嵌套字典的键集

Question

我有几行代码来填充set。你知道吗

x = {1: {2: 4, 3: 6}, 5: {2:6, 10: 25, 14: 12}}

keys = set()
for y in x:
    for z in x[y]:
        keys.add(z)

# keys is now `set([2, 3, 10, 14])`

我无法摆脱这种感觉，我可以做得更好，但我提出的似乎没有什么是伟大的。大多数实现首先构建list，这很烦人。在y中有大量的x，并且大多数y具有相同的z。你知道吗

# Builds a huuuuge list for large dicts.
# Adapted from https://stackoverflow.com/a/953097/241211
keys = set(itertools.chain(*x.values()))

# Still builds that big list, and hard to read as well.
# I wrote this one on my own, but it's pretty terrible.
keys = set(sum([x[y].keys() for y in x], []))

# Is this what I want?
# Finally got the terms in order from https://stackoverflow.com/a/952952/241211
keys = {z for y in x for z in x[y]}

原始代码是“最具python风格”还是其中一个行程序更好？还有别的吗？你知道吗

Answer 1

我将使用itertools模块，特别是chain类。你知道吗

>>> x = {1: {2: 4, 3: 6}, 5: {2:6, 10: 25, 14: 12}}
>>> from itertools import chain
>>> set(chain.from_iterable(x.itervalues()))
set([2, 3, 10, 14])

Answer 2

您可以使用dict.items()：

x = {1: {2: 4, 3: 6}, 5: {2:6, 10: 25, 14: 12}}
final_x = set(i for b in [b.keys() for i, b in x.items()] for i in b)

输出：

set([2, 3, 10, 14])

Answer 3

我会用

{k for d in x.itervalues() for k in d}

itervalues()（在python3中只是values()）不构建列表，并且这个解决方案不涉及字典查找（与{z for y in x for z in x[y]}相反）。你知道吗