search for: res2

I am doing cluster analysis [hclust(Dist, method="average")] on data that potentially contains redundant objects. As expected, the inclusion of redundant objects affects the clustering result, i.e., the data a1, = a2, = a3, b, c, d, e1, = e2 is likely to cluster differently from the same data without the redundancy, i.e., a1, b, c, d, e1. This is apparent when the outcome is visualized

Hi there, I have some data which are convenient to enter as lists. For example: t1<-list(fname="animal1",testname="hyla",dspkr="left",res1=39.7,res2=15.0) t2<-list(fname="animal1",testname="bufo",dspkr="left",res1=14.4,res2=56.1) t3<-list(fname="animal2",testname="hyla",dspkr="right",res1=22.6,res2=11.8) I would like to generate a dataframe, but *not* the way this approach works...

...mber, and 1 matrix) using optim() and BFGS method. Here is my code: # fonction to maximize g=function(x) { x1 = x[1:ncol(X)] x2 = x[(ncol(X)+1)] x3 = matrix(x[(ncol(X)+2):(ncol(X)+1+ncol(X)*ncol(Y))],nrow=ncol(X),ncol=ncol(Y)) x4 = x[(ncol(X)+1+ncol(X)*ncol(Y)+1):length(x)] res1=rep(0,nrow(X)) res2=matrix(0,nrow=nrow(X),ncol=ncol(Y)) log.res2=matrix(0,nrow=nrow(X),ncol=ncol(Y)) res2.b=rep(0,nrow(X)) res3 = rep(0,nrow(X)) res3.b = rep(0,nrow(X)) for (i in 1:nrow(X)) { res1[i]=1/(1+exp(-t(x1)%*%X[i,]-x2)) for (t in 1:ncol(Y)) { res2[i,t] = ((1-(1+exp(-t(x3[,t])%*%X[i,]-x4[t]))^(-1))^(abs(Y[i,t]...

...= select i1 %cmp, i32 %mul, i32 %mul2 Source code for above pattern: return (a*b) >=0 ? (a*b) : -a*b; Currently, llvm(-O3) can not recognize this as abs(a*b). I initially think we could do this in instCombine phase in opt. Below is what I think: %res = OP i32 %b, %a %sub = sub i32 0, %b %res2 = OP i32 %sub, %a We could do the transform: %res2 = OP i32 %sub, %a ==> %res2 = sub i32 0, %res Then we can get the advantage: 1: if %res2 is the only user of %sub, %sub can be eliminated; 2: if %res2 is not the only user of %sub, we could change some heavy instruction like div to sub; 3:...

Thank you very much for your reply. Your code work well with this example. I modified a little to fit my real data, I got an error massage. Error in split.default(x = seq_len(nrow(x)), f = f, drop = drop, ...) : Group length is 0 but data length > 0 On Thu, Jan 31, 2013 at 12:21 PM, arun kirshna [via R] < ml-node+s789695n4657196h87@n4.nabble.com> wrote: > Hi, > Try this: >

Hi, I have a program written in R which is good on the version 1.2, but for the fallowing versions of R, an error always is at the same place. That is at the level of the fallowing line: Sur<- getInitial(res2[m:M,2]~SSasymp(res2[m:M,1],Asymp,resp0,lrc),data=res2) Error in eval(expr,envir,enclos):numeric envir arg not of length one I don't know at all this langage for the instant. Thanks, Christelle ------------------------------------------------------------- NetCourrier, votre bureau virtue...

Hi Roman, Thanks for your good idea. I think it can solve the abs issue very well. I can continue with my work now^-^. But if it is not abs and there is no select, %res = OP i32 %b, %a %sub = sub i32 0, %b %res2 = OP i32 %sub, %a theoretically, we can still do the following transform for the above pattern: %res2 = OP i32 %sub, %a ==> %res2 = sub i32 0, %res Not sure whether we can do it in instCombine. Thanks. BRS// Chen Zheng Power Compiler Backend Developer From: Roman Lebedev <lebedev.ri...

...else res<-NA return(res)} Then, I tried to include some lines to take this "swap" function into account by "splitting" somehow my function in 2: myFunc<- function(x,y,z,swap) { res1= NULL res2=NULL if(z =="plus") res1=x+y else if(z=="minus") res1=x-y else if(z== "multiply") res1=x*y else if(z=="divide") res1=x/y else res1<-NA if(z ==...

Hello R community, I am trying to combine two CSV files that look like this: File A Row_ID_CR, Data1, Data2, Data3 1, aa, bb, cc 2, dd, ee, ff File B Row_ID_N, Src_Row_ID, DataN1 1a, 1, This is comment 1 2a, 1, This is comment 2 3a,

Dear R users, I have a data structure as follows: id two res1 res2 c1 c2 inter 1 -0.786093166 1 0 1 2 6 3 -0.308495749 1 0 0 1 2 5 -0.738033048 1 0 0 0 1 7 -0.52176252 1 0...

