away from the reference point the greater the radius. I've also calculated velocity labelled [GrA_Vel],[GrB_Vel]. The direction of each group is expressed as the orientation. This is labelled [GrA_Rotation],[GrB_Rotation]



Question on how I want the distribution to be adjusted for velocity and distance (radius):



I'm hoping to use SVD. Specifically, if I have the rotation angle of each scatter, this provides the direction. The velocity can be used to describe a scaling matrix [GrA_Scaling],[GrB_Scaling]. So this scaling matrix can be used to expand the radius in the x-direction and contract the radius in the y-direction. This expresses the COV matrix.



Finally, the distribution mean value is found by translating the groups location (x,y) by half the velocity.



Put simply: the radius is applied to each group's scatter point. The COV matrix is adjusted by the radius and velocity. So using the scaling matrix to expand the radius in x-direction and contract in y-direction. The direction is measured from the rotation angle. Then determine the distribution mean value by translating the groups location (x,y) by half the velocity.



Below is the df of these variables



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import matplotlib.animation as animation

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data = d)


I've made an animated plot of each xy coordinate.



GrA_X = [10,12,17,16,16,14,12,8]
GrA_Y = [10,12,13,7,6,7,8,8]

GrB_X = [5,8,13,16,19,15,13,5]
GrB_Y = [6,15,12,10,8,9,10,8]

Item_X = [6,8,14,18,13,11,16,15]
Item_Y = [10,12,8,12,15,12,10,8]

scatter_GrA = ax.scatter(GrA_X, GrA_Y)
scatter_GrB = ax.scatter(GrB_X, GrB_Y)
scatter_Item = ax.scatter(Item_X, Item_Y)

def animate(i) :
scatter_GrA.set_offsets([[GrA_X[0+i], GrA_Y[0+i]]])
scatter_GrB.set_offsets([[GrB_X[0+i], GrB_Y[0+i]]])
scatter_Item.set_offsets([[Item_X[0+i], Item_Y[0+i]]])

ani = animation.FuncAnimation(fig, animate, np.arange(0,9),
interval = 1000, blit = False)






python pandas matplotlib covariance gaussian






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Jan 17 at 1:43









tel

7,27621431




7,27621431










asked Nov 19 '18 at 4:19







user9410826















  • 2





    I've re-read the question several times and experimented with the code, but I'm afraid I still don't grasp the essence of the question. For example, what is Item? The animation shows that Item moves over time, along with GrA and GrB, but there's no discernible relationship between the three. Could you rephrase and/or simplify the question, and maybe provide specific output that would result from specific example input?

    – Peter Leimbigler
    Nov 23 '18 at 1:59











  • The put simply part is what I'm after. I've changed item to the reference point. I basically want to 1) apply the radius to each group. 2) Use the orientation to provide direction. 3) using the scaling factor to expand the radius in the x-direction and contract in the y-direction.

    – user9410826
    Nov 23 '18 at 11:14











  • Thanks for looking at this. The relationship between the 3 is distance between the two groups and that of the reference. That's what determines the radius. To provide a real world application, the two group scatters are people. The radius represents influence over a certain area. This influence should adjust for their velocity and distance to the reference point. Does this make sense?

    – user9410826
    Nov 23 '18 at 11:18






  • 1





    I also do not understand the question. It seems all the variables you talk about in the text do not even exist in the code?

    – ImportanceOfBeingErnest
    Nov 23 '18 at 18:30











  • @ImportanceOfBeingErnest. I haven't included how I worked them out in the code. I've just added them in the df. I didn't want to confuses readers with code that isn't pertinant to the question (something I've failed to do). But I wanted to provide context on how I calculated the variables. Should I strip the question right back?

    – user9410826
    Nov 24 '18 at 4:15














  • 2





    I've re-read the question several times and experimented with the code, but I'm afraid I still don't grasp the essence of the question. For example, what is Item? The animation shows that Item moves over time, along with GrA and GrB, but there's no discernible relationship between the three. Could you rephrase and/or simplify the question, and maybe provide specific output that would result from specific example input?

    – Peter Leimbigler
    Nov 23 '18 at 1:59











  • The put simply part is what I'm after. I've changed item to the reference point. I basically want to 1) apply the radius to each group. 2) Use the orientation to provide direction. 3) using the scaling factor to expand the radius in the x-direction and contract in the y-direction.

    – user9410826
    Nov 23 '18 at 11:14











  • Thanks for looking at this. The relationship between the 3 is distance between the two groups and that of the reference. That's what determines the radius. To provide a real world application, the two group scatters are people. The radius represents influence over a certain area. This influence should adjust for their velocity and distance to the reference point. Does this make sense?

    – user9410826
    Nov 23 '18 at 11:18






  • 1





    I also do not understand the question. It seems all the variables you talk about in the text do not even exist in the code?

    – ImportanceOfBeingErnest
    Nov 23 '18 at 18:30











  • @ImportanceOfBeingErnest. I haven't included how I worked them out in the code. I've just added them in the df. I didn't want to confuses readers with code that isn't pertinant to the question (something I've failed to do). But I wanted to provide context on how I calculated the variables. Should I strip the question right back?

    – user9410826
    Nov 24 '18 at 4:15








2




2





I've re-read the question several times and experimented with the code, but I'm afraid I still don't grasp the essence of the question. For example, what is Item? The animation shows that Item moves over time, along with GrA and GrB, but there's no discernible relationship between the three. Could you rephrase and/or simplify the question, and maybe provide specific output that would result from specific example input?

– Peter Leimbigler
Nov 23 '18 at 1:59





I've re-read the question several times and experimented with the code, but I'm afraid I still don't grasp the essence of the question. For example, what is Item? The animation shows that Item moves over time, along with GrA and GrB, but there's no discernible relationship between the three. Could you rephrase and/or simplify the question, and maybe provide specific output that would result from specific example input?

– Peter Leimbigler
Nov 23 '18 at 1:59













The put simply part is what I'm after. I've changed item to the reference point. I basically want to 1) apply the radius to each group. 2) Use the orientation to provide direction. 3) using the scaling factor to expand the radius in the x-direction and contract in the y-direction.

– user9410826
Nov 23 '18 at 11:14





The put simply part is what I'm after. I've changed item to the reference point. I basically want to 1) apply the radius to each group. 2) Use the orientation to provide direction. 3) using the scaling factor to expand the radius in the x-direction and contract in the y-direction.

– user9410826
Nov 23 '18 at 11:14













Thanks for looking at this. The relationship between the 3 is distance between the two groups and that of the reference. That's what determines the radius. To provide a real world application, the two group scatters are people. The radius represents influence over a certain area. This influence should adjust for their velocity and distance to the reference point. Does this make sense?

– user9410826
Nov 23 '18 at 11:18





Thanks for looking at this. The relationship between the 3 is distance between the two groups and that of the reference. That's what determines the radius. To provide a real world application, the two group scatters are people. The radius represents influence over a certain area. This influence should adjust for their velocity and distance to the reference point. Does this make sense?

– user9410826
Nov 23 '18 at 11:18




1




1





I also do not understand the question. It seems all the variables you talk about in the text do not even exist in the code?

– ImportanceOfBeingErnest
Nov 23 '18 at 18:30





I also do not understand the question. It seems all the variables you talk about in the text do not even exist in the code?

– ImportanceOfBeingErnest
Nov 23 '18 at 18:30













@ImportanceOfBeingErnest. I haven't included how I worked them out in the code. I've just added them in the df. I didn't want to confuses readers with code that isn't pertinant to the question (something I've failed to do). But I wanted to provide context on how I calculated the variables. Should I strip the question right back?

– user9410826
Nov 24 '18 at 4:15





@ImportanceOfBeingErnest. I haven't included how I worked them out in the code. I've just added them in the df. I didn't want to confuses readers with code that isn't pertinant to the question (something I've failed to do). But I wanted to provide context on how I calculated the variables. Should I strip the question right back?

– user9410826
Nov 24 '18 at 4:15












1 Answer
1






active

oldest

votes


















3





+100









Update



The question has been updated, and has gotten somewhat clearer. I've updated my code to match. Here's the latest output:



enter image description here



Aside from the styling, I think this matches what the OP described.



