replace every nth occurrence of a pattern using awk [duplicate] - linux

Question

This question already has answers here:
Printing with sed or awk a line following a matching pattern
(9 answers)
Closed 6 years ago.
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.