How do I calculate sum of numbers from different rows?

Question

I have a file containing numbers in many rows and 5 columns like this:

1  2  3  4  5   
6  7  8  9  10   
11 12 13 14 15   
16 17 18 19 20   
21 22 23 24 25   
...

If I want to get sums of the numbers in the 1st row and 3rd row in this way: 1+11, 2+12, 3+13, 4+14, 5+15 which means what I want to see finally is 12 14 16 18 20 What should I do?

Answer 1

Solution 1

#!/bin/bash
{ read -ra line1; read _; read -ra line3; }
for i in "${!line1[@]}"; do
  result+=("$(( ${line1[i]} + ${line3[i]} ))")
done
printf %s\\n "${result[*]}"

Example:

$ bash script < yourfilename
12 14 16 18 20
$ 

Script explanation

{ read -ra line1; read _; read -ra line3; } This part reads the first three lines of the file, keeping only the first two lines. It also splits them into arrays line1 and line3. It assumes a stream of data from stdin. See the example on how to do that.

for i in "${!line1[@]}"; do result+=("$(( ${line1[i]} + ${line3[i]} ))"); done This part loops over the elements of one of the two arrays, and sums the same elements from line1 and line2. The results are put into a new array called result

printf %s\\n "${result[*]}" Prints the computed array elements space-separated.

---

Solution 2

Another solution avoiding loops, assuming the input has 5 columns as requested:

#!/bin/bash
{ read a b c d e; read _; read v w x y z; }
echo "$((a+v)) $((b+w)) $((c+x)) $((d+y)) $((e+z))"

Example:

$ cat yourfilename
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
...
$ bash script < yourfilename
12 14 16 18 20
$ 

Script explanation

{ read a b c d e; read _; read v w x y z; } Reads the numbers from line 1 (resp. line 3) into variables a to e (resp. v to z)

echo "$((a+v)) $((b+w)) $((c+x)) $((d+y)) $((e+z))" Prints the sum of the variables space-separated as requested

Answer 2

$ sed -n '1p;3p' file | datamash -W sum 1-5
12      14      16      18      20

This first uses sed to only output lines 1 and 3. These are then fed into GNU datamash, which is told to sum each individual column from column 1 to 5. The -W instructs datamash to treat any run of whitespaces as field delimiters.

The output is tab-delimited. To get space-delimited output, use the --output-delimiter=' ' option with datamash:

$ sed -n '1p;3p' file | datamash -W --output-delimiter=' ' sum 1-5
12 14 16 18 20

Answer 3

awk 'NR==1 || NR==3 { for (i=1; i<=NF; i++) a[i]+=$(i)}END{for (i in a) printf "%s ", a[i]; print ""}'

Example:

$ awk 'NR==1 || NR==3 { for (i=1; i<=NF; i++) a[i]+=$(i)}END{for (i in a) printf "%s ", a[i]; print ""}'' yourfilename
12 14 16 18 20
$ 

EDIT: To stop reading lines from the file after the 3rd, we can exit early. You won't see a sensible difference with files containing less than 10k lines though.

awk 'NR==1 {split($0,a)} NR==3 {for (i in a) printf "%s ", a[i]+$i; print ""; exit}'

Proof that it works with your example:

$ cat yourfilename
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
21 22 23 24 25
...
$ awk 'NR==1 {split($0,a)} NR==3 {for (i in a) printf "%s ", a[i]+$i; print ""; exit}' yourfilename
12 14 16 18 20
$