How to find repetition of nested list if the element of list are not in order?

Question

#orignal list
list_one = [['a','b'],['a','c'],['b','c'],['b','a']]
li=[]
for i in list_one:
    for j in list_one:
        if i[0] == j[1] and j[0] == i[1]:
            li.append([i,j])
print(li)

#[[['a', 'b'], ['b', 'a']], [['b', 'a'], ['a', 'b']]]

I need the output to be [a,b] only the a and b can vary depending on the condition like a can be apple or anything

Does this answer your question? [finding duplicates in a list of lists](https://stackoverflow.com/questions/19811418/finding-duplicates-in-a-list-of-lists) — Carlo Zanocco, Aug 29 '20 at 16:44
@CarloZanocco "... if the elements of the list are not in order" — Thierry Lathuille, Aug 29 '20 at 16:45
@ThierryLathuille Why not order it? He don't ask the question with the constraint to not order the list. He just say they are not in order. — Carlo Zanocco, Aug 29 '20 at 16:47
@CarloZanocco I don't see any answer in the duplicate you suggest that can answer this question... — Thierry Lathuille, Aug 29 '20 at 16:50
Considering the @ThierryLathuille maybe you want to preserve the order of the list so check [here](https://stackoverflow.com/questions/480214/how-do-you-remove-duplicates-from-a-list-whilst-preserving-order) — Carlo Zanocco, Aug 29 '20 at 16:51

score 0 · Answer 1 · answered Aug 29 '20 at 16:52

When considering a sublist, you only need to look for a duplicate in the queue of the list, beyond the current sublist. With minor changes to your code, you could do:

list_one = [['a','b'],['a','c'],['b','c'],['b','a']]
li=[]
for index, i in enumerate(list_one):
    for index2 in range(index+1, len(list_one)):
        j = list_one[index2]                
        if i[0] == j[1] and j[0] == i[1]:
            li.append([i,j])
print(li)
# [[['a', 'b'], ['b', 'a']]]

score 0 · Answer 2 · answered Aug 29 '20 at 16:53

It seems you are looking for items which contain letters that have not been seen before. The simplest way to do this is by building a cache.

cache = set() # will contain all individual letters that have been seen
li = []
for item in list_one:
    if not any(letter in cache for letter in item):
        li.append(item)
    cache.update([letter for letter in item])
print(li)

[['a', 'b']]

score 0 · Accepted Answer · answered Aug 31 '20 at 09:51

One way could be:

Sort the list, find the duplicate nested.

#orignal list
data = [['a','b'],['a','c'],['b','c'],['b','a']]
seen = []
data = [sorted(x) for x in data]
data = [x if x in seen else seen.append(x) for x in data]
print([x for x in data if x])

OUTPUT:

[['a', 'b']]

How to find repetition of nested list if the element of list are not in order?

3 Answers3