Find sublist of second largest elements in a list (Python 3)

Question

Given a list:

lis = [37.21, 37.21, 37.2, 44, 44, 44, 101, 101]

What is a simple way to extract the second-largest elements?

[44, 44, 44]

My attempt

lis = [37.21, 37.21, 37.2, 44, 44, 44, 101, 101]
def sublist_of_second_largest(lis):
    maxx=max(lis)
    n=lis.count(maxx)
    for i in range(n):
        lis.remove(maxx)
    maxx=max(lis)
    n=lis.count(maxx)
    out=[]
    for i in range(n):
        out.append(maxx)
    return out

print(sublist_of_second_largest(lis))

Always try to use snake_case when naming variables and functions. see [PEP8](https://www.python.org/dev/peps/pep-0008/) — Zain Arshad, Jun 13 '20 at 12:43

Zain Arshad · Answer 1 · 2020-06-13T12:46:48.953

3

Simple pythonic way:

lis = [37.21, 37.21, 37.2, 44, 44, 44, 101, 101]
num = sorted(set(lis), reverse=True)[1]
print([i for i in lis if i == num])

Ouput:

[44, 44, 44]

edited Jun 13 '20 at 12:46

answered Jun 13 '20 at 12:35

Zain Arshad

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score 1 · Accepted Answer · answered Jun 13 '20 at 19:26

While Zain's answer is correct and consice, due to the sorting it has an O(n log n) runtime. This might not matter to you, but if it does here is an O(n) implementation:

def sublist_of_second_largest(num_lst):
    num_set = set(num_lst)
    num_set.remove(max(num_set))
    snd_largest_num = max(num_set)
    return [val for val in num_lst if val == snd_largest_num]

print(sublist_of_second_largest([37.21, 37.21, 37.2, 44, 44, 44, 101, 101]))

Prints:

[44, 44, 44]

Find sublist of second largest elements in a list (Python 3)

My attempt

2 Answers2