Calculate average of column in units of time (HH:MM)

thealmightyos · 05-05-2015, 11:05 AM

Here is an example of my data.

Code:

Group1,2015-04-21,05:05,PASS
Group1,2015-04-22,10:16,PASS
Group1,2015-04-23,05:25,PASS
Group1,2015-04-24,06:38,PASS
Group1,2015-04-25,06:27,PASS
Group1,2015-04-26,10:11,PASS
Group1,2015-04-27,05:48,PASS
Group1,2015-04-28,00:03,FAIL
Group1,2015-04-29,05:33,PASS
Group1,2015-04-30,05:30,PASS
Group1,2015-05-01,13:39,PASS
Group1,2015-05-02,06:53,PASS
Group1,2015-05-03,09:51,PASS
Group1,2015-05-04,06:01,PASS

What I want is an average of column 3 which should be easy, right?

Code:

awk -F',' '{ total += $3; count++ } END { print total/count }' test.txt
6.5

6.5 is not correct and is in decimal form. According to excel (yes, I am lazy) It should be 6 hours, 57 Minutes, 9 seconds. What do I need to do differently in order to get the correct answer in the format HH:MM?

schneidz · 05-05-2015, 11:27 AM

this mite help:
http://www.unix.com/unix-for-dummies...rage-time.html

pan64 · 05-05-2015, 11:33 AM

as excel will recognize date format and calculate average based on that you need to do the same thing. Convert hours:minutes into a usable format and calculate average. But I think you need to take into account the date too.
you may find this page useful to do the conversion: http://www.theunixschool.com/2013/01...functions.html

Samsonite2010 · 05-05-2015, 11:36 AM

If it really is just the hours and minutes you are interested in you probably want to multiply the hours by 60 then add the minutes then you will be dealing with one unit (minutes). As pan64 suggests - they date might be important?

schneidz · 05-05-2015, 11:42 AM

Code:

[schneidz@hyper density]$ date +s -d '2015-04-21 05:05'
1429607100
[schneidz@hyper density]$ date -d @1429607100
Tue Apr 21 05:05:00 EDT 2015
[schneidz@hyper density]$ man date

syg00 · 05-05-2015, 07:56 PM

The OP apparently is happy to ignore the date (based on excel example).
Using [,:] as FS busts the time up - the arithmetic becomes simple. If there is a lot of data, I'd be looking to avoid call-outs to date if I could.

thealmightyos · 05-06-2015, 05:38 PM

Correct. There is a lot of data. This is a tiny snip from a much larger group of files. The date is not important in this scenario, just part of the report that I have to maneuver around to get what I need which is averages for the hours and minutes it took to run. It looks like I am just failing maths

syg00 · 05-06-2015, 08:15 PM

I like to use things like this to expose the data - makes it easy to see what fields you need to use. The hours is one field, the minutes the next. As suggested above, totalling them should be trivial. In your END clause do the final division - or (better) use modulo 60.

Code:

awk -F"[,:]"  'NR < 3 {print "\nRecord: " NR ; for (i=1; i<=NF; i++) print "Field "i":\t",$i}' file