Python - sorted() vs sort()

Asked Jul 26 '19 at 17:57

Active Jul 26 '19 at 17:59

Viewed 56 times

    Python 3.6.7

    >>> a = [('a',1), ('b', 2), ('c',3), ('d',4)]
    >>> b = [('b',2), ('c', 3), ('a',1)]
    >>>
    >>>
    >>> a.sort() == b.sort()
    True

    >>> a = [('a',1), ('b', 2), ('c',3), ('d',4)]
    >>> b = [('b',2), ('c', 3), ('a',1)]
    >>>
    >>> sorted(a) == sorted(b)
    False
    >>>

I was expecting sort() to return false. What is the difference between sort and sorted causing this behaviour.

edited Jul 26 '19 at 17:59

asked Jul 26 '19 at 17:57

Anand

https://stackoverflow.com/a/22442440/11793921 I think this is the answer you are looking for. – VodkaBottle Jul 26 '19 at 17:59
3

`a.sort() == b.sort()` compares `None == None` – Andrej Kesely Jul 26 '19 at 18:00
1

`sort()` return None But `sorted()` return sorted list. – hmn Falahi Jul 27 '19 at 11:04

Python - sorted() vs sort()

0 Answers0