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?
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
对于Python 2和3:
import six
six.next(six.itervalues(d))
其他回答
对于Python 2和3:
import six
six.next(six.itervalues(d))
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。
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
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]