I have a dictionary:
{'three': 3, 'two': 2, 'one': 1}
and I need to make some operations with tuple of keys from this dictionary, so I'm trying to use this code:
list(D.keys()) # now I need to sort it
list(D.keys()).sort() # why such construction returns None? (Use print() to track it)
I'm using VS 2013 and IntelliSense prompts me sort method when I print list(D.keys()).
So what I'm missing?
p.s Such construction works well:
L = list(D.keys())
L.sort()
Thank You!
The .sort() method modifies the list in place and returns None. To use it on a list returned by a method you need to save the list in a variable.
L = list(D.keys())
L.sort()
print L
If you want to eliminate the temporary variable, use sorted:
print sorted(D.keys())
list.sort() is an in-place sort; you can use sorted to return a new one.
sorted(D)
(It also works on any iterable, so you don’t have to use list(D) [which would be equivalent to list(D.keys())].)
If the question is how to fix your code, John's and minitech's answers both cover it.
If the question is "why does list.sort return None?", then Guido explained his reasoning to the Python-Dev mailing list, sort() return value.
In summary, it is to prevent you writing a series of modifications of an object x as x.sort().compress().spiflicate()
You might not agree with Guido's preferred style, but AFAIK he honestly represented his reasons there, and has said similar things elsewhere. So that's the reason Python is how it is, there's no other motive.
.sort() # why such construction returns None?because read the documentation?