Grep matching cell of the csv file and index of that row

Question

I have csv file and I want to see numbers that start with 0.99 and their row's index which is the first cell of that row.

This is what I have so far:

cat fil.csv | grep '0\\.99'| tee > (cut -d, -f1) | tr , \\n |  grep '0\\.99'

input:

id, f1,f2,f3
f1,0.54,0.12,0.432
f2,0.1231,0.99999,0.99832
f3,0.121,nan,0.12321
f4,0.99712,0.121,0.434

desired output: ideally i want this but it would be too complex for oneliner:

f2,0.99999,0.99832
f4,0.99712

I can settle for this, which is what I want from the command that I wrote:

f2
0.99999
0.99832
f4
0.99712

Answer 1

With awk:

$ awk -F, '$0~"0\\.99*"{printf $1;for(i=1;i<=NF;i++){if($i~"0\\.99*"){printf ","$i}};printf "\n"}'
f2,0.99999,0.99832
f4,0.99712

In more readable form:

$ awk -F, '
      $0~"0\\.99*"{
          printf $1
          for(i=1; i<=NF; i++){
              if($i~"0\\.99*"){
                  printf ","$i
              }
          }
          printf "\n"
      }
  '

Answer 2

If you can use perl:

$ perl -F, -anle '
  BEGIN { $, = "," }
  @h = grep { /^0\.99/ } @F;
  print $F[0], @h if @h;
' file
f2,0.99999,0.99832
f4,0.99712