Select columns in numpy

Question

Assumes there is an index and a matrix L

>>> index
(array([0, 2, 3, 3]), array([0, 2, 2, 3]))
>>> L
array([[  1,  -1,  -5, -10],
   [-15,   0,  -1,  -5],
   [-10, -15,  10,  -1],
   [ -5, -10,   1,  15]])

I want to select the columns according to the index[1], I've tried:

>>> L[:,index[1]]
array([[  1,  -5,  -5, -10],
   [-15,  -1,  -1,  -5],
   [-10,  10,  10,  -1],
   [ -5,   1,   1,  15]])

but the result is not i expected, what I expected is:

>>> for i in index[1]:
...     print L[:,i]
[  1 -15 -10  -5]
[-5 -1 10  1]
[-5 -1 10  1]
[-10  -5  -1  15]

How can i get the expected result without for loop? and why this unexpected result comes out? Thanks.

willeM_ Van Onsem · Accepted Answer · 2017-08-02 19:56:21Z

3

You simply need to transpose it:

L[:,index[1]].T
#             ^ transpose

By using a transpose, the columns are rows and vice versa. So here (you can transpose before the selection, and then use L.T[index[1],:]) we first make the selection and then turn the columns into rows.

This produces:

>>> L[:,index[1]].T
array([[  1, -15, -10,  -5],
       [ -5,  -1,  10,   1],
       [ -5,  -1,  10,   1],
       [-10,  -5,  -1,  15]])

Note that of course behind the curtains there are still some loops that are done. But these are done outside Python and thus are more efficient.

edited Aug 2, 2017 at 19:56

answered Aug 2, 2017 at 19:47

willeM_ Van Onsem

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