Assume that I have a NumPy array A with n dimensions, which might be very large, and assume that I have k one-dimensional Boolean masks M1, ..., Mk
I would like to extract an n-dimensional array, B, from A which contains all the elements of A located at indices where the "outer-AND" of all the masks is True.
...but I would like to do this without first forming the (possibly very large) "outer-AND" of all the masks, and without having to extract the specified elements from each axis one axis at a time hence creating (possibly many) intermediate copies in the process.
The example below demonstrates the two ways of extracting the elements from A just described above:
from functools import reduce
import numpy as np
m = 100
for _ in range(m):
n = np.random.randint(0, 10)
k = np.random.randint(0, n + 1)
A_shape = tuple(np.random.randint(0, 10, n))
A = np.random.uniform(-1, 1, A_shape)
M_lst = [np.random.randint(0, 2, dim).astype(bool) for dim in A_shape]
# Creating shape of B:
B_shape = tuple(map(np.count_nonzero, M_lst)) + A_shape[len(M_lst):]
# size of B:
B_size = np.prod(B_shape)
# --- USING "OUTER-AND" OF ALL MASKS --- #
# Creating "outer-AND" of all masks:
M = reduce(np.bitwise_and, (np.expand_dims(M, tuple(np.r_[:i, i+1:n])) for i, M in enumerate(M_lst)), True)
# extracting elements from A and reshaping to the correct shape:
B1 = A[M].reshape(B_shape)
# Checking that the correct number of elements was extracted
assert B1.size == B_size
# THE PROBLEM WITH THIS METHOD IS THE POSSIBLY VERY LARGE OUTER-AND OF ALL THE MASKS!
# --- USING ONE MASK AT A TIME --- #
B2 = A
for i, M in enumerate(M_lst):
B2 = B2[tuple(slice(None) for _ in range(i)) + (M,)]
assert B2.size == np.prod(B_shape)
assert B2.shape == B_shape
# THE PROBLEM WITH THIS METHOD IS THE POSSIBLY LARGE NUMBER OF POSSIBLY LARGE INTERMEDIATE COPIES!
assert np.all(B1 == B2)
# EDIT 1:
# USING np.ix_ AS SUGGESTED BY Chrysophylaxs
i = np.ix_(*M_lst)
B3 = A[i]
assert B3.shape == B_shape
assert B3.size == B_size
assert np.prod(list(map(np.size, i))) == B_size
print(f'All three methods worked all {m} times')
Is there a smarter (more efficient) way to do this, possibly using an existing NumPy function?
