How to shift the entire numpy array, with wrapping












3















Not sure how to best title this question, but basically I would like to generate a new numpy array to be based on an existing array. The only difference is that the values have been shifted to an index I specify. Also assume that wrapping is required.



For simplicity, consider the base array:



[[0,1,2],

 [3,4,5],

 [6,7,8]]


If I want the zero (0) or the first element from the base array to be shifted at (0,1), it will be:



[[2,0,1],

 [5,3,4],

 [8,6,7]]


If I want the first element moved at (2,2) it will be:



[[4,5,3],

 [7,8,6],

 [1,2,0]]





python numpy







share|improve this question

























  • I assume the 5 in matrix_2 position [2, 0] should be an 8. Can't edit because the edit is "too short" per SO requirements.

    – ShlomiF
    Nov 26 '18 at 6:37











  • Yes, sorry typo. I updated the question. Thanks

    – user1179317
    Nov 26 '18 at 6:39











  • Depending on how often you do this, you might want to consider a ring buffer instead. The answer recommending np.roll is O(n).

    – Mateen Ulhaq
    Nov 26 '18 at 6:47


















3















Not sure how to best title this question, but basically I would like to generate a new numpy array to be based on an existing array. The only difference is that the values have been shifted to an index I specify. Also assume that wrapping is required.



For simplicity, consider the base array:



[[0,1,2],

 [3,4,5],

 [6,7,8]]


If I want the zero (0) or the first element from the base array to be shifted at (0,1), it will be:



[[2,0,1],

 [5,3,4],

 [8,6,7]]


If I want the first element moved at (2,2) it will be:



[[4,5,3],

 [7,8,6],

 [1,2,0]]





python numpy







share|improve this question

























  • I assume the 5 in matrix_2 position [2, 0] should be an 8. Can't edit because the edit is "too short" per SO requirements.

    – ShlomiF
    Nov 26 '18 at 6:37











  • Yes, sorry typo. I updated the question. Thanks

    – user1179317
    Nov 26 '18 at 6:39











  • Depending on how often you do this, you might want to consider a ring buffer instead. The answer recommending np.roll is O(n).

    – Mateen Ulhaq
    Nov 26 '18 at 6:47
















3












3








3








Not sure how to best title this question, but basically I would like to generate a new numpy array to be based on an existing array. The only difference is that the values have been shifted to an index I specify. Also assume that wrapping is required.



For simplicity, consider the base array:



[[0,1,2],

 [3,4,5],

 [6,7,8]]


If I want the zero (0) or the first element from the base array to be shifted at (0,1), it will be:



[[2,0,1],

 [5,3,4],

 [8,6,7]]


If I want the first element moved at (2,2) it will be:



[[4,5,3],

 [7,8,6],

 [1,2,0]]





python numpy







share|improve this question
















Not sure how to best title this question, but basically I would like to generate a new numpy array to be based on an existing array. The only difference is that the values have been shifted to an index I specify. Also assume that wrapping is required.



For simplicity, consider the base array:



[[0,1,2],

 [3,4,5],

 [6,7,8]]


If I want the zero (0) or the first element from the base array to be shifted at (0,1), it will be:



[[2,0,1],

 [5,3,4],

 [8,6,7]]


If I want the first element moved at (2,2) it will be:



[[4,5,3],

 [7,8,6],

 [1,2,0]]





python numpy




python numpy






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 26 '18 at 6:39







user1179317

















asked Nov 26 '18 at 6:32









user1179317user1179317

612818




612818













  • I assume the 5 in matrix_2 position [2, 0] should be an 8. Can't edit because the edit is "too short" per SO requirements.

    – ShlomiF
    Nov 26 '18 at 6:37











  • Yes, sorry typo. I updated the question. Thanks

    – user1179317
    Nov 26 '18 at 6:39











  • Depending on how often you do this, you might want to consider a ring buffer instead. The answer recommending np.roll is O(n).

    – Mateen Ulhaq
    Nov 26 '18 at 6:47





















  • I assume the 5 in matrix_2 position [2, 0] should be an 8. Can't edit because the edit is "too short" per SO requirements.

    – ShlomiF
    Nov 26 '18 at 6:37











  • Yes, sorry typo. I updated the question. Thanks

    – user1179317
    Nov 26 '18 at 6:39











  • Depending on how often you do this, you might want to consider a ring buffer instead. The answer recommending np.roll is O(n).

