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In programming languages I would typically measure the performance of something relatively fast (<1 second completion) by looping it a few thousand times, and then getting the average time per completion.
How would I do the same in bash to compare the execution performance of two very fast commands?
It's better to use something like time command.
Bash also has a time built-in, what I'm suggesting is using time binary located at: /usr/bin/time.
For one runtime:
command time -f "\nElapsed: %E \nUser: %U \nSystem: %S \nMemory: %M\n" \
./MyCommand 1> /dev/null
Which outputs:
Elapsed: 0:00.01
User: 0.00
System: 0.00
Memory: 2412
command: forces bash to use /usr/bin/time instead of time built-in.You can use time with a loop to get a "avrage", "min", "max" of a specific resource say memory:
This code will run ./COMMAND commands 1000 times, then prints out the "min, max, avg" of its total (data+stack+text) memory usage.
#!/bin/bash
tmpfile=`mktemp`
for i in {1..1000}; do command time -ao $tmpfile -f "%K" ./COMMAND 1>/dev/null; done;
awk 'NR == 1 {min = $0} $0 > max {max = $0} {total += $0} END {print total/NR, min, max}' $tmpfile
rm $tmpfile
Here is the output:
2436.89 2524 2324
You can change %K with:
%E: Elapsed real time%I: Number of file inputs%P: Percentage of the CPU that this job got%k: Number of signals delivered to the process%U: CPU usage in user mode%S: CPU usage in kernel modeSee man time
Thanks to muru for writing a more clear awk statement.
echo $(...) | tr ' ' '\n', you could just do for i in {1..1000}; do command time -f "%K\n" ./COMMAND 2>&1 1>/dev/null; done. And in awk, uninitialized variables get evaluated to 0, so awk '$0 < min {min = $0} $0 > max {max = $0} {total += $0} END {print total/NR; max, min}' The GNU time command also supports logging to a file, -o file, so you can use that, and then run awk on that file later, instead of redirecting stderr and stdout
-v min=foo, or NR == 1 {min = $0}. Sorry about that
sleep .1, is rounded down to 0.00, so no matter how many iterations it runs, the CPU total will always be zero. (the built-in bash time has millisecond resolution). Also, the Elapsed time printed by %E is in the form [hours:]minutes:seconds.xxx (i.e. 0:00.11), so you can't total it with the simple awk + operator.
You can use Python's timeit module (which is usually used for python code snippets), with the subprocess.call function to run external commands:
$ python3 -m timeit -s 'import subprocess' 'subprocess.call(["sleep", "0.1"])'
10 loops, best of 3: 103 msec per loop
You can change the number of loops with -n:
$ python3 -m timeit -n 5 -s 'import subprocess' 'subprocess.call(["sleep", "0.1"])'
5 loops, best of 3: 103 msec per loop
And since sleep 0.1 should sleep for 100 ms, it looks like the measurement is pretty accurate.
Also see:
Nevermind, got it:
N=10000
time(for i in $(eval echo "{1..$N}"); do
./mycommand &>/dev/null
done)
for i in $(seq 1 $N) instead of the eval echo would be simpler. Or for ((i = 0; i < N; i++))
In general the easy way is to do time ./myscript.sh but this will show you zero second as your script is really fast.
Instead use this answer that use GNU date (able to show nanoseconds) and PS4.
I was searching for something, and came across both this question and:
https://github.com/sharkdp/hyperfine - "A command-line benchmarking tool"
YMMV
