How do I iterate over a Python dictionary, ordered by values?

Question

I've got a dictionary like:

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

I'd like to iterate over it by value, not by key. In other words:

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

What's the most straightforward way?

Answer 1

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

for these of you that hate lambda :-)

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

However operator version requires CPython 2.5+

Answer 2

For non-Python 3 programs, you'll want to use iteritems to get the performance boost of generators, which yield values one at a time instead of returning all of them at once.

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

For even larger dictionaries, we can go a step further and have the key function be in C instead of Python as it is right now with the lambda.

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

Hooray!

Answer 3

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

Answer 4

The items method gives you a list of (key,value) tuples, which can be sorted using sorted and a custom sort key:

Python 2.5.1 (r251:54863, Jan 13 2009, 10:26:13) 

>>> a={ 'a': 6, 'b': 1, 'c': 2 }
>>> sorted(a.items(), key=lambda (key,value): value)
[('b', 1), ('c', 2), ('a', 6)]

In Python 3, the lambda expression will have to be changed to lambda x: x[1].