numpy slicing and indexing different results

Question

In numpy subarrays obtained through any of slicing, masking or fancy indexing operations are just views to the original array, which can be demonstrated as follows:

$ python3
Python 3.5.2 (default, Nov 23 2017, 16:37:01) 
[GCC 5.4.0 20160609] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> import numpy as np
>>> np.__version__
'1.11.0'

>>> a = np.arange(3); a[:2] = 111; a
array([111, 111,   2])

>>> a = np.arange(3); a[a<2] = 111; a
array([111, 111,   2])

>>> a = np.arange(3); a[[0,1]] = 111; a
array([111, 111,   2])

In the above example, the entire subarray was assigned to. However if we assign to an element of the subarray, the result of the slicing operation still behaves as a view, whereas the results of the masking and fancy indexing operations behave as independent copies:

>>> a = np.arange(3); a[:2][0] = 111; a
array([111,   1,   2])

>>> a = np.arange(3); a[a<2][0] = 111; a
array([0, 1, 2])

>>> a = np.arange(3); a[[0,1]][0] = 111; a
array([0, 1, 2])

Is this a bug in numpy, or is it by design? If it is by design, then what's the substantiation for such an inconsistency?

Answer 1

It's not a bug. As far as you pass a slice object to Numpy array the returned sub array is a view of the original items which means that even slice assignment or single item assignments will change the original array. But in other cases the returned result is not a view. It's, in fact, a shallow view (copy) of the chosen slice which only supports slice assignment like what other mutable objects in Python support.

It's also mentioned in documentation :

[...] As with index arrays, what is returned is a copy of the data, not a view as one gets with slices.