[Tutor] Re: Sorting a dictionary in one line?

[Derrick 'dman' Hudson]

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On Sun, Jan 19, 2003 at 11:13:50PM +0100, Magnus Lycka wrote:
| At 12:56 2003-01-19 -0800, Terry Carroll wrote:
[...]
| >1) Why doesn't this work? If I understand this right, dict.keys() returns
| >a list, and lists have a sort() method, which returns a list, which is a
| >sequence; so why is this an iteration over a non-sequence?
|=20
| No. .sort() returns None!
|=20
| >>> a =3D [3,4,2,1]
| >>> print a.sort()
| None
| >>> print a
| [1, 2, 3, 4]
|=20
| Lists can obviously be very big. For that reason it's important
| that we are able to perform sorts without having to duplicate
| the list. Thus .sort() modifies the original list, not a copy of
| it. The .sort() method could still have had a "return self" in
| the end as a convenience, but it doesn't. The thing is that if
| we wrote "sortedList =3D myList.sort()" it would be very confusing
| to discover that "myList" had become sorted in the operation. For
| that reason .sort() returns None. You need to do:
|=20
| sortedList =3D myList; sortedList.sort()
|=20
| Now there is no ambiguity.

Just be aware that in the above two lines of code, "both" lists are
sorted because the two names are really references to the same mutable
object.  If you want to have a sorted and unsorted version of the
list, make a copy :

    sortedList =3D myList[:]
    sortedList.sort()

-D

--=20
No harm befalls the righteous,
but the wicked have their fill of trouble.
        Proverbs 12:21
=20
http://dman.ddts.net/~dman/

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