How can I specify key argument in sort() function however complex the argument be?

Question

I have a nested list having elements of format ['name',float_val].

Nlist = [['naveen',31.5],['murali',44.1],['sai',23.1],['satya',23.1]]

I want to sort the above nested list by float object. How should I specify the 'key' argument in sort() function to fulfill my need?

Answer 1

Use this

sort(x, key=lambda x: x[1])

Answer 2

According the original Python documentation:

Both list.sort() and sorted() have a key parameter to specify a function to be called on each list element prior to making comparisons.

Note that you can use both list.sort() or sorted():

Python lists have a built-in list.sort() method that modifies the list in-place. There is also a sorted() built-in function that builds a new sorted list from an iterable.

You can find some useful examples here.

# take second element for sort
def takeSecond(elem):
    return elem[1]

# your list
Nlist = [['naveen',31.5],['murali',44.1],['sai',23.1],['satya',23.1]]

# sort list with key
Nlist.sort(key=takeSecond)

# print list
print('Sorted list:', Nlist)

Out:

Sorted list: [['sai', 23.1], ['satya', 23.1], ['naveen', 31.5], ['murali', 44.1]]