How to check if string contains a part of an array in Ruby?

Question

I have an array of banned words in Ruby:

bw = ["test"]

I want to check @nick against it.

For example

@nick = 'justatest'

would match.

I have tried:

if @nick.include?(bw)
 ##
end

but that doesn't seem to be the right way to do it as it doesn't catch it.

if bw.any? { |word| @nick =~ /#{word}/} should work. Example — engineersmnky
– engineersmnky, Commented Apr 29, 2015 at 13:20

Gagan Gami · Accepted Answer · 2015-04-29 13:58:41Z

2

If your concern is to check only (as your question's title suggests) then:

> bw.any? { |word| @nick[word] }
=> true

Though it might be faster to convert the array of strings into a Regexp:

> bw = ["test", "just", "gagan"]
> g = /#{bw.join("|")}/
> g === @nick 
 => true

edited Apr 29, 2015 at 13:58

answered Apr 29, 2015 at 13:31

Gagan Gami

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4 Comments

michael_wu Over a year ago

@nick.include?(word) could be written shorter as @nick[word]. Not sure if this is really an improvement, though.

Gagan Gami Over a year ago

@undur_gongor : thanks for your point for shorter trick. I have updated my answer :)

michael_wu Over a year ago

I assume, the regexp version is only faster if you can build it once and query it often. I just tried and (at least for small lists) the regexp creation is far more expensive than iterating through the array. In addition, you may have to think about escaping special characters as in mudasobwa's answer.

D-side Over a year ago

@undur_gongor yes, regexes are built into a slightly optimized form that takes some more time for building, but results in faster checks.

jerhinesmith · Accepted Answer · 2015-05-21 20:41:11Z

2

@nick =~ Regexp.new(bw.map { |w| Regexp.escape(w) }.join('|'))

Here we join all strings in bw into one regular expression, escaping possible symbols that have special meaning in a regular expression and check the input against it.

Another way to achieve this functionality is to check for every blacklisted word separately:

bw.any? { |w| @nick[Regexp.new(Regexp.escape(w))] }

Hope it helps.

edited May 21, 2015 at 20:41

jerhinesmith

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answered Apr 29, 2015 at 13:29

Aleksei Matiushkin

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3 Comments

engineersmnky Over a year ago

this is a much better solution and other than Regexp.escape I just added it as a suggestion in my comment below.

michael_wu Over a year ago

I fail to see the benefit of a regexp here (in the second solution).

Aleksei Matiushkin Over a year ago

@undur_gongor Frankly, me too.

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