How to rotate a matrix (nested list) counter clockwise by 90 degrees
up vote
4
down vote
favorite
I'm trying to rotate a matrix counter clockwise by 90 degrees.
For example, if:
m = [[1,2,3],
[2,3,3],
[5,4,3]]
then the result should be
m = [[3,3,3],
[2,3,4],
[1,2,5]]
So far, I found:
rez = [[m[j][i] for j in range(len(m))] for i in range(len(m[0]))]
for row in rez:
print(row)
This gives me
[1, 2, 5]
[2, 3, 4]
[3, 3, 3]
This is close, but the rows would need to be reverses. Does anyone know a simple way to rotate this matrix counter clockwise by 90 degrees?
python python-3.x
add a comment |
up vote
4
down vote
favorite
I'm trying to rotate a matrix counter clockwise by 90 degrees.
For example, if:
m = [[1,2,3],
[2,3,3],
[5,4,3]]
then the result should be
m = [[3,3,3],
[2,3,4],
[1,2,5]]
So far, I found:
rez = [[m[j][i] for j in range(len(m))] for i in range(len(m[0]))]
for row in rez:
print(row)
This gives me
[1, 2, 5]
[2, 3, 4]
[3, 3, 3]
This is close, but the rows would need to be reverses. Does anyone know a simple way to rotate this matrix counter clockwise by 90 degrees?
python python-3.x
Here you do not rotate, you transpose.
– Willem Van Onsem
Nov 12 at 20:34
You can here usenumpywithnp.rot90(m).
– Willem Van Onsem
Nov 12 at 20:35
1
list(zip(*m))[::-1]
– Warren Weckesser
Nov 12 at 20:44
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I'm trying to rotate a matrix counter clockwise by 90 degrees.
For example, if:
m = [[1,2,3],
[2,3,3],
[5,4,3]]
then the result should be
m = [[3,3,3],
[2,3,4],
[1,2,5]]
So far, I found:
rez = [[m[j][i] for j in range(len(m))] for i in range(len(m[0]))]
for row in rez:
print(row)
This gives me
[1, 2, 5]
[2, 3, 4]
[3, 3, 3]
This is close, but the rows would need to be reverses. Does anyone know a simple way to rotate this matrix counter clockwise by 90 degrees?
python python-3.x
I'm trying to rotate a matrix counter clockwise by 90 degrees.
For example, if:
m = [[1,2,3],
[2,3,3],
[5,4,3]]
then the result should be
m = [[3,3,3],
[2,3,4],
[1,2,5]]
So far, I found:
rez = [[m[j][i] for j in range(len(m))] for i in range(len(m[0]))]
for row in rez:
print(row)
This gives me
[1, 2, 5]
[2, 3, 4]
[3, 3, 3]
This is close, but the rows would need to be reverses. Does anyone know a simple way to rotate this matrix counter clockwise by 90 degrees?
python python-3.x
python python-3.x
asked Nov 12 at 20:33
user10642683
Here you do not rotate, you transpose.
– Willem Van Onsem
Nov 12 at 20:34
You can here usenumpywithnp.rot90(m).
– Willem Van Onsem
Nov 12 at 20:35
1
list(zip(*m))[::-1]
– Warren Weckesser
Nov 12 at 20:44
add a comment |
Here you do not rotate, you transpose.
– Willem Van Onsem
Nov 12 at 20:34
You can here usenumpywithnp.rot90(m).
– Willem Van Onsem
Nov 12 at 20:35
1
list(zip(*m))[::-1]
– Warren Weckesser
Nov 12 at 20:44
Here you do not rotate, you transpose.
– Willem Van Onsem
Nov 12 at 20:34
Here you do not rotate, you transpose.
