python assignment in array vs scalar

Question

I have a 2D array A of shape (4,3), and a 1D array a of shape (4,). I want to swap the first two rows of A, as well as the first two elements in a. I did the following:

A[0,:],A[1,:] = A[1,:],A[0,:]
a[0],a[1] = a[1],a[0]

Apparently, it works for a, but fails for A. Now, the second row becomes the first row, but the first row remains unchanged. If I do the following:

first_row_copy = A[0,:].copy()
A[0,:] = A[1,:]
A[1,:] = first_row_copy

Then, it seems to work. Why the first method doesn't work? (but works for a) Also, what's the difference between A_copy = A[0,:].copy() and A_copy = A[0,:]?

ShadowRanger · Accepted Answer · 2017-12-30 02:06:57Z

5

numpy slices are views of the underlying memory, they don't make independent copies by default (this is a performance/memory optimization). So:

A[0,:],A[1,:] = A[1,:],A[0,:]

Makes a view of A[1,:] and a view of A[0,:], then assigns the values of A[0,:] to equal what's in the view of A[1,:]. But when it gets to assigning A[1,:], A[0,:]'s view is now showing the post-copy data, so you get the incorrect result. Simply adding .copy to the second element here would be sufficient in this case:

A[0,:], A[1,:] = A[1,:], A[0,:].copy()

because the tuple on the right is always constructed completely before assignments to the left begin, so you can use the live view for the first assignment, and only need to make a copy to preserve the values for the second assignment.

answered Dec 30, 2017 at 2:06

ShadowRanger

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2 Comments

Physicist Over a year ago

then why does it work for a? Because it is not numpy slices and so it is making copy by default?

ShadowRanger Over a year ago

@Physicist: Plain indexing for a single value makes copies (there is nothing to be gained by avoiding copying the memory in that case).

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