Pairwise Set Intersection in Python

Question

If I have a variable number of sets (let\'s call the number n), which have at most m elements each, what\'s the most efficient way to calculate the pairwise in

Answer 1

this ought to do what you want

import random as RND
import string
import itertools as IT

mock some data

fnx = lambda: set(RND.sample(string.ascii_uppercase, 7))
S = [fnx() for c in range(5)]

generate an index list of the sets in S so the sets can be referenced more concisely below

idx = range(len(S))

get all possible unique pairs of the items in S; however, since set intersection is commutative, we want the combinations rather than permutations

pairs = IT.combinations(idx, 2)

write a function perform the set intersection

nt = lambda a, b: S[a].intersection(S[b])

fold this function over the pairs & key the result from each function call to its arguments

res = dict([ (t, nt(*t)) for t in pairs ])

the result below, formatted per the first option recited in the OP, is a dictionary in which the values are the set intersections of two sequences; each values keyed to a tuple comprised of the two indices of those sequences

this solution, is really just two lines of code: (i) calculate the permutations; (ii) then apply some function over each permutation, storing the returned value in a structured container (key-value) container

the memory footprint of this solution is minimal, but you can do even better by returning a generator expression in the last step, ie

res = ( (t, nt(*t)) for t in pairs )

notice that with this approach, neither the sequence of pairs nor the corresponding intersections have been written out in memory--ie, both pairs and res are iterators.