How to get execution time of a script effectively?

I would like to display the completion time of a script.

What I currently do is –

#!/bin/bash
date  ## echo the date at start
# the script contents
date  ## echo the date at end

This just show’s the time of start and end of the script. Would it be possible to display a fine grained output like processor time/ io time , etc?

Asked By: mtk

||

Just use time when you call the script:

time yourscript.sh
Answered By: Trudbert

Just call times without arguments upon exiting your script.

With ksh or zsh, you can also use time instead. With zsh, time will also give you the wall clock time in addition to the user and system CPU time.

To preserve the exit status of your script, you can make it:

ret=$?; times; exit "$ret"

Or you can also add a trap on EXIT:

trap times EXIT

That way, times will be called whenever the shell exits and the exit status will be preserved.

$ bash -c 'trap times EXIT; : {1..1000000}'
0m0.932s 0m0.028s
0m0.000s 0m0.000s
$ zsh -c 'trap time EXIT; : {1..1000000}'
shell  0.67s user 0.01s system 100% cpu 0.677 total
children  0.00s user 0.00s system 0% cpu 0.677 total

Also note that all of bash, ksh and zsh have a $SECONDS special variable that automatically gets incremented every second. In both zsh and ksh93, that variable can also be made floating point (with typeset -F SECONDS) to get more precision. This is only wall clock time, not CPU time.

Answered By: Stéphane Chazelas

If time isn’t an option,

start=`date +%s`
stuff
end=`date +%s`

runtime=$((end-start))

or, if you need sub-second precision and have bc installed,

start=`date +%s.%N`
stuff
end=`date +%s.%N`

runtime=$( echo "$end - $start" | bc -l )
Answered By: Rob Bos

I’m a bit late to the bandwagon, but wanted to post my solution (for sub-second precision) in case others happen to stumble upon this thread through searching. The output is in format of days, hours, minutes, and finally seconds:

res1=$(date +%s.%N)

# do stuff in here

res2=$(date +%s.%N)
dt=$(echo "$res2 - $res1" | bc)
dd=$(echo "$dt/86400" | bc)
dt2=$(echo "$dt-86400*$dd" | bc)
dh=$(echo "$dt2/3600" | bc)
dt3=$(echo "$dt2-3600*$dh" | bc)
dm=$(echo "$dt3/60" | bc)
ds=$(echo "$dt3-60*$dm" | bc)

LC_NUMERIC=C printf "Total runtime: %d:%02d:%02d:%02.4fn" $dd $dh $dm $ds

Hope someone out there finds this useful!

[edit] You need to count all characters in field definition in bash printf, if you want pad seconds to 2 digits before dot you have to define it as %07.4f (all digits and dot count too in to filed length) so the line should look like:
LC_NUMERIC=C printf "Total runtime: %d:%02d:%02d:%07.4fn" $dd $dh $dm $ds

Answered By: jwchew
#!/bin/bash
start=$(date +%s.%N)

# HERE BE CODE

end=$(date +%s.%N)    
runtime=$(python -c "print(${end} - ${start})")

echo "Runtime was $runtime"

Yes, this calls Python, but if you can live with that then this is quite a nice, terse solution.

Answered By: Alex

Here’s a variation of Alex’s answer. I only care about minutes and seconds, but I also wanted it formatted differently. So I did this:

start=$(date +%s)
end=$(date +%s)
runtime=$(python -c "print '%u:%02u' % ((${end} - ${start})/60, (${end} - ${start})%60)")
Answered By: mpontillo
#!/bin/bash
begin=$(date +"%s")

Script

termin=$(date +"%s")
difftimelps=$(($termin-$begin))
echo "$(($difftimelps / 60)) minutes and $(($difftimelps % 60)) seconds elapsed for Script Execution."
Answered By: Arun Binoy
#!/bin/csh
#PBS -q glean
#PBS -l nodes=1:ppn=1
#PBS -l walltime=10:00:00
#PBS -o a.log
#PBS -e a.err
#PBS -V
#PBS -M shihcheng.guo@gmail.com
#PBS -m abe
#PBS -A k4zhang-group
START=$(date +%s)
for i in {1..1000000}
do
echo 1
done
END=$(date +%s)
DIFF=$(echo "$END - $START" | bc)
echo "It takes DIFF=$DIFF seconds to complete this task..."
Answered By: Shicheng Guo

