I have a file in format like this:
method1|duration=300s
method2|duration=300s
method1|duration=500s
method1|duration=300s
method1|duration=700s
method3|duration=300s
...
How can I get the average duration time of method1 using Linux Shell? Thanks.
Use this:
awk -F'[=s]' '/method1/{t+=$2;c++}END{printf "%ss\n",t/c}' file
-F[=s] sets the field delimiter to = or s making it easy to extract the numerical value between them. t+=$2 will add the numerical value to the total t. c++ will count the lines containing the term method1. After the last line of input has been processed END, we print the total t divided by c, the number of lines containing method1.
Note: Initialization of the variables c and t is not necessary since awk initializes them with 0 on their first usage.
You can use awk for that. If duration is always in seconds something like this should work:
awk -F '\\||=' 'BEGIN { sum = 0; rows = 0 } { if ($1 == "method1") { sum += $3; rows += 1; } } END { print sum / rows "s" }' data_file
Using awk you can do it as:
awk -F '[=s]' 'BEGIN{count=0;sum=0} /method1/{sum+=$2;count+=1} END{print sum/count}' filename
All the duration must be of same time unit.
This will initially set count to 0 and sum to 0. During each iteration, sum will get added and count will be incremented if method1 is found in the line. At the end, sum/count will be printed out.
