Awk and calculating start time from end time and duration - time

Question

I have a file with date, end time and duration in decimal format and I need to calculate the start time. The file looks like:
20140101;1212;1.5
20140102;1515;1.58
20140103;1759;.69
20140104;1100;12.5
...
The duration 1.5 for the time 12:12 means one and a half hours and the start time would be 12:12 - 1:30 = 10:42 AM or 11:00 - 12.5 = 11:00 - 12:30 = 22:30 PM. Is there an easy way for calculating such time differences in Awk or is it the good ol' split-multiply-subtract-and-handle-the-day-break-yourself all over again?
Since the values are in hours and minutes, only the minutes matter and the seconds can be discarded, for example duration 1.58 means 1:34 and the leftover 0.8 seconds can be discarded.
I'm on GNU Awk 4.1.3

Answer 1

As you are using gawk take adventage of its native time functions:
gawk -F\; '{tmst=sprintf("%s %s %s %s %s 00",\
substr($1,1,4),\
substr($1,5,2),\
substr($1,7,2),\
substr($2,1,2),\
substr($2,3,2))
t1=mktime(tmst)
seconds=sprintf("%f",$3)+0
seconds*=60*60
difference=strftime("%H%M",t1-seconds)
print $0""FS""difference}' file
Results:
20140101;1212;1.5;1042
20140102;1515;1.58;1340
20140103;1759;.69;1717
20140104;1100;12.5;2230
Check: https://www.gnu.org/software/gawk/manual/html_node/Time-Functions.html
Explanation:
tmst=sprintf(..) :used to create a date string from the file
that conforms with the datespec of mktime function YYYY MM
DD HH MM SS [DST].
t1=mktime(tmst) :turn datespec into a timestamp than can be
handle by gawk (as the number of seconds elapsed since 1
January 1970)
seconds=sprintf("%f",$3)+0 : convert third field to float.
seconds*=60*60 : convert hours (in float) to seconds.
difference=strftime("%H%M",t1-seconds) : get the difference in
human maner, hours an minutes.

Answer 2

I highly recommend to use a programming language which supports datetime calculations, because the calculation can be tricky in detail because daylight saving shifts. You can use Python for example:
start_times.py:
import csv
from datetime import datetime, timedelta
with open('input.txt', 'rb') as csvfile:
reader = csv.reader(csvfile, delimiter=';', quotechar='|')
for row in reader:
end_day = row[0]
end_time = row[1]
# Create a datetime object
end = datetime.strptime(end_day + end_time, "%Y%m%d%H%M")
# Translate duration into minutes
duration=float(row[2])*60
# Calculate start time
start = end - timedelta(minutes=duration)
# Column 3 is the start day (can differ from end day!)
row.append(start.strftime("%Y%m%d"))
# Column 4 is the start time
row.append(start.strftime("%H%M"))
print ';'.join(row)
Run:
python start_times.py
Output:
20140101;1212;1.5;20140101;1042
20140102;1515;1.58;20140102;1340
20140103;1759;.69;20140103;1717
20140104;1100;12.5;20140103;2230 <-- you see, the day matters!
The above example is using the system's timezone. If the input data refers to a different timezone, Pyhon's datetime module allows to specify it.

Answer 3

I would do something like this:
awk 'BEGIN{FS=OFS=";"}
{ h=substr($2,0,2); m=substr($2,3,2); mins=h*60 + m; diff=mins - $3*60;
print $0, int(diff/60) ":" int(diff%60)
}' file
That is, convert everything to minutes and then back to hours/minutes.
Test
$ awk 'BEGIN{FS=OFS=";"}{h=substr($2,0,2); m=substr($2,3,2); mins=h*60 + m; diff=mins - $3*60; print $0, int(diff/60) ":" int(diff%60)}' a
20140101;1212;1.5;10:42
20140102;1515;1.58;13:40
20140103;1759;.69;17:17