I have two lists listA = [0, 4, 1, 3, 2] and listB = [0, 0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4] what I want is to get all those indexes where the element of listA appears in listB or that as a result would have listR = [0, 1, 10, 11, 12, 13, 2, 3, 7, 8, 9, 4, 5, 6] and try doing with a for but obviously it takes a long time.
An efficient method can be to build before a dictionary in which the keys are the numbers observed in B, and the values are lists with the indexes within B in which they appear.
This dictionary is easy to build:
from collections import defaultdict
listB = [0, 0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4]
dic = defaultdict(list)
for indice, elemento in enumerate(listB):
dic[elemento].append(indice)
When this loop ends, dic contains the following dictionary:
{0: [0, 1],
1: [2, 3],
2: [4, 5, 6],
3: [7, 8, 9],
4: [10, 11, 12]}
where you see, for example, that the data 3 appeared in indexes [7,8,9] .
From this dictionary it is easy to build the list that you ask, simply by going through A and adding to the result list the lists that come out of the dictionary, according to the value of the element of A:
listA = [0, 4, 1, 3, 2]
r = []
for elemento in listA:
r.extend(dic[elemento])
When finished, r has the list you were looking for:
[0, 1, 10, 11, 12, 2, 3, 7, 8, 9, 4, 5, 6]
Note that if any data that was not in B appears when traversing A, line dic[elemento] will cause an error because there is no such key in the dictionary. You can easily fix it if instead of dic[elemento] you put dic.get(elemento, [-1]) , for example, so you can -1 for cases where you can not find it (or [None] , or the value you want to put)