如果你想保留大部分键，同时删除一些键，另一种方法是:

{k: bigdict[k] for k in bigdict.keys() if k not in ['l', 'm', 'n']}

可能:

subdict=dict([(x,bigdict[x]) for x in ['l', 'm', 'n']])

Python 3甚至支持以下内容:

subdict={a:bigdict[a] for a in ['l','m','n']}

注意你可以在字典中检查是否存在，如下所示:

subdict=dict([(x,bigdict[x]) for x in ['l', 'm', 'n'] if x in bigdict])

分别地。对于python 3

subdict={a:bigdict[a] for a in ['l','m','n'] if a in bigdict}

你也可以使用map(这是一个非常有用的函数):

sd = dict(map(lambda k:(k, l.get(k, None))， l)))

例子:

large_dictionary = {'a1':123, 'a2':45, 'a3':344}
list_of_keys = ['a1', 'a3']
small_dictionary = dict(map(lambda key: (key, large_dictionary.get(key, None)), list_of_keys))

PS:我借用了.get(键，None)从以前的答案:)

此答案使用与所选答案类似的字典推导，但不会对缺失项进行省略。

Python 2版本:

{k:v for k, v in bigDict.iteritems() if k in ('l', 'm', 'n')}

Python 3版本:

{k:v for k, v in bigDict.items() if k in ('l', 'm', 'n')}

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。

如果你想保留大部分键，同时删除一些键，另一种方法是:

{k: bigdict[k] for k in bigdict.keys() if k not in ['l', 'm', 'n']}