Bash shell script for 100 iterations and log time

Question

How do I run 100 iterations using a bash shell script? I want to know how long it will take to execute one command (start and end time). I want to keep track which iteration is currently running. I want to log each iteration. I have one automated script I need to run and log it.

for i in 1 2 3
do
    command1
done

But I want to know how long it takes to complete one iteration - and write the information to a log file too!

Answer 1

You may use seq in iteration as well:

for i in `seq 1 100`; do ... done

Answer 2

for ((i = 1; i <= 100; i++)); do
    echo "--- Iteration #$i: $(date) ---"
    time command1
done 2>&1 | tee timing.log

There's a lot going on here. What's happening?

1. The for loop iterates from 1 to 100 using C-style syntax.
2. The $i in the echo printout prints the current iteration number.
3. $(date) inserts a timestamp into each printout.
4. The time command runs a command and prints how long it took to execute.
5. The output from everything inside of the loop is piped to tee, which saves a copy to timing.log.
6. The 2>&1 redirects stderr to stdout so that the log file will contain both regular output and error messages.

Answer 3

You can try with:

for i in {1..100}; do time some_script.sh; done 2>&1 | grep ^real | sed -e s/.*m// | awk '{sum += $1} END {print sum / NR}'

Explanation

• The for loop runs some_script.sh 100 times, measuring its execution time with time
• The stderr of the for loop is redirected to stdout, this is to capture the output of time so we can grep it
• grep ^real is to get only the lines starting with "real" in the output of time
• sed is to delete the beginning of the line up to minutes part (in the output of time)
• For each line, awk adds to the sum, so that in the end it can output the average, which is the total sum, divided by the number of input records (= NR)

Limitations

The snippet assumes that the running time of some_script.sh is less than 1 minute, otherwise it won't work at all. Depending on your system, the time builtin might work differently. Another alternative is to use the time command /usr/bin/time instead of the bash builtin.

Note: This script was extracted from here.

Answer 4

This script is based on the answer from @marian0, but there's no limitation to the running time. Name it timeit.sh and then do ./timeit.sh 10 ./script to run ./script 10 times and print the average time. Additional arguments can be added as desired.

#!/bin/bash

for i in `seq 1 $1`; do
    time "${@:2}"
done 2>&1 |\
    grep ^real |\
    sed -r -e "s/.*real\t([[:digit:]]+)m([[:digit:]]+\.[[:digit:]]+)s/\1 \2/" |\
    awk '{sum += $1 * 60 + $2} END {print sum / NR}'

Answer 5

The following script shows one way to do it.

#!/usr/bin/bash
for i in {1..100} ; do
    echo =============================
    echo "Number $i: $(date +%Y-%m-%d-%H:%M:%S)"
    ( time ( echo $i ; sleep 1 ) ) 2>&1 | sed 's/^/   /'
done | tee timing.log

It uses the bash range feature to run 100 iterations of the loop, outputting the loop counter and date.

It then times your command (echo $i ; sleep 1 in this case) and combines standard output and error before nicely formatting it, and sending it to both the terminal and a log file for later analysis.

A smaple run with five iterations:

pax> testprog.sh
=============================
Number 1: 2010-09-16-13:44:19
   1
   real 0m1.063s
   user 0m0.077s
   sys  0m0.015s
=============================
Number 2: 2010-09-16-13:44:20
   2
   real 0m1.056s
   user 0m0.030s
   sys  0m0.046s
=============================
Number 3: 2010-09-16-13:44:21
   3
   real 0m1.057s
   user 0m0.046s
   sys  0m0.030s
=============================
Number 4: 2010-09-16-13:44:22
   4
   real 0m1.057s
   user 0m0.061s
   sys  0m0.031s
=============================
Number 5: 2010-09-16-13:44:23
   5
   real 0m1.057s
   user 0m0.046s
   sys  0m0.015s