Python set intersection question

Question

I have three sets:

s0 = [set([16,9,2,10]), set([16,14,22,15]), set([14,7])]   # true, 16 and 14
s1 = [set([16,9,2,10]), set([16,14,22,15]), set([7,8])]    # false

I want a function that will return True if every set in the list intersects with at least one other set in the list. Is there a built-in for this or a simple list comprehension?

Answer 1

It's a little verbose but I think it's a pretty efficient solution. It takes advantage of the fact that when two sets intersect, we can mark them both as connected. It does this by keeping a list of flags as long as the list of sets. when set i and set j intersect, it sets the flag for both of them. It then loops over the list of sets and only tries to find a intersection for sets that haven't already been intersected. After reading the comments, I think this is what @Victor was talking about.

s0 = [set([16,9,2,10]), set([16,14,22,15]), set([14,7])]   # true, 16 and 14
s1 = [set([16,9,2,10]), set([16,14,22,15]), set([7,8])]    # false


def connected(sets):
    L = len(sets)

    if not L: return True
    if L == 1: return False

    passed = [False] * L
    i = 0
    while True:
        while passed[i]: 
            i += 1
            if i == L: 
                return True

        for j, s in enumerate(sets):
            if j == i: continue
            if sets[i] & s: 
                passed[i] = passed[j] = True
                break
        else:
            return False


print connected(s0)
print connected(s1)

I decided that an empty list of sets is connected (If you produce an element of the list, I can produce an element that it intersects ;). A list with only one element is dis-connected trivially. It's one line to change in either case if you disagree.

Answer 2

all(any(a & b for a in s if a is not b) for b in s)

Answer 3

This strategy isn't likely to be as efficient as @Victor's suggestion, but might be more efficient than jchl's answer due to increased use of set arithmetic (union).

s0 = [set([16,9,2,10]), set([16,14,22,15]), set([14,7])]
s1 = [set([16,9,2,10]), set([16,14,22,15]), set([7,8])]

def freeze(list_of_sets):
    """Transform a list of sets into a frozenset of frozensets."""
    return frozenset(frozenset(set_) for set_ in list_of_sets)

def all_sets_have_relatives(set_of_sets):
    """Check if all sets have another set that they intersect with.

    >>> all_sets_have_relatives(s0)   # true, 16 and 14
    True
    >>> all_sets_have_relatives(s1)   # false
    False
    """
    set_of_sets = freeze(set_of_sets)
    def has_relative(set_):
        return set_ & frozenset.union(*(set_of_sets - set((set_,))))
    return all(has_relative(set) for set in set_of_sets)

Answer 4

This may give better performance depending on the distribution of the sets.

def all_intersect(s):
   count = 0
   for x, a in enumerate(s):
      for y, b in enumerate(s):
         if a & b and x!=y:
            count += 1
            break
   return count == len(s)

Answer 5

Here's a more efficient (if much more complicated) solution, that performs a linear number of intersections and a number of unions of order O( n*log(n) ), where n is the length of s:

def f(s):
    import math
    j = int(math.log(len(s) - 1, 2)) + 1
    unions = [set()] * (j + 1)
    for i, a in enumerate(s):
        unions[:j] = [set.union(set(), *s[i+2**k:i+2**(k+1)]) for k in range(j)]
        if not (a & set.union(*unions)):
            return False
        j = int(math.log(i ^ (i + 1), 2))
        unions[j] = set.union(a, *unions[:j])
    return True

Note that this solution only works on Python >= 2.6.

Answer 6

Here's a very simple solution that's very efficient for large inputs:

def g(s):
    import collections
    count = collections.defaultdict(int)
    for a in s:
        for x in a:
            count[x] += 1
    return all(any(count[x] > 1 for x in a) for a in s)

Answer 7

As usual I'd like to give the inevitable itertools solution ;-)

from itertools import combinations, groupby
from operator import itemgetter


def any_intersects( sets ):
    # we are doing stuff with combinations of sets
    combined = combinations(sets,2) 
    # group these combinations by their first set
    grouped = (g for k,g in groupby( combined, key=itemgetter(0)))
    # are any intersections in each group
    intersected = (any((a&b) for a,b in group) for group in grouped)
    return all( intersected )


s0 = [set([16,9,2,10]), set([16,14,22,15]), set([14,7])]
s1 = [set([16,9,2,10]), set([16,14,22,15]), set([7,8])] 
print any_intersects( s0 ) # True
print any_intersects( s1 ) # False

This is really lazy and will only do the intersections that are required. It can also be a very confusing and unreadable oneliner ;-)

Answer 8

To answer your question, no, there isn't a built-in or simple list comprehension that does what you want. Here's another itertools based solution that is very efficient -- surprisingly about twice as fast as @THC4k's itertools answer using groupby() in timing tests using your sample input. It could probably be optimized a bit further, but is very readable as presented. Like @AaronMcSmooth, I arbitrarily decided what to return when there are no or is only one set in the input list.

from itertools import combinations

def all_intersect(sets):
    N = len(sets)
    if not N: return True
    if N == 1: return False

    intersected = [False] * N
    for i,j in combinations(xrange(N), 2):
        if not intersected[i] or not intersected[j]:
            if sets[i] & sets[j]:
                intersected[i] = intersected[j] = True
    return all(intersected)