Most efficient way to implement numpy.in1d for muliple arrays

Question

What is the best way to implement a function which takes an arbitrary number of 1d arrays and returns a tuple containing the indices of the matching values (if any).

Here is some pseudo-code of what I want to do:

a = np.array([1, 0, 4, 3, 2])
b = np.array([1, 2, 3, 4, 5])
c = np.array([4, 2])

(ind_a, ind_b, ind_c) = return_equals(a, b, c)
# ind_a = [2, 4]
# ind_b = [1, 3]
# ind_c = [0, 1]

(ind_a, ind_b, ind_c) = return_equals(a, b, c, sorted_by=a)
# ind_a = [2, 4]
# ind_b = [3, 1]
# ind_c = [0, 1]

def return_equals(*args, sorted_by=None):
    ...

Answer 1

You can use numpy.intersect1d with reduce for this:

def return_equals(*arrays):
    matched = reduce(np.intersect1d, arrays)
    return np.array([np.where(np.in1d(array, matched))[0] for array in arrays])

reduce may be little slow here because we are creating intermediate NumPy arrays here(for large number of input it may be very slow), we can prevent this if we use Python's set and its .intersection() method:

matched = np.array(list(set(arrays[0]).intersection(*arrays[1:])))

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Related GitHub ticket: n-array versions of set operations, especially intersect1d