Can numpy do better than this?

Ian Kelly ian.g.kelly at gmail.com
Thu Jan 8 20:27:46 CET 2015


On Thu, Jan 8, 2015 at 10:56 AM, Rustom Mody <rustompmody at gmail.com> wrote:
> Given a matrix I want to shift the 1st column 0 (ie leave as is)
> 2nd by one place, 3rd by 2 places etc.
>
> This code works.
> But I wonder if numpy can do it shorter and simpler.
>
> ---------------------
> def transpose(mat):
>      return([[l[i] for l in mat]for i in range(0,len(mat[0]))])
> def rotate(mat):
>      return([mat[i][i:]+mat[i][:i] for i in range(0, len(mat))])
> def shiftcols(mat):
>     return ( transpose(rotate(transpose(mat))))

Without using numpy, your transpose function could be:

def transpose(mat):
    return list(zip(*mat))

numpy provides the roll function, but it doesn't allow for a varying
shift per index. I don't see a way to do it other than to roll each
column separately:

>>> mat = np.array([[1,2,3,4,5,6],
...         [7,8,9,10,11,12],
...         [13,14,15,16,17,18],
...         [19,20,21,22,23,24],
...         [25,26,27,28,29,30],
...         [31,32,33,34,35,36],
...         [37,38,39,40,41,42]])
>>> res = np.empty_like(mat)
>>> for i in range(mat.shape[1]):
...     res[:,i] = np.roll(mat[:,i], -i, 0)
...
>>> res
array([[ 1,  8, 15, 22, 29, 36],
       [ 7, 14, 21, 28, 35, 42],
       [13, 20, 27, 34, 41,  6],
       [19, 26, 33, 40,  5, 12],
       [25, 32, 39,  4, 11, 18],
       [31, 38,  3, 10, 17, 24],
       [37,  2,  9, 16, 23, 30]])



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