Colin Holgate wrote:
>
> >One method that nobody mentioned in any of the replies to your
> >question is extracting the words from the string rather than
> >deleting the spaces:
>
> I thought of that too, but didn't post it because it would also get
> rid of returns. I'm pretty sure that he wants to get rid of returns
> as well, but he didn't ask for that, so I let it go.
Use the itemdelimiter and you can leave the returns:
This one is fastest by 40 to 50% in all cases I tried, I think that's
largely because it only counts the items in the text once:
on removespaces1 theText
the itemdelimiter = " "
newtext = ""
wordTotal = theText.item.count
repeat with x = 1 to wordTotal
newText = newText & theText.item[x]
end repeat
return newText
end
For short strings this is faster than the recursion but slower on large
chunks of text:
on removespaces2 theText
the itemdelimiter = " "
newtext = ""
repeat with x = 1 to theText.item.count
newText = newText & theText.item[x]
end repeat
return newText
end
The first of these recursive methods is a hair faster (~4%) than the
second, but both do too much item counting.
on killSpaces inStr
the itemdelimiter = " "
if inStr.item.count < 2 then return inStr.word[1]
return inStr.item[1] & killSpaces(inStr.item[2..the maxInteger])
end
on removespacesrecursively theText
the itemdelimiter = " "
if theText.item.count > 1 then
theText = theText.item[1] & removespacesrecursively
(theText.item[2..the maxinteger])
end if
return theText
end
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