Slicing array with numpy?

Question

import numpy as np
r = np.arange(36)
r.resize((6, 6))

print(r)
# prints:
# [[ 0  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]]

print(r[:,::7])
# prints:
# [[ 0]
#  [ 6]
#  [12]
#  [18]
#  [24]
#  [30]]

print(r[:,0])
# prints:
# [ 0  6 12 18 24 30]

The r[:,::7] gives me a column, the r[:,0] gives me a row, they both have the same numbers. Would be glad if someone could explain to me why?

Explain how r[:,::7] and r[:,0] have the same numbers? — Divakar
– Divakar, Commented Apr 9, 2017 at 17:08
@Divakar Why shouldn't they contain the same numbers? The r[:,::7] is just r[:, 0:6:7] for an array of shape (6, 6). — MSeifert
– MSeifert, Commented Apr 9, 2017 at 17:29
@MSeifert Ah yes! Just that the first one uses a range of indices, whereas the latter one uses a scalar for indexing. Thanks for pointing out! — Divakar
– Divakar, Commented Apr 9, 2017 at 17:30
r[:,[0]], r[:,0:1] also produce the column array. You get the 1d when indexing with a scalar; the other cases the result is 2d. — hpaulj
– hpaulj, Commented Apr 9, 2017 at 17:55

MSeifert · Accepted Answer · 2017-04-09 17:34:15Z

2

Because the step argument is greater than the corresponding shape so you'll just get the first "row". However these are not identical (even if they contain the same numbers) because the scalar index in [:, 0] flattens the corresponding dimension (so you'll get a 1D array). But [:, ::7] will keep the number of dimensions intact but alters the shape of the step-sliced dimension.

answered Apr 9, 2017 at 17:34

MSeifert

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