Here's the code that was used to produce the above plot:



dfake = ({    
'GrA_X' : [15,15],
'GrA_Y' : [15,15],
'Reference_X' : [15,3],
'Reference_Y' : [15,15],
'GrA_Rad' : [15,25],
'GrA_Vel' : [0,10],
'GrA_Scaling' : [0,0.5],
'GrA_Rotation' : [0,45]
})

dffake = pd.DataFrame(dfake)
fig,axs = plt.subplots(1, 2, figsize=(16,8))
fig.subplots_adjust(0,0,1,1)
plotone(dffake, 'A', 0, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[0])
plotone(dffake, 'A', 1, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[1])
plt.show()


and the complete implementation of the plotone function that I used is in the code block below. If you just want to know about the math used to generate and transform the 2D gaussian PDF, check out the mvpdf function (and the rot and getcov functions it depends on):



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

def mvpdf(x, y, xlim, ylim, radius=1, velocity=0, scale=0, theta=0):
"""Creates a grid of data that represents the PDF of a multivariate gaussian.

x, y: The center of the returned PDF
(xy)lim: The extent of the returned PDF
radius: The PDF will be dilated by this factor
scale: The PDF be stretched by a factor of (scale + 1) in the x direction, and squashed by a factor of 1/(scale + 1) in the y direction
theta: The PDF will be rotated by this many degrees

returns: X, Y, PDF. X and Y hold the coordinates of the PDF.
"""
# create the coordinate grids
X,Y = np.meshgrid(np.linspace(*xlim), np.linspace(*ylim))

# stack them into the format expected by the multivariate pdf
XY = np.stack([X, Y], 2)

# displace xy by half the velocity
x,y = rot(theta) @ (velocity/2, 0) + (x, y)

# get the covariance matrix with the appropriate transforms
cov = getcov(radius=radius, scale=scale, theta=theta)

# generate the data grid that represents the PDF
PDF = sts.multivariate_normal([x, y], cov).pdf(XY)

return X, Y, PDF

def plotmv(x, y, xlim=None, ylim=None, radius=1, velocity=0, scale=0, theta=0, xref=None, yref=None, fig=None, ax=None):
"""Plot an xy point with an appropriately tranformed 2D gaussian around it.
Also plots other related data like the reference point.
"""
if xlim is None: xlim = (x - 5, x + 5)
if ylim is None: ylim = (y - 5, y + 5)

if fig is None:
fig = plt.figure(figsize=(8,8))
ax = fig.gca()
elif ax is None:
ax = fig.gca()

# plot the xy point
ax.plot(x, y, '.', c='C0', ms=20)

if not (xref is None or yref is None):
# plot the reference point, if supplied
ax.plot(xref, yref, '.', c='w', ms=12)

# plot the arrow leading from the xy point
if velocity > 0:
ax.arrow(x, y, *rot(theta) @ (velocity, 0),
width=.4, length_includes_head=True, ec='C0', fc='C0')

# fetch the PDF of the 2D gaussian
X, Y, PDF = mvpdf(x, y, xlim=xlim, ylim=ylim, radius=radius, velocity=velocity, scale=scale, theta=theta)

# normalize PDF by shifting and scaling, so that the smallest value is 0 and the largest is 1
normPDF = PDF - PDF.min()
normPDF = normPDF/normPDF.max()

# plot and label the contour lines of the 2D gaussian
cs = ax.contour(X, Y, normPDF, levels=6, colors='w', alpha=.5)
ax.clabel(cs, fmt='%.3f', fontsize=12)

# plot the filled contours of the 2D gaussian. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normPDF, levels=50, cmap='viridis', vmin=-.9, vmax=1)

# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])

# add some labels
ax.grid()
ax.set_xlabel('X distance (M)')
ax.set_ylabel('Y distance (M)')

# ensure that x vs y scaling doesn't disrupt the transforms applied to the 2D gaussian
ax.set_aspect('equal', 'box')

return fig, ax

def fetchone(df, l, i, **kwargs):
"""Fetch all the needed data for one xy point
"""
keytups = (
('x', 'Gr%s_X'%l),
('y', 'Gr%s_Y'%l),
('radius', 'Gr%s_Rad'%l),
('velocity', 'Gr%s_Vel'%l),
('scale', 'Gr%s_Scaling'%l),
('theta', 'Gr%s_Rotation'%l),
('xref', 'Reference_X'),
('yref', 'Reference_Y')
)

ret = {k:df.loc[i, l] for k,l in keytups}
# add in any overrides
ret.update(kwargs)

return ret

def plotone(df, l, i, xlim=None, ylim=None, fig=None, ax=None, **kwargs):
"""Plot exactly one point from the dataset
"""
# look up all the data to plot one datapoint
xydata = fetchone(df, l, i, **kwargs)

# do the plot
return plotmv(xlim=xlim, ylim=ylim, fig=fig, ax=ax, **xydata)


Old answer -2



I've adjusted my answer to match the example the OP posted:



enter image description here



Here's the code that produced the above image:



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

def datalimits(*data, pad=.15):
dmin,dmax = min(d.min() for d in data), max(d.max() for d in data)
spad = pad*(dmax - dmin)
return dmin - spad, dmax + spad

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data=d)

limitpad = .5
clevels = 5
cflevels = 50

xmin,xmax = datalimits(df['GrA_X'], df['GrB_X'], pad=limitpad)
ymin,ymax = datalimits(df['GrA_Y'], df['GrB_Y'], pad=limitpad)

X,Y = np.meshgrid(np.linspace(xmin, xmax), np.linspace(ymin, ymax))

fig = plt.figure(figsize=(10,6))
ax = plt.gca()

Zs =
for l,color in zip('AB', ('red', 'yellow')):
# plot all of the points from a single group
ax.plot(df['Gr%s_X'%l], df['Gr%s_Y'%l], '.', c=color, ms=15, label=l)

Zrows =
for _,row in df.iterrows():
x,y = row['Gr%s_X'%l], row['Gr%s_Y'%l]

cov = getcov(radius=row['Gr%s_Rad'%l], scale=row['Gr%s_Scaling'%l], theta=row['Gr%s_Rotation'%l])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))
Zrows.append(Z)

Zs.append(np.sum(Zrows, axis=0))

# plot the reference points

# create Z from the difference of the sums of the 2D Gaussians from group A and group B
Z = Zs[0] - Zs[1]

# normalize Z by shifting and scaling, so that the smallest value is 0 and the largest is 1
normZ = Z - Z.min()
normZ = normZ/normZ.max()

# plot and label the contour lines
cs = ax.contour(X, Y, normZ, levels=clevels, colors='w', alpha=.5)
ax.clabel(cs, fmt='%2.1f', colors='w')#, fontsize=14)

# plot the filled contours. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normZ, levels=cflevels, cmap='viridis', vmin=0, vmax=1)
# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])


ax.set_aspect('equal', 'box')


Old answer -1



It's a little hard to tell exactly what you're after. It is possible to scale and rotate a multivariate gaussian distribution via its covariance matrix. Here's an example of how to do so based on your data:



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(scale, theta):
cov = np.array([
[1*(scale + 1), 0],
[0, 1/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data=d)
xmin,xmax = min(df['GrA_X'].min(), df['GrB_X'].min()), max(df['GrA_X'].max(), df['GrB_X'].max())
ymin,ymax = min(df['GrA_Y'].min(), df['GrB_Y'].min()), max(df['GrA_Y'].max(), df['GrB_Y'].max())

X,Y = np.meshgrid(
np.linspace(xmin - (xmax - xmin)*.1, xmax + (xmax - xmin)*.1),
np.linspace(ymin - (ymax - ymin)*.1, ymax + (ymax - ymin)*.1)
)

fig,axs = plt.subplots(df.shape[0], sharex=True, figsize=(4, 4*df.shape[0]))
fig.subplots_adjust(0,0,1,1,0,-.82)

for (_,row),ax in zip(df.iterrows(), axs):
for c in 'AB':
x,y = row['Gr%s_X'%c], row['Gr%s_Y'%c]

cov = getcov(scale=row['Gr%s_Scaling'%c], theta=row['Gr%s_Rotation'%c])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))

ax.contour(X, Y, Z)

ax.plot(row['Gr%s_X'%c], row['Gr%s_Y'%c], 'x')
ax.set_aspect('equal', 'box')


This outputs:



enter image description here






share|improve this answer


























  • @JPeter I've updated the answer again to reflect the updates in the question's description/example. Let me know if this matches what you're aiming for.

