Sum all positive integers in a bash array

Question

How can I sum all positive integers in a bash array.

Here is my code:

#!/bin/bash

arr=(2 43 -1 -33 24 12 -6)

for (( i = 0; i < ${#arr[@]}; ++i )); do
  if (( arr[i] > 0 )); then
    sum=`expr $sum + $i`
  fi
done
echo "$sum"

Answer 1

First, you need to select the positive numbers, rather than the negative ones

if (( arr[i] > 0 )); then

then you need to sum the array values rather than the indices

sum=`expr $sum + ${arr[i]}`

or (since you are already using the (( ... )) arithmetic evaluation syntax elsewhere)

sum=$((sum + arr[i]))

Answer 2

You could use a trick to replace all array elements starting (#) with a - with nothing.
This way you can skip the check in the loop.

$ arr=( 2 43 -1 -33 24 12 -6 )
$ echo "${arr[@]/#-*}"
2 43   24 12

And the modified script:

#!/bin/bash

arr=( 2 43 -1 -33 24 12 -6 )

sum=0
for i in "${arr[@]/#-*}"; do
  (( sum+=i ))
done

echo "$sum"

Answer 3

This is my new attempt:

# 2+43+24+12=81
sum=0
arr=( 2 43 -1 -33 24 12 -6 )
arr=("${arr[@]/-*/0}")
sum=$(IFS=+; echo "$(( ${arr[*]} ))" )

echo "$sum"
81

Answer 4

Just for fun, you can also use awk to do the looping for you

arr=(2 43 -1 -33 24 12 -6); 
sum=$(echo ${arr[*]} | awk '
    BEGIN{RS=" "}
    $0>0{sum+=$0}
    END{print sum}');
echo $sum
81

Handy if you want to do something more than integer math

arr=(2 43 -1 -33 24 12 -6); 
sqrt=$(echo ${arr[*]} | awk '
    BEGIN{RS=" "}
    $0>0{sum+=$0}
    END{printf "%.2f", sum^0.5}');
echo $sqrt
9.00