How to check a list of list present in list without order

Question

all_combination = [[['a', 'b'], ['c', 'd']], [['e', 'f'], ['g', 'h']], [['c', 'd'], ['a', 'b']]]

assume that, [['a', 'b'], ['c', 'd']] = [['c', 'd'], ['a', 'b']] I want to remove the duplicate list from list but the order doesn't matter. and more thing i need a if condition that checks that both [['a', 'b'], ['c', 'd']] is equal to [['c', 'd'], ['a', 'b']].

This is my code i have tried so far.

all_combination = [[['a', 'b'], ['c', 'd']], [['e', 'f'], ['g', 'h']], [['c', 'd'], ['a', 'b']]]

unique = []
new = []
status = False
for i in all_combination:
    for j in i:
        for k in range(len(all_combination)):
            l = k+1
            for m in range(l, len(i)):
                if j == all_combination[k][m]:
                    unique.append(i)
print(new)

Expected Answer:

[[['a', 'b'], ['c', 'd']], [['e', 'f'], ['g', 'h']]]

or

[[['e', 'f'], ['g', 'h']], [['c', 'd'], ['a', 'b']]]

`c = Counter([tuple(sorted(l)) for l in list1]) result = [list(k) for k,v in c.items() if v==1]` — azro, Jun 04 '20 at 07:43
Now you have a nested list with 2 levels...? Please make sure that the question details are clear from the beginning to avoid have answerers wasting time — yatu, Jun 04 '20 at 07:57
I fail to see how the expected output in your second case is related to the rules you described in the first part. — Thierry Lathuille, Jun 04 '20 at 07:58

score 1 · Answer 1 · answered Jun 04 '20 at 07:41

1

Something like this?

list2 = [e for e in list1 if set(e) == {"a", "b"}]

Comparison between sets is independent of order.

answered Jun 04 '20 at 07:41

Thomas

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The `set` part is right. But the answer is too specific. It should be generalised I guess. – paradocslover Jun 04 '20 at 07:42
I can't make it more general if the question doesn't describe any more general requirements. – Thomas Jun 04 '20 at 07:44
2

The question seems to be about removing duplicates, where order doesn't matter. I can't see how this helps, since it only works for this list – yatu Jun 04 '20 at 07:46
guys kindly have a look on my below expected answer. – Ali Hassan Ahmed Khan Jun 04 '20 at 07:53
really sorry guys for the misunderstanding kindly please remove the down votes. Thank You. – Ali Hassan Ahmed Khan Jun 04 '20 at 08:18

score 1 · Answer 2 · answered Jun 04 '20 at 07:46

1

use set

sets don't care about the order

but they also don't care about the count

list1 = [["a","b"], ["c","d"], ["b","a"]]
list1 = [set(i) for i in list1]
ret = [list(i) for i in list1 if list1.count(i) ==1 ]
print (ret)

answered Jun 04 '20 at 07:46

Kuldeep Singh Sidhu

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guys kindly have a look on my below expected answer. – Ali Hassan Ahmed Khan Jun 04 '20 at 07:53
really sorry guys for the misunderstanding kindly please remove the down votes. Thank You. – Ali Hassan Ahmed Khan Jun 04 '20 at 08:18

Fredrik Nilsson · Answer 3 · 2020-06-04T07:54:43.483

1

To remove the duplicates you can do this (independent of order):

list1 = [["a","b"], ["c","d"], ["b","a"]]

uniques = set([frozenset(x) for x in list1])
uniques = [list(x) for x in uniques]

one line version

uniques = [list(x) for x in set([frozenset(x) for x in list1])]

edited Jun 04 '20 at 07:54

answered Jun 04 '20 at 07:51

Fredrik Nilsson

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guys kindly have a look on my below expected answer. – Ali Hassan Ahmed Khan Jun 04 '20 at 07:52
really sorry guys for the misunderstanding kindly please remove the down votes. Thank You. – Ali Hassan Ahmed Khan Jun 04 '20 at 08:18
Thank you so much @Fredrik Nilsson, Thanks for this act of kindness. :') – Ali Hassan Ahmed Khan Jun 04 '20 at 08:20

Mario Ishac · Answer 4 · 2020-06-04T07:54:53.400

1

You can also use collections.Counter:

from collections import Counter

list1 = [["a", "b"], ["c", "d"], ["b", "a"]]

list1_counter = Counter([frozenset(sub_list1) for sub_list1 in list])
list2 = []

for sub_list1 in list1:
    if list1_counter[frozenset(sub_list1)] == 1:
        list2.append(sub_list1)

list2 after this would equal [["c", "d"]].

list2 also preserves the order of the lists within list1.

edited Jun 04 '20 at 07:54

answered Jun 04 '20 at 07:51

Mario Ishac

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guys kindly have a look on my below expected answer. – Ali Hassan Ahmed Khan Jun 04 '20 at 07:52
really sorry guys for the misunderstanding kindly please remove the down votes. Thank You. – Ali Hassan Ahmed Khan Jun 04 '20 at 08:18

score 1 · Accepted Answer · answered Jun 04 '20 at 08:03

1

Try this:

all_combination = [[['a', 'b'], ['c', 'd']], [['e', 'f'], ['g', 'h']], [['c', 'd'], ['a', 'b']]]

tmp = []
for data_list in all_combination:
    for data in data_list:
        if data not in tmp:
            tmp = tmp + [data]

print(tmp)

result = [['a', 'b'], ['c', 'd'], ['e', 'f'], ['g', 'h']]

answered Jun 04 '20 at 08:03

dhentris

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really sorry guys for the misunderstanding kindly please remove the down votes. Thank You. – Ali Hassan Ahmed Khan Jun 04 '20 at 08:17

How to check a list of list present in list without order

5 Answers5