count and index characters in a column

user_10242059 0
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I have a file system.xyz with several columns:

          43
  Built with Packmol
  O           37.536208       36.873149        9.514500
  C           37.768292       35.784076       10.014380
  N           37.749829       34.667899        9.235406
  C           38.014779       33.336113        9.750827
  C           37.921777       32.283049        8.635104
  C           38.203826       30.885654        9.187454

and need to change them to:

@atom:o1 @mol: $atom:O 0 37.536208 36.873149 9.514500
@atom:c1 @mol: $atom:C 0 37.768292 35.784076 10.014380
@atom:n1 @mol: $atom:N 0 37.749829 34.667899 9.235406
@atom:c2 @mol: $atom:C 0 38.014779 33.336113 9.750827
@atom:c3 @mol: $atom:C 0 37.921777 32.283049 8.635104
@atom:c4 @mol: $atom:C 0 38.203826 30.885654 9.187454

I have managed to use this grep -A43 Built system.xyz | awk '{print "@atom:"tolower($1), "@mol: $atom:"$1,"0",$2,$3,$4}'

@atom:built @mol: $atom:Built 0 with Packmol 
@atom:o @mol: $atom:O 0 37.536208 36.873149 9.514500
@atom:c @mol: $atom:C 0 37.768292 35.784076 10.014380
@atom:n @mol: $atom:N 0 37.749829 34.667899 9.235406
@atom:c @mol: $atom:C 0 38.014779 33.336113 9.750827
@atom:c @mol: $atom:C 0 37.921777 32.283049 8.635104
@atom:c @mol: $atom:C 0 38.203826 30.885654 9.187454

but I have to manually enter the indexes of each character of the first column. Is there a way to count and index the characters in the first column?

grep indexing awk bash
1 Answers

Try this:

awk '
    BEGIN {fmt = "@atom:%s%d @mol: $atom:%s 0"}
    {$1 = sprintf(fmt, tolower($1), ++count[tolower($1)], $1)}
    1
'
user_75520
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