如何基于字典中的元素创建列表？

Question

I have this list of dictionaries:

L = [{'code': 'UE', 'country': 'France'}, {'code': 'AM', 'country': 'canada'}, {'code': 'AF', 'country': 'morocco'}]

Is there any way to retrieve a list of codes sorted like this: codes = ['AF', 'AM', 'UE'] ?

Answer 1

You can pass a generator to sorted :

>>> codes = sorted( d['code'] for d in L )
>>> codes
['AF', 'AM', 'UE']

Another option is to pass a list comprehension to sorted , which @MartijnPieters showed to be faster in this case.

codes = sorted([ d['code'] for d in L ])

Answer 2

Use a list comprehension selecting the key you need, passing the values to the sorted() function :

codes = sorted([d['code'] for d in L])

Demo:

>>> L = [{'code': 'UE', 'country': 'France'}, {'code': 'AM', 'country': 'canada'}, {'code': 'AF', 'country': 'morocco'}]
>>> sorted([d['code'] for d in L])
['AF', 'AM', 'UE']

Here, a list comprehension is faster than a generator expression:

>>> from timeit import timeit
>>> timeit("sorted(d['code'] for d in L)", 'from __main__ import L')
2.1713640689849854
>>> timeit("sorted([d['code'] for d in L])", 'from __main__ import L')
0.9132740497589111

sorted() requires a list to sort, so either you give it a list or it'll build a list from the iterable. Building a list from a generator expression is (a lot) less efficient than giving it a list in the first place.

Answer 3

Use sorted() function with list comprehension:

>>> L = [{'code': 'UE', 'country': 'France'}, {'code': 'AM', 'country': 'canada'}, {'code': 'AF', 'country': 'morocco'}]
>>>
>>> sorted([d['code'] for d in L])
['AF', 'AM', 'UE']