I have a file that collects time stamps of HH:MM need to add one hour to the 1st column and convert the both columns from 24H to 12H then subtract them.
Output in the file I collected the times from:
Start End Total Run Time 00:00 05:39 01:31 06:02 00:48 06:24 23:46 04:50 00:05 05:12 00:06 05:04 00:04 05:10 00:10 05:10 00:00 04:51 00:10 05:33 23:41 04:15 I want to give the total run time, between these two columns.
so the 1st time stamps should be 01:00 05:39 before subtracting to get the total time run 01:00-05:39.
I tried with just this command to subtract but the 24H to 12H and adding one hour to the $1 1st column, just confuses me.
awk 'BEGIN { OFS = "\t" } { $3 = $2 - $1 } 1' 1 Answer
You can't simply subtract datetime objects like 01:00 and 05:39 in awk (although GNU awk has its own Time Functions that you could use to do it).
Having said that, formating a datetime difference as as datetime is somewhat questionable. Miller's builtin strptime/strftime implementations look like they will handle it - including correctly rolling over the day boundary.
Ex. given
$ cat file Start End 00:00 05:39 01:31 06:02 00:48 06:24 23:46 04:50 00:05 05:12 00:06 05:04 00:04 05:10 00:10 05:10 00:00 04:51 00:10 05:33 23:41 04:15 then
$ mlr --pprint put -S ' ${Total Run Time} = strftime(strptime(${End},"%H:%M") - 3600 - strptime(${Start},"%H:%M"),"%H:%M") ' file Start End Total Run Time 00:00 05:39 04:39 01:31 06:02 03:31 00:48 06:24 04:36 23:46 04:50 04:04 00:05 05:12 04:07 00:06 05:04 03:58 00:04 05:10 04:06 00:10 05:10 04:00 00:00 04:51 03:51 00:10 05:33 04:23 23:41 04:15 03:34 
