如何迭代按字符串排序的Python字典？

我有一本字典:

{ 'a': 6, 'b': 1, 'c': 2 }

我想按价值迭代它,而不是按键.换一种说法:

(b, 1)
(c, 2)
(a, 6)

什么是最直接的方式？

sorted(dictionary.items(), key=lambda x: x[1])

对于那些讨厌lambda的人:-)

import operator
sorted(dictionary.items(), key=operator.itemgetter(1))

但是operator版本需要CPython 2.5+

对于非Python 3程序,您将需要使用iteritems来提高生成器的性能,这样可以一次生成一个值,而不是一次返回所有生成器.

sorted(d.iteritems(), key=lambda x: x[1])

对于更大的字典,我们可以更进一步,将关键函数放在C而不是Python中,就像现在使用lambda一样.

import operator
sorted(d.iteritems(), key=operator.itemgetter(1))

万岁!

It can often be very handy to use namedtuple. For example, you have a dictionary of name and score and you want to sort on 'score':

import collections
Player = collections.namedtuple('Player', 'score name')
d = {'John':5, 'Alex':10, 'Richard': 7}

sorting with lowest score first:

worst = sorted(Player(v,k) for (k,v) in d.items())

sorting with highest score first:

best = sorted([Player(v,k) for (k,v) in d.items()], reverse=True)

The order of 'key' and 'value' in the listed tuples is (value, key), but now you can get the name and score of, let's say the second-best player (index=1) very Pythonically like this:

    player = best[1]
    player.name
        'Richard'
    player.score
         7