    – tel
    Nov 25 '18 at 9:46











  • This is truly brilliant @tel. Sorry for swapping the figure in the original question. I tried to simplify the question to get across the aim. In regards to your second output, do the contours provide a probability of influence ranging from 0-1. Could you provide a short description on how you calculated this. I'll also be applying this to numerous data points in a df. Do you foresee any issues with this? Once again thank you. This is great!

    – user9410826
    Nov 25 '18 at 10:55











  • Sorry @tel. I'm getting an error on both edit 2 and edit 3: it's on cfs = ax.contour(X,Y, normPDF, levels = 50, cmap = 'viridis', vmin = -.9, vmax = 1). The error is if self.filled and len(self.levels) < 2: TypeError: len() of unsized object

    – user9410826
    Nov 25 '18 at 23:35













  • It looks like you're using an old version of Matplotlib (the line of code raising the error looks different in the latest version. The first thing to try then is upgrading your Matplotlib package (they fix tons of bugs all the time).

    – tel
    Nov 26 '18 at 0:03











  • Thanks @tel. I just installed a new env and got it going. Just a quick one. Is the underlying math the same for for edit 2 and 3. You're plotting a distribution of probability. I'm finding it hard to conceptualise how the density of scatter points affects the normalised output probability

    – user9410826
    Nov 26 '18 at 5:40











Your Answer






StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53368263%2fplot-scaled-and-rotated-bivariate-distribution-using-matplotlib%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown
























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3





+100









Update



The question has been updated, and has gotten somewhat clearer. I've updated my code to match. Here's the latest output:



enter image description here



Aside from the styling, I think this matches what the OP described.



Here's the code that was used to produce the above plot:



dfake = ({    
'GrA_X' : [15,15],
'GrA_Y' : [15,15],
'Reference_X' : [15,3],
'Reference_Y' : [15,15],
'GrA_Rad' : [15,25],
'GrA_Vel' : [0,10],
'GrA_Scaling' : [0,0.5],
'GrA_Rotation' : [0,45]
})

dffake = pd.DataFrame(dfake)
fig,axs = plt.subplots(1, 2, figsize=(16,8))
fig.subplots_adjust(0,0,1,1)
plotone(dffake, 'A', 0, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[0])
plotone(dffake, 'A', 1, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[1])
plt.show()


and the complete implementation of the plotone function that I used is in the code block below. If you just want to know about the math used to generate and transform the 2D gaussian PDF, check out the mvpdf function (and the rot and getcov functions it depends on):



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

def mvpdf(x, y, xlim, ylim, radius=1, velocity=0, scale=0, theta=0):
"""Creates a grid of data that represents the PDF of a multivariate gaussian.

x, y: The center of the returned PDF
(xy)lim: The extent of the returned PDF
radius: The PDF will be dilated by this factor
scale: The PDF be stretched by a factor of (scale + 1) in the x direction, and squashed by a factor of 1/(scale + 1) in the y direction
theta: The PDF will be rotated by this many degrees

returns: X, Y, PDF. X and Y hold the coordinates of the PDF.
"""
# create the coordinate grids
X,Y = np.meshgrid(np.linspace(*xlim), np.linspace(*ylim))

# stack them into the format expected by the multivariate pdf
XY = np.stack([X, Y], 2)

# displace xy by half the velocity
x,y = rot(theta) @ (velocity/2, 0) + (x, y)

# get the covariance matrix with the appropriate transforms
cov = getcov(radius=radius, scale=scale, theta=theta)

# generate the data grid that represents the PDF
PDF = sts.multivariate_normal([x, y], cov).pdf(XY)

return X, Y, PDF

def plotmv(x, y, xlim=None, ylim=None, radius=1, velocity=0, scale=0, theta=0, xref=None, yref=None, fig=None, ax=None):
"""Plot an xy point with an appropriately tranformed 2D gaussian around it.
Also plots other related data like the reference point.
"""
if xlim is None: xlim = (x - 5, x + 5)
if ylim is None: ylim = (y - 5, y + 5)

if fig is None:
fig = plt.figure(figsize=(8,8))
ax = fig.gca()
elif ax is None:
ax = fig.gca()

# plot the xy point
ax.plot(x, y, '.', c='C0', ms=20)

if not (xref is None or yref is None):
# plot the reference point, if supplied
ax.plot(xref, yref, '.', c='w', ms=12)

# plot the arrow leading from the xy point
if velocity > 0:
ax.arrow(x, y, *rot(theta) @ (velocity, 0),
width=.4, length_includes_head=True, ec='C0', fc='C0')

# fetch the PDF of the 2D gaussian
X, Y, PDF = mvpdf(x, y, xlim=xlim, ylim=ylim, radius=radius, velocity=velocity, scale=scale, theta=theta)

# normalize PDF by shifting and scaling, so that the smallest value is 0 and the largest is 1
normPDF = PDF - PDF.min()
normPDF = normPDF/normPDF.max()

# plot and label the contour lines of the 2D gaussian
cs = ax.contour(X, Y, normPDF, levels=6, colors='w', alpha=.5)
ax.clabel(cs, fmt='%.3f', fontsize=12)

# plot the filled contours of the 2D gaussian. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normPDF, levels=50, cmap='viridis', vmin=-.9, vmax=1)

# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])

# add some labels
ax.grid()
ax.set_xlabel('X distance (M)')
ax.set_ylabel('Y distance (M)')

# ensure that x vs y scaling doesn't disrupt the transforms applied to the 2D gaussian
ax.set_aspect('equal', 'box')

return fig, ax

def fetchone(df, l, i, **kwargs):
"""Fetch all the needed data for one xy point
"""
keytups = (
('x', 'Gr%s_X'%l),
('y', 'Gr%s_Y'%l),
('radius', 'Gr%s_Rad'%l),
('velocity', 'Gr%s_Vel'%l),
('scale', 'Gr%s_Scaling'%l),
('theta', 'Gr%s_Rotation'%l),
('xref', 'Reference_X'),
('yref', 'Reference_Y')
)

ret = {k:df.loc[i, l] for k,l in keytups}
# add in any overrides
ret.update(kwargs)

return ret

def plotone(df, l, i, xlim=None, ylim=None, fig=None, ax=None, **kwargs):
"""Plot exactly one point from the dataset
"""
# look up all the data to plot one datapoint
xydata = fetchone(df, l, i, **kwargs)

# do the plot
return plotmv(xlim=xlim, ylim=ylim, fig=fig, ax=ax, **xydata)


Old answer -2



I've adjusted my answer to match the example the OP posted:



enter image description here



Here's the code that produced the above image:



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

def datalimits(*data, pad=.15):
dmin,dmax = min(d.min() for d in data), max(d.max() for d in data)
spad = pad*(dmax - dmin)
return dmin - spad, dmax + spad

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data=d)

limitpad = .5
clevels = 5
cflevels = 50

xmin,xmax = datalimits(df['GrA_X'], df['GrB_X'], pad=limitpad)
ymin,ymax = datalimits(df['GrA_Y'], df['GrB_Y'], pad=limitpad)

X,Y = np.meshgrid(np.linspace(xmin, xmax), np.linspace(ymin, ymax))

fig = plt.figure(figsize=(10,6))
ax = plt.gca()

Zs =
for l,color in zip('AB', ('red', 'yellow')):
# plot all of the points from a single group
ax.plot(df['Gr%s_X'%l], df['Gr%s_Y'%l], '.', c=color, ms=15, label=l)

Zrows =
for _,row in df.iterrows():
x,y = row['Gr%s_X'%l], row['Gr%s_Y'%l]

cov = getcov(radius=row['Gr%s_Rad'%l], scale=row['Gr%s_Scaling'%l], theta=row['Gr%s_Rotation'%l])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))
Zrows.append(Z)

Zs.append(np.sum(Zrows, axis=0))

# plot the reference points

# create Z from the difference of the sums of the 2D Gaussians from group A and group B
Z = Zs[0] - Zs[1]

# normalize Z by shifting and scaling, so that the smallest value is 0 and the largest is 1
normZ = Z - Z.min()
normZ = normZ/normZ.max()

# plot and label the contour lines
cs = ax.contour(X, Y, normZ, levels=clevels, colors='w', alpha=.5)
ax.clabel(cs, fmt='%2.1f', colors='w')#, fontsize=14)

# plot the filled contours. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normZ, levels=cflevels, cmap='viridis', vmin=0, vmax=1)
# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])


ax.set_aspect('equal', 'box')


Old answer -1



It's a little hard to tell exactly what you're after. It is possible to scale and rotate a multivariate gaussian distribution via its covariance matrix. Here's an example of how to do so based on your data:



