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Standard deviation in awk

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# 8  
Old 08-23-2011
Code:
awk 'NR==FNR{a[NR+2]=$1;b[NR+2]=$2;c[NR+2]=$3;next}NR>2{$5=$5" "a[FNR];$6=$6" "b[FNR];$7=$7" "c[FNR]}1' file2 file1

This User Gave Thanks to bartus11 For This Post:
gd9629
# 9  
Old 08-23-2011
My love knows no boundaries for you my friend. Just one final quick thing, for file2, the standard deviation code you gave me before, how would I amend it so that all values were to 4 decimal places?
# 10  
Old 08-23-2011
Code:
function std_dev(data, count) {
    sum=0;
    for( x=1; x <= count; x++) {
        sum += data[x];
    }
    avg = sum/count;
    sumsq=0;
    for( x=1; x <= count; x++) {
        sumsq += (data[x] - avg)^2;
    }
    return sprintf ("%.4f", sqrt(sumsq/count));
}
BEGIN {
    cnt = 0;
}
END {
    std_dev(array17, cnt); std_dev(array19, cnt); std_dev(array20, cnt);
}
NR>1{
   array17[cnt]=$17; array19[cnt]=$19; array20[cnt]=$20; cnt++;
   if (cnt==720) {
      print std_dev(array17, cnt) " " std_dev(array19, cnt) " " std_dev(array20, cnt);
      cnt=0;
   }
}

This User Gave Thanks to bartus11 For This Post:
gd9629
# 11  
Old 08-24-2011
I have an output like this now

Code:
Site: Ridge Hill
yyyy mm dd hh mm ____CO2________SD_______CH4_______SD_______H20_______SD__     
2011 06 09 15 38 4.65205e+02 71.0168 2.03040e+00 0.1782 8.55509e-01 0.0674
2011 06 09 16 38 4.28448e+02 24.3320 1.99402e+00 0.0875 8.19965e-01 0.0448
2011 06 09 17 38 4.20397e+02 19.5077 2.11204e+00 0.0826 8.88530e-01 0.0407
2011 06 09 18 38 4.13831e+02 15.8686 2.20515e+00 0.0852 9.22705e-01 0.0367
2011 06 09 19 38 4.09194e+02 15.5070 2.27311e+00 0.0851 9.36471e-01 0.0350

Problem is I want the data to start from the nearest whole hour (so starts at 14:00). With the averaging script and SD script is there a way of telling awk to start calculating from the nearest hour?

averaging script
Code:
NR==1{
  gsub(" +","\t")
  print
}
NR>1&&(NR-1)%720{
  for (i=3;i<=NF;i++){
    a[i]+=$i
  }
}
NR>1&&!((NR-2)%720){
  t=$1"\t"$2"\t"
}
NR>1&&!((NR-1)%720){
  printf t
  for (i=3;i<=NF;i++){
    printf "%.5e\t",a[i]/720
    a[i]=0
  }
  printf "\n"
}

SD script
Code:
function std_dev(data, count) {
    sum=0;
    for( x=1; x <= count; x++) {
        sum += data[x];
    }
    avg = sum/count;
    sumsq=0;
    for( x=1; x <= count; x++) {
        sumsq += (data[x] - avg)^2;
    }
    return sprintf ("%.4f", sqrt(sumsq/count));
}


BEGIN {
    cnt = 0;
}
END {
    std_dev(array17, cnt); std_dev(array19, cnt); std_dev(array20, cnt);
}
NR>1{
   array17[cnt]=$17; array19[cnt]=$19; array20[cnt]=$20; cnt++;
   if (cnt==720) {
      print std_dev(array17, cnt) " " std_dev(array19, cnt) " " std_dev(array20, cnt);
      cnt=0;
   }

Thanks
Last edited by gd9629; 08-24-2011 at 01:17 PM..
# 12  
Old 08-24-2011
I think you mean it should start averaging from 15:00:00, cause there is no data for 14:00 to 14:38 in your sample file. Anyway, try this on your file:
Code:
awk 'NR==1;$2~"15:00:00"{p=1}p' testfile.txt > testfile2.txt

Then just run previous code without any modifications on testfile2.txt.
This User Gave Thanks to bartus11 For This Post:
gd9629
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