    – Mateen Ulhaq
    Nov 26 '18 at 6:47



















I assume the 5 in matrix_2 position [2, 0] should be an 8. Can't edit because the edit is "too short" per SO requirements.

– ShlomiF
Nov 26 '18 at 6:37





I assume the 5 in matrix_2 position [2, 0] should be an 8. Can't edit because the edit is "too short" per SO requirements.

– ShlomiF
Nov 26 '18 at 6:37













Yes, sorry typo. I updated the question. Thanks

– user1179317
Nov 26 '18 at 6:39





Yes, sorry typo. I updated the question. Thanks

– user1179317
Nov 26 '18 at 6:39













Depending on how often you do this, you might want to consider a ring buffer instead. The answer recommending np.roll is O(n).

– Mateen Ulhaq
Nov 26 '18 at 6:47







Depending on how often you do this, you might want to consider a ring buffer instead. The answer recommending np.roll is O(n).

– Mateen Ulhaq
Nov 26 '18 at 6:47














1 Answer
1






active

oldest

votes


















6














Use numpy.roll.
For instance, for the first output you can roll 1 index to the right, meaning along axis 1: 



import numpy as np

x = np.array([[0,1,2], [3,4,5], [6,7,8]])

x_shifted = np.roll(x, shift=1, axis=1)


Due to commutativity you can roll twice (once along each dimension) for the two-directional cyclic permutation effect: 



x_double_shifted = np.roll(np.roll(x, shift=2, axis=1), shift=2, axis=0)


Obviously can be done more "pretty" ;-)

Good luck!






share|improve this answer


























  • This works great. Thanks!

    – user1179317
    Nov 26 '18 at 6:49











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6














Use numpy.roll.
For instance, for the first output you can roll 1 index to the right, meaning along axis 1: 



import numpy as np

x = np.array([[0,1,2], [3,4,5], [6,7,8]])

x_shifted = np.roll(x, shift=1, axis=1)


Due to commutativity you can roll twice (once along each dimension) for the two-directional cyclic permutation effect: 



x_double_shifted = np.roll(np.roll(x, shift=2, axis=1), shift=2, axis=0)


Obviously can be done more "pretty" ;-)

Good luck!






share|improve this answer


























  • This works great. Thanks!

    – user1179317
    Nov 26 '18 at 6:49
















6














Use numpy.roll.
For instance, for the first output you can roll 1 index to the right, meaning along axis 1: 



import numpy as np

x = np.array([[0,1,2], [3,4,5], [6,7,8]])

x_shifted = np.roll(x, shift=1, axis=1)


Due to commutativity you can roll twice (once along each dimension) for the two-directional cyclic permutation effect: 



x_double_shifted = np.roll(np.roll(x, shift=2, axis=1), shift=2, axis=0)


Obviously can be done more "pretty" ;-)

Good luck!






share|improve this answer


























  • This works great. Thanks!

    – user1179317
    Nov 26 '18 at 6:49














6












6








6







Use numpy.roll.
For instance, for the first output you can roll 1 index to the right, meaning along axis 1: 



import numpy as np

x = np.array([[0,1,2], [3,4,5], [6,7,8]])

x_shifted = np.roll(x, shift=1, axis=1)


Due to commutativity you can roll twice (once along each dimension) for the two-directional cyclic permutation effect: 



x_double_shifted = np.roll(np.roll(x, shift=2, axis=1), shift=2, axis=0)


Obviously can be done more "pretty" ;-)

Good luck!






share|improve this answer















Use numpy.roll.
For instance, for the first output you can roll 1 index to the right, meaning along axis 1: 



import numpy as np

x = np.array([[0,1,2], [3,4,5], [6,7,8]])

x_shifted = np.roll(x, shift=1, axis=1)


Due to commutativity you can roll twice (once along each dimension) for the two-directional cyclic permutation effect: 



x_double_shifted = np.roll(np.roll(x, shift=2, axis=1), shift=2, axis=0)


Obviously can be done more "pretty" ;-)

Good luck!







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 26 '18 at 6:59

























answered Nov 26 '18 at 6:43









ShlomiFShlomiF

839410




839410













  • This works great. Thanks!

    – user1179317
    Nov 26 '18 at 6:49



















  • This works great. Thanks!

    – user1179317
    Nov 26 '18 at 6:49

















This works great. Thanks!

– user1179317
Nov 26 '18 at 6:49





This works great. Thanks!

– user1179317
Nov 26 '18 at 6:49


















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