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()))

没有外部库，适用于Python 2.7和3.x:

>>> list(set({"a":1, "b": 2}.values()))[0]
1

对于任意键，省略.values()

>>> list(set({"a":1, "b": 2}))[0]
'a'

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()))

As others mentioned, there is no "first item", since dictionaries have no guaranteed order (they're implemented as hash tables). If you want, for example, the value corresponding to the smallest key, thedict[min(thedict)] will do that. If you care about the order in which the keys were inserted, i.e., by "first" you mean "inserted earliest", then in Python 3.1 you can use collections.OrderedDict, which is also in the forthcoming Python 2.7; for older versions of Python, download, install, and use the ordered dict backport (2.4 and later) which you can find here.

Python 3.7 现在字典是按插入顺序排列的。

在python3

list(dict.values())[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'