mjhjmtu

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=$3print $temp'

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

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

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="$i" 'BEGINsum=0sum=$i-sumENDprint sum' l.txt ; done| tr "n" " " | awk '$1="";$2="";print $0' | sed "s/^ //g"

output

8 6 4 2