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I have 2 arrays:

>>> a.shape
(9, 3, 11)
>>> b.shape
(9,)

I would like to compute the equivalent of c[i, j] = f(a[i, j, :], b[i]) where f(a0, b0) is a function that takes 2 parameters, with len(a0) == 11 and len(b0) == 9. Here, i is iterating on range(9) and j is iterating on range(3).

Is there a way to code this using numpy.vectorize? Or is it simpler with some clever broadcasting?

I have been trying for 2 hours and I just don't understand how to make it work... I tried to broadcast or to use signatures but to no avail.

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  • 1
    np.array([f(a[i,j,:], b[i]) for j in range(3)] for i in range(9)]). a + b[:,None,None] will work but the result is (9,3,11). Or maybe np.sum(a + b[:,None,None]).axis=2) if you want (9,3), but then you might as well np.sum(a, axis=2) first. Commented Nov 30, 2021 at 8:04
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    np.vectorize in default mode passes scalar values to the function. There is a signature option that makes it pass arrays, but it is trickier to use, and even slower. apply_along_axis can iterate on your i,j dimensions of a (slowly), but it can't, at the same time iterate on b. It's just a one array iterator. The nested list comprehension is probably your best option. Commented Nov 30, 2021 at 8:10
  • 1
    A sample of the function might help. Also you need to make it clear what the function returns. Is it as scalar, so c can be a numeric dtype array. That requires, I assume, some sort of reduction on the size 11 dimension. Commented Nov 30, 2021 at 16:45
  • @hpaulj, in the end I could find a solution: np.vectorize(f, signature="(k),(1)->()") and then I have to call it like f(a, b[:, None, None]. Do you think this would be much slower than a loop? Commented Nov 30, 2021 at 17:06

2 Answers 2

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In the end, I could make it work like this:

>>> f = np.vectorize(f, signature="(k),(1)->()")
>>> print(a.shape)
(9, 3, 11)
>>> print(b.shape)
(9,)
>>> print(f(a, b[:, None, None]).shape)
(9, 3)

This ensures that f gets called with the correct shapes and iterates properly. It is frankly not straightforward from the Numpy documentation to understand the trick to use a (1) in the signature for this purpose.

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numpy.apply_along_axis is what you need.

import numpy as np

a = np.ones( (9,3,11) )
b = np.ones( 9 )

def f(a0, b0):
    return sum(a0[:9]+b0)

c = np.apply_along_axis( f, 2, a, b )
print(c)

c's shape is (9,3).

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4 Comments

I get a (9, 3, 9) shaped c.
OK, f needs to return one value. not a list of 9. Now c is (9,3).
You could do just return np.sum(a0) + b0, so we don't need the index in the function definition! The important thing is to return the scalar value.
That won't work, because b0 is an array of 9. It will do a scalar addition and return a list, which is how I ended up with (9,3,9) in the first place. HOWEVER, I only made up that dummy function because you didn't provide yours. I presume yours is a bit more complicated.

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