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。

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.

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'

这个怎么样。这里还没有提到。

Py 2和Py 3

a = {"a":2,"b":3}
a[list(a)[0]] # the first element is here
>>> 2

在Python 3中，非破坏性和迭代性:

next(iter(mydict.values()))

在Python 2中，非破坏性和迭代性:

mydict.itervalues().next()

如果你想让它在python2和python3中都能工作，你可以使用six包:

six.next(six.itervalues(mydict))

虽然在这一点上，它是相当神秘的，我更喜欢你的代码。

如果你想删除任何项目，请执行以下操作:

key, value = mydict.popitem()

请注意，“first”在这里可能不是一个合适的术语，因为dict在Python < 3.6中不是有序类型。Python 3.6+字典是有序的。

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