Question

我有几行来填充set。

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])`

我无法摆脱我能做得更好的感觉，但我想出的任何事情似乎都很棒。大多数实现首先构建list，这很烦人。 x中有很多y，大多数y都有z。

# 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＆＃34;最pythonic＆＃34;或者是单线之一更好？还有别的吗？

Answer 1

我会用

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

itervalues()（仅在Python 3中values()）没有构建列表，此解决方案不涉及字典查找（与{z for y in x for z in x[y]}相反）。

Answer 2

我会使用itertools模块，特别是chain类。

>>> 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 3

您可以使用composeExtensions：

dict.items()

输出：

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)

嵌套字典的键集

3 个答案: