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?

Answer 1

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.