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I'm using numpy.

I have an ndarray with shape of [T, H, W, C] and I want to transpose it to become like: [T, C, H, W]. However, this array is huge and I want to be memory-efficient.

But I just found np.transpose to do this which is not in-place.

Why do operations like np.transpose don't have their in-place counterpart?

I used to think that any operation named np.Bar would have its in-place counterpart named np.Bar_, only to find that this is not the truth.

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From np.transpose docs

A view is returned whenever possible.

meaning no extra memory is allocated for the output array.

>>> import numpy as np

>>> A = np.random.rand(2, 3, 4, 5)
>>> B = np.transpose(A, axes=(0, 3, 1, 2))

>>> A.shape
(2, 3, 4, 5)
>>> B.shape
(2, 5, 3, 4)

You can use np.shares_memory to check if B is a view of A: 

>>> np.shares_memory(A, B)
True

So, you can safely transpose your data withnp.transpose.

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8 Comments

The idea was to demonstrate that the kind of transpose OP intends to perform does return a view rather than a copy.
so what happens if I do A = np.transpose(A,axes=xxx)? Does it need memory re-allocation?
@游凯超, If np.transpose(A,axes=xxx) is a view then A is just reassigned (both point to the same data structure).
is there any solution if we want the real transpose rather than a view?
@JingpengWu, a quick-and-dirty approach would be to copy the array and then transpose the copy.
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