Finding the index of elements in an array/list based on another list or array

Question

I have two lists/arrays, I want to find the index of elements in one list if the same number exists in another list. Here's an example

 list_A = [1,7,9,7,11,1,2,3,6,4,9,0,1]
 list_B = [9,1,7] 
 #output required : [0,1,2,3,5,10,12]

Any method to do this using hopefully numpy

Answer 1

Using a list-comprehension and enumerate():

>>> list_A = [1,7,9,7,11,1,2,3,6,4,9,0,1]
>>> list_B = [9,1,7]
>>> [i for i, x in enumerate(list_A) if x in list_B]
[0, 1, 2, 3, 5, 10, 12]

Using numpy:

>>> import numpy as np
>>> np.where(np.isin(list_A, list_B))
(array([ 0,  1,  2,  3,  5, 10, 12], dtype=int64),)

In addition, as @Chris_Rands points out, we could also convert list_B to a set first, as in is O(1) for sets as opposed to O(n) for lists.

Time comparison:

import random
import numpy as np
import timeit

list_A = [random.randint(0,100000) for _ in range(100000)]
list_B = [random.randint(0,100000) for _ in range(50000)]

array_A = np.array(A)
array_B = np.array(B)

def lists_enumerate(list_A, list_B):
    return [i for i, x in enumerate(list_A) if x in set(list_B)]

def listB_to_set_enumerate(list_A, list_B):
    set_B = set(list_B)
    return [i for i, x in enumerate(list_A) if x in set_B]

def numpy(array_A, array_B):
    return np.where(np.isin(array_A, array_B))

Results:

>>> %timeit lists_enumerate(list_A, list_B)
48.8 s ± 638 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
>>> %timeit listB_to_set_enumerate(list_A, list_B)
11.2 ms ± 856 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
>>> %timeit numpy(array_A, array_B)
23.3 ms ± 167 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

So clearly for larger lists the best solution is to either convert list_B to a set before applying the enumerate, or use numpy.