可以使用列表解析从dicts的词典中创建所有成员字典的列表吗？

Say I have a dict of dicts:

说我有一个dicts的词典：

foo = {
    'category': {
        'key1': 'bar',
        'key2': 'bar',
        'key3': 'bar',
    },
    'category2': {
        'key4': 'bar',
        'key5': 'bar',
    },
}

To get a single list of all keys in the member-dicts, I have a function as follows:

要获得member-dicts中所有键的单个列表，我有如下函数：

def _make_list():
    baz = list()
    for key,val in foo.items():
        baz += list(val.keys())
    return baz

The generated list looks like: ['key1', 'key2', 'key3', 'key4', 'key5', ].

生成的列表如下所示：['key1'，'key2'，'key3'，'key4'，'key5'，]。

This is simple enough, and it works, but I wonder: is there a way to accomplish this with a one-liner list comprehension? The keys of the member dicts will always be unique.

这很简单，并且它可以工作，但我想知道：有没有办法通过单行列表理解来实现这一点？成员dicts的键将始终是唯一的。

2 个解决方案

#1

Here's one way to do it:

这是一种方法：

>>> [k for d in foo.values() for k in d]
['key1', 'key2', 'key3', 'key4', 'key5']

#2

An approach would be using itertools.chain():

一种方法是使用itertools.chain（）：

import itertools

[k for k in itertools.chain(*(d.keys() for d in foo.values()))]

If what you want is just a one line of code, and not necessarily a list comprehension, you can also try (mentioned by @Duncan):

如果您想要的只是一行代码，而不一定是列表理解，您也可以尝试（由@Duncan提及）：

list(itertools.chain(*foo.values()))

Output:

输出：

>>> [k for k in itertools.chain(*(d.keys() for d in foo.values()))]
['key3', 'key2', 'key1', 'key5', 'key4']
>>>
>>> list(itertools.chain(*foo.values()))
['key3', 'key2', 'key1', 'key5', 'key4']

#1