How can I sort a list of dictionaries by a value in the dictionary?

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In Python how do I sort a list of dictionaries by values of the dictionary?

I'm writing a Python 3.2 app and I have a list of dictionaries containing the following:

teamlist = [{ "name":"Bears", "wins":10, "losses":3, "rating":75.00 },
            { "name":"Chargers", "wins":4, "losses":8, "rating":46.55 },
            { "name":"Dolphins", "wins":3, "losses":9, "rating":41.75 },
            { "name":"Patriots", "wins":9, "losses":3, "rating": 71.48 }]

I want the list to be sorted by the values in the rating key. How do I accomplish this?

I had a little bit of fun answering a similar question a while back ( http://stackoverflow.com/a/11732149/748858 ). I discussed sorting fairly in-depth so I'll leave a link here with the hopes that it will be helpful. — mgilson, Aug 15 '12 at 13:59

score 6 · Accepted Answer · answered Aug 15 '12 at 13:55

6

Use operator.itemgetter as the key:

sorted(teamlist, key=operator.itemgetter('rating'))

answered Aug 15 '12 at 13:55

ecatmur

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Do you know if there is any advantage to using `operator.itemgetter` over the `lambda` key that I use in my answer? – Brendan Wood Aug 15 '12 at 14:03
Thanks, this did it. Much easier than the old way I tried before using lambda and maps. – user338413 Aug 15 '12 at 14:04
@BrendanWood -- Some people don't like `lambda`. Also, `operator.itemgetter` is *slightly* faster -- But, probably not enough to make it worth worrying about in common situations. – mgilson Aug 15 '12 at 14:08
@mgilson -- Thanks, I always like learning faster ways of doing things. – Brendan Wood Aug 15 '12 at 14:11
@BrendanWood they're equivalent, although `itemgetter` has the advantage when you want to sort by multiple keys (you can write `itemgetter(key1, key2, ...)`. I assume it's mainly there because Guido doesn't like lambdas. – ecatmur Aug 15 '12 at 14:11

score 4 · Answer 2 · answered Aug 15 '12 at 13:55

4

You can use the sorted function with a sorting key referring to whichever field you wish.

teamlist_sorted = sorted(teamlist, key=lambda x: x['rating'])

answered Aug 15 '12 at 13:55

Brendan Wood

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I had used this in a previous app I did but I kept getting a syntax error this time. Don't know if that was because I'm using 3.2 for this app. – user338413 Aug 15 '12 at 14:05
@user338413: this is perfectly valid in 3.2. You must've done something slightly different. – DSM Aug 15 '12 at 14:09
Yup, I just tried it myself in 3.2 and it works fine for me. – Brendan Wood Aug 15 '12 at 14:09

score 1 · Answer 3 · answered Aug 15 '12 at 14:00

1

from operator import itemgetter
newlist = sorted(team_list, key=itemgetter('rating'))

answered Aug 15 '12 at 14:00

NIlesh Sharma

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How can I sort a list of dictionaries by a value in the dictionary?

3 Answers3