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I have some files in the format

Y15-SUB-B04-P17-BK_M02734_4_000000000-ANNUF_1_1111_24724_4878;size=1;
Y15-SUB-B05-P22-LM_M02734_4_000000000-ANNUF_1_1111_20624_14973;size=1;
Y15-SUB-B05-P22-LM_M02734_4_000000000-ANNUF_1_1103_11326_10379;size=1;

I wish to replace every occurrence of the underscore (_) with a colon (:) EXCEPT for the first one. I want an output like this:

Y15-SUB-B04-P17-BK_M02734:4:000000000-ANNUF:1:1111:24724:4878;size=1;
Y15-SUB-B05-P22-LM_M02734:4:000000000-ANNUF:1:1111:20624:14973;size=1;
Y15-SUB-B05-P22-LM_M02734:4:000000000-ANNUF:1:1103:11326:10379;size=1;

I know I can use sed -i '' 's/_/:/g' old_file to replace ALL (or sed 's/_/:/g' old_file > new_file), and that I could add numbers to replace only the 2nd, 4th or so occurrence:

sed 's/_/:/2' old_file > new_file

But how to replace every occurrence on each line BUT the first?

Using GNU sed (other versions may behave differently, thanks glenn jackman):

 sed -i'' 's/_/:/2g' file

This will change all _ to : skipping the first occurrence on each line.

Here is another simple awk script,(standar Linux gawk), no loops:

cat script.awk
match($0,/^[^_]*_/,a) # match current line to first _ (including) into a[0] variable
 sub(a[0],""); # remove a[0] from current line
 gsub("_",":"); # replace all _ to : in current line
 print a[0]""$0; # ouput a[0] and current line

run:

awk -f script.awk input.txt

or:

awk 'match($0,/^[^_]*_/,a)sub(a[0],"");gsub("_",":");print a[0]""$0;' input.txt

Is awk okay? You could use _ as the field separator, and print out:

<field 1>_<field 2>:<field n>:<field n+1>:...

Like this:

awk -F_ ' printf("%s_%s", $1, $2); for (x = 3; x <=NF; x++) printf(":%s", $x); ; printf("n"); '

If the structure is the same for each line you could hard-code the number of fields to avoid the loop (runs in about 2/3 of the time according to a very rough preliminary trial):

awk -F_ 'printf("%s_%s:%s:%s:%s:%s:%s:%sn", $1, $2, $3, $4, $5, $6, $7, $8);'

Simple sed command will work fine for this

command:sed "s/_/:/2g" filename

output

Y15-SUB-B04-P17-BK_M02734:4:000000000-ANNUF:1:1111:24724:4878;size=1;
Y15-SUB-B05-P22-LM_M02734:4:000000000-ANNUF:1:1111:20624:14973;size=1;
Y15-SUB-B05-P22-LM_M02734:4:000000000-ANNUF:1:1103:11326:10379;size=1;

Note:Suppose if you want to replace in same file use below command

sed -i "s/_/:/2g" filename

With perl, to match the character _ and replace from the first instance on-wards as below.

perl -pe '$n=0s_++$n > 1 ? ":" : $&;ge' file

The part s_ identifies the _ within the line and if its the 2^nd occurrence replace with : or replace with the same character($&)

Using Posix-sed constructs only we do like as:

$ sed -e '
 y/_/n/
 s/n/_/
 y/n/:/
' inp.file


$ perl -pe '1 while s/_.*?K_/:/g' inp.file

As per my comment below your question, here's the (very) small C program that accomplishes the described functionality, and probably an order-of-magnitude (or two) faster than sed...

#include <stdio.h>
#include <string.h>
int main ( int argc, char *argv[] ) 
 FILE *infp = (argc>1? fopen(argv[1],"r") : stdin ),
 *outfp = (argc>2? fopen(argv[2],"w") : stdout );
 char line[9999], *lptr=line, this = '_', that = ':';
 int nskip = 1, nfound = 0;
 if ( infp != NULL && outfp != NULL ) 
 while ( (lptr=fgets(line,9990,infp)) != NULL ) 
 nfound = 0;
 while ( (lptr=strchr(lptr,this)) != NULL ) 
 if ( ++nfound > nskip ) *lptr = that;
 lptr++; 
 if ( fputs(line,outfp) == EOF ) break;
 /* --- end-of-while(fgets()!=NULL) --- */
 if ( infp != stdin ) fclose(infp);
 if ( outfp != stdout ) fclose(outfp);
 /* --- end-of-if(in,outfp!=NULL) --- */
 /* --- end-of-function main() --- */

(P.S. If that's your boat, and if you're ever in-or-around NYC Harbor and need crew, email me. I'm a recreational sailor myself.)