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(scale, theta):
cov = np.array([
[1*(scale + 1), 0],
[0, 1/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data=d)
xmin,xmax = min(df['GrA_X'].min(), df['GrB_X'].min()), max(df['GrA_X'].max(), df['GrB_X'].max())
ymin,ymax = min(df['GrA_Y'].min(), df['GrB_Y'].min()), max(df['GrA_Y'].max(), df['GrB_Y'].max())

X,Y = np.meshgrid(
np.linspace(xmin - (xmax - xmin)*.1, xmax + (xmax - xmin)*.1),
np.linspace(ymin - (ymax - ymin)*.1, ymax + (ymax - ymin)*.1)
)

fig,axs = plt.subplots(df.shape[0], sharex=True, figsize=(4, 4*df.shape[0]))
fig.subplots_adjust(0,0,1,1,0,-.82)

for (_,row),ax in zip(df.iterrows(), axs):
for c in 'AB':
x,y = row['Gr%s_X'%c], row['Gr%s_Y'%c]

cov = getcov(scale=row['Gr%s_Scaling'%c], theta=row['Gr%s_Rotation'%c])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))

ax.contour(X, Y, Z)

ax.plot(row['Gr%s_X'%c], row['Gr%s_Y'%c], 'x')
ax.set_aspect('equal', 'box')


This outputs:



enter image description here






share|improve this answer


























  • @JPeter I've updated the answer again to reflect the updates in the question's description/example. Let me know if this matches what you're aiming for.

    – tel
    Nov 25 '18 at 9:46











  • This is truly brilliant @tel. Sorry for swapping the figure in the original question. I tried to simplify the question to get across the aim. In regards to your second output, do the contours provide a probability of influence ranging from 0-1. Could you provide a short description on how you calculated this. I'll also be applying this to numerous data points in a df. Do you foresee any issues with this? Once again thank you. This is great!

    – user9410826
    Nov 25 '18 at 10:55











  • Sorry @tel. I'm getting an error on both edit 2 and edit 3: it's on cfs = ax.contour(X,Y, normPDF, levels = 50, cmap = 'viridis', vmin = -.9, vmax = 1). The error is if self.filled and len(self.levels) < 2: TypeError: len() of unsized object

    – user9410826
    Nov 25 '18 at 23:35













  • It looks like you're using an old version of Matplotlib (the line of code raising the error looks different in the latest version. The first thing to try then is upgrading your Matplotlib package (they fix tons of bugs all the time).

    – tel
    Nov 26 '18 at 0:03











  • Thanks @tel. I just installed a new env and got it going. Just a quick one. Is the underlying math the same for for edit 2 and 3. You're plotting a distribution of probability. I'm finding it hard to conceptualise how the density of scatter points affects the normalised output probability

    – user9410826
    Nov 26 '18 at 5:40
















3





+100









Update



The question has been updated, and has gotten somewhat clearer. I've updated my code to match. Here's the latest output:



enter image description here



Aside from the styling, I think this matches what the OP described.



Here's the code that was used to produce the above plot:



dfake = ({    
'GrA_X' : [15,15],
'GrA_Y' : [15,15],
'Reference_X' : [15,3],
'Reference_Y' : [15,15],
'GrA_Rad' : [15,25],
'GrA_Vel' : [0,10],
'GrA_Scaling' : [0,0.5],
'GrA_Rotation' : [0,45]
})

dffake = pd.DataFrame(dfake)
fig,axs = plt.subplots(1, 2, figsize=(16,8))
fig.subplots_adjust(0,0,1,1)
plotone(dffake, 'A', 0, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[0])
plotone(dffake, 'A', 1, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[1])
plt.show()


and the complete implementation of the plotone function that I used is in the code block below. If you just want to know about the math used to generate and transform the 2D gaussian PDF, check out the mvpdf function (and the rot and getcov functions it depends on):



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

def mvpdf(x, y, xlim, ylim, radius=1, velocity=0, scale=0, theta=0):
"""Creates a grid of data that represents the PDF of a multivariate gaussian.

x, y: The center of the returned PDF
(xy)lim: The extent of the returned PDF
radius: The PDF will be dilated by this factor
scale: The PDF be stretched by a factor of (scale + 1) in the x direction, and squashed by a factor of 1/(scale + 1) in the y direction
theta: The PDF will be rotated by this many degrees

returns: X, Y, PDF. X and Y hold the coordinates of the PDF.
"""
# create the coordinate grids
X,Y = np.meshgrid(np.linspace(*xlim), np.linspace(*ylim))

# stack them into the format expected by the multivariate pdf
XY = np.stack([X, Y], 2)

# displace xy by half the velocity
x,y = rot(theta) @ (velocity/2, 0) + (x, y)

# get the covariance matrix with the appropriate transforms
cov = getcov(radius=radius, scale=scale, theta=theta)

# generate the data grid that represents the PDF
PDF = sts.multivariate_normal([x, y], cov).pdf(XY)

return X, Y, PDF

def plotmv(x, y, xlim=None, ylim=None, radius=1, velocity=0, scale=0, theta=0, xref=None, yref=None, fig=None, ax=None):
"""Plot an xy point with an appropriately tranformed 2D gaussian around it.
Also plots other related data like the reference point.
"""
if xlim is None: xlim = (x - 5, x + 5)
if ylim is None: ylim = (y - 5, y + 5)

if fig is None:
fig = plt.figure(figsize=(8,8))
ax = fig.gca()
elif ax is None:
ax = fig.gca()

# plot the xy point
ax.plot(x, y, '.', c='C0', ms=20)

if not (xref is None or yref is None):
# plot the reference point, if supplied
ax.plot(xref, yref, '.', c='w', ms=12)

# plot the arrow leading from the xy point
if velocity > 0:
ax.arrow(x, y, *rot(theta) @ (velocity, 0),
width=.4, length_includes_head=True, ec='C0', fc='C0')

# fetch the PDF of the 2D gaussian
X, Y, PDF = mvpdf(x, y, xlim=xlim, ylim=ylim, radius=radius, velocity=velocity, scale=scale, theta=theta)

# normalize PDF by shifting and scaling, so that the smallest value is 0 and the largest is 1
normPDF = PDF - PDF.min()
normPDF = normPDF/normPDF.max()

# plot and label the contour lines of the 2D gaussian
cs = ax.contour(X, Y, normPDF, levels=6, colors='w', alpha=.5)
ax.clabel(cs, fmt='%.3f', fontsize=12)

# plot the filled contours of the 2D gaussian. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normPDF, levels=50, cmap='viridis', vmin=-.9, vmax=1)

# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])

# add some labels
ax.grid()
ax.set_xlabel('X distance (M)')
ax.set_ylabel('Y distance (M)')

# ensure that x vs y scaling doesn't disrupt the transforms applied to the 2D gaussian
ax.set_aspect('equal', 'box')

return fig, ax

def fetchone(df, l, i, **kwargs):
"""Fetch all the needed data for one xy point
"""
keytups = (
('x', 'Gr%s_X'%l),
('y', 'Gr%s_Y'%l),
('radius', 'Gr%s_Rad'%l),
('velocity', 'Gr%s_Vel'%l),
('scale', 'Gr%s_Scaling'%l),
('theta', 'Gr%s_Rotation'%l),
('xref', 'Reference_X'),
('yref', 'Reference_Y')
)

ret = {k:df.loc[i, l] for k,l in keytups}
# add in any overrides
ret.update(kwargs)

return ret

def plotone(df, l, i, xlim=None, ylim=None, fig=None, ax=None, **kwargs):
"""Plot exactly one point from the dataset
"""
# look up all the data to plot one datapoint
xydata = fetchone(df, l, i, **kwargs)

# do the plot
return plotmv(xlim=xlim, ylim=ylim, fig=fig, ax=ax, **xydata)


Old answer -2



I've adjusted my answer to match the example the OP posted:



enter image description here



Here's the code that produced the above image:



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

def datalimits(*data, pad=.15):
dmin,dmax = min(d.min() for d in data), max(d.max() for d in data)
spad = pad*(dmax - dmin)
return dmin - spad, dmax + spad

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data=d)

limitpad = .5
clevels = 5
cflevels = 50

xmin,xmax = datalimits(df['GrA_X'], df['GrB_X'], pad=limitpad)
ymin,ymax = datalimits(df['GrA_Y'], df['GrB_Y'], pad=limitpad)

