CODEWITHSUNDEEP
pythonlistsortingdictionary
Eren Süleymanoğlu
Eren Süleymanoğlu

Reputation: 1254

How to sort a list of objects with python?

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?

Upvotes: 3

Views: 9936

Answers (5)

Preet Kukreti
Preet Kukreti

Reputation: 8607

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).

Upvotes: 0

Jon Clements
Jon Clements

Reputation: 142126

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')

Upvotes: 1

wladyslaw.czyzewski
wladyslaw.czyzewski

Reputation: 770

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}]

Upvotes: 3

Igor Chubin
Igor Chubin

Reputation: 64563

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

Upvotes: 0

Sven Marnach
Sven Marnach

Reputation: 601489

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

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

Upvotes: 12

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