– Willem Van Onsem
Nov 12 at 20:34
You can here use
numpy with np.rot90(m).– Willem Van Onsem
Nov 12 at 20:35
You can here use
numpy with np.rot90(m).– Willem Van Onsem
Nov 12 at 20:35
1
1
list(zip(*m))[::-1]– Warren Weckesser
Nov 12 at 20:44
list(zip(*m))[::-1]– Warren Weckesser
Nov 12 at 20:44
add a comment |
3 Answers
3
active
oldest
votes
up vote
2
down vote
You could do the following:
m = [[1, 2, 3],
[2, 3, 3],
[5, 4, 3]]
result = list(map(list, zip(*m)))[::-1]
print(result)
Output
[[3, 3, 3],
[2, 3, 4],
[1, 2, 5]]
With map(list, zip(*m)) you create an iterable of the columns, and with the expression list(...)[::-1] you convert that iterable into a list and reverse it.
edited Nov 12 at 20:41
jhpratt
3,07582232
answered Nov 12 at 20:38
Daniel Mesejo
11.2k1924
add a comment |
up vote
0
down vote
What you here basically do is map a matrix A to a matrix B such that:
Bi j=Aj i
In case you rotate elements, that means that if you rotate an n×m-matrix, then that means that:
Bi j=Aj n-i
So we can calculate this as:
rez = [[m[j][ni] for j in range(len(m))] for ni in range(len(m[0])-1, -1, -1)]which is thus the transpose, but than "reversed". Using indices is however typically not how you do such processing in Python, since now it works only for items that are subscriptable, so I advice you to look for a more elegant solution.
But that being said, numpy offers a numpy.rot90 function to rotate matrices:
>>> np.rot90(m)
array([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
answered Nov 12 at 20:42
Willem Van Onsem
142k16135227
add a comment |
up vote
0
down vote
Other option is to use scipy.ndimage.rotate
Rotate an array.
The array is rotated in the plane defined by the two axes given by the
axes parameter using spline interpolation of the requested order.
import numpy as np
from scipy import ndimage
m = np.matrix([[1,2,3],
[2,3,3],
[5,4,3]])
ndimage.rotate(m, 90.0) #angle as float.
Out:
array([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
Same result you can get by using the
zip()function to transpose rows and columns of a 5.1.4. Nested List then reverse the nested list with[::-1]+ put in a np.matrix :
matrix = [[1, 2, 3],
[2, 3, 3],
[5, 4, 3]]
np.matrix(list(zip(*matrix)))[::-1]
Out:
matrix([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
answered Nov 12 at 20:57
n1tk
1,09121226
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You could do the following:
m = [[1, 2, 3],
[2, 3, 3],
[5, 4, 3]]
result = list(map(list, zip(*m)))[::-1]
print(result)
Output
[[3, 3, 3],
[2, 3, 4],
[1, 2, 5]]
With map(list, zip(*m)) you create an iterable of the columns, and with the expression list(...)[::-1] you convert that iterable into a list and reverse it.
edited Nov 12 at 20:41
jhpratt
3,07582232
answered Nov 12 at 20:38
Daniel Mesejo
11.2k1924
add a comment |
up vote
2
down vote
You could do the following:
m = [[1, 2, 3],
[2, 3, 3],
[5, 4, 3]]
result = list(map(list, zip(*m)))[::-1]
print(result)
Output
[[3, 3, 3],
[2, 3, 4],
[1, 2, 5]]
With map(list, zip(*m)) you create an iterable of the columns, and with the expression list(...)[::-1] you convert that iterable into a list and reverse it.
edited Nov 12 at 20:41
jhpratt
3,07582232
answered Nov 12 at 20:38
Daniel Mesejo
11.2k1924
add a comment |
up vote
2
down vote
up vote
2
down vote
You could do the following:
m = [[1, 2, 3],
[2, 3, 3],
[5, 4, 3]]
result = list(map(list, zip(*m)))[::-1]
print(result)
Output
[[3, 3, 3],
[2, 3, 4],
[1, 2, 5]]
With map(list, zip(*m)) you create an iterable of the columns, and with the expression list(...)[::-1] you convert that iterable into a list and reverse it.
