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

Alright, let's make this last step a bit harder and do two things. Make a copy of favorite_

I made a copy of favorite _things and called the copy sorted_things[:] then appended .sort() tired of watching the movie, what am I doing wrong here.

Thanks Joe

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[-2:]
favorite_things = sorted_things[:].sort() # favorite_thingd = sorted_things[:] should make a copy of original list, right?
                                           # then append .sort() should sort the new list, right?
                                           # why does this code return Task 1 is no longer passing?

3 Answers

Chris Freeman
Chris Freeman
Treehouse Moderator 68,423 Points

There are two errors in your code. First it seems what you really wanted was to make a copy of favorite_things, then sort it and assign it to sorted_things. Something like:

sorted_things = favorite_things[:].sort()

The issue with this approach is that .sort() sorts a list in-place and returns None. This means that the new temporary list favorite_things[:] does get sorted in place, but it is the return value None that gets assigned to sorted_things.

To correct this, break the statement into two parts:

sorted_things = favorite_things[:]  # sorted_things get the copy
sorted_things.sort()  # sorted_things is sorted in-place

Post back if you need more help. Good Luck!!

Thank you sir. I had my head pointed the wrong way. But I bet I remember that kind hint. Thanks Joe

sorted_things = favorite_things[:] # sorted_things get the copy sorted_things.sort()