Break string into words using scan method + regexp, if word has `'` character, drop this character and everything after it

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sample_string = "let's could've they'll you're won't"
sample_string.scan(/\w+/)

Above gives me:

["let", "s", "could", "ve", "they", "ll", "you", "re", "won", "t"]

What I want:

["let", "could", "they", "you", "won"]

Been playing around in https://rubular.com/ and trying assertions like \w+(?<=') but no luck.

ruby parsing regex
2 Answers

Given:

> sample_string = "let's could've they'll you're won't"

You can do split and map:

> sample_string.split.map{|w| w.split(/'/)[0]}
=> ["let", "could", "they", "you", "won"]
user_2986070

You can use

sample_string.scan(/(?<![\w'])\w+/)
sample_string.scan(/\b(?<!')\w+/)

See the Rubular demo. The patterns (they are absolute synonyms) match

  • (?<![\w']) - a location in the string that is not immediately preceded with a word or ' char
  • \b(?<!') - a word boundary position which is not immediately preceded with a ' char
  • \w+ - one or more word chars.

See the Ruby demo:

sample_string = "let's could've they'll you're won't"
p sample_string.scan(/(?<![\w'])\w+/)
# => ["let", "could", "they", "you", "won"]
user_38329700
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