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 <[email protected]>
To: "[email protected]" <[email protected]>
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
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