比较一下所有提到的方法的速度:

更新于2020.07.13(谢谢@user3780389): 仅用于bigdict中的键。

 IPython 5.5.0 -- An enhanced Interactive Python.
Python 2.7.18 (default, Aug  8 2019, 00:00:00) 
[GCC 7.3.1 20180303 (Red Hat 7.3.1-5)] on linux2
import numpy.random as nprnd
  ...: keys = nprnd.randint(100000, size=10000)
  ...: bigdict = dict([(_, nprnd.rand()) for _ in range(100000)])
  ...: 
  ...: %timeit {key:bigdict[key] for key in keys}
  ...: %timeit dict((key, bigdict[key]) for key in keys)
  ...: %timeit dict(map(lambda k: (k, bigdict[k]), keys))
  ...: %timeit {key:bigdict[key] for key in set(keys) & set(bigdict.keys())}
  ...: %timeit dict(filter(lambda i:i[0] in keys, bigdict.items()))
  ...: %timeit {key:value for key, value in bigdict.items() if key in keys}
100 loops, best of 3: 2.36 ms per loop
100 loops, best of 3: 2.87 ms per loop
100 loops, best of 3: 3.65 ms per loop
100 loops, best of 3: 7.14 ms per loop
1 loop, best of 3: 577 ms per loop
1 loop, best of 3: 563 ms per loop

正如预期的那样:字典推导式是最好的选择。

还有一个问题(我更喜欢Mark Longair的答案)

di = {'a':1,'b':2,'c':3}
req = ['a','c','w']
dict([i for i in di.iteritems() if i[0] in di and i[0] in req])

interesting_keys = ('l', 'm', 'n')
subdict = {x: bigdict[x] for x in interesting_keys if x in bigdict}

比较一下所有提到的方法的速度:

更新于2020.07.13(谢谢@user3780389): 仅用于bigdict中的键。

 IPython 5.5.0 -- An enhanced Interactive Python.
Python 2.7.18 (default, Aug  8 2019, 00:00:00) 
[GCC 7.3.1 20180303 (Red Hat 7.3.1-5)] on linux2
import numpy.random as nprnd
  ...: keys = nprnd.randint(100000, size=10000)
  ...: bigdict = dict([(_, nprnd.rand()) for _ in range(100000)])
  ...: 
  ...: %timeit {key:bigdict[key] for key in keys}
  ...: %timeit dict((key, bigdict[key]) for key in keys)
  ...: %timeit dict(map(lambda k: (k, bigdict[k]), keys))
  ...: %timeit {key:bigdict[key] for key in set(keys) & set(bigdict.keys())}
  ...: %timeit dict(filter(lambda i:i[0] in keys, bigdict.items()))
  ...: %timeit {key:value for key, value in bigdict.items() if key in keys}
100 loops, best of 3: 2.36 ms per loop
100 loops, best of 3: 2.87 ms per loop
100 loops, best of 3: 3.65 ms per loop
100 loops, best of 3: 7.14 ms per loop
1 loop, best of 3: 577 ms per loop
1 loop, best of 3: 563 ms per loop

正如预期的那样:字典推导式是最好的选择。

使用地图(halfdanrump的答案)对我来说是最好的，尽管还没有计时……

但是如果你使用一个字典，如果你有一个big_dict:

一定要确保你遍历了要求。这是至关重要的，并且会影响算法的运行时间(大O, theta，你能想到的) 把它写得足够通用，以避免在没有键的情况下出现错误。

例如:

big_dict = {'a':1,'b':2,'c':3,................................................}
req = ['a','c','w']

{k:big_dict.get(k,None) for k in req )
# or 
{k:big_dict[k] for k in req if k in big_dict)

请注意，在相反的情况下，req很大，但my_dict很小，您应该通过my_dict进行循环。

一般来说，我们在做一个交集问题的复杂度是O(min(len(dict)) min(len(req)))Python自己的intersection实现考虑了两个集合的大小，所以它看起来是最优的。而且，作为c语言的核心库的一部分，可能比大多数未优化的python语句要快。 因此，我考虑的解决方案是:

dict = {'a':1,'b':2,'c':3,................................................}
req = ['a','c','w',...................]

{k:dic[k] for k in set(req).intersection(dict.keys())}

它将关键操作移到python的c代码中，并适用于所有情况。

解决方案

from operator import itemgetter
from typing import List, Dict, Union


def subdict(d: Union[Dict, List], columns: List[str]) -> Union[Dict, List[Dict]]:
    """Return a dict or list of dicts with subset of 
    columns from the d argument.
    """
    getter = itemgetter(*columns)

    if isinstance(d, list):
        result = []
        for subset in map(getter, d):
            record = dict(zip(columns, subset))
            result.append(record)
        return result
    elif isinstance(d, dict):
        return dict(zip(columns, getter(d)))

    raise ValueError('Unsupported type for `d`')

使用实例

# pure dict

d = dict(a=1, b=2, c=3)
print(subdict(d, ['a', 'c']))

>>> In [5]: {'a': 1, 'c': 3}

# list of dicts

d = [
    dict(a=1, b=2, c=3),
    dict(a=2, b=4, c=6),
    dict(a=4, b=8, c=12),
]

print(subdict(d, ['a', 'c']))

>>> In [5]: [{'a': 1, 'c': 3}, {'a': 2, 'c': 6}, {'a': 4, 'c': 12}]