For example, I have two list of sets:

list1 = [{'a','b'}, {'c','d'}, {'a','b','c'}, {'c','f'}]
list2 = [{'c','d','e'}, {'e','f'}, {'a','b','d'}, {'c','f'}]

I need to output a list of indices where list1[i] and list2[i] don't share common elements. (no intersection)

In this case, {'a','b'} has no common elements in {'c','d','e'}.

1. {'c','d'} has no common elements in {'e','f'}.
2. {'a','b','c'} has common elements 'a' and 'b' in {'a','b','d'}.
3. {'c','f'} has common elements 'c' and 'f' in {'c','f'}.

So list1[0] and list1[1] do not have the same element(s) in list2[0] and list2[1]

It will return a list of indices: list = [0,1]

My approach is:

for l1,l2 in zip(list1,list2):
    for i in l1:
        if i in l2:
            print(i)

This is clearly not correct. Any help is appreciated.

CodePudding user response：

You can enumerate over the zipped lists and filter the indices based on whether the pair of sets is disjoint:

list1 = [{'a','b'}, {'c','d'}, {'a','b','c'}, {'c','f'}]
list2 = [{'c','d','e'}, {'e','f'}, {'a','b','d'}, {'c','f'}]

indices = [i for i, (a, b) in enumerate(zip(list1, list2)) if a.isdisjoint(b)]

# [0, 1]

CodePudding user response：

I found a solution, I'm using a dictionary to output the common letters for each set

iter = [i for i in range(len(list1))]
list3 = []
dict = {}
for i in iter:
    for l1, l2 in zip(list1[i], list2[i]):
        letter = ""
        if l1 not in list2[i]:
            letter  = l1
        if l2 not in list1[i]:
            letter  = l2
        if letter != '':
        dict[f'Index  {str(i)}'] = letter
print(dict)