CODEWITHSUNDEEP
pythonpython-3.xlistlist-comprehension
kartik
kartik

Reputation: 686

Python: Return top n lists from sublist having maximum length

I have a list of lists, where each sublist is a list of objects. The task is to return a list of top n sublists in the descending order of their length.

Illustration:

[[{},{},{},{}],[{},{},{}],[{},{}],[{}],[]]

For above, I want to return a list of top 3 sublists as per length, which is

[[{},{},{},{}],[{},{},{}],[{},{}]]

as the sublists have largest length 3,2,1 respectively from the list.

Upvotes: 2

Views: 168

Answers (2)

Siddharth
Siddharth

Reputation: 536

>>> a = [[{},{},{},{}],[{},{},{}],[{},{}],[{}],[]]
>>> sorted(a, key=len, reverse=True)
[[{}, {}, {}, {}], [{}, {}, {}], [{}, {}], [{}], []]

You can always choose select top k after splicing the result of sorted by sorted(a, key=len, reverse=True)[0:k].

Main observation to note is argument key, which can also accept any crazy lambda function too.

Upvotes: 2

lmiguelvargasf
lmiguelvargasf

Reputation: 69725

You have two possible options, let's say:

l = [[{},{},{},{}],[{},{},{}],[{},{}],[{}],[]]

You can produce a completely new list and take the desired 3 elements:

sorted(l, key=len, reverse=True)[:3]

Or, you can sort your original list and them take the first 3 elements:

l.sort(key=len, reverse=True)
l[:3]

In terms of performance, the second option seems faster:

In [1]: %timeit sorted(l, key=len, reverse=True)[:3]
1.9 µs ± 28.1 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

In [2]: %%timeit
    ...: l.sort(key=len, reverse=True)
    ...: l[:3]
    ...: 
1.22 µs ± 33.8 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Upvotes: 4

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