How to find the second lowest lists into a nested list by their second value?

Here is a handy little function that incorporates heapq.nlargest:

Return a list with the n largest elements from the dataset

import heapq
num = heapq.nlargest(2, [key for item, key in nl])[-1]
print [item for item in nl if item[-1] == num] #[['Harsh', 20], ['Beria', 20]]

Your example can be simplified into this:

second_lowest = sorted(set(v[1] for v in nl))[1]
result = [v for v in nl if v[1] == second_lowest]
print(result) # [['Harsh', 20], ['Beria', 20]]

I use set of your values to find second, unique lowest value.
Than having this file, find elements in nl that match the found value.

I did it by finding the second lowest value using a set and then selecting elements from the list with the same value.

#ordering by value
nl.sort(key = lambda x: x[1])

values_set = set()
for value in nl:
    values_set.add(value[1])

values_list = list(values_set)
#ordering
values_list.sort()
#getting second lowest values
lowest_values = [lowest for lowest in nl if lowest[1] == values_list[1] ]

You can try below code. It works fine.

lst=[['Harry',37.21],['Berry',37.21],['Tina',37.2],['Akriti',41],['Harsh',39]]
names=[]
lowest = lst[0].__getitem__(1)
second_lowest=0
for l in lst:
    if l[1] < lowest:
        second_lowest = lowest
        lowest = l[1]
    elif l[1]<=second_lowest:
        second_lowest=l[1]
for l in lst:
    if l[1]==second_lowest:
        names.append(l[0])


print(lowest)
print(second_lowest)
print(names)

if __name__ == '__main__':
arr = []
for _ in range(int(input())):
    name = input()
    score = float(input())
    arr1 = [name, score]
    arr.append(arr1)
arr.sort(key=lambda x: x[1])
# print(arr)
# print(min(arr,key=lambda x:x[1]))
arr.remove(min(arr,key=lambda x:x[1]))
# print(arr)
minimum = min(arr,key=lambda x:x[1])
# print(minimum[1])
a=[]
minimum = minimum[1]
for i  in arr:
    if(i[1] == minimum):
        a.append(i[0])
a.sort()
for i in a:
    print(i)

I have come across similar kind of problem and after referring this page I got some ideas and able to solve it.

nl = [['Harsh', 20], ['Beria', 20], ['Varun', 19], ['Kakunami', 19], ['Vikas', 21]]

second = max(nl, key= lambda x: x[1])[1]
first = min(nl, key= lambda x: x[1])[1]
    
for i in range(len(nl)):
    if nl[i][1] <= second and nl[i][1] != first:
        second = nl[i][1]
    
for i in range(len(nl)):
    if second == nl[i][1]:
        print(nl[i], end=", ")

if __name__ == '__main__':

    n = []
    s = []
    for _ in range(int(input())):
        name = input()
        n.append(name)
        score = float(input())
        s.append(score)
        
data = [[x,y] for x,y in zip(n,s)]

min_marks = min([x[1] for x in data])

filtered_data = [d for d in data if d[1] != min_marks]

sec_min = min([x[1] for x in filtered_data])

students_with_sec_min_marks = sorted([d[0] for d in filtered_data if d[1] == sec_min])

for s in students_with_sec_min_marks:
    print(s)

if __name__ == '__main__':
    lst = []
    names = []
    for _ in range(int(input())):
        name = input()
        score = float(input())
        lst.append([name, score])
     
    lowest = max(lst, key=lambda x: x[1])[1]
    second_lowest=0
    for l in lst:
        if l[1] < lowest:
            second_lowest = lowest
            lowest = l[1]
        elif l[1] < second_lowest and l[1] != lowest:
            second_lowest = l[1]
    for l in lst:
        if l[1] == second_lowest:
            names.append(l[0])  
    
    names.sort()
    for nm in names:
        print(nm)

I used the below code and it worked good:

records = []

if __name__ == '__main__':
    for i in range(int(input())):
        name = input()
        score = float(input())
        records.append([name,score])
        records.sort(key=lambda x:x[1])
    minimum = min(records,key=lambda x:x[1])
    minimum = minimum[1]
    args = records.copy()
    
    for i in args:
        if(i[1] == minimum):
            records.remove(i)
        else:
            continue
    
    minimum = min(records,key=lambda x:x[1])
    minimum = minimum[1]
    a=[]
    for i  in records:
        if(i[1] == minimum):
            a.append(i[0])

    a.sort()
    for i in a:
        print(i)

This way, you will also be able to remove multiple same lowest scores if they exist. Running it without copy() and working on the original list causes the loop to miss the several same lowest scores.

students=[]
if __name__ == '__main__':
    for _ in range(int(input())):
        name = input()
        score = float(input())
        students.append([name,score])
    
            
students.sort(key=lambda x:x[1]  )  
students.remove(min(students,key=lambda x:x[1] ))
names=[]
score=students[0][1]
for student in students:
    if student[1]==score:
        names.append(student[0])
names.sort()
for each in names:
    print(each)