X,Y = np.meshgrid(np.linspace(xmin, xmax), np.linspace(ymin, ymax))

fig = plt.figure(figsize=(10,6))
ax = plt.gca()

Zs =
for l,color in zip('AB', ('red', 'yellow')):
# plot all of the points from a single group
ax.plot(df['Gr%s_X'%l], df['Gr%s_Y'%l], '.', c=color, ms=15, label=l)

Zrows =
for _,row in df.iterrows():
x,y = row['Gr%s_X'%l], row['Gr%s_Y'%l]

cov = getcov(radius=row['Gr%s_Rad'%l], scale=row['Gr%s_Scaling'%l], theta=row['Gr%s_Rotation'%l])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))
Zrows.append(Z)

Zs.append(np.sum(Zrows, axis=0))

# plot the reference points

# create Z from the difference of the sums of the 2D Gaussians from group A and group B
Z = Zs[0] - Zs[1]

# normalize Z by shifting and scaling, so that the smallest value is 0 and the largest is 1
normZ = Z - Z.min()
normZ = normZ/normZ.max()

# plot and label the contour lines
cs = ax.contour(X, Y, normZ, levels=clevels, colors='w', alpha=.5)
ax.clabel(cs, fmt='%2.1f', colors='w')#, fontsize=14)

# plot the filled contours. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normZ, levels=cflevels, cmap='viridis', vmin=0, vmax=1)
# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])


ax.set_aspect('equal', 'box')


Old answer -1



It's a little hard to tell exactly what you're after. It is possible to scale and rotate a multivariate gaussian distribution via its covariance matrix. Here's an example of how to do so based on your data:



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(scale, theta):
cov = np.array([
[1*(scale + 1), 0],
[0, 1/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data=d)
xmin,xmax = min(df['GrA_X'].min(), df['GrB_X'].min()), max(df['GrA_X'].max(), df['GrB_X'].max())
ymin,ymax = min(df['GrA_Y'].min(), df['GrB_Y'].min()), max(df['GrA_Y'].max(), df['GrB_Y'].max())

X,Y = np.meshgrid(
np.linspace(xmin - (xmax - xmin)*.1, xmax + (xmax - xmin)*.1),
np.linspace(ymin - (ymax - ymin)*.1, ymax + (ymax - ymin)*.1)
)

fig,axs = plt.subplots(df.shape[0], sharex=True, figsize=(4, 4*df.shape[0]))
fig.subplots_adjust(0,0,1,1,0,-.82)

for (_,row),ax in zip(df.iterrows(), axs):
for c in 'AB':
x,y = row['Gr%s_X'%c], row['Gr%s_Y'%c]

cov = getcov(scale=row['Gr%s_Scaling'%c], theta=row['Gr%s_Rotation'%c])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))

ax.contour(X, Y, Z)

ax.plot(row['Gr%s_X'%c], row['Gr%s_Y'%c], 'x')
ax.set_aspect('equal', 'box')


This outputs:



enter image description here






share|improve this answer


























  • @JPeter I've updated the answer again to reflect the updates in the question's description/example. Let me know if this matches what you're aiming for.

    – tel
    Nov 25 '18 at 9:46











  • This is truly brilliant @tel. Sorry for swapping the figure in the original question. I tried to simplify the question to get across the aim. In regards to your second output, do the contours provide a probability of influence ranging from 0-1. Could you provide a short description on how you calculated this. I'll also be applying this to numerous data points in a df. Do you foresee any issues with this? Once again thank you. This is great!

    – user9410826
    Nov 25 '18 at 10:55











  • Sorry @tel. I'm getting an error on both edit 2 and edit 3: it's on cfs = ax.contour(X,Y, normPDF, levels = 50, cmap = 'viridis', vmin = -.9, vmax = 1). The error is if self.filled and len(self.levels) < 2: TypeError: len() of unsized object

    – user9410826
    Nov 25 '18 at 23:35













  • It looks like you're using an old version of Matplotlib (the line of code raising the error looks different in the latest version. The first thing to try then is upgrading your Matplotlib package (they fix tons of bugs all the time).

    – tel
    Nov 26 '18 at 0:03











  • Thanks @tel. I just installed a new env and got it going. Just a quick one. Is the underlying math the same for for edit 2 and 3. You're plotting a distribution of probability. I'm finding it hard to conceptualise how the density of scatter points affects the normalised output probability

    – user9410826
    Nov 26 '18 at 5:40














3





+100







3





+100



3




+100





Update



The question has been updated, and has gotten somewhat clearer. I've updated my code to match. Here's the latest output:



enter image description here



Aside from the styling, I think this matches what the OP described.



Here's the code that was used to produce the above plot:



dfake = ({    
'GrA_X' : [15,15],
'GrA_Y' : [15,15],
'Reference_X' : [15,3],
'Reference_Y' : [15,15],
'GrA_Rad' : [15,25],
'GrA_Vel' : [0,10],
'GrA_Scaling' : [0,0.5],
'GrA_Rotation' : [0,45]
})

dffake = pd.DataFrame(dfake)
fig,axs = plt.subplots(1, 2, figsize=(16,8))
fig.subplots_adjust(0,0,1,1)
plotone(dffake, 'A', 0, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[0])
plotone(dffake, 'A', 1, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[1])
plt.show()


and the complete implementation of the plotone function that I used is in the code block below. If you just want to know about the math used to generate and transform the 2D gaussian PDF, check out the mvpdf function (and the rot and getcov functions it depends on):



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

def mvpdf(x, y, xlim, ylim, radius=1, velocity=0, scale=0, theta=0):
"""Creates a grid of data that represents the PDF of a multivariate gaussian.

x, y: The center of the returned PDF
(xy)lim: The extent of the returned PDF
radius: The PDF will be dilated by this factor
scale: The PDF be stretched by a factor of (scale + 1) in the x direction, and squashed by a factor of 1/(scale + 1) in the y direction
theta: The PDF will be rotated by this many degrees

returns: X, Y, PDF. X and Y hold the coordinates of the PDF.
"""
# create the coordinate grids
X,Y = np.meshgrid(np.linspace(*xlim), np.linspace(*ylim))

# stack them into the format expected by the multivariate pdf
XY = np.stack([X, Y], 2)

# displace xy by half the velocity
x,y = rot(theta) @ (velocity/2, 0) + (x, y)

# get the covariance matrix with the appropriate transforms
cov = getcov(radius=radius, scale=scale, theta=theta)

# generate the data grid that represents the PDF
PDF = sts.multivariate_normal([x, y], cov).pdf(XY)

return X, Y, PDF

def plotmv(x, y, xlim=None, ylim=None, radius=1, velocity=0, scale=0, theta=0, xref=None, yref=None, fig=None, ax=None):
"""Plot an xy point with an appropriately tranformed 2D gaussian around it.
Also plots other related data like the reference point.
"""
if xlim is None: xlim = (x - 5, x + 5)
if ylim is None: ylim = (y - 5, y + 5)

if fig is None:
fig = plt.figure(figsize=(8,8))
ax = fig.gca()
elif ax is None:
ax = fig.gca()

# plot the xy point
ax.plot(x, y, '.', c='C0', ms=20)

if not (xref is None or yref is None):
# plot the reference point, if supplied
ax.plot(xref, yref, '.', c='w', ms=12)

# plot the arrow leading from the xy point
if velocity > 0:
ax.arrow(x, y, *rot(theta) @ (velocity, 0),
width=.4, length_includes_head=True, ec='C0', fc='C0')

# fetch the PDF of the 2D gaussian
X, Y, PDF = mvpdf(x, y, xlim=xlim, ylim=ylim, radius=radius, velocity=velocity, scale=scale, theta=theta)

# normalize PDF by shifting and scaling, so that the smallest value is 0 and the largest is 1
normPDF = PDF - PDF.min()
normPDF = normPDF/normPDF.max()

# plot and label the contour lines of the 2D gaussian
cs = ax.contour(X, Y, normPDF, levels=6, colors='w', alpha=.5)
ax.clabel(cs, fmt='%.3f', fontsize=12)

# plot the filled contours of the 2D gaussian. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normPDF, levels=50, cmap='viridis', vmin=-.9, vmax=1)

# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])

