Pairwise addition of elements from two arrays with different lengths

Question

I'm trying to figure out a way to add each element of two arrays together.

For example:

list1=(1 2 3 4 5)    
list2=(32 45 5443 543 332 9889797)

Output:

33
47
5446
547
337
9889797

Note that one array may be longer than the other.

This is where I have got to, however, it fails when the second array is longer than the first.

list=(1 2 3 4 5)

list2=(32 45 5443 543 332 9889797)

x=0

for i in "${list[@]}"
do
    value1=$i
    value2="${list2[x]}"
    echo $((value1 + value2))
    x=$(($x + 1))
done

Answer 1

If you don't know a priori which array will be longer, iterate directly over indices from both:

# generate sum for all indices in a *or* b
declare -A finished=( )                # track which indices we've already added
for idx in "${!a[@]}" "${!b[@]}"; do   # iterate over indices from *both* arrays
  [[ ${finished[$idx]} ]] && continue  # skip anything tracked as finished
  printf '%s\n' "$idx => $(( ${a[$idx]:-0} + ${b[$idx]:-0} ))"
  finished[$idx]=1
done

This works with all kinds of arrays, including sparse arrays or associative arrays, so long as the indices used in arrays a and b match.

Let's take an example:

 #    0  1  2  3  4  5  6  7  8   ## indices
 a=(  1  2  3  4  5  6  7  8  9 )
 b=( 10 20 30 40 50 60 70 80 90 )

 # remove "3", at index 2, from array a
 unset a[2]

Now, after that unset a[2] was run, the highest index in a is still 8, but ${#a[@]} is 8 instead of 7; the rule that the highest index is one less than the number of entries in the array is broken.

However, this answer still works:

0 => 11
1 => 22
3 => 44
4 => 55
5 => 66
6 => 77
7 => 88
8 => 99
2 => 30

We're still aligning the 3 and the 30 by their index values, and treating the absent element 2 from a as a 0.

---

When run with:

declare -A a=( [foo]=1 [bar]=1 [baz]=1 ) b=( [bar]=1 [baz]=1 [qux]=1 )

...this emits:

bar => 2
baz => 2
foo => 1
qux => 1

---

To give a concrete example of a case where this works and many of the other answers don't:

a=( [10234]=20 [25432]=30 )
b=( [10234]=1  [25432]=2  )

...and the result is properly:

10234 => 21
25432 => 32

As another:

declare -A a=( [bob]=20 [jim]=30 )
declare -A b=( [bob]=1  [jim]=2  )

and the proper result:

bob => 21
jim => 32

---

I'm printing indices in the above to demonstrate more about what the code is doing under-the-hood, but of course you can just remove $idx => from the format strings to exclude them.

Answer 2

a=(4 5 3 2 6 2)
b=(7 5 3 4 6)

for ((i=0; i<${#a[@]} || i<${#b[@]}; i++)); do echo $((a[i] + b[i])); done

yields:

11
10
6
6
12
2

With b=(7 5 3 4 6 9 9 9 9) it yields:

11
10
6
6
12
11
9
9
9

Answer 3

If you iterate over the longer array, that should work:

list=(1 2 3 4 5)
list2=(32 45 5443 543 332 9889797)
x=0

for i in "${list2[@]}"
do
 value1=$i
 value2="${list[x]}"
 echo $((value1 + value2))
 x=$(($x + 1))
done

Answer 4

I am not sure I follow why this is so difficult?? Simply get the length of each array and use the indices from the shorter one.

list1=(1 2 3 4 5)
list2=(32 45 5443 543 332 9889797)

a1_len=${#list1[*]}
a2_len=${#list2[*]}

if (( a1_len > a2_len ))
then
    short_arr=("${list2[@]}")
    long_arr=("${list1[@]}")
    (( diff = a2_len - 1 ))
else
    short_arr=("${list1[@]}")
    long_arr=("${list2[@]}")
    (( diff = a1_len - 1 ))
fi

for i in ${!short_arr[*]}
do
    out_arr+=($((${short_arr[i]} + ${long_arr[i]})))
done

out_arr+=(${long_arr[@]:$diff})

echo "${out_arr[@]}"