How can I place multiple unrelated graphs on the same axes in ggplot2?
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I am trying to recreate an image found in a textbook in R, the original of which was built in MATLAB:
I have generated each of the graphs seperately, but what would be best practice them into an image like this in ggplot2?
Edit: Provided code used. This is just a transformation of normally distributed data.
library(ggplot2)
mean <- 6
sd <- 1
X <- rnorm(100000, mean = mean, sd = sd)
Y <- dnorm(X, mean = mean, sd = sd)
Y_p <- pnorm(X, mean = mean, sd = sd)
ch_vars <- function(X){
nu_vars <- c()
for (x in X){
nu_vars <- c(nu_vars, (1/(1 + exp(-x + 5))))
}
return(nu_vars)
}
nu_X <- ch_vars(X)
nu_Y <- ch_vars(Y)
data <- data.frame(x = X, y = Y, Y_p = Y_p, x = nu_X, y = nu_Y)
# Cumulative distribution
ggplot(data = data) +
geom_line(aes(x = X, y = Y_p))
# Distribution of initial data
ggplot(data = data_ch, aes(x = X)) +
geom_histogram(aes(y = ..density..), bins = 25, fill = "red", color = "black")
# Distribution of transformed data
ggplot(data = data, aes(x = nu_X)) +
geom_histogram(aes(y = ..density..), bins = 25, fill = "green", color = "black")
r ggplot2
asked Nov 22 '18 at 9:26
FindingIKKFindingIKK
4217
add a comment |
I am trying to recreate an image found in a textbook in R, the original of which was built in MATLAB:
I have generated each of the graphs seperately, but what would be best practice them into an image like this in ggplot2?
Edit: Provided code used. This is just a transformation of normally distributed data.
library(ggplot2)
mean <- 6
sd <- 1
X <- rnorm(100000, mean = mean, sd = sd)
Y <- dnorm(X, mean = mean, sd = sd)
Y_p <- pnorm(X, mean = mean, sd = sd)
ch_vars <- function(X){
nu_vars <- c()
for (x in X){
nu_vars <- c(nu_vars, (1/(1 + exp(-x + 5))))
}
return(nu_vars)
}
nu_X <- ch_vars(X)
nu_Y <- ch_vars(Y)
data <- data.frame(x = X, y = Y, Y_p = Y_p, x = nu_X, y = nu_Y)
# Cumulative distribution
ggplot(data = data) +
geom_line(aes(x = X, y = Y_p))
# Distribution of initial data
ggplot(data = data_ch, aes(x = X)) +
geom_histogram(aes(y = ..density..), bins = 25, fill = "red", color = "black")
# Distribution of transformed data
ggplot(data = data, aes(x = nu_X)) +
geom_histogram(aes(y = ..density..), bins = 25, fill = "green", color = "black")
r ggplot2
asked Nov 22 '18 at 9:26
FindingIKKFindingIKK
4217
3
Can you add example data and code of what you have tried?
– PoGibas
Nov 22 '18 at 9:38
Yourch_vars()function is unnecessary and very inefficient. You can achieve the same effect by taking advantage of R's built-in vectorization, for example:nu_X <- 1/(1 + exp(-X + 5))
– jdobres
Nov 22 '18 at 13:27
Thanks and very true, but there's more to come in the function. This is just the starting point.
– FindingIKK
Nov 22 '18 at 14:11
add a comment |
I am trying to recreate an image found in a textbook in R, the original of which was built in MATLAB:
I have generated each of the graphs seperately, but what would be best practice them into an image like this in ggplot2?
Edit: Provided code used. This is just a transformation of normally distributed data.
library(ggplot2)
mean <- 6
sd <- 1
X <- rnorm(100000, mean = mean, sd = sd)
Y <- dnorm(X, mean = mean, sd = sd)
Y_p <- pnorm(X, mean = mean, sd = sd)
ch_vars <- function(X){
nu_vars <- c()
for (x in X){
nu_vars <- c(nu_vars, (1/(1 + exp(-x + 5))))
}
return(nu_vars)
}
nu_X <- ch_vars(X)
nu_Y <- ch_vars(Y)
data <- data.frame(x = X, y = Y, Y_p = Y_p, x = nu_X, y = nu_Y)
# Cumulative distribution
ggplot(data = data) +
geom_line(aes(x = X, y = Y_p))
# Distribution of initial data
ggplot(data = data_ch, aes(x = X)) +
geom_histogram(aes(y = ..density..), bins = 25, fill = "red", color = "black")
# Distribution of transformed data
ggplot(data = data, aes(x = nu_X)) +
geom_histogram(aes(y = ..density..), bins = 25, fill = "green", color = "black")
r ggplot2
asked Nov 22 '18 at 9:26
FindingIKKFindingIKK
4217
I am trying to recreate an image found in a textbook in R, the original of which was built in MATLAB:
I have generated each of the graphs seperately, but what would be best practice them into an image like this in ggplot2?
