Remove spaces (apostrophes) around quotes with regex in ruby

TLDR:

I assume the spaces were inserted by some automation software and there can only be single spaces around the words.

s = "l' avant ou l 'apres ou encore ' maintenant' ou bien 'ceci ' et ' encore de l ' huile ' d 'accord d' accord d ' accord Je n' en ai pas .... s ' entendre Je m'appelle Victor"
first_rx = /(?<=\b[b-df-hj-np-tv-z]) ' ?(?=\p{L})|(?<=\b[b-df-hj-np-tv-z]) ?' (?=\p{L})/i
# If you find it overmatches, replace [b-df-hj-np-tv-z] with [dlnsmtc], 
# i.e. first letters of word that are usually contracted
second_rx = /\b'\b\K|' *((?:\b'\b|[^'])+)(?<=\S) *'/
puts s.gsub(first_rx, "'")
      .gsub(second_rx) { $~[1] ? "'#{$~[1]}'" : "" }    

Output:

l'avant ou l'apres ou encore 'maintenant' ou bien 'ceci' et 'encore de l'huile' d'accord d'accord d'accord Je n'en ai pas .... s'entendre Je m'appelle Victor

Explanation

The problem is really complex. There are several words that can be abbreviated and used with an apostrophe in French, de, le/la, ne, se, me, te, ce to name a few, but these are all consonants. You may remove all spaces between a single, standalone consonant, apostrophe and the next word using

s.gsub(/(?<=\b[b-df-hj-np-tv-z]) ' ?(?=\p{L})|(?<=\b[b-df-hj-np-tv-z]) ?' (?=\p{L})/i, "'")

If you find it overmatches, replace [b-df-hj-np-tv-z] with [dlnsmtc], i.e. first letters of word that are usually contracted. See the regex demo.

Next step is to remove spaces after initial and before trailing apostrophes. This is tricky:

s.gsub(/\b'\b\K|' *((?:\b'\b|[^'])+)(?<=\S) *'/) { $~[1] ? "'#{$~[1]}'" : "" }

where \b'\b is meant to match all apsotrophes in between word chars, those that we fixed at the previous step. See this regex demo. As there is no (*SKIP)(*F) support in Onigmo regex, the regex is a bit simplified but the replacement is a conditional one: if Group 1 matched, replace with ' + Group 1 value ($1) + ', else, replace with an empty string (since \K reset the match, dropped all text from the match memory buffer).

NOTE: this approach can be extended to handle some specific cases like aujourd'hui, too.

To remove all whitespace around the ', use gsub!, applied in several steps for proper whitespace removal:

str = "l' avant ou l 'apres ou encore ' maintenant'"
str.gsub!(/\b'\s+\b/, "'").gsub!(/\b\s+'\b/, "'").gsub!(/\b(\s+')\s+\b/, '\1')
puts str
# l'avant ou l'apres ou encore 'maintenant'

Here,
\b : word boundary,
\s+ : 1 or more whitespace,
string.gsub!(regex, replacement_string) : replace in the string argument regex with specified replacement_string (during this, the original string is changed),
\1 : in the replacement string, this refers to the first group captured in parenthesis in the regex: (...).

I believe the following should work for you

s.gsub(/'.*?'/){ |e| "'#{e[1...-1].strip}'" }

The regex portion lazy matches all text within single quotes (including quotes). Then, for each match you substitute for the quoted text with leading and trailing whitespace removed, and return this text in quotes.

So if you have alot of data like this, all the answers I have seen are wrong, and will not work. No regex can guess wether the preceding word should have a space or not. Unless you came up with a list of words (or patterns) that either do or don't.

The problem is, sometimes a space should be left, sometimes not. The only way to script that is to find a pattern which describes when the space should be there, or when not. You must teach your regex French grammar. It may be possible lol. But probably not, or difficult.

If this is a one off, my advice is to create regexes for 2 or 3 different situations, and use something like vim, to go through the data, and select manually yes or no to substitute each occurrence.

There may be some cases you can run - eg remove all spaces to the right of quotes? - but unfortunately I don't think you can automate this process.