I am trying to delete all the spaces that aren't sandwiched by a letter. So this

 I am trying; but ; I can't find a way    ;

should be this

I am trying;but;I can't find a way;

I tried this but it doesn't work.

:%s/\([A-Za-z]\)\@<! *\([A-Za-z]\)\@!//g

2 Answers 2


"Aren't sandwiched by a letter" means does not have: a letter before it and a letter after it, when expressed through De Morgan's laws: does not have a letter before it OR does not have a letter after it. What your regular expression is matching is does not have a letter before it AND does not have a letter after it; so it would only match between something like # ; $ (which your example does not have). Maybe it becomes clearer when we condense the collection to /\a and remove the superfluous grouping:

/\a\@<! \+\a\@!/

I've also changed * to \+; though matching no sandwiched spaces and replacing them with nothing doesn't do harm, it's imprecise (and may be a bit more inefficient).

To arrive at the correct pattern, we need to turn the AND to OR, by introducing a branch with \| (:help /\bar):

/\a\@<! \+\| \+\a\@!/
  • 1
    Thank you for the detailed explanation. What if some of the letters are accented characters like Š or é? What should I use instead of \a?
    – arty
    Oct 31, 2019 at 12:55
  • 3
    That's unfortunately still a weak spot in Vim; all built-in classes and sets do not contain them; you have to include them yourself; see stackoverflow.com/questions/19385458/… Oct 31, 2019 at 13:13

You can make use of \zs and \ze.


:s/\v<pat1>|<pat2>|<pat3>|<pat4>//g --- delete all matches using "very magic" mode

^\s+ --- leading spaces

\s+$ --- trailing spaces

[^[:alpha:][:space:]]\zs\s+ --- matches anything except alpha or space followed by space(s), but selects space(s) only

\s+\ze[^[:alpha:][:space:]] --- matches space(s) followed by anything except alpha or space, but selects space(s) only

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