replace every nth occurrence of a pattern using awk

Question

I'm trying to replace every nth occurrence of a string in a text file.

background: I have a huge bibtex file (called in.bib) containing hundreds of entries beginning with "@". But every entry has a different amount of lines. I want to write a string (e.g. "#") right before every (let's say) 6th occurrence of "@" so, in a second step, I can use csplit to split the huge file at "#" into files containing 5 entries each.

The problem is to find and replace every fifth "@".

Since I need it repeatedly, the suggested answer in printing with sed or awk a line following a matching pattern won't do the job. Again, I do not looking for just one matching place but many of it.

What I have so far:

awk '/^@/ && v++%5 {sub(/^@/, "\n#\n@")} {print > "out.bib"}' in.bib

replaces 2nd until 5th occurance (and no more). (btw, I found and adopted this solution here: "Sed replace every nth occurrence". Initially, it was meant to replace every second occurence--which it does.)

And, second:

awk -v p="@" -v n="5" '$0~p{i++}i==n{sub(/^@/, "\n#\n@")}{print > "out.bib"}' in.bib

replaces exactly the 5th occurance and nothing else. (adopted solution from here: "Display only the n'th match of grep"

What I need (and not able to write) is imho a loop. Would a for loop do the job? Something like:

for (i = 1; i <= 200; i * 5)
   <find "@"> and <replace with "\n#\n@">
then print

The material I have looks like this:

@article{karamanic_jedno_2007,
    title = {Jedno Kosova, Dva Srbije},
    journal = {Ulaznica: Journal for Culture, Art and Social Issues},
    author = {Karamanic, Slobodan},
    year = {2007}
}

@inproceedings{blome_eigene_2008,
    title = {Das Eigene, das Andere und ihre Vermischung. Zur Rolle von Sexualität und Reproduktion im Rassendiskurs des 19. Jahrhunderts},
    comment = {Rest of lines snippet off here for usability -- as in following entries. All original entries may have a different amount of lines.}
}

@book{doring_inter-agency_2008,
    title = {Inter-agency coordination in United Nations peacebuilding}
}

@book{reckwitz_subjekt_2008,
    address = {Bielefeld},
    title = {Subjekt}
}

What I want is every sixth entry looking like this:

#
@book{reckwitz_subjekt_2008,
    address = {Bielefeld},
    title = {Subjekt}
}

Thanks for your help.

Answer 1

Your code is almost right, i modified it.

To replace every nth occurrence, you need a modular expression.

So for better understanding with brackets, you need an expression like ((i % n) == 0)

awk -v p="@" -v n="5" ' $0~p { i++ } ((i%n)==0) { sub(/^@/, "\n#\n@") }{ print }' in.bib > out.bib

Answer 2

you can do the splitting in awk easily in one step.

awk -v RS='@' 'NR==1{next} (NR-1)%5==1{c++} {print RT $0 > FILENAME"."c}' file

will create file.1, file.2, etc with 5 records each, where the record is defined by the delimiter @.

Answer 3

Instead of doing this in multiple steps with multiple tools, just do something like:

awk '/@/ && (++v%5)==1{out="out"++c} {print > out}' file

Untested since you didn't provide any sample input/output.

If you don't have GNU awk and your input file is huge you'll need to add a close(out) right before the out=... to avoid having too many files open simultaneously.