[EMAIL PROTECTED] wrote:
> Problem:
>
> You have a list of unknown length, such as this: list =
> [X,X,X,O,O,O,O]. You want to extract all and only the X's. You know
> the X's are all up front and you know that the item after the last X is
> an O, or that the list ends with an X. There are never O's between
> X's.
>
> I have been using something like this:
> _____________________
>
> while list[0] != O:
> storage.append(list[0])
> list.pop(0)
> if len(list) == 0:
> break
> _____________________
>
> But this seems ugly to me, and using "while" give me the heebies. Is
> there a better approach?
It seems few suggested solutions actually do the same thing as
your code shows. I think this does. (Untested)
try:
ix = aList.index(O)
storage, aList = aList[:ix], aList[ix:]
except ValueError:
# No O's
storage, aList = aList, []
The main advantage with my approach over yours is that
all iterations, all the looping, takes place in fast C
code, unless your list elements are Python classes that
implement __cmp__ etc. If testing 'aList[0] == O'
involves Python, things might slow down a bit...
.index() takes linear time, but it's fairly fast. As
Alex suggested, you might want to replace it with a
O(log n) solution for big lists. You might need rather
big lists for the Python based O(log n) to get faster
than the C based O(n) though. Measure.
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