为什么（dictionary.keys（））。sort（）在python中不起作用？

Question

I'm new to Python and can't understand why a thing like this does not work. I can't find the issue raised elsewhere either.

toto = {'a':1, 'c':2 , 'b':3}
toto.keys().sort()           #does not work (yields none)
(toto.keys()).sort()         #does not work (yields none)
eval('toto.keys()').sort()   #does not work (yields none)

Yet if I inspect the type I see that I invoke sort() on a list, so what is the problem..

toto.keys().__class__     # yields <type 'list'>

The only way I have this to work is by adding some temporary variable, which is ugly

temp = toto.keys()
temp.sort()

What am I missing here, there must be a nicer way to do it.

Answer 1

sort() sorts the list in place . It returns None to prevent you from thinking that it's leaving the original list alone and returning a sorted copy of it.

Answer 2

sorted(toto.keys())

Should do what you want. The sort method you're using sorts in place and returns None.

Answer 3

sort() method sort in place, returning none . You have to use sorted(toto.keys()) which returns a new iterable, sorted.