edited Nov 12 at 20:41
jhpratt
3,07582232
answered Nov 12 at 20:38
Daniel Mesejo
11.2k1924
You could do the following:
m = [[1, 2, 3],
[2, 3, 3],
[5, 4, 3]]
result = list(map(list, zip(*m)))[::-1]
print(result)
Output
[[3, 3, 3],
[2, 3, 4],
[1, 2, 5]]
With map(list, zip(*m)) you create an iterable of the columns, and with the expression list(...)[::-1] you convert that iterable into a list and reverse it.
edited Nov 12 at 20:41
jhpratt
3,07582232
answered Nov 12 at 20:38
Daniel Mesejo
11.2k1924
edited Nov 12 at 20:41
jhpratt
3,07582232
edited Nov 12 at 20:41
jhpratt
3,07582232
edited Nov 12 at 20:41
jhpratt
3,07582232
3,07582232
answered Nov 12 at 20:38
Daniel Mesejo
11.2k1924
answered Nov 12 at 20:38
Daniel Mesejo
11.2k1924
answered Nov 12 at 20:38
Daniel Mesejo
11.2k1924
11.2k1924
add a comment |
add a comment |
up vote
0
down vote
What you here basically do is map a matrix A to a matrix B such that:
Bi j=Aj i
In case you rotate elements, that means that if you rotate an n×m-matrix, then that means that:
Bi j=Aj n-i
So we can calculate this as:
rez = [[m[j][ni] for j in range(len(m))] for ni in range(len(m[0])-1, -1, -1)]which is thus the transpose, but than "reversed". Using indices is however typically not how you do such processing in Python, since now it works only for items that are subscriptable, so I advice you to look for a more elegant solution.
But that being said, numpy offers a numpy.rot90 function to rotate matrices:
>>> np.rot90(m)
array([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
answered Nov 12 at 20:42
Willem Van Onsem
142k16135227
add a comment |
up vote
0
down vote
What you here basically do is map a matrix A to a matrix B such that:
Bi j=Aj i
In case you rotate elements, that means that if you rotate an n×m-matrix, then that means that:
Bi j=Aj n-i
So we can calculate this as:
rez = [[m[j][ni] for j in range(len(m))] for ni in range(len(m[0])-1, -1, -1)]which is thus the transpose, but than "reversed". Using indices is however typically not how you do such processing in Python, since now it works only for items that are subscriptable, so I advice you to look for a more elegant solution.
But that being said, numpy offers a numpy.rot90 function to rotate matrices:
>>> np.rot90(m)
array([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
answered Nov 12 at 20:42
Willem Van Onsem
142k16135227
add a comment |
up vote
0
down vote
up vote
0
down vote
What you here basically do is map a matrix A to a matrix B such that:
Bi j=Aj i
In case you rotate elements, that means that if you rotate an n×m-matrix, then that means that:
Bi j=Aj n-i
So we can calculate this as:
rez = [[m[j][ni] for j in range(len(m))] for ni in range(len(m[0])-1, -1, -1)]which is thus the transpose, but than "reversed". Using indices is however typically not how you do such processing in Python, since now it works only for items that are subscriptable, so I advice you to look for a more elegant solution.
But that being said, numpy offers a numpy.rot90 function to rotate matrices:
>>> np.rot90(m)
array([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
answered Nov 12 at 20:42
Willem Van Onsem
142k16135227
What you here basically do is map a matrix A to a matrix B such that:
Bi j=Aj i
In case you rotate elements, that means that if you rotate an n×m-matrix, then that means that:
Bi j=Aj n-i
So we can calculate this as:
rez = [[m[j][ni] for j in range(len(m))] for ni in range(len(m[0])-1, -1, -1)]which is thus the transpose, but than "reversed". Using indices is however typically not how you do such processing in Python, since now it works only for items that are subscriptable, so I advice you to look for a more elegant solution.
But that being said, numpy offers a numpy.rot90 function to rotate matrices:
>>> np.rot90(m)
array([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
answered Nov 12 at 20:42
Willem Van Onsem
142k16135227
edited Nov 12 at 20:47
answered Nov 12 at 20:42
Willem Van Onsem
142k16135227
answered Nov 12 at 20:42
Willem Van Onsem
142k16135227
answered Nov 12 at 20:42
Willem Van Onsem
142k16135227
142k16135227
add a comment |
add a comment |
up vote
0
down vote
Other option is to use scipy.ndimage.rotate
Rotate an array.
