Bash shell script for 100 iterations and log time taken - bash

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

Answer 3

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

Answer 4

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 5

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}'