查找列表中第二大元素的子列表（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))

Answer 1

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]

Answer 2

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]