Python-将单键字典的列表转换为单个字典 [英] Python - Convert list of single key dictionaries into a single dictionary

问题描述

我有一个单键词典列表.例如:

I have a list of single-key dictionaries. For example:

lst = [
    {'1': 'A'},
    {'2': 'B'},
    {'3': 'C'}
]

我想简单地将其转换为普通词典:

I'd like to simply convert this into a normal dictionary:

dictionary = {
    '1': 'A',
    '2': 'B',
    '3': 'C'
}

最简洁/最有效的方法是什么?

What's the most concise/efficient way to do this?

推荐答案

您可以使用reduce:

You can use reduce:

reduce(lambda r, d: r.update(d) or r, lst, {})

演示:

>>> lst = [
...     {'1': 'A'},
...     {'2': 'B'},
...     {'3': 'C'}
... ]
>>> reduce(lambda r, d: r.update(d) or r, lst, {})
{'1': 'A', '3': 'C', '2': 'B'}

或者您可以链接项目调用(Python 2):

or you could chain the items calls (Python 2):

from itertools import chain, imap
from operator import methodcaller

dict(chain.from_iterable(imap(methodcaller('iteritems'), lst)))

Python 3版本:

Python 3 version:

from itertools import chain
from operator import methodcaller

dict(chain.from_iterable(map(methodcaller('items'), lst)))

演示:

>>> from itertools import chain, imap
>>> from operator import methodcaller
>>> 
>>> dict(chain.from_iterable(map(methodcaller('iteritems'), lst)))
{'1': 'A', '3': 'C', '2': 'B'}

或使用字典理解:

{k: v for d in lst for k, v in d.iteritems()}

演示:

>>> {k: v for d in lst for k, v in d.iteritems()}
{'1': 'A', '3': 'C', '2': 'B'}

在这三种方法中，对于简单的3字典输入，对dict的理解最快:

Of the three, for the simple 3-dictionary input, the dict comprehension is fastest:

>>> import timeit
>>> def d_reduce(lst):
...     reduce(lambda r, d: r.update(d) or r, lst, {})
... 
>>> def d_chain(lst):
...     dict(chain.from_iterable(imap(methodcaller('iteritems'), lst)))
... 
>>> def d_comp(lst):
...     {k: v for d in lst for k, v in d.iteritems()}
... 
>>> timeit.timeit('f(lst)', 'from __main__ import lst, d_reduce as f')
2.4552760124206543
>>> timeit.timeit('f(lst)', 'from __main__ import lst, d_chain as f')
3.9764280319213867
>>> timeit.timeit('f(lst)', 'from __main__ import lst, d_comp as f')
1.8335261344909668

当增加输入列表中的项目数为1000时，chain方法就会赶上:

When you increase the number of items in the inputlist to 1000, then the chain method catches up:

>>> import string, random
>>> lst = [{random.choice(string.printable): random.randrange(100)} for _ in range(1000)]
>>> timeit.timeit('f(lst)', 'from __main__ import lst, d_reduce as f', number=10000)
5.420135974884033
>>> timeit.timeit('f(lst)', 'from __main__ import lst, d_chain as f', number=10000)
3.464245080947876
>>> timeit.timeit('f(lst)', 'from __main__ import lst, d_comp as f', number=10000)
3.877490997314453

从现在开始，进一步增加输入列表似乎无关紧要； chain()方法快了一个很小的百分比，但从未获得明显的优势.

Increasing the input list further doesn't appear to matter from here on out; the chain() approach is a small percentage faster but never gains a clear advantage.

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