# add some labels
ax.grid()
ax.set_xlabel('X distance (M)')
ax.set_ylabel('Y distance (M)')

# ensure that x vs y scaling doesn't disrupt the transforms applied to the 2D gaussian
ax.set_aspect('equal', 'box')

return fig, ax

def fetchone(df, l, i, **kwargs):
"""Fetch all the needed data for one xy point
"""
keytups = (
('x', 'Gr%s_X'%l),
('y', 'Gr%s_Y'%l),
('radius', 'Gr%s_Rad'%l),
('velocity', 'Gr%s_Vel'%l),
('scale', 'Gr%s_Scaling'%l),
('theta', 'Gr%s_Rotation'%l),
('xref', 'Reference_X'),
('yref', 'Reference_Y')
)

ret = {k:df.loc[i, l] for k,l in keytups}
# add in any overrides
ret.update(kwargs)

return ret

def plotone(df, l, i, xlim=None, ylim=None, fig=None, ax=None, **kwargs):
"""Plot exactly one point from the dataset
"""
# look up all the data to plot one datapoint
xydata = fetchone(df, l, i, **kwargs)

# do the plot
return plotmv(xlim=xlim, ylim=ylim, fig=fig, ax=ax, **xydata)


Old answer -2



I've adjusted my answer to match the example the OP posted:



enter image description here



Here's the code that produced the above image:



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

def datalimits(*data, pad=.15):
dmin,dmax = min(d.min() for d in data), max(d.max() for d in data)
spad = pad*(dmax - dmin)
return dmin - spad, dmax + spad

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data=d)

limitpad = .5
clevels = 5
cflevels = 50

xmin,xmax = datalimits(df['GrA_X'], df['GrB_X'], pad=limitpad)
ymin,ymax = datalimits(df['GrA_Y'], df['GrB_Y'], pad=limitpad)

X,Y = np.meshgrid(np.linspace(xmin, xmax), np.linspace(ymin, ymax))

fig = plt.figure(figsize=(10,6))
ax = plt.gca()

Zs =
for l,color in zip('AB', ('red', 'yellow')):
# plot all of the points from a single group
ax.plot(df['Gr%s_X'%l], df['Gr%s_Y'%l], '.', c=color, ms=15, label=l)

Zrows =
for _,row in df.iterrows():
x,y = row['Gr%s_X'%l], row['Gr%s_Y'%l]

cov = getcov(radius=row['Gr%s_Rad'%l], scale=row['Gr%s_Scaling'%l], theta=row['Gr%s_Rotation'%l])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))
Zrows.append(Z)

Zs.append(np.sum(Zrows, axis=0))

# plot the reference points

# create Z from the difference of the sums of the 2D Gaussians from group A and group B
Z = Zs[0] - Zs[1]

# normalize Z by shifting and scaling, so that the smallest value is 0 and the largest is 1
normZ = Z - Z.min()
normZ = normZ/normZ.max()

# plot and label the contour lines
cs = ax.contour(X, Y, normZ, levels=clevels, colors='w', alpha=.5)
ax.clabel(cs, fmt='%2.1f', colors='w')#, fontsize=14)

# plot the filled contours. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normZ, levels=cflevels, cmap='viridis', vmin=0, vmax=1)
# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])


ax.set_aspect('equal', 'box')


Old answer -1



It's a little hard to tell exactly what you're after. It is possible to scale and rotate a multivariate gaussian distribution via its covariance matrix. Here's an example of how to do so based on your data:



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(scale, theta):
cov = np.array([
[1*(scale + 1), 0],
[0, 1/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data=d)
xmin,xmax = min(df['GrA_X'].min(), df['GrB_X'].min()), max(df['GrA_X'].max(), df['GrB_X'].max())
ymin,ymax = min(df['GrA_Y'].min(), df['GrB_Y'].min()), max(df['GrA_Y'].max(), df['GrB_Y'].max())

X,Y = np.meshgrid(
np.linspace(xmin - (xmax - xmin)*.1, xmax + (xmax - xmin)*.1),
np.linspace(ymin - (ymax - ymin)*.1, ymax + (ymax - ymin)*.1)
)

fig,axs = plt.subplots(df.shape[0], sharex=True, figsize=(4, 4*df.shape[0]))
fig.subplots_adjust(0,0,1,1,0,-.82)

for (_,row),ax in zip(df.iterrows(), axs):
for c in 'AB':
x,y = row['Gr%s_X'%c], row['Gr%s_Y'%c]

cov = getcov(scale=row['Gr%s_Scaling'%c], theta=row['Gr%s_Rotation'%c])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))

ax.contour(X, Y, Z)

ax.plot(row['Gr%s_X'%c], row['Gr%s_Y'%c], 'x')
ax.set_aspect('equal', 'box')


This outputs:



enter image description here






share|improve this answer















Update



The question has been updated, and has gotten somewhat clearer. I've updated my code to match. Here's the latest output:



enter image description here



Aside from the styling, I think this matches what the OP described.



Here's the code that was used to produce the above plot:



dfake = ({    
'GrA_X' : [15,15],
'GrA_Y' : [15,15],
'Reference_X' : [15,3],
'Reference_Y' : [15,15],
'GrA_Rad' : [15,25],
'GrA_Vel' : [0,10],
'GrA_Scaling' : [0,0.5],
'GrA_Rotation' : [0,45]
})

dffake = pd.DataFrame(dfake)
fig,axs = plt.subplots(1, 2, figsize=(16,8))
fig.subplots_adjust(0,0,1,1)
plotone(dffake, 'A', 0, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[0])
plotone(dffake, 'A', 1, xlim=(0,30), ylim=(0,30), fig=fig, ax=axs[1])
plt.show()


and the complete implementation of the plotone function that I used is in the code block below. If you just want to know about the math used to generate and transform the 2D gaussian PDF, check out the mvpdf function (and the rot and getcov functions it depends on):



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

def mvpdf(x, y, xlim, ylim, radius=1, velocity=0, scale=0, theta=0):
"""Creates a grid of data that represents the PDF of a multivariate gaussian.

x, y: The center of the returned PDF
(xy)lim: The extent of the returned PDF
radius: The PDF will be dilated by this factor
scale: The PDF be stretched by a factor of (scale + 1) in the x direction, and squashed by a factor of 1/(scale + 1) in the y direction
theta: The PDF will be rotated by this many degrees

returns: X, Y, PDF. X and Y hold the coordinates of the PDF.
"""
# create the coordinate grids
X,Y = np.meshgrid(np.linspace(*xlim), np.linspace(*ylim))

# stack them into the format expected by the multivariate pdf
XY = np.stack([X, Y], 2)

# displace xy by half the velocity
x,y = rot(theta) @ (velocity/2, 0) + (x, y)

# get the covariance matrix with the appropriate transforms
cov = getcov(radius=radius, scale=scale, theta=theta)

# generate the data grid that represents the PDF
PDF = sts.multivariate_normal([x, y], cov).pdf(XY)

return X, Y, PDF

def plotmv(x, y, xlim=None, ylim=None, radius=1, velocity=0, scale=0, theta=0, xref=None, yref=None, fig=None, ax=None):
"""Plot an xy point with an appropriately tranformed 2D gaussian around it.
Also plots other related data like the reference point.
"""
if xlim is None: xlim = (x - 5, x + 5)
if ylim is None: ylim = (y - 5, y + 5)

if fig is None:
fig = plt.figure(figsize=(8,8))
ax = fig.gca()
elif ax is None:
ax = fig.gca()

# plot the xy point
ax.plot(x, y, '.', c='C0', ms=20)

if not (xref is None or yref is None):
# plot the reference point, if supplied
ax.plot(xref, yref, '.', c='w', ms=12)

# plot the arrow leading from the xy point
if velocity > 0:
ax.arrow(x, y, *rot(theta) @ (velocity, 0),
width=.4, length_includes_head=True, ec='C0', fc='C0')

# fetch the PDF of the 2D gaussian
X, Y, PDF = mvpdf(x, y, xlim=xlim, ylim=ylim, radius=radius, velocity=velocity, scale=scale, theta=theta)

# normalize PDF by shifting and scaling, so that the smallest value is 0 and the largest is 1
normPDF = PDF - PDF.min()
normPDF = normPDF/normPDF.max()

# plot and label the contour lines of the 2D gaussian
cs = ax.contour(X, Y, normPDF, levels=6, colors='w', alpha=.5)
ax.clabel(cs, fmt='%.3f', fontsize=12)

# plot the filled contours of the 2D gaussian. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normPDF, levels=50, cmap='viridis', vmin=-.9, vmax=1)

# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])

# add some labels
ax.grid()
ax.set_xlabel('X distance (M)')
ax.set_ylabel('Y distance (M)')

# ensure that x vs y scaling doesn't disrupt the transforms applied to the 2D gaussian
ax.set_aspect('equal', 'box')

return fig, ax

def fetchone(df, l, i, **kwargs):
"""Fetch all the needed data for one xy point
"""
keytups = (
('x', 'Gr%s_X'%l),
('y', 'Gr%s_Y'%l),
('radius', 'Gr%s_Rad'%l),
('velocity', 'Gr%s_Vel'%l),
('scale', 'Gr%s_Scaling'%l),
('theta', 'Gr%s_Rotation'%l),
('xref', 'Reference_X'),
('yref', 'Reference_Y')
)

ret = {k:df.loc[i, l] for k,l in keytups}
# add in any overrides
ret.update(kwargs)

return ret

def plotone(df, l, i, xlim=None, ylim=None, fig=None, ax=None, **kwargs):
"""Plot exactly one point from the dataset
"""
# look up all the data to plot one datapoint
xydata = fetchone(df, l, i, **kwargs)

# do the plot
return plotmv(xlim=xlim, ylim=ylim, fig=fig, ax=ax, **xydata)


Old answer -2



I've adjusted my answer to match the example the OP posted:



enter image description here



Here's the code that produced the above image:



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(radius=1, scale=1, theta=0):
cov = np.array([
[radius*(scale + 1), 0],
[0, radius/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

def datalimits(*data, pad=.15):
dmin,dmax = min(d.min() for d in data), max(d.max() for d in data)
spad = pad*(dmax - dmin)
return dmin - spad, dmax + spad

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data=d)

limitpad = .5
clevels = 5
cflevels = 50

xmin,xmax = datalimits(df['GrA_X'], df['GrB_X'], pad=limitpad)
ymin,ymax = datalimits(df['GrA_Y'], df['GrB_Y'], pad=limitpad)

X,Y = np.meshgrid(np.linspace(xmin, xmax), np.linspace(ymin, ymax))

fig = plt.figure(figsize=(10,6))
ax = plt.gca()

Zs =
for l,color in zip('AB', ('red', 'yellow')):
# plot all of the points from a single group
ax.plot(df['Gr%s_X'%l], df['Gr%s_Y'%l], '.', c=color, ms=15, label=l)

Zrows =
for _,row in df.iterrows():
x,y = row['Gr%s_X'%l], row['Gr%s_Y'%l]

cov = getcov(radius=row['Gr%s_Rad'%l], scale=row['Gr%s_Scaling'%l], theta=row['Gr%s_Rotation'%l])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))
Zrows.append(Z)

Zs.append(np.sum(Zrows, axis=0))

# plot the reference points

# create Z from the difference of the sums of the 2D Gaussians from group A and group B
Z = Zs[0] - Zs[1]

# normalize Z by shifting and scaling, so that the smallest value is 0 and the largest is 1
normZ = Z - Z.min()
normZ = normZ/normZ.max()

# plot and label the contour lines
cs = ax.contour(X, Y, normZ, levels=clevels, colors='w', alpha=.5)
ax.clabel(cs, fmt='%2.1f', colors='w')#, fontsize=14)

# plot the filled contours. Set levels high for smooth contours
cfs = ax.contourf(X, Y, normZ, levels=cflevels, cmap='viridis', vmin=0, vmax=1)
# create the colorbar and ensure that it goes from 0 -> 1
cbar = fig.colorbar(cfs, ax=ax)
cbar.set_ticks([0, .2, .4, .6, .8, 1])


ax.set_aspect('equal', 'box')


Old answer -1



It's a little hard to tell exactly what you're after. It is possible to scale and rotate a multivariate gaussian distribution via its covariance matrix. Here's an example of how to do so based on your data:



import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import scipy.stats as sts

def rot(theta):
theta = np.deg2rad(theta)
return np.array([
[np.cos(theta), -np.sin(theta)],
[np.sin(theta), np.cos(theta)]
])

def getcov(scale, theta):
cov = np.array([
[1*(scale + 1), 0],
[0, 1/(scale + 1)]
])

r = rot(theta)
return r @ cov @ r.T

d = ({
'Time' : [1,2,3,4,5,6,7,8],
'GrA_X' : [10,12,17,16,16,14,12,8],
'GrA_Y' : [10,12,13,7,6,7,8,8],
'GrB_X' : [5,8,13,16,19,15,13,5],
'GrB_Y' : [6,15,12,7,8,9,10,8],
'Reference_X' : [6,8,14,18,13,11,16,15],
'Reference_Y' : [10,12,8,12,15,12,10,8],
'GrA_Rad' : [8.3,8.25,8.2,8,8.15,8.15,8.2,8.3],
'GrB_Rad' : [8.3,8.25,8.3,8.4,8.6,8.4,8.3,8.65],
'GrA_Vel' : [0,2.8,5.1,6.1,1.0,2.2,2.2,4.0],
'GrB_Vel' : [0,9.5,5.8,5.8,3.16,4.12,2.2,8.2],
'GrA_Scaling' : [0,0.22,0.39,0.47,0.07,0.17,0.17,0.31],
'GrB_Scaling' : [0,0.53,0.2,0.2,0.06,0.1,0.03,0.4],
'GrA_Rotation' : [0,45,23.2,-26.56,-33.69,-36.86,-45,-135],
'GrB_Rotation' : [0,71.6,36.87,5.2,8.13,16.70,26.57,90],
})

df = pd.DataFrame(data=d)
xmin,xmax = min(df['GrA_X'].min(), df['GrB_X'].min()), max(df['GrA_X'].max(), df['GrB_X'].max())
ymin,ymax = min(df['GrA_Y'].min(), df['GrB_Y'].min()), max(df['GrA_Y'].max(), df['GrB_Y'].max())

X,Y = np.meshgrid(
np.linspace(xmin - (xmax - xmin)*.1, xmax + (xmax - xmin)*.1),
np.linspace(ymin - (ymax - ymin)*.1, ymax + (ymax - ymin)*.1)
)

fig,axs = plt.subplots(df.shape[0], sharex=True, figsize=(4, 4*df.shape[0]))
fig.subplots_adjust(0,0,1,1,0,-.82)

for (_,row),ax in zip(df.iterrows(), axs):
for c in 'AB':
x,y = row['Gr%s_X'%c], row['Gr%s_Y'%c]

cov = getcov(scale=row['Gr%s_Scaling'%c], theta=row['Gr%s_Rotation'%c])
mnorm = sts.multivariate_normal([x, y], cov)
Z = mnorm.pdf(np.stack([X, Y], 2))

ax.contour(X, Y, Z)

ax.plot(row['Gr%s_X'%c], row['Gr%s_Y'%c], 'x')
ax.set_aspect('equal', 'box')


This outputs:



enter image description here







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 25 '18 at 9:51

























answered Nov 23 '18 at 20:08









teltel

7,27621431




7,27621431













  • @JPeter I've updated the answer again to reflect the updates in the question's description/example. Let me know if this matches what you're aiming for.

    – tel
    Nov 25 '18 at 9:46











  • This is truly brilliant @tel. Sorry for swapping the figure in the original question. I tried to simplify the question to get across the aim. In regards to your second output, do the contours provide a probability of influence ranging from 0-1. Could you provide a short description on how you calculated this. I'll also be applying this to numerous data points in a df. Do you foresee any issues with this? Once again thank you. This is great!

    – user9410826
    Nov 25 '18 at 10:55











  • Sorry @tel. I'm getting an error on both edit 2 and edit 3: it's on cfs = ax.contour(X,Y, normPDF, levels = 50, cmap = 'viridis', vmin = -.9, vmax = 1). The error is if self.filled and len(self.levels) < 2: TypeError: len() of unsized object

    – user9410826
    Nov 25 '18 at 23:35













  • It looks like you're using an old version of Matplotlib (the line of code raising the error looks different in the latest version. The first thing to try then is upgrading your Matplotlib package (they fix tons of bugs all the time).

    – tel
    Nov 26 '18 at 0:03











  • Thanks @tel. I just installed a new env and got it going. Just a quick one. Is the underlying math the same for for edit 2 and 3. You're plotting a distribution of probability. I'm finding it hard to conceptualise how the density of scatter points affects the normalised output probability

    – user9410826
    Nov 26 '18 at 5:40



















  • @JPeter I've updated the answer again to reflect the updates in the question's description/example. Let me know if this matches what you're aiming for.

