Numpy comparing vectors of length m and n resulting in boolean matrix of size m,n

Question

I am wondering if there is a more efficient way to run the comparison (or really many other functions) using numpy.

a = np.array([1,2,5,7])
b = np.array([0,4,6])

np.repeat(a, len(b)).reshape(-1, len(b)) > b

> array([[ True, False, False],
       [ True, False, False],
       [ True,  True, False],
       [ True,  True,  True]])

Basically each output (m, n) is the comparison if A_m > B_n

score 3 · Accepted Answer · answered Feb 22 '22 at 12:45

3

You can use broadcasting to make this operation more efficient. Indeed, repeat creates a new temporary array. Here is the resulting code:

a.reshape(-1, 1) > b

answered Feb 22 '22 at 12:45

Jérôme Richard

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3

`A[:, None]` is a common alternative – Mad Physicist Feb 22 '22 at 12:47
Can you cast with higher dimensions, say `a` has length `m` but now `b` has size `n, o` – qwertylpc Feb 22 '22 at 13:04
For others who might stumble across this I did something like this `a[:, None, None] > b[None, :, :]` – qwertylpc Feb 22 '22 at 13:20

Numpy comparing vectors of length m and n resulting in boolean matrix of size m,n

1 Answers1