在bash和awk中查找和排序数据点

Question

First of all, let me clarify that I am unfortunately still quite inexperienced in programming, so I really need some help.

What I have:

I have a data file containing 3 columns: $1=(Energy1) , $2=(Energy2) , $3=(intensity of their frequency in combination) . If I plot these data eg in gnuplot by doing spl "datafile.dat" u 1:2:3 I obtain a surface plot with my 2D-spectrum.

What I want:

Now, I would like to select only certain data points, for which my ($1-$2)=5.7 give this specific value, thus obtaining a line spectrum along a diagonal, with all possible combinations of $1 and $2 yielding this value.

The new data-file should then contain the $1 -value and the intensity (stored in $3 ) corresponding to the selected line, which contained the correct values of $1 and $2 yielding 5.7.

I have tried do do this in bash using awk, but unfortunately until now I failed. PLEASE help me!!! thank you very much in advance.

Answer 1

You do not need awk for this, gnuplot can do it.

admissible(x,y,value,epsilon)=(abs(x-y-value)<epsilon)
plot 'datafile.dat' using (admissible($1,$2,5.7,1e-5)?$1:1/0):3 with points

Function admissible is tested for each line of data file, if it returns true then the point ($1,$3) is plotted, else the x-coordinate is set to undefined (1/0) and thus the point is not plotted. The only shortcoming is that you cannot use the lines style with this, since lines will be interrupted by non-admissible datapoints.

Answer 2

Maybe I don't understand all the issues, or maybe you are having a floating-equal problem as others have noted, but why doesn't a simple filter through the data work?:

awk -v s=5.7 -v e=.01 '{d=$1-$2-$s}d<e&&d>-e{print $1,$3}'

Tack on a sort if you want/need:

| sort -n

Or, is it possible that your data is too sparse, and you're looking for some value interpolation solution?

Answer 3

If you want to compare every $1 against every $2, you need to take 2 passes through the file, once to collect all the $1,$3 pairs, the next to do all the comparisons:

awk -v diff=5.7 '
    NR == FNR {
        # this is the first trip through
        val[$1] = $3
        next
    }
    {
        for (v1 in val) {
            if ( (v1 - $2) == diff ) {
                print v1, val[v1]
            }
        }
    }
' file file   # yes, give the same filename twice.

To address @Baruchel's comment about floating point precision, try this:

awk -v diff=5.7 -v epsilon=0.0001'
    NR == FNR {val[$1] = $3; next}
    {
        for (v1 in val) {
            delta = v1 - $2 - diff
            if (-epsilon <= delta && delta <= epsilon) 
                print v1, val[v1]
        }
    }
' file file