Re: Trimming down a list
Re: Trimming down a list
- Subject: Re: Trimming down a list
- From: Paul Berkowitz <email@hidden>
- Date: Wed, 10 Apr 2002 09:37:08 -0700
On 4/10/02 8:58 AM, "Steve Thompson" <email@hidden> wrote:
>
Now I'm really confused. Thank you for your example, Paul. I've created a
>
sub called RemoveDuplicates and I've passed my long list to it. The script
>
makes sense. The loop steps through each item of listRef to see if it's in
>
k. If it isn't it adds it to the end.
>
>
My sample file that I'm testing my script on contains 11,768 items. With
>
this script that should work as you've written it (ie, it makes sense to me)
>
I now get the first item in the list returned 11,768 times.
>
>
> if thisItem is not in k then set end of k to thisItem
>
>
Makes sense to me. What's going on?
>
Is it possible that you wrote
item 1
instead of
item i
in the line:
set thisItem to item i of RemoveDups's listRef
?
>
> if thisItem is not in k then set end of k to thisItem
>
>
I changed it to:
>
>
if thisItem is not in k then set k to k & thisItem
>
>
And it works. Is that a bug or does list not understand end of?
I've heard tell of limitations with very long lists, but i don't believe it
applies to 'set the end of' to.
Reduce your list to under 4500 items and try again. Does it work now? (There
is another limit around 4600, but it shouldn't apply to 'end of'.) Can you
find a point at which it breaks?
If it does, you would be better to break the list into sublists of about
4000 items each, do the 'end of' technique, and then concatenate (&) the
sublists together if you can do it in a way that doesn't give you
duplicates. I'll get back to this a little later. In the meantime, as I
already pointed out the first time, you should not do:
if thisItem is not in k then set k to k & thisItem
If you _have_ to concatenate, do it this way:
if thisItem is not in k then set k to k & {thisItem}
But that really should not be necessary.
--
Paul Berkowitz
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