search for: peak2

...ot;) #Subset dat2<-dat1[1:5,] res1<-do.call(cbind,lapply(seq_len(nrow(dat2)),function(i) do.call(rbind,lapply(split(rbind(dat2[i,],dat2[-i,]),1:nrow(rbind(dat2[i,],dat2[-i,]))), function(x) {x1<-rbind(dat2[i,],x); abs((x1$Peak1.v.[1]-x1$Peak1.v.[2])*(x1$Peak1.t.[1]-x1$Peak1.t.[2]))+abs((x1$Peak2.v.[1]-x1$Peak2.v.[2])*(x1$Peak2.t.[1]-x1$Peak2.t.[2]))+abs((x1$Npeak1.v.[1]-x1$Npeak1.v.[2])*(x1$Npeak1.t.[1]-x1$Npeak1.t.[2]))+abs((x1$Npeak2.v.[1]-x1$Npeak2.v.[2])*(x1$Npeak2.t.[1]-x1$Npeak2.t.[2]))})))) res2<-do.call(cbind,lapply(seq_len(ncol(res1)),function(i) c(c(tail(res1[seq(1,i,1),i],-1)...

...he current rate, it will take a few weeks to finish running. I'm looking for suggestions on performing the calculations more efficiently. Here's what I currently have: for (j in 1:dim(genes2)[1]){ for(i in 1:dim(genes1)[1]){ peak1<-as.vector(t(expression.data[genes1[i,1],])) peak2<-as.vector(t(expression.data[genes2[j,1],])) Cor.matrix[i,j]<-cor(peak1,peak2,method='s') } } Thanks so much. Alayne

...000 plot(eval(h), type = 'l') # however this yields an error h <- h/sqrt(sum(h^2)) Error in h^2 : non-numeric argument to binary operator what shall I do to get the matlab: h = h/norm(h) implemented in R? thanks for a hint christoph ---- # parameters peak1 <- 5.4 fwhm1 <- 5.2 peak2 <- 10.8 fwhm2 <- 7.35 dip <- 0.35 b1 <- 0.9 # dispersion b2 <- 0.9 #dispersion a1 <- peak1/b1 a2 <- peak2/b2 d1 <- a1*b1 d2 <- a2*b2 c <- dip

...re are only two peaks: CnBdens<-density(Ys/Xs) #probability density function for ratio of Ys to Xs for(p in 1:512) ifelse(CnBdens$y[p]>CnBdens$y[p-1],peak1<-p,break) #identifies first peak in probability distribution for(p in 1:512) ifelse(CnBdens$y[512-p]>CnBdens$y[512-p +1],peak2<-512-p,break) #identifies second peak in probability distribution but the simulation sometimes produces a small third peak at one end of the distribution. Is there any simple way to identify the two highest maxima in a trimodal distribution? Thanks for any help Rob Knell School of B...

...h) beforehand. Andy > Error in h^2 : non-numeric argument to binary operator > > what shall I do to get the matlab: h = h/norm(h) implemented in R? > > thanks for a hint > > christoph > > > ---- > # parameters > peak1 <- 5.4 > fwhm1 <- 5.2 > peak2 <- 10.8 > fwhm2 <- 7.35 > dip <- 0.35 > > b1 <- 0.9 # dispersion > b2 <- 0.9 #dispersion > a1 <- peak1/b1 > a2 <- peak2/b2 > d1 <- a1*b1 > d2 <- a2*b2 > c <- dip > > ______________________________________________ > R-help at sta...