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Python Python Collections (2016, retired 2019) Slices Back and Forth

Posted by cleancut
cleancut
1,218 Points

Calling .sort() method after creating a copy of a list using [:] in Python

I know that I can break sorted_things=favorite_things[:].sort() into 2 steps to make it pass the test:

sorted_things = favorite_things[:] sorted_things.sort()

However, why does doing both steps at once using sorted_things=favorite_things[:].sort() result in an empty list?

Any help would be much appreciated!

slices.py 
favorite_things = ['raindrops on roses', 'whiskers on kittens', 'bright copper kettles',
                   'warm woolen mittens', 'bright paper packages tied up with string',
                   'cream colored ponies', 'crisp apple strudels']

slice1 = favorite_things[1:4]
slice2 = favorite_things[5:]
sorted_things=favorite_things[:].sort()

1 Answer

Alexander Davison
Alexander Davison
65,469 Points

This is because .sort() sorts a variable in place, and there's another function called sorted() that just sorts a list without changing the variable in place. For example; in this code both a and b will be the list [1, 2, 3, 4, 5].

mixed = [3, 4, 2, 5, 1]

a = sorted(mixed)  # sorted() doesn't change a variable; it just returns the sorted version of a list

b = mixed[:]
b.sort()   # .sort() changes a variable in place, and returns nothing

# Both a and b are the same list, and the mixed list didn't change

sorted_things = favorite_things[:].sort()  # Remember, .sort() returns nothing, so sorted_things will be None

:point_right: I think the challenge is expecting you to use .sort().

I hope you get it. ~Alex

cleancut
cleancut
1,218 Points

Ah, I see! Thanks for reminding me that .sort() returns nothing!

Alexander Davison
Alexander Davison
65,469 Points

No problem :)