floating point math in bash with variables [duplicate] - bash

Question

This question already has answers here:
How do I use floating-point arithmetic in bash?
(23 answers)
Bash - Calculate the Average of Numbers Inputted
(4 answers)
Closed 1 year ago.
How do you perform floating-point math over variables in bash
the output I get is an integer like number
#! /bin/bash
# finding the average of n numbers
avg=0
read -p "" n
for (( i = 0; i < $n; i++ ))
do
read x
((avg = $avg + $x ))
done
#printf %.3f "$(( avg / n )) "
the goal is to show up to 3 decimal places
3
4
5
6
./avg.sh: line 22: printf: 5 : invalid number
5,000
I tried using | bc but I am missing sth

Answer 1

You can replace the entire loop with awk, which does floating-point math natively:
read -p "" n
awk -v n="$n" '{sum+=$1}; NR==n {printf "%.3f\n", sum/n; exit}'
Explanation: -v n="$n" copies the shell variable n into an awk variable with the same name. {avg+=$1} adds each line of input (well, the first "field" of each line) to a running sum. NR==n {printf "%.3f\n", sum/n; exit} means when it hits the nth line, it prints sum/n to three decimal places and exits.

Answer 2

Found this useful question
Added this section after the for loop
var=$(echo "scale=3; $avg / $n" | bc -l)
echo $var

Answer 3

Here is how you can compute the average of n numbers with POSIX shell arithmetic and have 3 or more decimals:
#!/usr/bin/env sh
decimals=3
# The decimal precision is one digit more than the display decimals,
# so it can have an accurate rounding when formatting.
# The precision factor is 10^(decimals + 1).
precision_factor=$(printf '1%0*d' $((decimals + 1)) 0)
average () {
# Multiply the sum of the arguments by the precision factor,
# and divide by the number of arguments.
# With IFS=+ $* expands to the sum of the arguments $1+$2+$3...
# Example, arguments 5 8 1 7 will expand to 5+8+1+7 because IFS is +
IFS=+ average=$(((($*)*precision_factor)/$#))
# The integer part is the average divided by the precision factor.
integer_part=$((average/precision_factor))
# The decimal part is the average head-stripped from the integer part.
decimal_part=${average#$integer_part}
# Assemble a C-locale (decimal point) string for floating-point average.
float_average="$integer_part.$decimal_part"
# Bash consider floating point arguments to printf must be formatted
# to the current locale. So specify we use the C-locale.
LC_NUMERIC=C printf 'Average: %.*f\n' "$decimals" "$float_average"
}
if [ $# -eq 0 ]; then
printf 'Enter a set of numbers: '
read -r input
average $input
else
average "$#"
fi