从词典列表中创建一个不相矛盾的词典

问题描述

此问题的灵感来自于。我想从一个字典列表中找到一本字典，它应该包含所有字典中只包含一次的所有键/值对，或者所有字典都同意关联值。示例（取自上述发布）：

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)

应该生成

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

我目前的实现如下所示：

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}

所以我的问题： >这可以更优雅吗？

So my question: Can this be done more elegantly?

Sidequestion：可以 next（iter（b））更好地完成，而不修改底层字典（即不是 b.pop（））？

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

推荐答案

到目前为止，所有的解决方案都假定所有的字典值是可以散列的。由于代码不会变慢，只有一点更复杂，没有这个假设，我会放弃它。这是一个适用于支持！= 的所有值的版本：

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

设置冲突将只包含字典键，这将始终是可以哈希的。

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

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