My method for bash:

# Reset BASH time counter
SECONDS=0
    # 
    # do stuff
    # 
ELAPSED="Elapsed: $(($SECONDS / 3600))hrs $((($SECONDS / 60) % 60))min $(($SECONDS % 60))sec"
Answered By: Mark

A small shell function that can be added before commands to measure their time:

tm() {
  local start=$(date +%s)
  $@
  local exit_code=$?
  echo >&2 "took ~$(($(date +%s)-${start})) seconds. exited with ${exit_code}"
  return $exit_code
}

Then use it in your script, or on your command line like so:

tm the_original_command with all its parameters
Answered By: Evgeny

Using only bash it is also possible to measure and calculate the time duration for a portion of the shell script (or the elapsed time for the entire script):

start=$SECONDS

... # do time consuming stuff

end=$SECONDS

you can now either just print the difference:

echo "duration: $((end-start)) seconds."

if you only need incremental duration do:

echo "duration: $((SECONDS-start)) seconds elapsed.."

You can also store the duration in a variable:

let diff=end-start
Answered By: gauteh

This question is quite old but in trying to find my favorite way of doing it this thread came up high… and I’m surprised no one mentioned it:

perf stat -r 10 -B sleep 1

‘perf’ is a performance analyzing tool included in the kernel under ‘tools/perf’ and often available to install as a separate package (‘perf’ in CentOS and ‘linux-tools’ on Debian/Ubuntu). The Linux Kernal perf Wiki has much more information about it.

Running ‘perf stat’ gives quite a bit of details including average execution time right at the end:

1.002248382 seconds time elapsed                   ( +-  0.01% )
Answered By: zbateson

Personally, I like to wrap all my script code in some “main” function like so:

main () {
 echo running ...
}

# stuff ...

# calling the function at the very end of the script
time main

Notice how easy is to use the time command in this scenario. Obviously you’re not measuring the precise time including script parse time, but I find it accurate enough in most situations.

Answered By: LeZuse

Timing function based on SECONDS, limited to second-level granularity only, doesn’t use any external commands:

time_it() {
  local start=$SECONDS ts ec
  printf -v ts '%(%Y-%m-%d_%H:%M:%S)T' -1
  printf '%sn' "$ts Starting $*"
  "$@"; ec=$?
  printf -v ts '%(%Y-%m-%d_%H:%M:%S)T' -1
  printf '%sn' "$ts Finished $*; elapsed = $((SECONDS-start)) seconds"
  # make sure to return the exit code of the command so that the caller can use it
  return "$ec"
}

For example:

time_it sleep 5

gives

2019-03-30_17:24:37 Starting sleep 5 2019-03-30_17:24:42 Finished
sleep 5; elapsed = 5 seconds
Answered By: codeforester
  1. Just use time [any command]. Ex: time sleep 1 will sleep for a real time (ie: as timed by a stop watch) of ~1.000 to ~1.020 sec, as shown here:

     $ time sleep 1
    
     real    0m1.011s
     user    0m0.004s
     sys 0m0.000s
    

    What a beautiful thing. You can put any command after it, and it outputs the result in a nice, human-readable form. I really like to use it for timing builds. Ex:

     # time your "make" build
     time make
    
     # time your "Bazel" build
     time bazel build //path/to/some:target
    

    …or for git operations which can potentially be really long, so I can develop realistic mental expectations:

     # time how long it takes to pull from a massive repo when
     # I'm working from home during COVID-19. NB: `git pull`
     # is sooooo much slower than just pulling the one branch
     # you need with `git pull origin <branch>`, so just fetch
     # or pull what you need!
     time git pull origin master
    
  2. For more-customized timing needs where you may need to manipulate the output or convert it to other forms, in bash, use the internal $SECONDS variable. Here’s a demo, including converting these seconds to other units, such as floating point minutes:

    Note that dt_min gets rounded from 0.01666666666... (1 second = that many minutes) to 0.017 in this case since I’m using the printf function to round. The sleep 1; part below is where you’d call your script to run and time, but I’m just sleeping for 1 second instead for the sake of this demo.

