python - Efficiently find overlap of a lists of tuples -

i have bunch of lists, each composed of tuples.

a = [(1,2,3),(4,5,7),(8,9,10),(5,6,2)] b = [(1,3,6),(4,2,8),(3,6,7),(5,2,8)] c = [(6,2,3),(1,7,2),(5,7,2),(7,2,7)] 

i need find group of tuples such first element of tuple appears in every list. (i know confusing) above example, overlap be:

overlap = [(1,2,3),(1,3,6),(1,7,2),(5,6,2),(5,2,8),(5,7,2)] 

this because tuple number '1' in first element of tuple appears in every list. same number '5'.

what best way this?

i have working code, feel there better way this.

big_list = [a,b,c] overlap = [] all_points = list(set([item item in big_list])) (f,s,t) in all_points:     in_all = true     lest in big_list:         present = false         (first, second, third) in lest:             if first == f:                 present = true         if not present:             in_all = false     if in_all:         overlap.append((f,s,t)) 

you can use set intersection this:

>>> itertools import chain >>> def get_first(seq):                                                return (x[0] x in seq) >>> common = set(get_first(a)).intersection(get_first(b), get_first(c)) 

now common contains:

>>> common set([1, 5]) 

now can loop on individual items a, b , c , choose tuples first item found in common:

>>> [x x in chain(a, b, c) if x[0] in common] [(1, 2, 3), (5, 6, 2), (1, 3, 6), (5, 2, 8), (1, 7, 2), (5, 7, 2)] 

sort first item:

>>> operator import itemgetter >>> sorted((x x in chain(a, b, c) if x[0] in common), key=itemgetter(0)) [(1, 2, 3), (1, 3, 6), (1, 7, 2), (5, 6, 2), (5, 2, 8), (5, 7, 2)]