Want to get average value for each hour Post: 302941292

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Top Forums UNIX for Advanced & Expert Users Want to get average value for each hour Post 302941292 by Don Cragun on Tuesday 14th of April 2015 08:17:34 PM

04-14-2015

Registered User

Quote:

Originally Posted by Saravanan_0074

I have received hourly average for using below script, but i need daily average value.

Code:

awk -F'|' 'NR>2 {hour=" " substr($2,1,2); array[$1 hour] += $3 count[$1 hour]++ } END { for (a in array) {print a, array[a]/count[a] } } ' filec | sort 
03/02/2015 00 26.2439
03/02/2015 01 25.3431
03/02/2015 02 24.7481
03/02/2015 03 26.6714
03/02/2015 04 10.4339
03/02/2015 05 17.3723
03/02/2015 06 17.5681
03/02/2015 07 23.8506
03/02/2015 08 14.3764
03/02/2015 09 15.7681
03/02/2015 10 18.7348
03/02/2015 11 20.1648
03/02/2015 12 22.8531
03/02/2015 13 22.2181
03/02/2015 14 16.4239
03/02/2015 15 25.3914
03/02/2015 16 20.2731
03/02/2015 17 12.3039
03/02/2015 18 17.9764
03/02/2015 19 22.6573
03/02/2015 20 15.7089
03/02/2015 21 32.6081
03/02/2015 22 17.2506
03/02/2015 23 21.6848

As mjf already said, why gather each hour's data and calculate an average of each input line if you're trying to calculate a daily average.

But, in addition to that, there aren't any | characters in your input. So, with -F'|' in your code, there is nothing in $2 nor in $3 in your input. And, there is no header in this data, so, unless you want to skip data from the midnight hour and the 1am hour data in your daily averages, you don't want the NR>2. And, as has been said before, if you're going to use a single line scrunched up awk script, you MUST separate statements from each other with semicolons. And, since you data appears to be in month/day/year format instead of year/month/day, you need to modify your sort if the goal is to print the output in increasing date order when you run this code with data from December in one year and January in the next year...

Perhaps you wanted something more like:

Code:

awk '{array[$1]+=$3;count[$1]++}END{for(a in array){print a,array[a]/count[a]}}' filec|sort -k1.7,1.10 -k1.1,1.2 -k1.4,1.5

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NAME

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