Edit: Provided code used. This is just a transformation of normally distributed data.
library(ggplot2)
mean <- 6
sd <- 1
X <- rnorm(100000, mean = mean, sd = sd)
Y <- dnorm(X, mean = mean, sd = sd)
Y_p <- pnorm(X, mean = mean, sd = sd)
ch_vars <- function(X){
nu_vars <- c()
for (x in X){
nu_vars <- c(nu_vars, (1/(1 + exp(-x + 5))))
}
return(nu_vars)
}
nu_X <- ch_vars(X)
nu_Y <- ch_vars(Y)
data <- data.frame(x = X, y = Y, Y_p = Y_p, x = nu_X, y = nu_Y)
# Cumulative distribution
ggplot(data = data) +
geom_line(aes(x = X, y = Y_p))
# Distribution of initial data
ggplot(data = data_ch, aes(x = X)) +
geom_histogram(aes(y = ..density..), bins = 25, fill = "red", color = "black")
# Distribution of transformed data
ggplot(data = data, aes(x = nu_X)) +
geom_histogram(aes(y = ..density..), bins = 25, fill = "green", color = "black")
r ggplot2
r ggplot2
asked Nov 22 '18 at 9:26
FindingIKKFindingIKK
4217
asked Nov 22 '18 at 9:26
FindingIKKFindingIKK
4217
edited Nov 22 '18 at 13:03
FindingIKK
asked Nov 22 '18 at 9:26
FindingIKKFindingIKK
4217
asked Nov 22 '18 at 9:26
FindingIKKFindingIKK
4217
asked Nov 22 '18 at 9:26
FindingIKKFindingIKK
4217
4217
3
Can you add example data and code of what you have tried?
– PoGibas
Nov 22 '18 at 9:38
Yourch_vars()function is unnecessary and very inefficient. You can achieve the same effect by taking advantage of R's built-in vectorization, for example:nu_X <- 1/(1 + exp(-X + 5))
– jdobres
Nov 22 '18 at 13:27
Thanks and very true, but there's more to come in the function. This is just the starting point.
– FindingIKK
Nov 22 '18 at 14:11
add a comment |
3
Can you add example data and code of what you have tried?
– PoGibas
Nov 22 '18 at 9:38
Yourch_vars()function is unnecessary and very inefficient. You can achieve the same effect by taking advantage of R's built-in vectorization, for example:nu_X <- 1/(1 + exp(-X + 5))
– jdobres
Nov 22 '18 at 13:27
Thanks and very true, but there's more to come in the function. This is just the starting point.
– FindingIKK
Nov 22 '18 at 14:11
3
3
Can you add example data and code of what you have tried?
– PoGibas
Nov 22 '18 at 9:38
Can you add example data and code of what you have tried?
– PoGibas
Nov 22 '18 at 9:38
Your
ch_vars() function is unnecessary and very inefficient. You can achieve the same effect by taking advantage of R's built-in vectorization, for example: nu_X <- 1/(1 + exp(-X + 5))– jdobres
Nov 22 '18 at 13:27
Your
ch_vars() function is unnecessary and very inefficient. You can achieve the same effect by taking advantage of R's built-in vectorization, for example: nu_X <- 1/(1 + exp(-X + 5))– jdobres
Nov 22 '18 at 13:27
Thanks and very true, but there's more to come in the function. This is just the starting point.
– FindingIKK
Nov 22 '18 at 14:11
Thanks and very true, but there's more to come in the function. This is just the starting point.
– FindingIKK
Nov 22 '18 at 14:11
add a comment |
1 Answer
1
active
oldest
votes
In short, you can't, or rather, you shouldn't.
ggplot is a high-level plotting packaging. More than a system for drawing shapes and lines, it's fairly "opinionated" about how data should be represented, and one of its opinions is that a plot should express a clear relationship between its axes and marks (points, bars, lines, etc.). The axes essentially define a coordinate space, and the marks are then plotted onto the space in a straightforward and easily interpretable manner.
The plot you show breaks that relationship -- it's a set of essentially arbitrary histograms all drawn onto the same box, where the axis values become ambiguous. The x-axis represents the values of 1 histogram and the y-axis represents another (and thus neither axis represents the histograms' heights).
It is of course technically possible to force ggplot to render something like your example, but it would require pre-computing the histograms, normalizing their values and bin heights to a common coordinate space, converting these into suitable coordinates for use with geom_rect, and then re-labeling the plot axes. It would be a very large amount of manual effort and ultimately defeats the point of using a high-level plotting grammar like ggplot.
answered Nov 22 '18 at 14:10
jdobresjdobres
5,1031623
add a comment |
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In short, you can't, or rather, you shouldn't.
ggplot is a high-level plotting packaging. More than a system for drawing shapes and lines, it's fairly "opinionated" about how data should be represented, and one of its opinions is that a plot should express a clear relationship between its axes and marks (points, bars, lines, etc.). The axes essentially define a coordinate space, and the marks are then plotted onto the space in a straightforward and easily interpretable manner.
