How to calculate anomaly using awk - awk

Question

A have a file:
file.txt
1 32
2 34
3 32
4 43
5 25
6 34
7 65
8 34
9 23
10 44
I would like to find anomaly on send column:
my below script printing anomalies considering row-2 to row-10 values. It is not considering row-1 values.
awk 'FNR==NR{
f=1;
if($1 >= 1 && $1 <= 10){
count++;
SUM+=$2;
};
next
}
FNR==1 && f==1{
AVG=SUM/count;
next
}
($1 >= 1 && $1 <= 10){
print $1, $2-AVG
}
' file.txt file.txt
My desire output:
1 -4.6
2 -2.6
3 -4.6
4 6.4
5 -11.6
6 -2.6
7 28.4
8 -2.6
9 -13.6
10 7.4
I got a solution of it:
awk '{f=$1>=1 && $1<=10}f && NR==FNR{sum+=$2; c++; next}f{ print $1, $2-(sum/c) }' file.txt file.txt
I am still wondering why the first script is not giving correct answer.

Answer 1

Since this is just 2 columns file, this can be done in a single pass awk also:
awk '{map[$1] = $2; s += $2}
END {mean = s/NR; for (i in map) print i, map[i] - mean}' file
1 -4.6
2 -2.6
3 -4.6
4 6.4
5 -11.6
6 -2.6
7 28.4
8 -2.6
9 -13.6
10 7.4

Answer 2

The first script in the OP is not giving the correct value, because you skip the first line in the second pass of your file. This is seen in the statement (FNR==1 && f==1) { AVG=sum/count; next }. Due to the next statement, you skip the computation of the deviation from the mean value for the first record.
This is an efficient computation of the deviation from the mean in a double pass:
awk '(NR==FNR){s+=$2;c++;next}
(FNR==1){s/=c}
{print $1,$2-s}' file file
If file contains values bigger than 10 or smaller than 1 in the first, column, but you only want to see this for values in the range of [0,10], then you can do:
awk '($1<1 || $1>10) {next}
(NR==FNR){s+=$2;c++;next}
(FNR==1){s/=c}
{print $1,$2-s}' file file
There are still other optimizations that can be done, but these only become beneficial when working with extremely large files (many millions of lines).