From: Josh Cheek <josh.cheek@...>
Date: 2010-10-10T10:39:32+09:00
Subject: Re: Better way to remove all occurrence of matches on a String

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On Sat, Oct 9, 2010 at 9:09 AM, Benoit Daloze <eregontp@gmail.com> wrote:

> Hi Rubyists !
>
> I remember many times I searched for a better way to "remove all
> occurrence of matches on a String".
>
> I usually do:
>  str.gsub! /\bword\b/, ''
>
> I remembered the String#[]= from tryruby.org, but that only replace
> the first occurrence.
> String#delete "Uses the same rules for building the set of characters
> as String#count", and does not work with Regexp.
> String#tr follow the same rules, and would need a second '' parameter.
>
> What feels a bit "awkward" to me in the first approach is to specify I
> want no replacement (so nothing, so I should not have to specify it).
>
> I think for a default second parameter to (g)sub(!) (without block),
> but this case is supposed to return an Enumerator (is it sometimes
> useful?).
> Also the "sub" part of the name clearly indicate it is substituting
> sth for a new thing (the new thing should not be empty).
>
> I would then go for #delete, adding a special case for RegExp
> arguments, but that is not consistent with current behavior with a
> String.
> However, String#delete seems very rarely used, and could get some
> interest back. (and "str".delete(/\bword\b/) do feel right to me)
>
> What do you think ?
>
> Benoit Daloze
>
>
I think gsub with an empty string is ugly too, but I am pretty confident
there is not a better way.

You need to be careful, though, delete and tr don't work the way you think
they work. They are based on characters not words. They don't work with
regex because they consider the input string to be a set of characters, and
that doesn't make sense with regex.

"abcde".delete("bd") # => "ace"
"abcde".tr("bd","12") # => "a1c2e"

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