python(or numpy) equivalent of match in R

>>> a = [5,4,3,2,1]>>> b = [2,3]>>> [ b.index(x) if x in b else None for x in a ][None, None, 1, 0, None]

Add 1 if you really need position "one based" instead of "zero based".

>>> [ b.index(x)+1 if x in b else None for x in a ][None, None, 2, 1, None]

You can make this one-liner reusable if you are going to repeat it a lot:

>>> match = lambda a, b: [ b.index(x)+1 if x in b else None for x in a ]>>> match<function <lambda> at 0x04E77B70>>>> match(a, b)[None, None, 2, 1, None]

A faster approach building on Paulo Scardine's answer (difference becomes more meaningful as the size of the arrays increases).If you don't mind losing the one-liner:

from typing import Hashable, Listdef match_list(a: List[Hashable], b: List[Hashable]) -> List[int]:    return [b.index(x) if x in b else None for x in a]def match(a: List[Hashable], b: List[Hashable]) -> List[int]:    b_dict = {x: i for i, x in enumerate(b)}    return [b_dict.get(x, None) for x in a]import randoma = [random.randint(0, 100) for _ in range(10000)]b = [i for i in range(100) if i % 2 == 0]%timeit match(a, b)>>> 580 µs ± 15.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)%timeit match_list(a, b)>>> 6.13 ms ± 146 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)match(a, b) == match_list(a, b)>>> True