I have two arrays:
a = np.array([[1, 2], [3, 4], [5, 6]])
b = np.array([[1, 1, 1, 3, 3],
[1, 2, 4, 5, 9],
[1, 2, 3, 4, 5]])
The expected output would match the shape of array 'a' and would be:
array([True, False], [False, True], [True, False])
The first dimension size of arrays a and b always matches (in this case 3).
What I wish to calculate is for each index of each array (0 to 2 as there are 3 dimensions here) is if each number in the array 'a' exists in the corresponding second dimension of array 'b'.
I can solve this in a loop with the following code, but I would like to vectorise it to gain a speed boost but having sat here for several hours, I cannot figure it out:
output = np.full(a.shape, False)
assert len(a) == len(b)
for i in range(len(a)):
output[i] = np.isin(a[i], b[i])
Thank you for any guidance! Anything would be very appreciated :)
Properly reshape the arrays so they can broadcast correctly while comparing:
(a[...,None] == b[:,None]).any(2)
#[[ True False]
# [False True]
# [ True False]]
a[...,None] adds an extra dimension to the end, with shape (3, 2, 1);b[:,None] inserts a dimension as 2nd axis, with shape (3, 1, 5);(3, 2, 5) so essentially you compare each element in the row of a to each element in the corresponding row of b;a;