1

I have some text like this.

    a x="true" i q={1}

That is, a line of text prefixed with 4 space characters. Then letters with spaces in between.

I'd like to transform the text to this

    a
x="true"
i
q={1}

(So keep the 4 spaces at the beginning and replace every next space with a line break.)

First, I tried s/ /\r/g, but then I realized that substitution also replaces the leading 4 spaces. However, I want to keep the 4 leading spaces.

Is there a way to use substitutions on parts of lines, instead of whole lines?

  • For that specific example this seems like an easy route: :s/a /a\r/g...or if the letters can be other than 'a'... `:s/\w /&\r/g' – B Layer Feb 4 at 4:52
  • Ooh, cool. Didn't know about \w and &! – 425nesp Feb 4 at 5:09
  • 1
    They're both pretty common, especially \w...do yourself a favor and read sections 4,5,6 in the pattern.txt reference (:h pattern). So many goodies. – B Layer Feb 4 at 5:12
2

Jump to the end for the simplest solution. I decided to thumb my nose at the KISS principle for a while before snapping out of it and posting the easy answer. :P

At least three ways come to mind to handle this.

The first uses the %v pattern "atom" (see :h /\%v). This is used to pin the search after/before a column number. For example this matches all text up to column 17: /^.*\%17v

We'll use it to skip over the opening four spaces and then substitute line feeds for every space afterwards:

s/\%>4v /\r/g

Another option is to use "look arounds". Specifically a "negative lookbehind" (:h /\@<!). We want to change all spaces to line feed except for those that are at the beginning of the line or separated from the beginning of the line by only whitespace.

s/\%(^\s*\)\@<!\s/\r/g

So that says to match any whitespace, \s, as long as there is no match preceding it (\@<!) made up of the beginning of the line plus optional whitespace characters (\%(^\s*\)). This actually will work for any length of opening indent.

Note that lookaround expressions don't actually consume any of the target string. This explains: /zero-width

These answers are specifically targeting your stated use case (four-space indent followed by the "sentence" to modify). But the subject asks a more general question ("apply substitution on parts of lines"). That question has been asked before and you can find some more generalized discussion there: How to run a substitute command on only a certain part of the line

In fact, to understand my third solution I recommend you read my answer there. This approach is probably overkill for the relatively simple problem we're looking at here (though I maintain this isn't as complicated as some folks make it out to be). For the sake of thoroughness I give you the sub-replace expression solution:

s/^\s\+\zs\(.*\)/\=substitute(submatch(0), ' ', '\r', 'g')/

Update: Silly me, I left out the easiest solution. This despite the fact that I posted something similar in a comment above (though that was targeting an earlier, easier version of the question)...

s/\(\S\+\) /\1\r/g

That is, match every sequence of non-whitespace characters (\S\+) followed by a space (). Surround all but the space in a capturing group (\(...\)). Each occurrence of this pattern in the line will be replaced with the captured characters (\1) and then a line feed (\r).

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