用于方差百分比计算的Shell脚本 [英] Shell script for variance percentage calculation

问题描述

我需要计算2个数字之间的方差百分比差异，这可以帮助我在Unix Shell脚本中进行的工作.另外，我喜欢将输出作为abs值(总是+ ve).

I need to calculate variance percentage difference between 2 numbers could some one help me how i can do in unix shell scripting. Also the I like to have the output as abs value (always +ve).

谢谢

推荐答案

使用 pure

Pseudo floating point using pure bash

I wrote this some years ago:

# 
# Bash source file for percent computing
#
# (C) 2011-2012 Felix Hauri - [e-mail]
# Licensed under terms of LGPL v3. www.gnu.org

# after sourcing script:
# syntaxe: percent <amount> <total> [varname]

percent() {
    local p=000$((${1}00000/$2))
    printf ${3+-v} $3 "%.2f%%" ${p:0:${#p}-3}.${p:${#p}-3}
}

export -f percent

可以这样使用:

percent 10 50
20.00%

或设置一个变量:

percent 10 50 result
echo $result
20.00%

伪 abs() 使用错误的负值

percent() {
    local p=000$((${1#-}00000/$2));
    printf ${3+-v} $3 "%.2f%%" ${p:0:${#p}-3}.${p:${#p}-3};
}

这会在您的第一个参数中删除所有minux符号:

This will drop any minux sign in your 1st argument:

value1=3947
value2=5853
percent $((value1-value2)) $value1 result
echo $result
48.29%

或更精确地说:

percent() {
    local p;
    printf -v p 00000%u $((${1#-}0000000/$2));
    printf ${3+-v} $3 "%.4f%%" ${p:0:${#p}-5}.${p:${#p}-5};
}

可以计算:

value1=3947
value2=5853
percent $((value1-value2)) $value1 result
echo $result
48.2898%

当然，由于它使用bash的64位整数，因此仅适用于较小的值:第一个参数不能大于 922337203685 ！

Of course, as this use bash's 64bits integers, this will only work with small values: 1st argument could not be bigger than 922337203685!

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