How to sort a list of objects with python?

Question

I have a list of objects look like below:

[{'id': 17L,
  'price': 0,
  'parent_count': 2},
 {'id': 39L,
  'price': 0,
  'parent_count': 1},
 {'id': 26L,
  'price': 2.0,
  'parent_count': 4},
 {'id': 25L,
  'price': 2.0,
  'parent_count': 3}]

I want to sort the objects by 'parent_count' in order to look like this:

 [{'id': 39L,
   'price': 0,
   'parent_count': 1},
  {'id': 17L,
   'price': 0,
   'parent_count': 2},
  {'id': 25L,
   'price': 2.0,
   'parent_count': 3},
  {'id': 26L,
   'price': 2.0,
   'parent_count': 4}]

Does anyone know a function?

Is 'parent_count' an optional key? The first object above has 'parent_say', not 'parent_count'? — Rob Cowie
– Rob Cowie, Commented Jul 27, 2012 at 10:22
i have edited the question. im sorry i wrote it wrong. there is no 'parent_say' key. — Eren Süleymanoğlu
– Eren Süleymanoğlu, Commented Jul 27, 2012 at 10:55

Sven Marnach · Accepted Answer · 2012-07-27 10:18:57Z

12

Use operator.itemgetter("parent_count") as key parameter to list.sort():

from operator import itemgetter
my_list.sort(key=itemgetter("parent_count"))

answered Jul 27, 2012 at 10:18

Sven Marnach

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6 Comments

Rob Cowie Over a year ago

This will raise a KeyError on the object that has 'parent_say' in place of 'parent_count'. If 'parent_count' is optional itemgetter() will need to be replaced with a callable that will return a suitable default value instead of raising KeyError.

Sven Marnach Over a year ago

@RobCowie: I will assume for now this was a typo – let's wait what the OP says.

Jon Clements Over a year ago

@RobCowie It's a consistent typo if it is (being both in source and output data)

Eren Süleymanoğlu Over a year ago

sorry man i wrote it wrong. all of them are 'parent_count'. there is no 'parent_say'.

Sven Marnach Over a year ago

@ErenSüleymanoğlu: As do all mutating methods in Python. It sorts the list in place.

|

wladyslaw.czyzewski · Accepted Answer · 2012-07-27 10:28:22Z

3

Also, you can use this method:

a = [{'id': 17L, 'price': 0, 'parent_count': 2}, {'id': 18L, 'price': 3, 'parent_count': 1}, {'id': 39L, 'price': 1, 'parent_count': 4}]
sorted(a, key=lambda o: o['parent_count'])

Result:

[{'parent_count': 1, 'price': 3, 'id': 18L}, {'parent_count': 2, 'price': 0, 'id': 17L}, {'parent_count': 4, 'price': 1, 'id': 39L}]

answered Jul 27, 2012 at 10:28

wladyslaw.czyzewski

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Comments

Jon Clements · Accepted Answer · 2012-07-27 11:14:28Z

1

Do you actually have "parent_say" and "parent_count"?

def get_parent(item):
    return item.get('parent_count', item['parent_say'])
    # return item.get('parent_count', item.get('parent_say')) if missing keys should just go to the front and not cause an exception

my_list.sort(key=get_parent)

or a bit more generic

def keygetter(obj, *keys, **kwargs):
    sentinel = object()
    default = kwargs.get('default', sentinel)
    for key in keys:
        value = obj.get(key, sentinel)
        if value is not sentinel:
            return value
    if default is not sentinel:
        return default
    raise KeyError('No matching key found and no default specified')

edited Jul 27, 2012 at 11:14

answered Jul 27, 2012 at 10:23

Jon Clements

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Comments

Igor Chubin · Accepted Answer · 2012-07-27 10:21:27Z

0

my_list.sort(key=lambda x:x["parent_count"])

answered Jul 27, 2012 at 10:21

Igor Chubin

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Comments

Eran · Accepted Answer · 2014-01-24 21:43:04Z

0

You can also do:

my_list.sort(key=lambda x: x.get('parent_count'))

which doesn't require operator.itemgetter and doesn't cause an error if the key doesn't exist (those that don't have the key get put at the start).

edited Jan 24, 2014 at 21:43

Eran

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answered Jul 27, 2012 at 10:22

Preet Kukreti

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1 Comment

Preet Kukreti Over a year ago

@ErenSüleymanoğlu it is an in-place sort. if you want a sorted copy, use the sorted builtin with the same args (but with the list my_list as the first arg)

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