Most efficient way to index a numpy array by the indices of another numpy array

Question

I have a numpy array A with shape (a, b, c), and another integer array L with shape (a, b). I want to make an array B such that B[i, j, k] = A[L[i, j],j, k] (assume shapes and values of L permit this).

What is the most efficient way to do this? The only ways I can think to do it involve creating meshgrids or new tiled arrays to expand L to the same shape as A which feels inefficient and not very nice. Surely this is quite a standard thing to do with an efficient implementation somewhere (is there an einops way?)?

I used einops einops.rearrange(A, i j k -> (i j) k) and flattened L similarly, indexed the first coordinate by L and reshaped back. This seems kind of stupid -- I am doing it like this because I don't really understand numpy indexing.

You need L of size (a,b,2) then B[i, j, k] = A[L[i, j,0], L[i, j,1], k] — Johnny Cheesecutter
– Johnny Cheesecutter, Commented Jun 4, 2024 at 23:09
If L is of shape (a,b) then call to L[a,b] will return single value float/int` and so you can't broadcast one value to two dimensions. — Johnny Cheesecutter
– Johnny Cheesecutter, Commented Jun 4, 2024 at 23:11
A: (a,b,c) -> float then L: (i,j) -> (a,b) this means that L[i,j] stores two values and it is actually and L[i,j,2] — Johnny Cheesecutter
– Johnny Cheesecutter, Commented Jun 4, 2024 at 23:13
Give a minimal reproducible example, and some code that does the job, even it isn't the most efficient. That will ensure that your specification is complete. — hpaulj
– hpaulj, Commented Jun 5, 2024 at 0:43

Chrysophylaxs · Accepted Answer · 2024-06-05 17:33:49Z

1

a, b, c = 5, 7, 4
A = np.random.rand(a, b, c)

# L indexes the first axis of A,
# so pick random indices that are lower than `a`:
L = np.random.randint(a, size=(a, b))

# Solution: create some indexers for the other axes:
j, k = np.indices((b, c), sparse=True)
# The broadcasting works in this case, because `L` indexes the first axis.
# In other situations, the following might be more robust:
# _, j, k = np.indices((a, b, c), sparse=True)

# Finally do the indexing:
B = A[L[:, :, None], j, k]

Checking:

B2 = np.zeros_like(A)
for i in range(a):
    for j in range(b):
        for k in range(c):
            B2[i, j, k] = A[L[i, j], j, k]

assert np.array_equal(B, B2)

answered Jun 5, 2024 at 17:33

Chrysophylaxs

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2 Comments

mozway Over a year ago

You can also expand the indices: B = A[L[:, :, None], *np.indices((b, c), sparse=True)] :)

mozway Over a year ago

Or A[L[:, :, None], *np.indices((a, b, c), sparse=True)[1:]] for the variant

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