    Command:

     start=$SECONDS; sleep 1; end=$SECONDS; dt_sec=$(( end - start )); 
     dt_min=$(printf %.3f $(echo "$dt_sec/60" | bc -l)); 
     echo "dt_sec = $dt_sec; dt_min = $dt_min"
    

    Output:

     dt_sec = 1; dt_min = 0.017
    

Related:

  1. Read more about bc and printf in my answer here: https://stackoverflow.com/questions/12722095/how-do-i-use-floating-point-division-in-bash/58479867#58479867
  2. I don’t remember where I first learned about the time command anymore, but it may have been from @Trudbert’s answer right here.
Answered By: Gabriel Staples

Use bash time builtin?

time: time [-p] PIPELINE
    Execute PIPELINE and print a summary of the real time, user CPU time,
    and system CPU time spent executing PIPELINE when it terminates.
    The return status is the return status of PIPELINE.  The `-p' option
    prints the timing summary in a slightly different format.  This uses
    the value of the TIMEFORMAT variable as the output format.

Example:

TIMEFORMAT="The command took %Rs"
time {
    sleep 0.1
}

Output:

The command took 0.108s
Answered By: brablc

The accepted solution using time writes to stderr.
The solution using times writes to stdout.
The solutions using $SECONDS are missing sub-second precision.
The other solutions involve calling external programs like date or perf which is not efficient.
If you are fine with any of these, use it.

But if you need an efficient solution to get the times with millisecond precision and need them into variables such that the original output remains undisturbed you may combine process substitution with some redirections around time which is much faster than calling external programs and allows redirections around the timing wrapper script as on the original command/script.

# Preparations:
Cmd=vgs  # example of a program to be timed which is writing to stdout and stderr
Cmd="eval { echo stdout; echo stderr >&2; sleep 0.1; }"  # other example; replace with your own
TIMEFORMAT="%3R %3U %3S"  # make time output easy to parse

Select one of the following variants parsing the output of time appropriate to you needs:
Shortest variant where stdout of $Cmd is written to stderr and nothing to stdout:

read Elapsed User System < <({ time $Cmd 2>&3; } 3>&2 2>&1 >&3)

Longer variant that keeps original stdout and stderr separate of each other:

{ read Elapsed User System < <({ time $Cmd 2>&4; } 4>&2 2>&1 >&3); } 3>&1

Most complicated variant includes closing the extra file descriptors such that $Cmd is called as if without this timing wrapper around it and lvm commands like vgs do not complain about leaked file descriptors:

{ read Elapsed User System < <({ time $Cmd 2>&4 4>&-; } 4>&2 2>&1 >&3 3>&-); } 3>&1

You can even fake a floating point addition in bash without calling bc which would be much slower:

CPU=`printf %04d $((10#${User/.}+10#${System/.}))`  # replace with your own postprocessing
echo CPU ${CPU::-3}.${CPU: -3} s, Elapsed $Elapsed s >&2  # redirected independent of $Cmd

Possible outputs with the two examples of $Cmd on a slow CPU:

File descriptor 3 (/dev/pts/1) leaked on vgs invocation. Parent PID 10756: bash
File descriptor 4 (/dev/pts/1) leaked on vgs invocation. Parent PID 10756: bash
  VG   #PV #LV #SN Attr   VSize VFree
  b3     3  24   0 wz--n- 1.31t 1.19t
CPU 0.052 s, Elapsed 0.056 s

Or:

stdout
stderr
CPU 0.008 s, Elapsed 0.109 s
Answered By: Juergen

Another buildin solution (according to: https://stackoverflow.com/a/385422/1350091) is:

/usr/bin/time -v command
Answered By: baziorek

similar to @LeZuse’s response but supports script arguments:

#!/bin/bash

main () {
   echo "my first argument is $1"
   # script content here
}

# "$@" will expand to the script's arguments
time main "$@"
Answered By: marmor

An alternative if you want to measure multiple parts of a script:

#!/bin/bash
mts=$(date +%s%3N);mtl=$mts

sleep 1
mtc=$(date +%s%3N);printf "line $LINENO: %.3fs [+%.3fs]\n" "$((mtc - mts))e-3" "$((mtc - mtl))e-3";mtl=$mtc

sleep 1.5
mtc=$(date +%s%3N);printf "line $LINENO: %.3fs [+%.3fs]\n" "$((mtc - mts))e-3" "$((mtc - mtl))e-3";mtl=$mtc

output:

line 5: 1.007s [+1.007s]
line 8: 2.514s [+1.507s]

So it returns the total execution time and the time since the last measured line.

Answered By: mgutt
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