The plot you show breaks that relationship -- it's a set of essentially arbitrary histograms all drawn onto the same box, where the axis values become ambiguous. The x-axis represents the values of 1 histogram and the y-axis represents another (and thus neither axis represents the histograms' heights).
It is of course technically possible to force ggplot to render something like your example, but it would require pre-computing the histograms, normalizing their values and bin heights to a common coordinate space, converting these into suitable coordinates for use with geom_rect, and then re-labeling the plot axes. It would be a very large amount of manual effort and ultimately defeats the point of using a high-level plotting grammar like ggplot.
answered Nov 22 '18 at 14:10
jdobresjdobres
5,1031623
add a comment |
In short, you can't, or rather, you shouldn't.
ggplot is a high-level plotting packaging. More than a system for drawing shapes and lines, it's fairly "opinionated" about how data should be represented, and one of its opinions is that a plot should express a clear relationship between its axes and marks (points, bars, lines, etc.). The axes essentially define a coordinate space, and the marks are then plotted onto the space in a straightforward and easily interpretable manner.
The plot you show breaks that relationship -- it's a set of essentially arbitrary histograms all drawn onto the same box, where the axis values become ambiguous. The x-axis represents the values of 1 histogram and the y-axis represents another (and thus neither axis represents the histograms' heights).
It is of course technically possible to force ggplot to render something like your example, but it would require pre-computing the histograms, normalizing their values and bin heights to a common coordinate space, converting these into suitable coordinates for use with geom_rect, and then re-labeling the plot axes. It would be a very large amount of manual effort and ultimately defeats the point of using a high-level plotting grammar like ggplot.
answered Nov 22 '18 at 14:10
jdobresjdobres
5,1031623
add a comment |
In short, you can't, or rather, you shouldn't.
ggplot is a high-level plotting packaging. More than a system for drawing shapes and lines, it's fairly "opinionated" about how data should be represented, and one of its opinions is that a plot should express a clear relationship between its axes and marks (points, bars, lines, etc.). The axes essentially define a coordinate space, and the marks are then plotted onto the space in a straightforward and easily interpretable manner.
The plot you show breaks that relationship -- it's a set of essentially arbitrary histograms all drawn onto the same box, where the axis values become ambiguous. The x-axis represents the values of 1 histogram and the y-axis represents another (and thus neither axis represents the histograms' heights).
It is of course technically possible to force ggplot to render something like your example, but it would require pre-computing the histograms, normalizing their values and bin heights to a common coordinate space, converting these into suitable coordinates for use with geom_rect, and then re-labeling the plot axes. It would be a very large amount of manual effort and ultimately defeats the point of using a high-level plotting grammar like ggplot.
answered Nov 22 '18 at 14:10
jdobresjdobres
5,1031623
In short, you can't, or rather, you shouldn't.
ggplot is a high-level plotting packaging. More than a system for drawing shapes and lines, it's fairly "opinionated" about how data should be represented, and one of its opinions is that a plot should express a clear relationship between its axes and marks (points, bars, lines, etc.). The axes essentially define a coordinate space, and the marks are then plotted onto the space in a straightforward and easily interpretable manner.
The plot you show breaks that relationship -- it's a set of essentially arbitrary histograms all drawn onto the same box, where the axis values become ambiguous. The x-axis represents the values of 1 histogram and the y-axis represents another (and thus neither axis represents the histograms' heights).
It is of course technically possible to force ggplot to render something like your example, but it would require pre-computing the histograms, normalizing their values and bin heights to a common coordinate space, converting these into suitable coordinates for use with geom_rect, and then re-labeling the plot axes. It would be a very large amount of manual effort and ultimately defeats the point of using a high-level plotting grammar like ggplot.
answered Nov 22 '18 at 14:10
jdobresjdobres
5,1031623
edited Nov 22 '18 at 18:17
answered Nov 22 '18 at 14:10
jdobresjdobres
5,1031623
answered Nov 22 '18 at 14:10
jdobresjdobres
5,1031623
answered Nov 22 '18 at 14:10
jdobresjdobres
5,1031623
5,1031623
add a comment |
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3
Can you add example data and code of what you have tried?
– PoGibas
Nov 22 '18 at 9:38
Your
ch_vars()function is unnecessary and very inefficient. You can achieve the same effect by taking advantage of R's built-in vectorization, for example:nu_X <- 1/(1 + exp(-X + 5))– jdobres
Nov 22 '18 at 13:27
Thanks and very true, but there's more to come in the function. This is just the starting point.
– FindingIKK
Nov 22 '18 at 14:11