如果mydict不为空,则访问任意元素,如下:

mydict[mydict.keys()[0]]

还有什么更好的办法吗?


当前回答

为了得到钥匙

next(iter(mydict))

得到一个值

next(iter(mydict.values()))

两者兼得

next(iter(mydict.items())) # or next(iter(mydict.viewitems())) in python 2

前两个是Python 2和3。后两个在python3中是lazy,但在python2中不是。

其他回答

first_key, *rest_keys = mydict

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'

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.

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。

对于Python 2和3:

import six

six.next(six.itervalues(d))