Reputation: 2077
I have a datafile with 10 columns as given below
ifile.txt
2 4 4 2 1 2 2 4 2 1
3 3 1 5 3 3 4 5 3 3
4 3 3 2 2 1 2 3 4 2
5 3 1 3 1 2 4 5 6 8
I want to add 11th column which will show the standard deviation of each rows along 10 columns. i.e. STDEV(2 4 4 2 1 2 2 4 2 1) and so on. I am able to do by taking tranpose, then using the following command and again taking transpose
awk '{x[NR]=$0; s+=$1} END{a=s/NR; for (i in x){ss += (x[i]-a)^2} sd = sqrt(ss/NR); print sd}'
Can anybody suggest a simpler way so that I can do it directly along each row.
Upvotes: 1
Views: 534
Reputation: 67567
You can do the same with one pass as well.
awk '{for(i=1;i<=NF;i++){s+=$i;ss+=$i*$i}m=s/NF;$(NF+1)=sqrt(ss/NF-m*m);s=ss=0}1' ifile.txt
Upvotes: 3
Reputation: 11246
Do you mean something like this ?
awk '{for(i=1;i<=NF;i++)s+=$i;M=s/NF;
for(i=1;i<=NF;i++)sd+=(($i-M)^2);$(NF+1)=sqrt(sd/NF);M=sd=s=0}1' file
2 4 4 2 1 2 2 4 2 1 1.11355
3 3 1 5 3 3 4 5 3 3 1.1
4 3 3 2 2 1 2 3 4 2 0.916515
5 3 1 3 1 2 4 5 6 8 2.13542
You just use the fields instead of transposing and using the rows.
Upvotes: 2