Bash: arithmetic addressed by line number and column - bash

Question

I have normally done this with Excel, but as I am trying to learn bash, I'd like to ask for advice here on how to do so. My input file resembles:
# s0 legend "1001"
# s1 legend "1002"
#target G0.S0
#type xy
2.0 -1052.7396157664
2.5 -1052.7330560932
3.0 -1052.7540013664
3.5 -1052.7780321236
4.0 -1052.7948229060
4.5 -1052.8081313831
5.0 -1052.8190310613
&
#target G0.S1
#type xy
2.0 -1052.5384564253
2.5 -1052.7040374678
3.0 -1052.7542803612
3.5 -1052.7781686744
4.0 -1052.7948927247
4.5 -1052.8081704241
5.0 -1052.8190543049
&
where the above only shows two data sets: s0 and s1. In reality I have 17 data sets and will combine them arbitrarily. By combine, I mean I would like to:
For two data sets, extract the second column of each separately.
Subtract these two columns row by row.
Multiply the difference by a constant, $C.
Note: $C multiplies very small numbers and the only way I could get it to not divide by zero was to take a massive scale.
Edit: After requests, I was apparently not entirely clear what I was going for. Take for example:
set0
2 x
3 y
4 z
set1
2 r
3 s
4 t
I also have defined a constant C.
I would like to perform the following operation:
C*(r - x)
C*(s - y)
C*(t - z)
I will be doing this for sets > 1, up to 16, for example (set 10) minus (set 0). Therefore, I need the flexibility to target a value based on its line number and column number, and preferably acting over a range of line numbers to make it efficient.
So far this works:
C=$(echo "scale=45;x=(small numbers)*(small numbers); x" | bc -l)
sed -n '5,11p' input.in | cut -c 5-20 > tmp1.in
sed -n '15,21p' input.in | cut -c 5-20 > tmp2.in
pr -m -t -s tmp1.in tmp2.in > tmp3.in
awk '{printf $2-$1 "\n"}' tmp3.in > tmp4.in
but the multiplication failed:
awk '{printf "%11.2f\n", "$C"*$1 }' tmp4.in > tmp5.in
returning:
0.00
0.00
0.00
0.00
0.00
0.00
0.00
I have a feeling the whole thing can be accomplished more elegantly with awk. I also tried this:
for (( i=0; i<=6; i++ ))
do
n=5+$i
m=10+n
awk 'NR==n{a=$2};NR==m{b=$2} {printf "%d\n", $b-$a}' input.in > temp.in
done
but all I get in temp.in is a long column of 0s.
I also tried
awk 'NR==5,NR==11{a=$2};NR==15,NR==21{b=$2} {printf "%d\n", $b-$a}' input.in > temp.in
but got the error
awk: (FILENAME=input.in FNR=20) fatal: attempt to access field -1052
Any idea how to formulate this with awk, and if that doesn't work, then why I cannot multiply with awk above? Thank you!

Answer 1

this does the math in one go
$ awk -v c=1 '/^&/ {s++}
s==1 {a[$1]=$2}
s==3 {print $1,a[$1],$2,c*(a[$1]-$2)}
/#type/ {s++}' file
2.0 -1052.7396157664 -1052.5384564253 -0.201159
2.5 -1052.7330560932 -1052.7040374678 -0.0290186
3.0 -1052.7540013664 -1052.7542803612 0.000278995
3.5 -1052.7780321236 -1052.7781686744 0.000136551
4.0 -1052.7948229060 -1052.7948927247 6.98187e-05
4.5 -1052.8081313831 -1052.8081704241 3.9041e-05
5.0 -1052.8190310613 -1052.8190543049 2.32436e-05
you can remove the decorations and add print formatting easily. The magic numbers 1=g1 and 3=2*g2-1 correspond to data groups 1 and 2 as the order presented in the data file, can be converted to awk variables as well.
The counter s keeps track of whether you're in a set or not, Odd numbers correspond to sets and even numbers between sets. The increment is done both at the start pattern and end pattern. The order of increment statements were set in such a way they, they are not printed following the pattern (unset first, print set values, reset last}. You can change the order and observe the effects.

Answer 2

This might be what you're looking for:
$ cat tst.awk
/^[#&]/ { lineNr=0; next }
{
++lineNr
if (lineNr in prev) {
print $1, c * ($2 - prev[lineNr])
}
prev[lineNr] = $2
}
$ awk -v c=100000 -f tst.awk file
2.0 20115.9
2.5 2901.86
3.0 -27.8995
3.5 -13.6551
4.0 -6.98187
4.5 -3.9041
5.0 -2.32436

Answer 3

In your first try, you should replace that line:
awk '{printf "%11.2f\n", "$C"*$1 }' tmp4.in > tmp5.in
with that one:
awk -v C=$C '{printf "%11.2f\n", C*$1 }' tmp4.in > tmp5.in
You are mixing notations of bash shell with notation with awk.
in shell you define variable without $, and you use them with $.
Here you are in awk script, there is no $ to use variables. Yet there are some special variables : $1 $2 ...
You have put single quote ' around your awk script, so the shell variables cant be used. I mean you have written $C, but the shell can not see it inside single-quote. That is why you have to write awk -v C=$C so that the shell variable $C is transferred to an awk variable called C.
In your other tries with awk, we can see such errors also. Now I think you'll make it.