[FIXED] Is there a function that can apply NumPy's broadcasting rules to a list of shapes and return the final shape?

Issue

This is not a question about how broadcasting works (i.e., it's not a duplicate of these questions).

I would just like find a function that can apply NumPy's broadcasting rules to a list of shapes and return the final shape, for example:

>>> broadcast_shapes([6], [4, 2, 3, 1], [2, 1, 1])
[4, 2, 3, 6]

Thanks!

Solution

Here is another direct implementation which happens to beat the others on the example. Honorable mention goes to @hpaulj's with @Warren Weckesser's hack which is almost as fast and much more concise:

def bs_pp(*shapes):
    ml = max(shapes, key=len)
    out = list(ml)
    for l in shapes:
        if l is ml:
            continue
        for i, x in enumerate(l, -len(l)):
            if x != 1 and x != out[i]:
                if out[i] != 1:
                    raise ValueError
                out[i] = x
    return (*out,)

def bs_mq1(*shapes):
    max_rank = max([len(shape) for shape in shapes])
    shapes = [[1] * (max_rank - len(shape)) + shape for shape in shapes]
    final_shape = [1] * max_rank
    for shape in shapes:
        for dim, size in enumerate(shape):
            if size != 1:
                final_size = final_shape[dim]
                if final_size == 1:
                    final_shape[dim] = size
                elif final_size != size:
                    raise ValueError("Cannot broadcast these shapes")
    return (*final_shape,)

import numpy as np

def bs_mq2(*shapes):
    max_rank = max([len(shape) for shape in shapes])
    shapes = np.array([[1] * (max_rank - len(shape)) + shape
                      for shape in shapes])
    shapes[shapes==1] = -1
    final_shape = shapes.max(axis=0)
    final_shape[final_shape==-1] = 1
    return (*final_shape,)

def bs_hp_ww(*shapes):
    return np.broadcast(*[np.empty(shape + [0,], int) for shape in shapes]).shape[:-1]

L = [6], [4, 2, 3, 1], [2, 1, 1]

from timeit import timeit

print('pp:       ', timeit(lambda: bs_pp(*L), number=10_000)/10)
print('mq 1:     ', timeit(lambda: bs_mq1(*L), number=10_000)/10)
print('mq 2:     ', timeit(lambda: bs_mq2(*L), number=10_000)/10)
print('hpaulj/ww:', timeit(lambda: bs_hp_ww(*L), number=10_000)/10)

assert bs_pp(*L) == bs_mq1(*L) and bs_pp(*L) == bs_mq2(*L) and bs_pp(*L) == bs_hp_ww(*L)

Sample run:

pp:        0.0021552839782088993
mq 1:      0.00398325570859015
mq 2:      0.01497043427079916
hpaulj/ww: 0.003267909213900566

Answered By - Paul Panzer