CODEWITHSUNDEEP
pythonsorting
JasonBeni
JasonBeni

Reputation: 51

Python 3.6 -- sorted set of tuple

I need to sort my list in alphabetical order but ignore the case of my letters.

my_list = [('CAT', 1), ('Crop', 1), ('PoP', 4), ('anTs', 1), ('apple', 1), ('cAt', 1), ('can', 1), ('dog', 1), ('long', 1), ('poeT', 1), ('toe', 2)]

I would like to receive :

my_list_sorted = [('anTs', 1), ('apple', 1), ('can', 1), ('CAT', 1), ('cAt', 1), ('Crop', 1), ('dog', 1), ('long', 1), ('poeT', 1), ('PoP', 4), ('toe', 2)]

I tried to do :

sorted(set(my_list),key=str.lower())

But I received this error : 

TypeError: descriptor 'lower' of 'str' object needs an argument

Someone can help me please ?

Thank you, Jason

Upvotes: 2

Views: 1234

Answers (3)

Maiki Bodhisattva
Maiki Bodhisattva

Reputation: 218

If you don't need to keep an original version of the list it's quicker to sort in place by calling a list method:

my_list.sort(key=lambda tup: tup[0].lower())

this method doesn't return anything because the original list object now becomes sorted.

Upvotes: 2

Christian Callau
Christian Callau

Reputation: 1140

This should work:

sorted(my_list, key=lambda x: x[0].casefold())

The key parameter of the function sorted is the value that will be used for comparison in the sorting process. In this case the key we want is the string part of the tuples in lowercase, so that's what we return using a one-line lambda function.

In python str is a basic built in function that converts the value passed to a string, that's why the error.

See str.casefold() to understand why is prefered over str.lower()

Upvotes: 8

Stephen.W
Stephen.W

Reputation: 2127

You need to provide a key function:

sorted(my_list, key=lambda item : item[0].lower())

Upvotes: 0

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