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

Here is given a nested list:

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

Now I have to find the second lowest lists into the nested list by their second value. And append the second lowest lists into another list.

So the output should be:

```['Harsh', 20], ['Beria', 20]
```

I wrote the following code but it doesn’t work:

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

result=[]

temp=max(nl, key=lambda x: x[1])

largest, larger = temp[1], temp[1]
for num in nl:
if num[1] < largest:
largest, larger = num[1], largest
elif num[1] < larger:
larger = num[1]
result.append(larger)
print(result)
```

Get the `min` of the total elements, filter using that valid then get min of remaining and keep elements equal to min of remaining:

```from operator import itemgetter
# min of all elements
mn = min(nl, key=itemgetter(1))[1]

# remove elements equal to min
filtered = [x for x in nl if x[1] != mn]

# get min of remaining
mn_fil = min(filtered,key=itemgetter(1))[1]

# filter remaining
out = [x for x in filtered if x[1] == mn_fil]
print(out)

[['Harsh', 20], ['Beria', 20]]
```

```In [19]: nl = [['Prashant', 32], ['Pallavi', 36], ['Dheeraj', 39], ['Shivam', 40]]
In [20]: from operator import itemgetter
In [21]: mn = min(nl, key=itemgetter(1))[1]
In [22]: filtered = [x for x in nl if x[1] != mn]
In [23]: mn_fil = min(filtered,key=itemgetter(1))[1]
In [24]: out = [x for x in filtered if x[1] == mn_fil]
In [25]: out
Out[25]: [['Dheeraj', 36]]
```

Using a single for loop we remove all elements from the temp list if we find a lower element, if we find and equally lower one we append it:

```mn = min(nl, key=itemgetter(1))[1]
temp = []
best = float("inf")
for ele in nl:
if mn < ele[1] < best:
best = ele[1]
temp = []
out.append(ele)
elif ele[1] == best:
temp.append(ele)
print(temp)
```