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])])) Not exactly pretty. My hunch is it can be improved??... -- https://mail.python.org/mailman/listinfo/python-list
