For each row of y I would like to get element whose indexes are specified in m.
>>> y = np.arange(15).reshape(3,5)
[[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]]
>>> m = np.array([[0, 1], [1, 2], [2, 3]])
Expected output:
[[0, 1]
[6, 7]
[12, 13]]
Solution with for cycle
>>> np.stack([y[i, cols] for i, cols in enumerate(m)])
Is there a way how to do it without a for cycle?
Using values from one array as index from another is called 'fancy indexing', however that indexing operation will be repeated for all rows:
y = numpy.arange(15).reshape(3,5)
y[:, [0, 2, 3]]
# array([[ 0, 2, 3],
# [ 5, 7, 8],
# [10, 12, 13]])
If you want to individually "use one index value per row", you need to give that row-to-index relation as another index:
y[[0, 1, 2], [0, 2, 3]]
# array([ 0, 7, 13])
Since your index array m is 2D, you need to tell NumPy which of these two dimension in m corresponds to your row index. You do this by adding another empty axis to the ascending index, (keyword: broadcasting), and you get
y = numpy.arange(15).reshape(3,5)
m = numpy.array([[0, 1], [1, 2], [2, 3]])
y[numpy.arange(len(m))[:, None], m]
# array([[ 0, 1],
# [ 6, 7],
# [12, 13]])
One line, although not much nicer than your own proposal using the for loop:
y[..., m][np.identity(3, dtype=bool)]
Though it will give you some insights in numpy indexing.
