Subtracting same column between two rows in awk

Question

I want to subtract the second line with the first line. The file is like this

tmptxt

A B 1 2 3 4 
C D 9 8 7 6

The desired output is

8 6 4 2

How to do this in awk?

I managed to output for a single column only:

awk '{$temp=$3-prev3; prev3=$3}{print $temp}'

Answer 1

A solution with awk

awk '
    NR==1 {split($0,a)}
    NR==2 {split($0,b)}
    END {for(i=1;i<=NF;i++) printf "%d ", b[i]-a[i]}
' input.txt

gives a result of

0 0 8 6 4 2

Since awk interpret strings without valid numbers as 0 during arithmetic operations, in case you want to remove the results in which the source field contains non-numeric values, you can do this by adding an additional condition.

awk '
    NR==1 {split($0,a)}
    NR==2 {split($0,b)}
    END {
        for(i=1;i<=NF;i++)
        if(a[i] ~ /^[0-9]+$/ && b[i] ~ /^[0-9]+$/)
        printf "%d ", b[i]-a[i]
    }
' input.txt

gives a result of

8 6 4 2

Answer 2

Alternative Python solution:

python -c 'import sys; f=open(sys.argv[1],"r"); 
print(" ".join(str(int(d2)-int(d1)) 
for d1,d2 in zip(next(f).split()[2:], next(f).split()[2:]))); f.close()' file

The output:

8 6 4 2

Answer 3

below command also provides same output. l.txt contains the input which you have provided in question

for i in {1..6}; do awk -v i="$i" 'BEGIN{sum=0}{sum=$i-sum}END{print sum}' l.txt ; done| tr "\n" " " | awk '{$1="";$2="";print $0}' | sed "s/^ //g"

output

8 6 4 2