Using awk to interpolate data column based in a data file with date and time

筅森魡賤 提交于 2019-12-01 14:54:17

Here is one solution in Gnu awk. It runs twice for the first given data file, remembers first and last data points (y1, y2) and their timestamps (x2, x2), computes slopes of the points (k=(y2-y1)/(x2-x1)) and inter- and extrapolates values for empty elements ((y=(x1-x)+y1).

It's not fool proof, it doesn't check for division by zeroes or if there are two points for the slopes or any other checks whatsoever.

$ cat inexpolator.awk
BEGIN {
    FS=OFS=";"
    ARGC=3; ARGV[2]=ARGV[1]        # run it twice for first file
}
BEGINFILE {                        # on the second round
        for(i in p)                # compute the slopes
            k[i]=(y2[i]-y1[i])/(x2[i]-x1[i])
}
{
    split($1,a,"[:.]")             # reformat the timestamp
    ts=mktime(a[3] " " a[2] " " a[1] " " a[4] " " a[5] " " a[6])
}
NR==FNR {                          # remember first and last points for slopes
    for(i=2;i<=NF;i++) {
        p[i]
        if(y1[i]=="") { y1[i]=$i; x1[i]=ts }
        if($i!="") { y2[i]=$i; x2[i]=ts }
    }
    next                           # only on the first round
}
{                                  # reformat ts again for output
    printf "%s", strftime("%d.%m.%Y:%H:%M:%S",ts) OFS  # print ts
    for(i=2;i<=NF;i++) {
        if($i=="") $i=k[i]*(ts-x1[i])+y1[i]            # compute missing points
        printf "%.1f%s", $i, (i<NF?OFS:ORS)            # print points
    }
}

Run it:

$ awk -f inexpolator.awk Matrix.txt
13.09.2016:23:44:10;0.0;4.0
13.09.2016:23:44:20;10.0;5.0
13.09.2016:23:44:30;20.0;6.0
13.09.2016:23:44:40;30.0;7.0
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