bash - count number of patterns -

i have csv file, , wanna remove columns have less 5 different values. e.g

a b c; 1 1 1; 1 2 2; 1 3 4; 2 4 5; 1 6 7; 

then wanna remove column a since has 2 different values (1,2). how this?

a solution using arrays:

infile="infile.txt"  different=5 rows=0  while  read -a line ;   data+=( ${line[@]/;/} )                        # remove semicolons   ((rows++)) done < "$infile"  cols=$(( ${#data[@]}/rows ))                     # calculate number of rows  result=() (( cntr1=0; cntr1<cols; cntr1+=1 ));   cnt=()   save=( ${data[cntr1]} )                        # add column header   (( cntr2=cols; cntr2<${#data[@]}; cntr2+=cols ));     cnt[${data[cntr1+cntr2]}]=1     save+=( ${data[cntr1+cntr2]} )               # add column data   done   if [ ${#cnt[@]} -eq $different ] ;        # choose column?     result+=( ${save[@]} )                       # add column result   fi done  cols=$((${#result[@]}/rows))                     # recalculate number of columns  (( cntr1=0; cntr1<rows; cntr1+=1 ));   (( cntr2=0; cntr2<${#result[@]}; cntr2+=rows ));     printf " %s" "${result[cntr1+cntr2]}"   done   printf ";\n" done 

the output:

 b c;  1 1;  2 2;  3 4;  4 5;  6 7;