most efficient way to index a numpy array with a number of 1d boolean arrays

Assume that I have a numpy array A with n dimensions, which might be very large, and assume that I have k 1-dimensional boolean masks M1, ..., Mk

I would like to extract from A an n-dimensional array B 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?.

EDIT 1: I added the solution suggested by Chrysophylaxs
EDIT 2: Accepted answer from Chrysophylaxs as the index used has the same size as the resulting B which is exactly what I wanted.

>Solution :

IIUC, you’re looking for np.ix_; an example:

import numpy as np

arr = np.arange(60).reshape(3, 4, 5)

x = [True, False, True]
y = [False, True, True, False]
z = [False, True, False, True, False]

out = arr[np.ix_(x, y, z)]

out:

array([[[ 6,  8],
        [11, 13]],

       [[46, 48],
        [51, 53]]])