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
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.
O(N).
list_B to a set firstin operation is O(N) for lists and O(1) for sets, so if both lists are large, makes a difference
For arrays.