06-19-2008
Average in awk
Hi
I am looking for an awk script which can compute average of all the fields every 5th line. The file looks:
A B C D E F G H I J K L M
1 18 13 14 12 14 13 11 12 12 15 15 15
2 17 17 13 13 13 12 12 11 12 14 15 14
3 16 16 12 12 12 11 11 12 11 16 14 13
4 15 15 11 11 11 12 11 12 11 15 14 16
5 14 14 10 12 11 12 11 13 12 14 16 16
6 13 13 11 11 13 11 10 12 12 14 15 15
7 12 12 27 24 12 12 11 11 15 16 15 14
8 12 11 26 23 11 13 11 12 14 15 15 14
9 12 11 25 22 11 12 11 11 13 14 15 14
10 15 12 24 21 11 12 10 13 13 15 16 14
I need to compute the average of the individual fields (A to M) for every 5th line (line 1 to 5: 6-10 etc).
I greatly appreciate any help.
Thanks
10 More Discussions You Might Find Interesting
1. UNIX for Dummies Questions & Answers
Suppose I have 500 files in a directory and I need to Use awk to calculate average of column 3 for each of the file, how would I do that? (6 Replies)
Discussion started by: grossgermany
6 Replies
2. Shell Programming and Scripting
Hi,
I have the data like this
$1 $2
1 12
2 13
3 14
4 12
5 12
6 12
7 13
8 14
9 12
10 12
i want to compute average of $1 and $2 every 5th line (1-5 and 6-10)
Please help me with awk
Thank you (4 Replies)
Discussion started by: saint2006
4 Replies
3. UNIX for Dummies Questions & Answers
Hi
I am looking for an awk script which can compute the average of the last column based on the date and time. The file looks:
site1,"2000-01-01 00:00:00", "2000-01-01 00:59:00",0.013
site2,"2000-02-01 01:00:00", "2000-02-01 01:59:00",0.035
site1,"2000-02-01 02:00:00", "2000-02-01... (15 Replies)
Discussion started by: kathy wang
15 Replies
4. Shell Programming and Scripting
Hi,
I have the following data in a file for example:
P1 XXXXXXX.1 YYYYYYY.1 ZZZ.1
P1 XXXXXXX.2 YYYYYYY.2 ZZZ.2
P1 XXXXXXX.3 YYYYYYY.3 ZZZ.3
P1 XXXXXXX.4 YYYYYYY.4 ZZZ.4
P1 XXXXXXX.5 YYYYYYY.5 ZZZ.5
P1 XXXXXXX.6 YYYYYYY.6 ZZZ.6
P1 XXXXXXX.7 YYYYYYY.7 ZZZ.7
P1 XXXXXXX.8 YYYYYYY.8 ZZZ.8
P2... (6 Replies)
Discussion started by: alex2005
6 Replies
5. Shell Programming and Scripting
Hi guys, I am not an expert in shell and I need help with awk command. I have a file with values like
200 1 1
200 7 2
200 6 3
200 5 4
300 3 1
300 7 2
300 6 3
300 4 4
I need resulting file with averages of... (3 Replies)
Discussion started by: saif
3 Replies
6. Shell Programming and Scripting
I want to calculate the average line by line of some files with several lines on them, the files are identical, just want to average the 3rd columns of those files.:wall:
Example file:
File 1
001 0.046 0.667267
001 0.047 0.672028
001 0.048 0.656025
001 0.049 ... (2 Replies)
Discussion started by: AriasFco
2 Replies
7. Shell Programming and Scripting
I need to find the average from a file like:
data => BW:123 M:30 RTD:0 1 0 1 0 0 1 1 1 1 0 0 1 1 0'
data => BW:123 N:30 RTD:0 1 0 1 0 0 1 1 1 1 0 0 1 1 0'
data => BW:123 N:30 RTD:0 1 0 1 0 0 1 1 1 1 0 0 1 1 0'
data => BW:123 N:30 RTD:0 1 0 1 0 0 1 1 1 1 0 0 1 1 0'
data => BW:123 N:30 RTD:0 1... (4 Replies)
Discussion started by: Slagle
4 Replies
8. Shell Programming and Scripting
I am trying to modify the awk below to include the gene name ($5) for each target and can not seem to do so. Also, I'm not sure the calculation is right (average of all targets that are the same is $4 using the values in $7)? Thank you :).
awk '{if((NR>1)&&($4!=last)){printf("%s\t%f\t%s\n",... (1 Reply)
Discussion started by: cmccabe
1 Replies
9. Shell Programming and Scripting
In the below awk I am trying to combine all matching $4 into a single $5 (up to the -), and count the lines in $6 and average all values in $7. The awk is close but it seems to only be using the last line in the file and skipping all others. The posted input is a sample of the file that is over... (3 Replies)
Discussion started by: cmccabe
3 Replies
10. Shell Programming and Scripting
Hi, I'm using awk to try and get a moving average for the second column of numbers ($2) in the below example broken out by unique identifier in column 1 ($1) :
H1,1.2
H1,2.3
H1,5.5
H1,6.6
H1,8.7
H1,4.1
H1,6.4
H1,7.8
H1,9.6
H1,3.2
H5,50.1
H5,54.2
H5,58.8
H5,60.9
H5,54.3
H5,52.7... (8 Replies)
Discussion started by: theflamingmoe
8 Replies
LEARN ABOUT REDHAT
logfile
LOGFILE(1) mrtg LOGFILE(1)
NAME
logfile - description of the mrtg-2 logfile format
SYNOPSIS
This document provides a description of the contents of the mrtg-2 logfile.
OVERVIEW
The logfile consists of two main sections. A very short one at the beginning:
The first Line
It stores the traffic counters from the most recent run of mrtg
The rest of the File
Stores past traffic rate averates and maxima at increassing intervals
The first number on each line is a unix time stamp. It represents the number of seconds since 1970.
DETAILS
The first Line
The first line has 3 numbers which are:
A (1st column)
A timestamp of when MRTG last ran for this interface. The timestamp is the number of non-skip seconds passed since the standard UNIX
"epoch" of midnight on 1st of January 1970 GMT.
B (2nd column)
The "incoming bytes counter" value.
C (3rd column)
The "outgoing bytes counter" value.
The rest of the File
The second and remaining lines of the file 5 numbers which are:
A (1st column)
The Unix timestamp for the point in time the data on this line is relevant. Note that the interval between timestamps increases as you
prograss through the file. At first it is 5 minutes and at the end it is one day between two lines.
This timestamp may be converted in EXCEL by using the following formula:
=(x+y)/86400+DATE(1970,1,1)
you can also ask perl to help by typing
perl -e 'print scalar localtime(x),"
"'
x is the unix timestamp and y is the offset in seconds from UTC. (Perl knows y).
B (2nd column)
The average incoming transfer rate in bytes per second. This is valid for the time between the A value of the current line and the A
value of the previous line.
C (3rd column)
The average outgoing transfer rate in bytes per second since the previous measurement.
D (4th column)
The maximum incoming transfer rate in bytes per second for the current interval. This is calculated from all the updates which have
occured in the current interval. If the current interval is 1 hour, and updates have occured every 5 minutes, it will be the biggest 5
minute transferrate seen during the hour.
E (5th column)
The maximum outgoing transfer rate in bytes per second for the current interval.
AUTHOR
Butch Kemper <kemper@bihs.net> and Tobias Oetiker <oetiker@ee.ethz.ch>
3rd Berkeley Distribution 2.9.17 LOGFILE(1)