python - Numpy: views vs copy by slicing

Question

When I am doing the slicing, an unexpected thing happened that seems the first to be view but the second is copy.

First

First slice of row, then slice of column. It seems is a view.

>>> a = np.arange(12).reshape(3, 4)   
>>> a[0:3:2, :][:, [0, 2]] = 100
>>> a
array([[100,   1, 100,   3],
       [  4,   5,   6,   7],
       [100,   9, 100,  11]])

Second

But if I first slice of column, then slice of row, it seems a copy:

>>> a[:, [0, 2]][0:3:2, :] = 0
>>> a
array([[100,   1, 100,   3],
       [  4,   5,   6,   7],
       [100,   9, 100,  11]])

I am confused because the two methods finally will cause seem position to change, but why the second actually doesn't change the number?

Answer 1

All that matters is whether you slice by rows or by columns. Slicing by rows can return a view because it is a contiguous segment of the original array. Slicing by column must return a copy because it is not a contiguous segment. For example:

A1 A2 A3
B1 B2 B3
C1 C2 C3

By default, it is stored in memory this way:

A1 A2 A3 B1 B2 B3 C1 C2 C3

So if you want to choose every second row, it is:

[A1 A2 A3] B1 B2 B3 [C1 C2 C3]

That can be described as {start: 0, size: 3, stride: 6}.

But if you want to choose every second column:

[A1] A2 [A3 B1] B2 [B3 C1] C2 [C3]

And there is no way to describe that using a single start, size, and stride. So there is no way to construct such a view.

If you want to be able to view every second column instead of every second row, you can construct your array in column-major aka Fortran order instead:

np.array(a, order='F')

Then it will be stored as such:

A1 B1 C1 A2 B2 C2 A3 B3 C3

Answer 2

The accepted answer by John Zwinck is actually false (I just figured this out the hard way!). The problem in the question is a combination of doing "lvalue indexing" with numpy's fancy indexing. The following doc explains exactly this case

https://scipy-cookbook.readthedocs.io/items/ViewsVsCopies.html

in the section "But fancy indexing does seem to return views sometimes, doesn't it?"

Answer 3

This is my understanding, for your reference

a[0:3:2, :]                     # basic indexing, a view
... = a[0:3:2, :][:, [0, 2]]    # getitme version, a copy,
                                # because you use advanced
                                # indexing [:,[0,2]]
a[0:3:2, :][:, [0, 2]] = ...    # howver setitem version is
                                # like a view, setitem version
                                # is different from getitem version,
                                # this is not c++
a[:, [0, 2]]                    # getitem version, a copy,
                                # because you use advanced indexing
a[:, [0, 2]][0:3:2, :] = 0      # the copy is modied,
                                # but a keeps unchanged.

If I have any misunderstanding, please point it out.