filter list of list by matching with lists in other list regardless of their position

0
votes

I have 2 lists of lists and I want to filter list1 based on matching with elements in list2:

list1 = [[1,2,3], [4,5,16]]
list2 = [[9,8,50], [7,10,3]]

the last element in each list is unique id (example list[0][-1] is unique) I want to get the elements from list1 (list) that have the same last element in elements in list2

In my example

list1[0][-1] is similar to list2[1][-1]

I try this:

[x for x, y in zip(list1, list2) if x[-1] == y[-1]]

but it filter by position (check elements that have the same position in both lists) I want to check them regardless of their position

EDIT expected result:

[1,2,3] from the list 1 (because hit last element (list[-1]) is the same as  last element in a list inside list2)
2
python
Why don't you show us the expected result? – Kelly Bundy
done, I added it – Ravanelli

2 Answers

3
votes

you can use a set to first store all the ids from list2, and then filter list1 using a list comprehension, like this:

list2_ids = {x[-1] for x in list2}
result = [x for x in list1 if x[-1] in list2_ids]
-1
votes
list1 = [[1,2,3], [4,5,16]]
list2 = [[9,8,50], [7,10,3]]

lst = [x for x in list1 if x[-1] in set([y[-1] for y in list2])]
#Out[3]: [[1, 2, 3]]