You can use this one, expected to be fast enough. Extra lines in your files, besides the tables, are not expected to be a problem.
grep "#$" *.dlg | sort -rk11 | awk '!seen[$1]++'
grep
fetches all the histogram lines which are then sorted in reverse order by last field, that means lines with most #
on the top, and finally awk
removes the duplicates. Note that when grep
is parsing more than one file, it has -H
by default to print the filenames at the beginning of the line, so if you test it for one file, use grep -H
.
Result should be like this:
file1.dlg: 3 | -5.47 | 17 | -5.44 | 2 |##########
file2.dlg: 3 | -5.47 | 17 | -5.44 | 2 |####
file3.dlg: 3 | -5.47 | 17 | -5.44 | 2 |#######
Here is a modification to get the first appearence in case of many equal max lines in a file:
grep "#$" *.dlg | sort -k11 | tac | awk '!seen[$1]++'
We replaced the reversed parameter in sort, with the 'tac' command which is reversing the file stream, so now for any equal lines, initial order is preserved.
Second solution
Here using only awk:
awk -F"|" '/#$/ && $NF > max[FILENAME] {max[FILENAME]=$NF; row[FILENAME]=$0}
END {for (i in row) print i ":" row[i]}' *.dlg
Update: if you execute it from different directory and want to keep only the basename of every file, to remove the path prefix:
awk -F"|" '/#$/ && $NF > max[FILENAME] {max[FILENAME]=$NF; row[FILENAME]=$0}
END {for (i in row) {sub(".*/","",i); print i ":" row[i]}}'
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…