[Tutor] creating the equivalent of string.strip()

[Ricardo Aráoz]

Ian Witham wrote:
> 
> On 10/3/07, *Alan Gauld* <alan.gauld at btinternet.com
> <mailto:alan.gauld at btinternet.com>> wrote:
> 
> 
>     "Ricardo Aráoz" <ricaraoz at gmail.com <mailto:ricaraoz at gmail.com>> wrote
> 
>     >sorry, forgot a piece of the code :
>     >
>     >s = list(s)
>     >while s[0].isspace() :
>     >     while s[-1].isspace() :
>     >         del s[-1]
>     >     del s[0]
>     > s = ''.join(s)
> 
>     It still won't work. Strings are immutable, you can't use del
>     to delete a character. Try it.
> 
> 
> Alan, he does convert to a list first and then rejoin to a string
> afterwards, so using del is not the problem.
> 
> However, the script is still flawed.
> The loop to delete trailing whitespace is nested within the loop to
> delete the leading whitespace. The trailing whitespace loop only needs
> to be run once, but it is currently being processed once for each
> leading space and not at all if there are no leading spaces!
> For instance, "  Myriad Harbor  " strips OK, but "Myriad Harbor  " does not.
> 
> The solution of course is to keep the two loops separate:
> 
> s = list(s)
> while s[0].isspace() :
>     del s[0]
> while s[-1].isspace() :
>     del s[-1]
> s = ''.join(s)
> 

You are right Ian. It's an extra "s[-1].isspace()"  (and a jump) for
each leading whitespace, so though it works yours is still better. Don't
know why didn't see it :(, perhaps because I took code from the previous
post and did a quick fix.  :( (and blushing in shame)