Re: A question on technique
Re: A question on technique
- Subject: Re: A question on technique
- From: Deivy Petrescu <email@hidden>
- Date: Sun, 15 Mar 2015 19:14:31 -0400
> On Mar 15, 2015, at 17:33 , Gil Dawson <email@hidden> wrote:
>
> Hi--
>
> I wish to translate a sequence of codes into corresponding mnemonics.
>
> Two same-length properties encode the correspondence.
>
> property Codes : {"9420", "94ae", "94d0", "942f", "942C", "9470"}
> property Mnemonics : {"RCL", "ENM", "ReW", "EOC", "EDM", "RfW"}
>
> Eventually I expect the two lists to be much larger.
>
> The following code works well enough...
>
> on MnemonicOf(aCode)
> set Found to false
> repeat with k from 1 to the length of Codes
> if item k of Codes is equal to aCode then
> set ReturnResult to item k of Mnemonics
> set Found to true
> exit repeat
> end if
> end repeat
> if not Found...
>
> ...but is there another, perhaps faster, way to do this?
>
> I think of constructs like...
>
> if aCode is in Codes then...
>
> ...but I don't see how to index the Mnemonics list.
>
> I think of maybe using records, but I don't see quite how.
>
> Any suggestions?
>
> --Gil
>
>
I don’t know the OS nor the idea of the whole script.
So two suggestions:
1. AppleScript
<script>
property Codes : {"9420", "94ae", "94d0", "942f", "942C", "9470"}
property Mnemonics : {"RCL", "ENM", "ReW", "EOC", "EDM", "RfW"}
tid(space)
set TCodes to Codes as string
set TMnemonics to Mnemonics as string
set k to round (offset of "9470" in TCodes) / 5 rounding up
set j to round (offset of "ENM" in TMnemonics) / 4 rounding up
return {k, j}
—>{6,2}
</script>
The script above assumes all the items are of the same length (4 in Codes and 3 in Mnemonics) you can make minor changes to accommodate the variation.
2. Javascript
<script>
Codes = ["9420", "94ae", "94d0", "942f", "942C", "9470"]
Mnemonics = ["RCL", "ENM", "ReW", "EOC", "EDM", "RfW"]
Mnemonics.indexOf("EOC")
</script>
—>3
Deivy Petrescu
email@hidden
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