...,26,28,18,20,34,28,16,10,18,20,14,34,22,30,44,30) Stv.df <- data.frame(rt=data,subj=factor(rep(paste("subj",1:5,sep=""),4)),drug=factor( rep(paste("drug",1:4,sep=""),c(5,5,5,5)))) res0 <- aov(rt~drug,data=Stv.df) res1 <- aov(rt~drug+subj,data=Stv.df) res2 <- aov(rt~drug+Error(subj/drug),data=Stv.df) res0$resid works resid(res0) works res0$df.residual works res1$resid works resid(res1) works res1$df.residual work res2$resid doesn't work and reports NULL resid(res2) doesn't work and reports NULL res2$df.residual doesn't work and repo...

...t2[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),0),res1[-c(1:i),i]))) row.names(res2)<-1:nrow(res2) ?res2 #????? [,1]???? [,2]???? [,3]???? [,4]???? [,5] #1?? 0.0000?? 0.0000?? 0.0000 379.1364?? 0.0000 #2?? 0.0000?? 0.0000?? 0.0000 312.8267?? 0.0000 #3??...

Hi, (Please use ?dput() to share the example dataset. Avoid using images to show dataset. Also, please read the posting guide esp. regarding home work, assignments etc.) res <- sapply(Gene[,-1],function(x) tapply(x,list(Gene$Genotype),mean)) #or res2 <-? aggregate(.~Genotype, data=Gene,mean) #or library(plyr) ?res3 <- ddply(Gene,.(Genotype),numcolwise(mean)) identical(res2,res3) #[1] TRUE resNew <- data.frame(Genotype=rownames(res),res,stringsAsFactors=FALSE) ?attr(resNew,"row.names") <- attr(res2,"row.names&quo...

Hi, I use lme to fit models like R> res1 <- lme(y~A+B, data=mydata, random=~1|subject) R> res2 <- lme(y~B+A, data=mydata, random=~1|subject) (only difference between these two models are the sequence in which the indep variables are written in formula) where y is continuous and A, B, and subject are factors. To get ANOVA table I used R> anova(res1) R> anova(res2) and found ANOVA...

Hi everybody, I am beginer in R and I need your precious help. I want to create a small function in R as in sas to retrieve date. I have a file with data that import in R. DATE PAYS nb_pays.ILI. 1 24/04/2009 usa 0 2 24/04/2009 usa 0 3 24/04/2009 Mexique 0 4 24/04/2009

Hello I am very new to R.  I have an R task to complete that I have not been able to find a straightforward answer to as of yet.  I have a list of values. I would like to count the number of values that are in one bin, the number that fall in the next bin, etc. For example My input file is:  123 48 342 442 43 232 32 129 191 147 I would like the output to be similar to:  0-100 3 100-200 4

..., b=1, c=1)) call.fitfunc <- c(list(fitfunc), as.name("a"), as.name("b"), as.name("c"), as.name("x")) call.fitfunc <- as.call(call.fitfunc) frml <- as.formula("y ~ eval(call.fitfunc)") #"computed" call of nls: res2 <- nls(frml, start = c(a=1, b=1, c=1)) list(res1 = res1, res2 = res2) } #----------------------------------------- the argument `n' defines the number of (simulated) data points x/y which are going to be fitted by some model ( here y ~ a*exp(-b*x)+c ) the first call to `nls' i...

Using R version 2.4.1 (2006-12-18) on Windows, I have a dataset which resembles this: id att1 att2 att3 1 1 1 0 2 1 0 0 3 0 1 1 4 1 1 1 ratings <- data.frame(id = c(1,2,3,4), att1 = c(1,1,0,1), att2 = c(1,0,0,1), att3 = c(0,1,1,1)) I would like to get a cross tab of counts of co-ocurrence, which might resemble this:

Hi. Since a customer requested us that feature, I wrote this little patch for app_dial to allow to play an announcement to the called party, as soon he answers. you can define the file to play in the dial() option, using A(filename). for example: exten => blah,1,Dial(Zap/blah,30,rA(/my/own/announce)Tt) that doesn't break anything ... feel free to blame me for anything bad this patch

Dear R-users, I was wondering if someone could give me some advices on the following problem. I tried to apply moran’s test to a small dataset and couldn’t succeed, here is the error message:   mor <- moran.test(x, res2)   Avis dans moran.test(x, res2) :   Out-of-range p-value: reconsider test arguments   mor     Moran's I test under randomization data:  x  weights: res2   Moran I statistic standard deviate = NaN, p-value = NA alternative hypothesis: greater sample estimates: Moran I statistic ...