bash – 在多次运行中的平均运行时间

这是一个有趣的问题 so I took a few minute and wrote a script for it.下面是脚本的简易版本,它接受一个具有运行time命令输出的文件. The full script还有更多选项,直接接受命令直接运行平均值.

码：

#!/bin/bash
# calculate the mean average of wall clock time from multiple /usr/bin/time results.

file=${1}
cnt=0

if [ ${#file} -lt 1 ]; then
    echo "you must specify a file containing output of /usr/bin/time results"
    exit 1
elif [ ${#file} -gt 1 ]; then
    samples=(`grep --color=never  real ${file} | awk '{print $2}' | cut -dm -f2 | cut -ds -f1`)

    for sample in `grep --color=never real ${file} | awk '{print $2}' | cut -dm -f2 | cut -ds -f1`; do
        cnt=$(echo ${cnt}+${sample} | bc -l)
    done

    # Calculate the 'Mean' average (sum / samples).
    mean_avg=$(echo ${cnt}/${#samples[@]} | bc -l)
    mean_avg=$(echo ${mean_avg} | cut -b1-6)

    printf "tSamples:t%s ntMean Avg:t%snn" ${#samples[@]} ${mean_avg}
fi

示例(这里我命名脚本timeit.sh和chmod 775 timeit.sh和chown jon timeit.sh *)：

[ 09:22 jon@hozbox.com ~ ]$/usr/bin/time -a -o times.log -p sleep 3
[ 09:23 jon@hozbox.com ~ ]$/usr/bin/time -a -o times.log -p sleep 1
[ 09:23 jon@hozbox.com ~ ]$/usr/bin/time -a -o times.log -p sleep 2
[ 09:23 jon@hozbox.com ~ ]$/usr/bin/time -a -o times.log -p sleep 2
[ 09:23 jon@hozbox.com ~ ]$/usr/bin/time -a -o times.log -p sleep 7
[ 09:23 jon@hozbox.com ~ ]$/usr/bin/time -a -o times.log -p sleep 0.5
[ 09:23 jon@hozbox.com ~ ]$./timeit.sh times.log
        Samples:        6
        Mean Avg:       2.5833

[ 09:23 jon@hozbox.com ~ ]$cat times.log
real 3.00
user 0.00
sys 0.00
real 1.00
user 0.00
sys 0.00
real 2.00
user 0.00
sys 0.00
real 2.00
user 0.00
sys 0.00
real 7.00
user 0.00
sys 0.00
real 0.50
user 0.00
sys 0.00

* chown不是必需的,但我无法帮助它！

（编辑：李大同）

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