如果你不想使用random.choice()，你可以尝试这样做:

>>> list(myDictionary)[i]
'VENEZUELA'
>>> myDictionary = {'VENEZUELA':'CARACAS', 'IRAN' : 'TEHRAN'}
>>> import random
>>> i = random.randint(0, len(myDictionary) - 1)
>>> myDictionary[list(myDictionary)[i]]
'TEHRAN'
>>> list(myDictionary)[i]
'IRAN'

因为最初的帖子想要一对:

import random
d = {'VENEZUELA':'CARACAS', 'CANADA':'TORONTO'}
country, capital = random.choice(list(d.items()))

(python 3风格)

下面是一个字典类的Python代码，它可以在O(1)时间内返回随机键。(为了可读性，我在代码中包含了myypy类型):

from typing import TypeVar, Generic, Dict, List
import random

K = TypeVar('K')
V = TypeVar('V')
class IndexableDict(Generic[K, V]):
    def __init__(self) -> None:
        self.keys: List[K] = []
        self.vals: List[V] = []
        self.dict: Dict[K, int] = {}

    def __getitem__(self, key: K) -> V:
        return self.vals[self.dict[key]]

    def __setitem__(self, key: K, val: V) -> None:
        if key in self.dict:
            index = self.dict[key]
            self.vals[index] = val
        else:
            self.dict[key] = len(self.keys)
            self.keys.append(key)
            self.vals.append(val)

    def __contains__(self, key: K) -> bool:
        return key in self.dict

    def __len__(self) -> int:
        return len(self.keys)

    def random_key(self) -> K:
        return self.keys[random.randrange(len(self.keys))]

我需要遍历字典中的键的范围，而不是每次都对它排序，并找到了Sorted Containers库。我发现这个库可以通过索引随机访问字典项，直观地解决了这个问题，而不用每次都遍历整个字典:

>>> import sortedcontainers
>>> import random
>>> d = sortedcontainers.SortedDict({1: 'a', 2: 'b', 3: 'c'})
>>> random.choice(d.items())
(1, 'a')
>>> random.sample(d.keys(), k=2)
[1, 3]

我通过寻找一个相当类似的解决方案找到了这篇文章。如果要从字典中选取多个元素，可以使用:

idx_picks = np.random.choice(len(d), num_of_picks, replace=False) #(Don't pick the same element twice)
result = dict ()
c_keys = [d.keys()] #not so efficient - unfortunately .keys() returns a non-indexable object because dicts are unordered
for i in idx_picks:
    result[c_keys[i]] = d[i]

从字典集dict_data中选择50个随机键值:

sample = random.sample(set(dict_data.keys()), 50)