CODEWITHSUNDEEP
python
Jack Lee
Jack Lee

Reputation: 13

Sorting list dict by index and get top n elements

What I have is a list of dict:

x = [{25: 16}, {40: 32}, {11: 50}]

What i am trying is to sort it by value and get the top n elements:

where n = 2

what i did:

sorted_lst = x.sort(key=lambda d: d.values)
print(sorted_lst)
filtered = sorted_lst[:n]

but this did not work.

the output that I am trying to get is:

[{11: 50}, {40: 32}]

Upvotes: 0

Views: 86

Answers (3)

RoadRunner
RoadRunner

Reputation: 26325

You could use collections.Counter.most_common to get the top n elements:

>>> from collections import Counter
>>> lst = [{25: 16}, {40: 32}, {11: 50}]
>>> [{k: v} for k, v in Counter(dict(tuple(*d.items()) for d in lst)).most_common(2)]
[{11: 50}, {40: 32}]

Upvotes: 1

Christian Sloper
Christian Sloper

Reputation: 7510

Here is a slightly different solution to @mikksu's variant, using iter and next. Iter turns the view of values into an iterator and next gets the first value.

sorted(x, key = lambda x: next(iter(x.values())), reverse=True)[:2]

Upvotes: 2

Michael Szczesny
Michael Szczesny

Reputation: 5026

Without iterator in lambda

x = [{25: 16}, {40: 32}, {11: 50}]
sorted(x, key = lambda i: list(i.values()), reverse = True)[:2]
>>> [{11: 50}, {40: 32}]

Upvotes: 2

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