[R] Distance calculation

[arun]

HI,
Try this:


dat1<- read.csv("DQP.csv",sep="\t")
res<- do.call(cbind,lapply(seq_len(nrow(dat1)),function(i) do.call(rbind,lapply(split(rbind(dat1[i,],dat1[-i,]),1:nrow(rbind(dat1[i,],dat1[-i,]))), function(x) {x1<-rbind(dat1[i,],x);colnames(x1)<-gsub("[.]","",colnames(x1));
 if({indx<- colSums(x1[,2:3]==0);indx[1]==0 & indx[2]==1 }) #2 peaks 1 peak comparison
{x2<- x1[order(x1$Peak2t,x1$Npeak2t),];
with(x2,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak1v[1]-Peak2v[2])*((Peak1t[1]+12)-Peak2t[2]))+ abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak1v[1]-Npeak2v[2])*((Npeak1t[1]+12)-Npeak2t[2])) })
}
else #cases where peaks are similar 
{with(x1,{abs((Peak1v[1]-Peak1v[2])*(Peak1t[1]-Peak1t[2]))+abs((Peak2v[1]-Peak2v[2])*(Peak2t[1]-Peak2t[2])) + abs((Npeak1v[1]-Npeak1v[2])*(Npeak1t[1]-Npeak1t[2]))+abs((Npeak2v[1]-Npeak2v[2])*(Npeak2t[1]-Npeak2t[2]))})
}

}))))

res2<-do.call(cbind,lapply(seq_len(ncol(res)),function(i) c(c(tail(res[seq(1,i,1),i],-1),0),res[-c(1:i),i])))
row.names(res2)<-1:nrow(res2)

 dim(res2)
#[1] 124 124
 res2[1:5,1:5]
#    [,1]   [,2]  [,3]  [,4]   [,5]
#1  0.000  1.512 7.031 3.662 13.030
#2  1.512  0.000 7.109 4.880 18.731
#3  7.031  7.109 0.000 0.056  1.280
#4  3.662  4.880 0.056 0.000  0.584
#5 13.030 18.731 1.280 0.584  0.000


A.K.



________________________________
From: eliza botto <eliza_botto at hotmail.com>
To: "smartpink111 at yahoo.com" <smartpink111 at yahoo.com> 
Sent: Wednesday, March 27, 2013 8:58 AM
Subject: RE: Distance calculation



Dear Arun,
I would like to ask a small question.
In the distance calculation procedure, if there are only 2 peaks and 1 peaks stations and there are no 3 and 4 peaks stations, like the file i attached. How to modify the script below to eliminate 3 peaks and 4 peaks scripts??
I hope you mind my hasty questioning
Thanks in advance
Elisa