Reputation: 49188
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?
Upvotes: 30
Views: 27832
Reputation: 21175
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
Upvotes: 5
Reputation: 134581
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+
Upvotes: 44
Reputation: 107754
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]
.
Upvotes: 3
Reputation: 10228
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!
Upvotes: 7