I have some files with the following contents.I would like to calculate average of fifth column. How can I do this with awk?
I used the following code to calculate average. But the output is wrong.
Saying "the output is wrong" doesn't help us much. What, in the output you are getting, is wrong? With your sample input, what output are you trying to get? The output you said you want does not show the sum of values for each animal nor the average value for each animal (except for animals that only appear in one of your input files). Are you trying to prints animal sums in the output, or animal averages?
What values are you trying to average in your final line of output? Are you trying to calculate the average of all of the input values (which seems to be what your code is doing)? Are you trying to calculate the average of the averages for each animal? Are you trying to calculate the average of the sums for each animal?
This is the cronjob
----------------------
root@a7germ:/home/paxtemp > crontab -l|grep test
57 * * * * /home/paxtemp/test_1.sh
0,5,10,15,20,25,30,35,40,45,50,55 * * * * /home/paxtemp/test.sh
root@a7germ:/home/paxtemp >
This is the contents of test.sh script... (2 Replies)
Hi All
I like to know how can we calculate the number of rows and the average of the values present in the file. I will not know what will be the rowcount, which will be dynamic in nature of the file.
eg.
29
33
48
30
28 (6 Replies)
Dear all,
i have 200 values in a file. How can i calculate a weighted average and output into a new file avg.dat?
INPUT:
file1.dat
1.3453
2.434
2.345
.....
OUTPUT:
avg.dat
file1: 1.762
Thanks.
Po (3 Replies)
Hey guys.....
I have many files (lets say 100 or more) of same size, and I want to create a new output file and calculate the average of first row fifth column in all files and print it in first row of output file, then 2nd row fifth col in all 100 files and print it in 2nd row of output... (1 Reply)
Is there an awk script that can easily perform the following operation?
I have a data file that is in the format of
1944-12,5.6
1945-01,9.8
1945-02,6.7
1945-03,9.3
1945-04,5.9
1945-05,0.7
1945-06,0.0
1945-07,0.0
1945-08,0.0
1945-09,0.0
1945-10,0.2
1945-11,10.5
1945-12,22.3... (3 Replies)
Hi
I have file like below
111,victor,48,12,36
342,Peter,54,58,30
476,Scott,25,36,48
567,Patty,74,17,95
I have written below code to calcualte avereage for every id
Victor = 48+12+36/3
#!/bin/ksh
/usr/xpg4/bin/awk '
BEGIN {FS=","} {sum=0; n=0;i=3 (1 Reply)
Can anyone explain what each line of the code does and how it works? I have no experience with python so I am not sure how the arrays and such work. I found this code while looking through the forums.
f = open("exams","r")
l = f.readline()
while l:
l = l.split(" ")
values = l
... (22 Replies)
Discussion started by: totoro125
22 Replies
LEARN ABOUT SUNOS
uptime
uptime(1) User Commands uptime(1)NAME
uptime - show how long the system has been up
SYNOPSIS
uptime
DESCRIPTION
The uptime command prints the current time, the length of time the system has been up, and the average number of jobs in the run queue over
the last 1, 5 and 15 minutes. It is, essentially, the first line of a w(1) command.
EXAMPLES
Below is an example of the output uptime provides:
example% uptime
10:47am up 27 day(s), 50 mins, 1 user, load average: 0.18, 0.26, 0.20
ATTRIBUTES
See attributes(5) for descriptions of the following attributes:
+-----------------------------+-----------------------------+
| ATTRIBUTE TYPE | ATTRIBUTE VALUE |
|Availability |SUNWcsu |
+-----------------------------+-----------------------------+
SEE ALSO w(1), who(1), whodo(1M), attributes(5)NOTES
who -b gives the time the system was last booted.
SunOS 5.10 18 Mar 1994 uptime(1)