Substitute the n-th occurrence of a word in vim

Question

I saw other questions dealing with the finding the n-th occurrence of a word/pattern, but I couldn't find how you would actually substitute the n-th occurrence of a pattern in vim. There's the obvious way of hard coding all the occurrences like

:s/.*\(word\).*\(word\).*\(word\).*/.*\1.*\2.*newWord.*/g 

Is there a better way of doing this?

Answer 1

For information,

s/\%(\(pattern\).\{-}\)\{41}\zs\1/2/

also works to replace 42th occurrence. However, I prefer the solution given by John Kugelman which is more simple -- even if it will not limit itself to the current line.

Answer 2

You can do this a little more simply by using multiple searches. The empty pattern in the :s/pattern/repl/ command means replace the most recent search result.

:/word//word//word/ s//newWord/
or
:/word//word/ s/word/newWord/

You could then repeat this multiple times by doing @:, or even 10@: to repeat the command 10 more times.

Alternatively, if I were doing this interactively I would do something like:

3/word
:s//newWord/r

That would find the third occurrence of word starting at the cursor and then perform a substitution.

Answer 3

Replace each N^th occurrence of PATTERN in a line with REPLACE.

:%s/\(\zsPATTERN.\{-}\)\{N}/REPLACE/

Answer 4

To replace the nth occurrence of PATTERN in a line in vim, in addtion to the above answer I just wanted to explain the pattern matching i.e how it is actually working for easy understanding.

So I will be discussing the \(.\{-}\zsPATTERN\)\{N} solution,

The example I will be using is replacing the second occurrence of more than 1 space in a sentence(string). According to the pattern match code->

1. According to the zs doc,

   \zs - Scroll the text horizontally to position the cursor at the start (left side) of the screen.

2. .\{-} 0 or more as few as possible (*)

   Here . is matching any character and {} the number of times. e.g ab{2,3}c here it will match where b comes either 2 or 3 times.

   In this case, we can also use .* which is 0 or many as many possible. According to vim non-greedy docs, "{-}" is the same as "*" but uses the shortest match first algorithm.

3. \{N} -> Matches n of the preceding atom

   /\<\d\{4}\> search for exactly 4 digits, same as /\<\d\d\d\d>

   **ignore these \<\> they are for exact searching, like search for fred -> \<fred\> will only search fred not alfred.

4. \( \) combining the whole pattern.

5. PATTERN here is your pattern you are matching -> \s\{1,} (\s - space and {1,} as explained just above, search for 1 or more space)

"abc subtring def"

    :%s/\(.\{-}\zs\s\{1,}\)\{2}/,/

OUTPUT -> "abc subtring,def"

# explanation: first space would be between abc and substring and second 
# occurence of the pattern would be between substring and def, hence that 
# will be replaced by the "," as specified in replace command above. 

Answer 5

This answers your actual question, but not your intent.

You asked about replacing the nth occurrence of a word (but seemed to mean "within a line"). Here's an answer for the question as asked, in case someone finds it like I did =)

For weird tasks (like needing to replace every 12th occurrence of "dog" with "parrot"), I like to use recursive recordings.

First blank the recording in @q

qqq

Now start a new recording in q

qq

Next, manually do the thing you want to do (using the example above, replace the 12th occurrence of "dog" with "parrot"):

/dog
nnnnnnnnnnn

delete "dog" and get into insert

diwi

type parrot

parrot

Now play your currently empty "@q" recording

@q

which does nothing.

Finally, stop recording:

q

Now your recording in @q calls itself at the end. But because it calls the recording by name, it won't be empty anymore. So, call the recording:

@q

It will replay the recording, then at the end, as the last step, replay itself again. It will repeat this until the end of the file.

TLDR;

qq
q
/dog
nnnnnnnnnnndiwiparrot<esc>
@q
q
@q

Answer 6

Well, if you do /gc then you can count the number of times it asks you for confirmation, and go ahead with the replacement when you get to the nth :D

Answer 7

Two regex do the job as initially intended:

%s/^\(.\{-}word\)\{3}\zs/* for each line, OR

%s/\(word\(\_.\{-}word\)\{15}\)\zs\(\_.*\)/*\3 for whole document, followed by:

%s/word\*/newword

Note that can change \{3} and \{15} for desired occurrence number, as well as the match word.