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If there is maximum of 1 entry per file:
awk '++A[$0]>=ARGC-1' file*

This would then be a bit more robust:
awk '{$1=$1} ++A[$0]>=ARGC-1' file*

But 1000 files is probably going to be too many...

Try:-
awk -F'\t' 'FNR==1{ofile="NEW" FILENAME}$10=="S"{print > ofile}' *.subVCF

You forgot to mention which OS and shell you are using. With bourne type shells, this might work:ls *.geno | while read FN; do awk -f example.awk $FN final.map > ${FN/geno/dat}; done

Would this come close: awk 'FNR==NR {T[$1]=$3 FS $4; next} {print $0, T[$1]}' file2 file1
#file1-header1
file1-header2
111 qwe rtz uio
198 asd fgh jkl nko1 vhu1
165 yxc...

If the pattern is always 4 characters then you can replace the loop
{for (t in T) if ($2 ~ t) $(NF+1)=T[t]}
1by the efficient
{print $0, T[substr($2,1,4)]}

Well, try awk '
FNR==NR {X=$1
sub ($1 FS, "")
T[X]=$0
next
}
{for (t in T) if ($2 ~ t) $(NF+1)=T[t]}...

I have a comma (,) delimited file.

106232145,"medicare","medicare,medicaid",789


I would like to count the number of fields in each line.
I tried the below code

awk -F ',' '{print NF-1}'...

awk -F, '{gsub(/"[^"]*"/,x);print NF}'

You were on the right track. Here is an approach with two-dimensional arrays :
awk '{split($2,F,/,/)} NR==FNR{for(i in F) A[$1,F[i]]; next} {for(i in F) if(($1,F[i]) in A) $3=$3 "*"}1' FS='\t'...