sorting list of dictionaries in python

Question

I have this list:

L = [{'status': 1, 'country': 'France'}, {'status': 1, 'country': 'canada'}, {'status': 1, 'country': 'usa'}]

How to sort this list by country (or by status) elements, ASC/DESC.

If these dictionaries are all matching (ie. they all have a 'status' key and a 'country' key, and no other keys), consider using instances of a class created by collections.namedtuple instead. It will make a lot of your code, including sorting, much simpler to read and write. See this link for details — kampu
– kampu, Commented May 14, 2013 at 9:55

Ashwini Chaudhary · Accepted Answer · 2013-05-14 10:11:07Z

9

Use list.sort() to sort the list in-place or sorted to get a new list:

>>> L = [{'status': 1, 'country': 'France'}, {'status': 1, 'country': 'canada'}, {'status': 1, 'country': 'usa'}]
>>> L.sort(key= lambda x:x['country'])
>>> L
[{'status': 1, 'country': 'France'}, {'status': 1, 'country': 'canada'}, {'status': 1, 'country': 'usa'}]

You can pass an optional key word argument reverse = True to sort and sorted to sort in descending order.

As a upper-case alphabet is considered smaller than it's corresponding smaller-case version(due to their ASCII value), so you may have to use str.lower as well.

>>> L.sort(key= lambda x:x['country'].lower())
>>> L
[{'status': 1, 'country': 'canada'}, {'status': 1, 'country': 'France'}, {'status': 1, 'country': 'usa'}]

edited May 14, 2013 at 10:11

answered May 14, 2013 at 9:45

Ashwini Chaudhary

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jamylak · Accepted Answer · 2013-05-14 09:45:59Z

7

>>> from operator import itemgetter
>>> L = [{'status': 1, 'country': 'France'}, {'status': 1, 'country': 'canada'}, {'status': 1, 'country': 'usa'}]
>>> sorted(L, key=itemgetter('country'))
[{'status': 1, 'country': 'France'}, {'status': 1, 'country': 'canada'}, {'status': 1, 'country': 'usa'}]
>>> sorted(L, key=itemgetter('country'), reverse=True)
[{'status': 1, 'country': 'usa'}, {'status': 1, 'country': 'canada'}, {'status': 1, 'country': 'France'}]
>>> sorted(L, key=itemgetter('status'))
[{'status': 1, 'country': 'France'}, {'status': 1, 'country': 'canada'}, {'status': 1, 'country': 'usa'}]

answered May 14, 2013 at 9:45

jamylak

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3 Comments

Ashwini Chaudhary Over a year ago

I think we have to use str.lower here as dictionary wise canada must come before France.

jamylak Over a year ago

@AshwiniChaudhary if that's required then your answer should be accepted ;) I just followed what the question asked so I'll leave this here for other people visiting

Drwhite Over a year ago

@jamylak: AshwiniChaudhary has right, canada should come before france in the first short, actually the short is correct with str.lower

John La Rooy · Accepted Answer · 2013-05-14 10:11:19Z

2

Pulling the key out to a named function is a good idea for real code, because now you can write tests for it explicitly

def by_country(x):
    # For case insensitive ordering by country
    return x['country'].lower()

L.sort(key=by_country)

Of course you can adapt to use sorted(L, key=...) or reverse=True etc.

answered May 14, 2013 at 10:11

John La Rooy

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