How to sort a list of tuples, by the second tuple element?

Question

I have a list like this:

[('TIPE_ORG', 'Corea'), ('TIPE_ORG', 'United Kingdom'), ('TIPE_ORG', 'Russia'), ('TIPE_ORG', 'Germany'),('TIPE_PER', 'Pepe Martínez')]

I want it to be sorted by text length from largest to smallest of the second parameter Let it be like this:

[('TIPE_ORG', 'United Kingdom'),('TIPE_PER', 'Pepe Martínez'), ('TIPE_ORG', 'Germany'), ('TIPE_ORG', 'Russia'),('TIPE_ORG', 'Corea')]

I have tried to do this, but having two parameters does not order it for the second, but for the first TIPE_ORG:

x.sort(key=len, reverse=True)

Answer 1

this can be done by sorting inplace with sort method or sorting and return anew sorted list with sorted method. for the equality decition, we use lambda function. see below:

my_list=[('TIPE_ORG', 'Corea'), ('TIPE_ORG', 'United Kingdom'),('TIPE_PER', 'Pepe Martínez')]

my_new_list = sorted(my_list, key=lambda this_tup:len(this_tup[1]), reverse=True)

or

sort(my_list, key=lambda this_tup:len(this_tup[1]), reverse=True)

you can refer to this link they have good examples about lambda expressions and how to use it.

lambda

Answer 2

You can use sorted method with a lambda function as below.

lst=[('TIPE_ORG', 'Corea'), ('TIPE_ORG', 'United Kingdom'), ('TIPE_ORG', 'Russia'), ('TIPE_ORG', 'Germany'),('TIPE_PER', 'Pepe Martínez')]

sorted(lst,key=lambda key:len((key[1])),reverse=True)

Output will be

[('TIPE_ORG', 'United Kingdom'),
('TIPE_PER', 'Pepe Martínez'),
('TIPE_ORG', 'Germany'),
('TIPE_ORG', 'Russia'),
('TIPE_ORG', 'Corea')]