The array is rotated in the plane defined by the two axes given by the
axes parameter using spline interpolation of the requested order.
import numpy as np
from scipy import ndimage
m = np.matrix([[1,2,3],
[2,3,3],
[5,4,3]])
ndimage.rotate(m, 90.0) #angle as float.
Out:
array([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
Same result you can get by using the
zip()function to transpose rows and columns of a 5.1.4. Nested List then reverse the nested list with[::-1]+ put in a np.matrix :
matrix = [[1, 2, 3],
[2, 3, 3],
[5, 4, 3]]
np.matrix(list(zip(*matrix)))[::-1]
Out:
matrix([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
answered Nov 12 at 20:57
n1tk
1,09121226
add a comment |
up vote
0
down vote
Other option is to use scipy.ndimage.rotate
Rotate an array.
The array is rotated in the plane defined by the two axes given by the
axes parameter using spline interpolation of the requested order.
import numpy as np
from scipy import ndimage
m = np.matrix([[1,2,3],
[2,3,3],
[5,4,3]])
ndimage.rotate(m, 90.0) #angle as float.
Out:
array([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
Same result you can get by using the
zip()function to transpose rows and columns of a 5.1.4. Nested List then reverse the nested list with[::-1]+ put in a np.matrix :
matrix = [[1, 2, 3],
[2, 3, 3],
[5, 4, 3]]
np.matrix(list(zip(*matrix)))[::-1]
Out:
matrix([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
answered Nov 12 at 20:57
n1tk
1,09121226
add a comment |
up vote
0
down vote
up vote
0
down vote
Other option is to use scipy.ndimage.rotate
Rotate an array.
The array is rotated in the plane defined by the two axes given by the
axes parameter using spline interpolation of the requested order.
import numpy as np
from scipy import ndimage
m = np.matrix([[1,2,3],
[2,3,3],
[5,4,3]])
ndimage.rotate(m, 90.0) #angle as float.
Out:
array([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
Same result you can get by using the
zip()function to transpose rows and columns of a 5.1.4. Nested List then reverse the nested list with[::-1]+ put in a np.matrix :
matrix = [[1, 2, 3],
[2, 3, 3],
[5, 4, 3]]
np.matrix(list(zip(*matrix)))[::-1]
Out:
matrix([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
answered Nov 12 at 20:57
n1tk
1,09121226
Other option is to use scipy.ndimage.rotate
Rotate an array.
The array is rotated in the plane defined by the two axes given by the
axes parameter using spline interpolation of the requested order.
import numpy as np
from scipy import ndimage
m = np.matrix([[1,2,3],
[2,3,3],
[5,4,3]])
ndimage.rotate(m, 90.0) #angle as float.
Out:
array([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
Same result you can get by using the
zip()function to transpose rows and columns of a 5.1.4. Nested List then reverse the nested list with[::-1]+ put in a np.matrix :
matrix = [[1, 2, 3],
[2, 3, 3],
[5, 4, 3]]
np.matrix(list(zip(*matrix)))[::-1]
Out:
matrix([[3, 3, 3],
[2, 3, 4],
[1, 2, 5]])
answered Nov 12 at 20:57
n1tk
1,09121226
edited Nov 13 at 5:02
answered Nov 12 at 20:57
n1tk
1,09121226
answered Nov 12 at 20:57
n1tk
1,09121226
answered Nov 12 at 20:57
n1tk
1,09121226
1,09121226
add a comment |
add a comment |
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Here you do not rotate, you transpose.
– Willem Van Onsem
Nov 12 at 20:34
You can here use
numpywithnp.rot90(m).– Willem Van Onsem
Nov 12 at 20:35
1
list(zip(*m))[::-1]– Warren Weckesser
Nov 12 at 20:44