Indexing Numpy array with list and tuple gives different results?

Question

Sample code:

import numpy as np
a = np.zeros((5,5))
a[[0,1]] = 1     #(list of indices)
print('results with list based indexing\n', a)

a = np.zeros((5,5))
a[(0,1)] = 1   #(tuple of indices)
print('results with tuple based indexing\n',a)

Result:

results with list based indexing
 [[ 1.  1.  1.  1.  1.]
  [ 1.  1.  1.  1.  1.]
  [ 0.  0.  0.  0.  0.]
  [ 0.  0.  0.  0.  0.]
  [ 0.  0.  0.  0.  0.]]

results with tuple based indexing
[[  0.   1.   0.   0.   0.]
 [  0.   0.   0.   0.   0.]
 [  0.   0.   0.   0.   0.]
 [  0.   0.   0.   0.   0.]
 [  0.   0.   0.   0.   0.]]

As you must have noticed, indexing array with list gave a different result than with tuple of same indices. I'm using python3 with numpy version 1.13.3

What is the fundamental difference in indexing a numpy array with list and tuple?

wim · Accepted Answer · 2018-02-26 16:39:12Z

7

By design. Numpy's getitem and setitem syntax does not duck-type, because the different types are used to support different features. This is just a plain old __setitem__:

a[(0,1)] = 1

It's the same as doing a[0,1] = 1. In both cases, ndarray's setitem receives two arguments: a tuple for the index (0, 1) and the value 1.

a[[0,1]] = 1

This is a special case of broadcasting. It would usually be written a[0:2] = 1, but you could also slice/mutate other rows for example a[[0,1,3]]. The scalar 1 gets "stretched" across all columns of rows 0 and 1 in the assignment.

edited Feb 26, 2018 at 16:39

answered Feb 26, 2018 at 16:27

wim

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2 Comments

Tejas Over a year ago

Got it thanks, I also thought of it by another way, writing it down as a answer

hpaulj Over a year ago

The list case could also be written a a[0,1], :]. This makes it clearer, to humans, that the list is selecting two whole rows.

Tejas · Accepted Answer · 2018-02-26 17:20:56Z

0

I also tried following cases in terminal.

a[[0],[1]] = 1
a[(0),(1)] = 1
a[(0,),(1,)] = 1

All of them are equivalent to a[0,1] = 1. What I understood from this is that, Numpy is expecting N different sequences of integers for indexing N dimensions of array.

Meaning, indexing array as a[[ 0,1 ]] results in a[ [0,1], :].

[0,1] is taken as a one sequence of integers to access first dimension of the array and since no indexing is mentioned for remaining dimension it is taken as:by default which results in a[[0,1],:].

Answer by wim gave me a direction for this thought.

Cheers!

answered Feb 26, 2018 at 17:20

Tejas

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