print 3 consecutive column after specific string from CSV - shell

Question

I need to print 2 columns after specific string (in my case it is 64). There can be multiple instances of 64 within same CSV row, however next instance will not occur within 3 columns of previous occurrence. Output of each instance should be in next line and unique. The problem is, the specific string does not fall in same column for all rows. All row is having kind of dynamic data and there is no header for CSV. Let say, below is input file (its just a sample, actual file is having approx 300 columns & 5 Million raws):
00:TEST,123453103279586,ABC,XYZ,123,456,65,906,06149,NIL TS21,1,64,906,06149,NIL TS22,1,64,916,06149,NIL BS20,1,64,926,06149,NIL BS30,1,64,906,06149,NIL CAML,1,ORIG,0,TERM,1,1,1,6422222222
00:TEST,123458131344169,ABC,XYZ,123,456,OCCF,1,1,1,64,857,19066,NIL TS21,1,64,857,19066,NIL TS22,1,64,857,19066,NIL BS20,1,64,857,19067,NIL BS30,1,64,857,19068,NIL PSS,1,E2 EPSDATA,GRANTED,NONE,1,N,N,256000,5
00:TEST,123458131016844,ABC,XYZ,123,456,HOLD,,1,64,938,36843,NIL TS21,1,64,938,36841,NIL TS22,1,64,938,36823,NIL BS20,1,64,938,36843,NIL BS30,1,64,938,36843,NIL CAML,1,ORIG,0,TERM,00,50000,N,N,N,N
00:TEST,123453102914690,ABC,XYZ,123,456,HOLD,,1,PBS,TS11,64,938,64126,NIL TS21,1,64,938,64126,NIL TS22,1,64,938,64126,NIL BS20,1,64,938,64226,NIL BS30,1,64,938,64326,NIL CAML,1,ORIG,0,TERM,1,1,1,6422222222,2222,R
Output required(only unique entries):
64,906,06149
64,857,19066
64,857,19067
64,857,19068
64,938,36843
64,938,36841
64,938,36823
64,938,36843
64,938,36843
64,938,64326
There is no performance related concerns. I have tried to search many threads but could not get anything near related. Please help.

Answer 1

We can use a pipe of two commands... first to put the 64's leading on a line and a second to print first three columns if we see a leading 64.
sed 's/,64[,\n]/\n64,/g' | awk -F, '/^64/ { print $1 FS $2 FS $3 }'
There are ways of doing this with a single awk command, but this felt quick and easy to me.
Though the sample data from the question contains redundant lines, karakfa (see below) reminds me that the question speaks of a "unique data" requirement. This version uses the keys of an associative array to keep track of duplicate records.
sed 's/,64[,\n]/\n64,/g' | awk -F, 'BEGIN { split("",a) } /^64/ && !((x=$1 FS $2 FS $3) in a) { a[x]=1; print x }'

Answer 2

gawk:
awk -F, '{for(i=0;++i<=NF;){if($i=="64")a=4;if(--a>0)s=s?s","$i:$i;if(a==1){print s;s=""}}}' file

Answer 3

Sed for fun
sed -n -e 's/$/,n,n,n/' -e ':a' -e 'G;s/[[:blank:],]\(64,.*\)\(\n\)$/\2\1/;s/.*\(\n\)\(64\([[:blank:],][^[:blank:],]\{1,\}\)\{2\}\)\([[:blank:],][^[:blank:],]\{1,\}\)\{3\}\([[:blank:],].*\)\{0,1\}$/\1\2\1\5/;s/^.*\n\(.*\n\)/\1/;/^64.*\n/P;s///;ta' YourFile | sort -u
assuming column are separated by blank space or comma
need a sort -u for uniq (possible in sed but a new "simple" action of the same kind to add in this case)

Answer 4

awk to the rescue!
$ awk -F, '{for(i=1;i<=NF;i++)
if($i==64)
{k=$i FS $(++i) FS $(++i);
if (!a[k]++)
print k
}
}' file
64,906,06149
64,916,06149
64,926,06149
64,857,19066
64,857,19067
64,857,19068
64,938,36843
64,938,36841
64,938,36823
64,938,64126
64,938,64226
64,938,64326
ps. your sample output doesn't match the given input.