Why (dictionary.keys()).sort() is not working in python?

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.

I'm not sure about the solution but aren't dictionary keys unordered? — fauxCoder
– fauxCoder, Commented Jan 31, 2010 at 5:17
@Shraptnel: you are correct. The OP is placing the keys into a list and sorting that list. — mechanical_meat
– mechanical_meat, Commented Jan 31, 2010 at 5:30

Robert Munteanu · Accepted Answer · 2010-02-07 21:29:23Z

6

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.

edited Feb 7, 2010 at 21:29

Robert Munteanu

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answered Jan 31, 2010 at 5:16

Tyler

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Comments

CTT · Accepted Answer · 2010-01-31 05:18:09Z

6

sorted(toto.keys())

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

answered Jan 31, 2010 at 5:18

CTT

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2 Comments

Chris Lutz Over a year ago

Alternatively, use sorted(toto.iterkeys()) in case the dict is large to prevent the creation and destruction of the unsorted keys list.

jathanism Over a year ago

In Python 2.6 and 3.x you can just use sorted(toto) to get the same result. In fact, the use of .keys() is deprecated.

Stefano Borini · Accepted Answer · 2010-01-31 05:18:54Z

1

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

answered Jan 31, 2010 at 5:18

Stefano Borini

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4 Comments

nicolas Over a year ago

Is there any way to retain a similar postfix notation ? Like for len, one can use alist.__len__() instead of len(alist)

Stefano Borini Over a year ago

if you are calling l.__len__() it means that you are doing it wrong.

nicolas Over a year ago

Ok thanks for the advice. I have a hard time not using postfix notation though like alist.map().filter().sort().map().len() it puts the operation close to its arguments, so its really handy, instead of operation1(operation2(operation3(arg, arg3), arg2),arg1)

Stefano Borini Over a year ago

what you want to do is called method chaining. it's a pattern which is not implemented by any standard python type.

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