Re: finding the index of an item in a list?
Re: finding the index of an item in a list?
- Subject: Re: finding the index of an item in a list?
- From: Yuma Decaux <email@hidden>
- Date: Wed, 01 Feb 2012 19:44:13 +1300
Hi Stan,
Thanks for that part. It works fine though the values come with e power symbols. Guess i'll round them up or something.
Oh, and thanks to Zak as well for your suggestion. For some reason applescript told me access denied to 1 of itemlist. Weird, it;s perhaps some coercion issue?
Best regards,
Yuma
On 1/02/2012, at 5:15 PM, Stan Cleveland wrote:
> On Jan 31, 2012, at 7:20 PM, Yuma Decaux wrote:
>
>> Hi List pun intended
>>
>> Been questing at this for a few hours now, almost at the end of my script and i am stuck with finding the index number of an item in a list. I would like to placehold that and use it in an expression such as:
>>
>> Set list1 to get value of every cell of column 4 --that's in the numbers tell block)
>> Set list1 to {}
>> Repeat with i in list1
>> Set nextitem to index of item i +1 --this to increment the item at each pass of the repeat loop
>> Set differential to item i - item nextitem
>> Copy differential to the end of list2
>> End repeat
>>
>> Etc etc
>>
>> The issue i have here is with getting the index of i in the list list1. It just says "can't get index of 0.0054323 --which is the value of that item not its index
>
> Hi Yuma,
>
> It seems that you're trying to find differences between adjacent items, so the following may get you started:
> set list1 to {23, 453, 675, 123, 39, 54, 2345, 679, 4563, 8265}
> set list2 to {}
> set lastitem to 0 -- starting value
> repeat with i from 1 to (length of list1)
> set nextitem to item i of list1
> set differential to lastitem - nextitem
> set end of list2 to differential
> set lastitem to nextitem
> end repeat
> return list2
>
> HTH,
> Stan C.
>
>
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