Awk printing out smallest and highest number, in a time format - bash

Question

I'm fairly new to linux/bash shell and I'm really having trouble printing two values (the highest and lowest) from a particular column in a text file. The file is formatted like this:
Geoff Audi 2:22:35.227
Bob Mercedes 1:24:22.338
Derek Jaguar 1:19:77.693
Dave Ferrari 1:08:22.921
As you can see the final column is a timing, I'm trying to use awk to print out the highest and lowest timing in the column. I'm really stumped, I've tried:
awk '{print sort -n < $NF}' timings.txt
However that didn't even seem to sort anything, I just received an output of:
1
0
1
0
...
Repeating over and over, it went on for longer but I didn't want a massive line of it when you get the point after the first couple iterations.
My desired output would be:
Min: 1:08:22.921
Max: 2:22:35.227

Answer 1

After question clarifications: if the time field always has a same number of digits in the same place, e.g. h:mm:ss.ss, the solution can be drastically simplified. Namely, we don't need to convert time to seconds to compare it anymore, we can do a simple string/lexicographical comparison:
$ awk 'NR==1 {m=M=$3} {$3<m&&m=$3; $3>M&&M=$3} END {printf("min: %s\nmax: %s",m,M)}' file
min: 1:08:22.921
max: 2:22:35.227
The logic is the same as in the (previous) script below, just using a simpler string-only based comparison for ordering values (determining min/max). We can do that since we know all timings will conform to the same format, and if a < b (for example "1:22:33" < "1:23:00") we know a is "smaller" than b. (If values are not consistently formatted, then by using the lexicographical comparison alone, we can't order them, e.g. "12:00:00" < "3:00:00".)
So, on first value read (first record, NR==1), we set the initial min/max value to the timing read (in the 3rd field). For each record we test if the current value is smaller than the current min, and if it is, we set the new min. Similarly for the max. We use short circuiting instead if to make expressions shorter ($3<m && m=$3 is equivalent to if ($3<m) m=$3). In the END we simply print the result.
Here's a general awk solution that accepts time strings with variable number of digits for hours/minutes/seconds per record:
$ awk '{split($3,t,":"); s=t[3]+60*(t[2]+60*t[1]); if (s<min||NR==1) {min=s;min_t=$3}; if (s>max||NR==1) {max=s;max_t=$3}} END{print "min:",min_t; print "max:",max_t}' file
min: 1:22:35.227
max: 10:22:35.228
Or, in a more readable form:
#!/usr/bin/awk -f
{
split($3, t, ":")
s = t[3] + 60 * (t[2] + 60 * t[1])
if (s < min || NR == 1) {
min = s
min_t = $3
}
if (s > max || NR == 1) {
max = s
max_t = $3
}
}
END {
print "min:", min_t
print "max:", max_t
}
For each line, we convert the time components (hours, minutes, seconds) from the third field to seconds which we can later simply compare as numbers. As we iterate, we track the current min val and max val, printing them in the END. Initial values for min and max are taken from the first line (NR==1).

Answer 2

Given your statements that the time field is actually a duration and the hours component is always a single digit, this is all you need:
$ awk 'NR==1{min=max=$3} {min=(min<$3?min:$3); max=(max>$3?max:$3)} END{print "Min:", min ORS "Max:", max}' file
Min: 1:08:22.921
Max: 2:22:35.227

Answer 3

You don't want to run sort inside of awk (even with the proper syntax).
Try this:
sed 1d timings.txt | sort -k3,3n | sed -n '1p; $p'
where
the first sed will remove the header
sort on the 3rd column numerically
the second sed will print the first and last line