    – tel
    Nov 25 '18 at 9:46











  • This is truly brilliant @tel. Sorry for swapping the figure in the original question. I tried to simplify the question to get across the aim. In regards to your second output, do the contours provide a probability of influence ranging from 0-1. Could you provide a short description on how you calculated this. I'll also be applying this to numerous data points in a df. Do you foresee any issues with this? Once again thank you. This is great!

    – user9410826
    Nov 25 '18 at 10:55











  • Sorry @tel. I'm getting an error on both edit 2 and edit 3: it's on cfs = ax.contour(X,Y, normPDF, levels = 50, cmap = 'viridis', vmin = -.9, vmax = 1). The error is if self.filled and len(self.levels) < 2: TypeError: len() of unsized object

    – user9410826
    Nov 25 '18 at 23:35













  • It looks like you're using an old version of Matplotlib (the line of code raising the error looks different in the latest version. The first thing to try then is upgrading your Matplotlib package (they fix tons of bugs all the time).

    – tel
    Nov 26 '18 at 0:03











  • Thanks @tel. I just installed a new env and got it going. Just a quick one. Is the underlying math the same for for edit 2 and 3. You're plotting a distribution of probability. I'm finding it hard to conceptualise how the density of scatter points affects the normalised output probability

    – user9410826
    Nov 26 '18 at 5:40

















@JPeter I've updated the answer again to reflect the updates in the question's description/example. Let me know if this matches what you're aiming for.

– tel
Nov 25 '18 at 9:46





@JPeter I've updated the answer again to reflect the updates in the question's description/example. Let me know if this matches what you're aiming for.

– tel
Nov 25 '18 at 9:46













This is truly brilliant @tel. Sorry for swapping the figure in the original question. I tried to simplify the question to get across the aim. In regards to your second output, do the contours provide a probability of influence ranging from 0-1. Could you provide a short description on how you calculated this. I'll also be applying this to numerous data points in a df. Do you foresee any issues with this? Once again thank you. This is great!

– user9410826
Nov 25 '18 at 10:55





This is truly brilliant @tel. Sorry for swapping the figure in the original question. I tried to simplify the question to get across the aim. In regards to your second output, do the contours provide a probability of influence ranging from 0-1. Could you provide a short description on how you calculated this. I'll also be applying this to numerous data points in a df. Do you foresee any issues with this? Once again thank you. This is great!

– user9410826
Nov 25 '18 at 10:55













Sorry @tel. I'm getting an error on both edit 2 and edit 3: it's on cfs = ax.contour(X,Y, normPDF, levels = 50, cmap = 'viridis', vmin = -.9, vmax = 1). The error is if self.filled and len(self.levels) < 2: TypeError: len() of unsized object

– user9410826
Nov 25 '18 at 23:35







Sorry @tel. I'm getting an error on both edit 2 and edit 3: it's on cfs = ax.contour(X,Y, normPDF, levels = 50, cmap = 'viridis', vmin = -.9, vmax = 1). The error is if self.filled and len(self.levels) < 2: TypeError: len() of unsized object

– user9410826
Nov 25 '18 at 23:35















It looks like you're using an old version of Matplotlib (the line of code raising the error looks different in the latest version. The first thing to try then is upgrading your Matplotlib package (they fix tons of bugs all the time).

– tel
Nov 26 '18 at 0:03





It looks like you're using an old version of Matplotlib (the line of code raising the error looks different in the latest version. The first thing to try then is upgrading your Matplotlib package (they fix tons of bugs all the time).

– tel
Nov 26 '18 at 0:03













Thanks @tel. I just installed a new env and got it going. Just a quick one. Is the underlying math the same for for edit 2 and 3. You're plotting a distribution of probability. I'm finding it hard to conceptualise how the density of scatter points affects the normalised output probability

– user9410826
Nov 26 '18 at 5:40





Thanks @tel. I just installed a new env and got it going. Just a quick one. Is the underlying math the same for for edit 2 and 3. You're plotting a distribution of probability. I'm finding it hard to conceptualise how the density of scatter points affects the normalised output probability

– user9410826
Nov 26 '18 at 5:40


















draft saved

draft discarded




















































Thanks for contributing an answer to Stack Overflow!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53368263%2fplot-scaled-and-rotated-bivariate-distribution-using-matplotlib%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

鏡平學校

ꓛꓣだゔៀៅຸ໢ທຮ໕໒ ,ໂ'໥໓າ໼ឨឲ៵៭ៈゎゔit''䖳𥁄卿' ☨₤₨こゎもょの;ꜹꟚꞖꞵꟅꞛေၦေɯ,ɨɡ𛃵𛁹ޝ޳ޠ޾,ޤޒޯ޾𫝒𫠁သ𛅤チョ'サノބޘދ𛁐ᶿᶇᶀᶋᶠ㨑㽹⻮ꧬ꧹؍۩وَؠ㇕㇃㇪ ㇦㇋㇋ṜẰᵡᴠ 軌ᵕ搜۳ٰޗޮ޷ސޯ𫖾𫅀ल, ꙭ꙰ꚅꙁꚊꞻꝔ꟠Ꝭㄤﺟޱސꧨꧼ꧴ꧯꧽ꧲ꧯ'⽹⽭⾁⿞⼳⽋២៩ញណើꩯꩤ꩸ꩮᶻᶺᶧᶂ𫳲𫪭𬸄𫵰𬖩𬫣𬊉ၲ𛅬㕦䬺𫝌𫝼,,𫟖𫞽ហៅ஫㆔ాఆఅꙒꚞꙍ,Ꙟ꙱エ ,ポテ,フࢰࢯ𫟠𫞶 𫝤𫟠ﺕﹱﻜﻣ𪵕𪭸𪻆𪾩𫔷ġ,ŧآꞪ꟥,ꞔꝻ♚☹⛵𛀌ꬷꭞȄƁƪƬșƦǙǗdžƝǯǧⱦⱰꓕꓢႋ神 ဴ၀க௭எ௫ឫោ ' េㇷㇴㇼ神ㇸㇲㇽㇴㇼㇻㇸ'ㇸㇿㇸㇹㇰㆣꓚꓤ₡₧ ㄨㄟ㄂ㄖㄎ໗ツڒذ₶।ऩछएोञयूटक़कयँृी,冬'𛅢𛅥ㇱㇵㇶ𥄥𦒽𠣧𠊓𧢖𥞘𩔋цѰㄠſtʯʭɿʆʗʍʩɷɛ,əʏダヵㄐㄘR{gỚṖḺờṠṫảḙḭᴮᵏᴘᵀᵷᵕᴜᴏᵾq﮲ﲿﴽﭙ軌ﰬﶚﶧ﫲Ҝжюїкӈㇴffצּ﬘﭅﬈軌'ffistfflſtffतभफɳɰʊɲʎ𛁱𛁖𛁮𛀉 𛂯𛀞నఋŀŲ 𫟲𫠖𫞺ຆຆ ໹້໕໗ๆทԊꧢꧠ꧰ꓱ⿝⼑ŎḬẃẖỐẅ ,ờỰỈỗﮊDžȩꭏꭎꬻ꭮ꬿꭖꭥꭅ㇭神 ⾈ꓵꓑ⺄㄄ㄪㄙㄅㄇstA۵䞽ॶ𫞑𫝄㇉㇇゜軌𩜛𩳠Jﻺ‚Üမ႕ႌႊၐၸဓၞၞၡ៸wyvtᶎᶪᶹစဎ꣡꣰꣢꣤ٗ؋لㇳㇾㇻㇱ㆐㆔,,㆟Ⱶヤマފ޼ޝަݿݞݠݷݐ',ݘ,ݪݙݵ𬝉𬜁𫝨𫞘くせぉて¼óû×ó£…𛅑הㄙくԗԀ5606神45,神796'𪤻𫞧ꓐ㄁ㄘɥɺꓵꓲ3''7034׉ⱦⱠˆ“𫝋ȍ,ꩲ軌꩷ꩶꩧꩫఞ۔فڱێظペサ神ナᴦᵑ47 9238їﻂ䐊䔉㠸﬎ffiﬣ,לּᴷᴦᵛᵽ,ᴨᵤ ᵸᵥᴗᵈꚏꚉꚟ⻆rtǟƴ𬎎

Why https connections are so slow when debugging (stepping over) in Java?