Can numpy do better than this?

[Rustom Mody]

On Friday, January 9, 2015 at 6:43:45 AM UTC+5:30, Rustom Mody wrote:
> On Friday, January 9, 2015 at 12:58:52 AM UTC+5:30, Ian wrote:
> > On Thu, Jan 8, 2015 at 10:56 AM, Rustom Mody 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]])
> 
> Thanks Ian!
> With that I came up with the expression
> 
> transpose(array([list(roll(mat[:,i],i,0)) for i in range(mat.shape[1])]))

That is numpy transpose of course (following a 'from numpy import *')
Not a vanilla (list) transpose