对于Python 2和3:

import six

six.next(six.itervalues(d))

忽略字典排序的问题，这样可能会更好:

next(dict.itervalues())

这样就避免了项查找和生成不使用的键的列表。

Python3

next(iter(dict.values()))

另一种方法是在一行中做到这一点，同时保持字典的完整性:

arbitrary_value = mydict.setdefault(*mydict.popitem())

popitem() returns a tuple of (key, value) for the last item that was added into the dictionary and this pair is passed into setdefault as positional arguments. The setdefault tries to insert key into mydict with value value if it doesn't already exist, but does nothing if does exist; and then returns the value of that key to the caller. Because we already popped the (key, value) pair out of the dictionary, we insert it back into it via setdefault and then proceed to return value, which is what we want.

如果你只需要访问一个元素(碰巧是第一个元素，因为字典不保证顺序)，你可以在Python 2中简单地这样做:

my_dict.keys()[0]    # key of "first" element
my_dict.values()[0]  # value of "first" element
my_dict.items()[0]   # (key, value) tuple of "first" element

请注意(据我所知)Python不能保证连续两次调用这些方法中的任何一个都会返回相同顺序的list。Python3不支持这一点。

在Python 3中:

list(my_dict.keys())[0]    # key of "first" element
list(my_dict.values())[0]  # value of "first" element
list(my_dict.items())[0]   # (key, value) tuple of "first" element

在python3中，方法是:

dict.keys() 

返回一个类型为:dict_keys()的值，当获取dict的key的第一个成员时，我们将得到一个错误:

dict.keys()[0]
TypeError: 'dict_keys' object does not support indexing

最后，我将dict.keys()转换为list @1st，并通过列表拼接方法获得了第一个成员:

list(dict.keys())[0]

子类化字典是一种方法，但效率不高。这里如果你提供一个整数，它将返回d[list(d)[n]]，否则按预期访问字典:

class mydict(dict):
    def __getitem__(self, value):
        if isinstance(value, int):
            return self.get(list(self)[value])
        else:
            return self.get(value)

d = mydict({'a': 'hello', 'b': 'this', 'c': 'is', 'd': 'a',
            'e': 'test', 'f': 'dictionary', 'g': 'testing'})

d[0]    # 'hello'
d[1]    # 'this'
d['c']  # 'is'