Question

I want to remove characters &%*@()!{} from a string. I tried this code:

keyword.gsub!(/[\&\%\*\@\(\)\!\{\}]/, '')`

but it failed.

Answer 1

Your regex isn’t right. It appears you don’t understand how [...] works in a regex.

You can use:

gsub(/[&%*@()!{}]+/, '')

For instance:

'foo&%*@()!{}bar'.gsub(/[&%*@()!{}]+/, '') # => "foobar"

An alternate way to do this, without using a regex is to use the tr method:

'foo&%*@()!{}bar'.tr('&%*@()!{}', '') # => "foobar"

The benefit of using tr, AKA “translate” is it doesn’t take a regex, and can do deletions, like here, or translations from one character to another. It’s also very fast.

require 'benchmark'

n = 1_000_000
Benchmark.bm() do |b|

  b.report { n.times { 'foo&%*@()!{}bar'.gsub(/[&%*@()!{}]+/, '') } }
  b.report { n.times { 'foo&%*@()!{}bar'.tr('&%*@()!{}', '')   } }

end

Returns on my machine running 1.9.3-p362:

   user     system      total        real
4.120000   0.010000   4.130000 (  4.125929)
1.280000   0.000000   1.280000 (  1.282932)