strip chars

[-= JB =-]

That is interesting and nice.

I have only run one test but it was mixed pretty good &
my result was 94 67 with the repeat for each being the
fastest method.

Thanks to both Mark & Richard for the valuable help.

-=>JB<=-



On Jul 25, 2008, at 7:06 AM, Richard Gaskin wrote:

> RegEx solutions are conveniently compact to write, but given the  
> complexity of the RegEx subsystem they tend to be slower than more  
> brute-force solutions.
>
> Being a benchmarking fan I couldn't resist testing this latest  
> case, extracting numerals from an alphanumeric string.
>
> Below is a test handler which calls two functions, the first using  
> RegEx and the second using "repeat for each".  To run just put a  
> button and three fields on a card, and put some alphanumeric test  
> into the first field. It takes the text from field 1 and runs it  
> through each of the two functions, putting the results into flds 2  
> and 3 and then displaying in the Message Box the execution times  
> and whether the results matched (useful to make sure the algos are  
> doing the same job).
>
> In my tests here on strings long and short the "repeat for each"  
> method seems roughly twice as fast as the RegEx one.
>
>
> on mouseup
>   put fld 1 into tSrc
>   put 1000 into n
>   --
>   -- ReplaceText:
>   --
>   put the millisecs into t
>   repeat n
>     get rplText(tSrc)
>   end repeat
>   put the millisecs - t into t1
>   put it into fld 2
>   --
>   -- Repeat for Each:
>   --
>   put the millisecs into t
>   repeat n
>     get rptText(tSrc)
>   end repeat
>   put the millisecs - t into t2
>   put it into fld 3
>   --
>   -- Results:
>   put t1 && t2 &cr& (fld 2 = fld 3)
> end mouseup
>
>
> function rplText pSrc
>   return replacetext( pSrc ,"[^0-9]","")
> end rplText
>
>
> function rptText pSrc
>   put empty into s
>   repeat for each char k in pSrc
>     if k is in "0123456789" then put k after s
>   end repeat
>   return s
> end rptText
>
>
> -- 
>  Richard Gaskin
>  Managing Editor, revJournal
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