从字典中返回一组具有共同值的键对

Question

How can I write a function, that would take a dictionary and return me a set that would consist of pairs of keys that have at least one common value?

Example:

I have the following dictionary:

dict = {
'C': {'123'}, 
'A': {'123', '456'}, 
'D': {'123'}, 
'B': {'789', '456'}, 
'E': {'789'}}

MyFunction(dict) should return me:

{("A", "B"), ("A", "C"), ("A", "D"), ("B", "E"), ("C", "D")}

Answer 1

Using itertools.combinations :

from itertools import combinations

d = {
    'C': {'123'}, 
    'A': {'123', '456'}, 
    'D': {'123'}, 
    'B': {'789', '456'}, 
    'E': {'789'}
}

def MyFunction(d):
    out = set()
    for i, j in combinations(d, 2):
        if d[j].intersection(d[i]) and (i, j) not in out and (j, i) not in out:
            out.add((i, j))
    return set(tuple(sorted(i)) for i in out)

print(MyFunction(d))
print(MyFunction(d) == {("A", "B"), ("A", "C"), ("A", "D"), ("B", "E"), ("C", "D")})

Output is:

{('A', 'D'), ('A', 'B'), ('B', 'E'), ('A', 'C'), ('C', 'D')}
True

If you consider ('A', 'C') and ('C', 'A') same, you can replace

return set(tuple(sorted(i)) for i in out)

with just

return out

Answer 2

defaultdict + combinations

For a brute force solution, you can invert your dictionary of sets, then use a set comprehension:

from collections import defaultdict
from itertools import combinations

d = {'C': {'123'}, 'A': {'123', '456'}, 
     'D': {'123'}, 'B': {'789', '456'}, 
     'E': {'789'}}

dd = defaultdict(set)

for k, v in d.items():
    for w in v:
        dd[w].add(k)

res = {frozenset(i) for v in dd.values() if len(v) >= 2 for i in combinations(v, 2)}

print(res)

{frozenset({'A', 'D'}), frozenset({'C', 'D'}),
 frozenset({'B', 'E'}), frozenset({'B', 'A'}),
 frozenset({'C', 'A'})}

As you can see the items in res are frozenset objects, ie they aren't depending on sorting within tuples. frozenset is required instead of set since set is not hashable.

Answer 3

A more efficient one-pass solution would be to use a seen dict that keeps track of a list of keys that have "seen" a given value so far:

pairs = set()
seen = {}
for key, values in d.items():
    for value in values:
        if value in seen:
            for seen_key in seen[value]:
                pairs.add(frozenset((key, seen_key)))
        seen.setdefault(value, []).append(key)

pairs would become:

{frozenset({'D', 'A'}), frozenset({'B', 'E'}), frozenset({'B', 'A'}), frozenset({'C', 'D'}), frozenset({'C', 'A'})}

You can then easily transform it to a set of lexicographically sorted tuples if you want:

{tuple(sorted(p)) for p in pairs}

which returns:

{('A', 'C'), ('B', 'E'), ('C', 'D'), ('A', 'D'), ('A', 'B')}