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How to sort a list of dicts in Python

I'm using the MongoDB group function (it's similar to SQL's GROUP BY) to aggregate some results for my live-log-analyzer project. This function is pretty cool, but it does not sort the grouped data. Here is how to sort the data. (It is only one line of Python, but I have a hard time remembering how to do this.)

DATA is the output of the mongoDB group function. I want to sort this list of dicts by 'ups_ad'.

from pprint import pprint

DATA = [
    {u'avg': 2.9165000000000001,
     u'count': 10.0,
     u'total': 29.165000000000003,
     u'ups_ad': u'10.194.154.49:80'},
    {u'avg': 2.6931000000000003,
     u'count': 10.0,
     u'total': 26.931000000000001,
     u'ups_ad': u'10.194.155.176:80'},
    {u'avg': 1.9860909090909091,
     u'count': 11.0,
     u'total': 21.847000000000001,
     u'ups_ad': u'10.195.71.146:80'},
    {u'avg': 1.742818181818182,
     u'count': 11.0,
     u'total': 19.171000000000003,
     u'ups_ad': u'10.194.155.48:80'}
    ]

data_sorted = sorted(DATA, key=lambda item: item['ups_ad'])
pprint(data_sorted)

Results:

[{u'avg': 2.9165000000000001,
  u'count': 10.0,
  u'total': 29.165000000000003,
  u'ups_ad': u'10.194.154.49:80'},
 {u'avg': 2.6931000000000003,
  u'count': 10.0,
  u'total': 26.931000000000001,
  u'ups_ad': u'10.194.155.176:80'},
 {u'avg': 1.742818181818182,
  u'count': 11.0,
  u'total': 19.171000000000003,
  u'ups_ad': u'10.194.155.48:80'},
 {u'avg': 1.9860909090909091,
  u'count': 11.0,
  u'total': 21.847000000000001,
  u'ups_ad': u'10.195.71.146:80'}]
References:

Update 2010-04-28: Apparently I didn't use Google properly when I first wrote this post. Searching today produced several sources for doing exactly this.

1 Comment — feed icon Comments feed for this post


#1 Beni Cherniavsky-Paskin commented on 2010-08-31:

You can also do sorted(DATA, key=operator.itemgettr('ups_ad'))

See also operator.attrgetter() and operator.methodcaller().

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