Shell script for variance percentage calculation

Question

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).

Thanks

Answer 1

Pseudo floating point using pure bash

I wrote this some years ago:

# 
# Bash source file for percent computing
#
# (C) 2011-2012 Felix Hauri - felix@f-hauri.ch
# 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

This could be used this way:

percent 10 50
20.00%

or to set a variable:

percent 10 50 result
echo $result
20.00%

Pseudo abs() for using wrong negative values

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

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

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

Or with more precision:

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

Could compute:

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

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