I need to do the follwing with bash what is the elegant way ? to get the final $sum value
worker_machine=32
executors_per_node=3
executer=$worker_machine/$executors_per_node-1
spare=$executer X 0.07
sum=$executer-$spare ( with round the number to down )
example:
32/3 = 10 - 1 = 9
9 X 0.7 = 0.6
9 – 0.6 = 8 ( with round the number to down )
Using awk, taking values from shell variables:
awk -v n="$worker_machine" -v m="$executors_per_node" \
'BEGIN { printf("%d\n", 0.93 * (n / m - 1)) }' /dev/null
The awk script doesn't get any input as usual, so we use /dev/null as input file and do our calculation and output in a BEGIN block.
Using bc:
sum=$( printf '0.93 * (%d / %d - 1)\n' "$worker_machine" "$executors_per_node" | bc )
printf '%.0f\n' "$sum"
Using dc:
sum=$( printf '%d\n%d\n/\n1\n-\n0.93\n*\np\n' "$worker_machine" "$executors_per_node" | dc )
printf '%.0f\n' "$sum"
BEGIN block, it would wait for input from the terminal if not provided a file.
The shell is able to to the math (as integers, which round down):
$ sum=$(( ( worker_machine/executors_per_node-1 ) * 93 / 100 ))
$ echo "$sum"
8
The default scale for bc is 0, so the results of division would be integers:
$ sum=$(bc <<<"($worker_machine / $executors_per_node - 1)*93/100")
$ echo "$sum"
8
As we want an integer, we may ask awk for one
$ sum=$(awk -v m="$worker_machine" -v n="$executors_per_node" 'BEGIN{ print( int((m/n-1)*93/100) )}' /dev/null)
$ echo "$sum"
8
$(( ... )). The best option would be to useawkIMHO.