Implementing `sumproduct` in UNIX shell - bash

Question

I have some output from a script thescript which reads:
202 1 0 1 0 0 0
Now I want to selectively sum this number with awk, depending on the value of a ${SUM_MASK}:
SUM_MASK=1,1,0,0,0,0,0
I would like to have something like:
thescript | awk <SOMETHING>
where the each number output of thescript gets multiplied by the corresponding number in ${SUM_MASK}, obtaining:
203
as result of:
203 = 202 * 1 + 1 * 1 + 0 * 0 + 1 * 0 + 0 * 0 + 0 * 0 + 0 * 0
This would be similar to the sumproduct function in spreadsheet software.
The following code snipets do the trick, but I would like to avoid using process substitution:
SUM_MASK="1,1,0,0,0,0,0"; paste <(thescript) <(echo ${SUM_MASK} | tr ',' '\n') | awk '{ SUM += $1 * $2 } END { print SUM }'
and named pipes:
SUM_MASK="1,1,0,0,0,0,0"; mkfifo fA; mkfifo fB; thescript > fA & echo ${SUM_MASK} | tr ',' '\n' > fB & paste fA fB | awk '{ SUM += $1 * $2 } END { print SUM }' > result.text; rm -f fA fB
how could I achieve that?

Answer 1

echo "202 1 0 1 0 0 0" |
awk -v summask="1,1,0,0,0,0,0" '
BEGIN {split(summask, mask, /,/)}
{ sumproduct=0
for (i=1; i<=NF; i++) {
sumproduct += $i * mask[i]
}
print sumproduct
}
'
203

Answer 2

There's no need for external tools such as awk here -- bash is capable of resolving this with built-in capabilities only. Consider the below implementation as a function:
sumproduct() {
local -a sum_inputs sum_mask
local idx result
# read your sum_inputs into an array from stdin
IFS=', ' read -r -a sum_inputs # this could be <<<"$1" to use the first argument
# and your sum_mask from the like-named variable
IFS=', ' read -r -a sum_mask <<<"$SUM_MASK" # or <<<"$2" for the second argument
# ...iterate over array elements in sum_inputs; find the corresponding sum_mask; math.
result=0
for idx in "${!sum_inputs[#]}"; do
(( result += ${sum_mask[$idx]} * ${sum_inputs[$idx]} ))
done
echo "$result"
}
To test this:
echo "202 1 0 1 0 0 0" | SUM_MASK=1,1,0,0,0,0,0 sumproduct
...correctly yields:
203

Answer 3

You do not actually need sum product, but masked summation, for example this should be faster if you have a lot of masked columns.
$ awk -v mask='1,1,0,0,0,0,0' 'BEGIN {n=split(mask,m,",");
for(i=1; i<=n; i++) if(m[i]) ix[i]}
{sum=0;
for(i in ix) sum += $i;
print sum}' file
203

Answer 4

With the one-digit multipliers you can make a simple loop
SUM_MASK=1,1,0,0,0,0,0
offset=0
sum=0;
for i in 202 1 0 1 0 0 0; do
j="${SUM_MASK:$offset:1}"
((sum += i * j ))
((offset+=2))
done
echo "${sum}"
This solution can be used in a script prodsum that can be called with thescript | prodsum :
offset=0
sum=0;
for i ; do
j="${SUM_MASK:$offset:1}"
((sum += i * j ))
((offset+=2))
done
echo "${sum}"
EDIT: When the SUB_MASK can have numbers>9, use the following:
SUM_MASK=1,10,0,0,0,0,0
sum=0;
remaining_mask="$SUM_MASK"
for i in 202 1 0 1 0 0 0; do
j="${remaining_mask%%,*}"
remaining_mask="${remaining_mask#*,}"
((sum += i * j ))
done
echo "${sum}"