How to find sublist which present one list not another list in python? [duplicate]

Question

I need to compare two lists which are basically list-of-list find out the sublists which are present in one list but not other. Also the arrangement of the sublists does not consider i.e. ['a','b'] = ['b,'a']. The two lists are

List_1 = [['T_1','T_2'],['T_2','T_3'],['T_1','T_3']]
List_2 = [['T_1','T_2'],['T_3','T_1']]

The output list should be

out_list = [['T_2','T_3']]

Answer 1

For two element sublists, this should suffice:

[x for x in List_1 if x not in List_2 and x[::-1] not in List_2]

Code:

List_1 = [['T_1','T_2'],['T_2','T_3'],['T_1','T_3']]
List_2 = [['T_1','T_2'],['T_3','T_1']]

print([x for x in List_1 if x not in List_2 and x[::-1] not in List_2])

Answer 2

Here's a little messy functional solution that uses sets and tuples in the process (sets are used because what you're trying to calculate is the symmetric difference, and tuples are used because unlike lists, they're hashable, and can be used as set elements):

List_1 = [['T_1','T_2'],['T_2','T_3'],['T_1','T_3']]
List_2 = [['T_1','T_2'],['T_3','T_1']]

f = lambda l : tuple(sorted(l))

out_list = list(map(list, set(map(f, List_1)).symmetric_difference(map(f, List_2))))

print(out_list)

Output:

[['T_2', 'T_3']]

Answer 3

I'd say frozensets are more appropiate for such task:

fs2 = set(map(frozenset,List_2))
out = set(map(frozenset,List_1)).symmetric_difference(fs2)

print(out)                                 
# {frozenset({'T_2', 'T_3'})}

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The advantage of using frozensets here is that they can be hashed, hence you can simply map both lists and take the set.symmetric_difference.

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If you want a nested list from the output, you can simply do:

list(map(list, out))

Note that some sublists might appear in a different order, though given the task should not be a problem

Answer 4

You can convert lists to sets for equality comparison and use any() to add into list only items which doesn't exists in second list:

List_1 = [['T_1', 'T_2'], ['T_2', 'T_3'], ['T_1', 'T_3']]
List_2 = [['T_1', 'T_2'], ['T_3', 'T_1']]
out_list = [l1 for l1 in List_1 if not any(set(l1) == set(l2) for l2 in List_2)]

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For better understanding resources consumption and efficiency of each answer I've done some tests. Hope it'll help to choose best.

Results on data from question:

• Olvin Roght's answer - 12.963876624000001;
• yatu's answer - 8.218290244000002;
• rusu_ro1's answer - 8.857162503000001;
• MrGeek's answer - 11.631234766000002;
• Austin's answer - 3.452045860999995;
• GZ0's answer - 7.037438627.

Results on bigger data:

• Olvin Roght's answer - 83.452110953;
• yatu's answer - 0.1939603360000035;
• rusu_ro1's answer - 0.24479892000000802;
• MrGeek's answer - 0.32636319700000627;
• Austin's answer - 5.052051797000004;
• GZ0's answer - 0.20400504799999908.

Answer 5

if you do not have duplicates in your lists you can use:

 set(frozenset(e) for e in List_1).symmetric_difference({frozenset(e) for e in List_2})

output:

{frozenset({'T_2', 'T_3'}), frozenset({1, 2})}

if you need a list of lists as output you can use:

[list(o) for o in output]

ouptut:

[['T_2', 'T_3']]