bash- find average of numbers in line

Question

I am trying to read a file line by line and find the average of the numbers in each line. I am getting the error: expr: non-numeric argument

I have narrowed the problem down to sum=expr $sum + $i, but I'm not sure why the code doesn't work.

while read -a rows
do
    for i in "${rows[@]}"
    do
        sum=`expr $sum + $i`
        total=`expr $total + 1`
    done
    average=`expr $sum / $total`
done < $fileName

The file looks like this (the numbers are separated by tabs):

1       1       1       1       1
9       3       4       5       5
6       7       8       9       7
3       6       8       9       1
3       4       2       1       4
6       4       4       7       7

Answer 1

With some minor corrections, your code runs well:

while read -a rows
do
    total=0
    sum=0
    for i in "${rows[@]}"
    do
        sum=`expr $sum + $i`
        total=`expr $total + 1`
    done
    average=`expr $sum / $total`
    echo $average
done <filename

With the sample input file, the output produced is:

1
5
7
5
2
5

Note that the answers are what they are because expr only does integer arithmetic.

Using sed to preprocess for expr

The above code could be rewritten as:

$ while read row; do expr '(' $(sed 's/  */ + /g' <<<"$row") ')' / $(wc -w<<<$row); done < filename
1
5
7
5
2
5

Using bash's builtin arithmetic capability

expr is archaic. In modern bash:

while read -a rows
do
    total=0
    sum=0
    for i in "${rows[@]}"
    do
        ((sum += $i))
        ((total++))
    done
    echo $((sum/total))
done <filename

Using awk for floating point math

Because awk does floating point math, it can provide more accurate results:

$ awk '{s=0; for (i=1;i<=NF;i++)s+=$i; print s/NF;}' filename
1
5.2
7.4
5.4
2.8
5.6