Get counts of exact matches for tuple in a nested list of tuples

Question

I have a list that contains multiple lists of tuples. I want to compare each tuple to every other tuple in the lists and return a count of the exact matches.

    foo = [[('a',),('a','b'),('a','b','c'),('b','c'),('c',)],[('a',),('a','b')]]

Expected results:

('a',)        2
('a','b')     2
('a','b','c') 1 
('b','c')     1
('c',)        1

Any help appreciated.

Answer 1

Using collections.Counter :

>>> import itertools
>>> import collections
>>>
>>> foo = [[('a',),('a','b'),('a','b','c'),('b','c'),('c',)],[('a',),('a','b')]]
>>> for x, cnt in collections.Counter(itertools.chain.from_iterable(foo)).most_common():
...     print(x, cnt)
...
('a',) 2
('a', 'b') 2
('a', 'b', 'c') 1
('c',) 1
('b', 'c') 1

Answer 2

You need to merge all the lists into one, and then you can use Collections.Counter , like this

foo = [[('a',),('a','b'),('a','b','c'),('b','c'),('c',)],[('a',),('a','b')]]
from collections import Counter
print Counter(item for items in foo for item in items)

Output

Counter({('a', 'b'): 2, ('a',): 2, ('b', 'c'): 1, ('c',): 1, ('a', 'b', 'c'): 1})

The same result can be achieved with normal dict as well

result = {}
for items in foo:
    for item in items:
        result[item] = result.get(item, 0) + 1
print result

Output

{('b', 'c'): 1, ('c',): 1, ('a', 'b'): 2, ('a',): 2, ('a', 'b', 'c'): 1}