[Tutor] (no subject)

[Magnus Lycka]

At 10:48 2002-10-02 +0200, Annika Scheffler wrote:
>Hi Rob,
>
>Thanks for your help! In fact, it does work that way for me, too. But when
>writing a function like this:
>
>def sortList(someList):
>
>
>           return someList.sort()
>
>print sortList("What have we here".split())
>
>I get "none" as a result. Why's that?

Same reason as

a =3D someList.sort()
print a
None

The statement "return someList.sort()" means:
"Please python, would you return the return
value of the sort() operation on the someList
object to the calling function?"

.sort() changes the list it sorts. It doesn't
return any value. You need to return the sorted
list, not the non-existing return value of the
sort operation.

You need

def sortList(someList):
     someList.sort()
     return someList

Contrast this with

a =3D someList.index(5)

or

return someList.count('x')

In these cases we use list operations that actually
return a value we care about.

The reason why sort() and reverse() chages the list
instead of returning a sorted/reversed list, leaving
the original unchaged, is that the list might be very
big, and the return value will be equally big. This
might use up more memory than you care for.

If you need to keep the sorted list intact, you can
change your function to:

def sortList(someList):
     sorted =3D someList[:]
     sorted.sort()
     return sorted

Please note that the first version changes the
"original" list.

If you type

a =3D [1,2,3,4]
b =3D a

then both a and b will refer to the _same_ list.
Not two identical.

b.reverse()
print a
[4, 3, 2, 1]



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Magnus Lyck=E5, Thinkware AB
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