Question

This question is inspired by this question. I’d like to get a dictionary from a list of dictionaries that should contain all key/value pairs from all dictionaries that are either only contained once, or where all dictionaries agree on the associated value. Example (taken from the aforementioned posting):

dicts = [dict(a=3, b=89, d=2), dict(a=3, b=89, c=99), dict(a=3, b=42, c=33)]
print dict_itersection(dicts)

should yield

{'a': 3, 'd': 2}

My current implementation looks like this:

import collections

def dict_intersection(dicts):
        c=collections.defaultdict(set)
        for d in dicts:
                for a, b in d.iteritems():
                        c[a].add(b)
        return {a: next(iter(b)) for a, b in c.iteritems() if len(b) == 1}

So my question: Can this be done more elegantly?

Sidequestion: can next(iter(b)) be done better without modification of the underlying dictionary (i.e. not b.pop())?

Answer 1

All solutions so far assume that all dictionary values are hashable. Since the code won’t get slower and only little more complex without this assumption, I’d drop it. Here’s a version that works for all values that support !=:

def dict_intersection(dicts):
    result = {}
    conflicting = set()
    for d in dicts:
        for k, v in d.iteritems():
            if k not in conflicting and result.setdefault(k, v) != v:
                del result[k]
                conflicting.add(k)
    return result

The set conflicting will only contain dictionary keys, which will always be hashable.