你可以做:

for k in sorted(d.keys()):
    print d[k]

这将为您提供一个一致排序的键集(我猜是相对于builtin.hash())，如果排序对您有任何意义，您可以对其进行处理。这意味着，例如，即使扩展字典，数值类型也会保持一致排序。

例子

# lets create a simple dictionary
d = {1:1, 2:2, 3:3, 4:4, 10:10, 100:100}
print d.keys()
print sorted(d.keys())

# add some other stuff
d['peter'] = 'peter'
d['parker'] = 'parker'
print d.keys()
print sorted(d.keys())

# some more stuff, numeric of different type, this will "mess up" the keys set order
d[0.001] = 0.001
d[3.14] = 'pie'
d[2.71] = 'apple pie'
print d.keys()
print sorted(d.keys())

请注意，字典在打印时是排序的。但是键集本质上是一个hashmap!

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'

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'

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。

first_key, *rest_keys = mydict