Removing characters

[jmatro]

How do I remove all letters (a-zA-Z) from a string?

For example, given "1abc2def14xyz9" I want to obtain "12149".

Given all the tools in REBOL, it should be easy, but I can't see it.  Hope I'm not missing something obvious.

Thanks.

[Guest_jmatro]

Found one way to do it!!

    string: "1abc2def14xyz9"

    remove-each v string [find "abcdefghijklmnopqrstuvwxyz" v]

But is there a simpler way?

Thanks.

[CarlRead]

I doubt you could find much shorter code to do it, but depending on how long your strings are, there may be more efficient methods.  You're using a FIND for every character in the string, which might make the process slow with long strings.

Alternative approaches could be a routine to loop through the string comparing the characters, (note you'd need to compare #"a" as well as #"A"), or to use PARSE.

If you need examples for those just ask.

[Vincent]

Not simpler, but faster:

"trim/with" removes all specified caracters from a string.

trim/with string "abcdefghijklmnopqrstuvwyzABCDEFGHIJKLMNOPQRSTUVWYZ"

[CarlRead]

Not simpler, but faster:

Doh!

Pity there's not more monitoring this forum - that answer should've arrived a lot quicker.