Using vim, I know that I can replace all occurrences of a string across multiple lines by specifying a count or a range, as documented in :h subst.

I have a markdown table which contains C function prototypes, as well as English prose. I would like to replace, for example, the word 'float', but only the first three occurrences.

If I were writing a program to do this, many languages have a .replace method on strings that will accept a howmany parameter that limits the number of times the replacement is made.

Can I do something like that with vim? Is there any way to say, for example, "replace float with long double at most 3 times"?


3 Answers 3


Taking a straight programmatic approach, we can simply implement this as a for loop:

:for x in range(3) | :%s/float/long double/ | endfor

Update: @PeterRincker offers this user command that you can wrap around the above for easy invocation:

command! -count=1 -nargs=+ -complete=command Times for _ in range(<count>) | execute '<args>' | endfor.

Then for the above example we'd just need...

:3Times %s/float/long double/ 
  • 2
    In case this is a common thing: command! -count=1 -nargs=+ -complete=command Times for _ in range(<count>) | execute '<args>' | endfor. Now they can do :3Times s/foo/bar/ May 6, 2021 at 13:18
  • Good stuff @PeterRincker. I added it to my answer. Cheers.
    – B Layer
    May 7, 2021 at 4:26
  • 1
    This is nicely pragmatic. And the extra Times command is just whipped cream on top of the cake. If I could give another +1 I would.
    – aghast
    May 7, 2021 at 17:21
  • @aghast LOL. I'm glad you're enjoying the dessert we served. ;)
    – B Layer
    May 7, 2021 at 18:38

Because I love macros, here's a simple macro based-solution:

  1. Record a macro to replace one instance of float on the current line:

     qq:s/float/long double<CR>q
                           Press Return here
  2. (If necessary) Undo the change you just made by hitting u

  3. Play the macro three times on each line in your file:

     :%norm! 3@q
  • I like this approach. It fits well with "normal" use and seems likely to be something one could stumble upon while floundering around looking for a better way. I had never thought of :%norm as a valid option before now, which means I learned something from this. Thanks!
    – aghast
    May 7, 2021 at 17:23

We can also prepare the repeated text with

:let list = repeat(['long double'], 3)

and then replace with:

:s/float/\=empty(l) ? submatch(0) : remove(list,0)/g

Alas, I'm not inspired at this time for solving the resetting of the list for each new line with a :%s/.... Well we could work with repeat([repeat(['long double'], 3)], line('$')) and then remove from list[line('.')]. It's a bit overkill, but quite possibly much more faster than :for based solutions with vimscript < 9.

Note: It can certainly be built into a command like (in other similar Q/A I've provided multiple examples of such encapsulation of :substitute (SO, ...)

:SubstituteFirst/3/float/long double
  • Perhaps use a :global to reset it? :g/float/ let list = repeat(['long double'], 3) | s//\=empty(l) ? submatch(0) : remove(list,0)/g
    – filbranden
    May 6, 2021 at 5:05
  • 1
    In a 5900 line text file a for loop substitution with a pattern that has 20K matches completes with no discernible delay on the two systems I tried it on.
    – B Layer
    May 6, 2021 at 6:54

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