How to compare two decimal numbers in bash/awk?

Question

You can do it using Bash's numeric context:

if (( $(echo "$result1 > $result2" | bc -l) )); then

bc will output 0 or 1 and the (( )) will interpret them as false or true respectively.

The same thing using AWK:

if (( $(echo "$result1 $result2" | awk '{print ($1 > $2)}') )); then

if awk 'BEGIN{exit ARGV[1]>ARGV[2]}' "$z" "$y"
then
  echo z not greater than y
else
  echo z greater than y
fi

if [[ `echo "$result1 $result2" | awk '{print ($1 > $2)}'` == 1 ]]; then
  echo "$result1 is greater than $result2"
fi

Following up on Dennis's reply:

Although his reply is correct for decimal points, bash throws (standard_in) 1: syntax error with floating point arithmetic.

result1=12
result2=1.27554e-05


if (( $(echo "$result1 > $result2" | bc -l) )); then
    echo "r1 > r2"
else
    echo "r1 < r2"
fi

This returns incorrect output with a warning although with an exit code of 0.

(standard_in) 1: syntax error
r1 < r2

While there is no clear solution to this (discussion thread 1 and thread 2), I used following partial fix by rounding off floating point results using awk followed by use of bc command as in Dennis's reply and this thread

Round off to a desired decimal place: Following will get recursive directory space in TB with rounding off at the second decimal place.

result2=$(du -s "/home/foo/videos" | tail -n1 | awk '{$1=$1/(1024^3); printf "%.2f", $1;}')

You can then use bash arithmetic as above or using [[ ]] enclosure as in following thread.

if (( $(echo "$result1 > $result2" | bc -l) )); then
    echo "r1 > r2"
else
    echo "r1 < r2"
fi

or using -eq operator where bc output of 1 is true and 0 is false

if [[ $(bc <<< "$result1 < $result2") -eq 1 ]]; then
    echo "r1 < r2"
else
    echo "r1 > r2"
fi