search for: res3

Recent sensible changes to the dynload mechanism have made an old problem resurface: how should we deal with packages which contain Fortran code and may need to be linked against additional libraries such as -lf2c? The current consensus is that extra Fortran libraries maybe needed are handled via the make variable FLIBS, and that `-lf2c' or `-lg2c' are added by default if g77 is used.

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,

...t;,"y1","m","n","x","y")? >>>>attr(res2,"out.attrs")<-NULL? >>>>tail(res2) >>>> >>>>#install.packages(pkgs="plyr") >>>>library(plyr)? >>>> >>>>res3<- join(res2,d4,by=c("m1","n1","x1","y1"),type="inner")? >>>>head(res3) >>>>tail(res3) >>>> >>>>res3<-res3[,c(1:16)] >>>>head(res3) >>>>tail(res3) >>>> >>...

...) { 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]-Yb[i])))*(((1+exp(-t(x3[,t])%*%X[i,]-x4[t]))^(-1))^(1-abs(Y[i,t]-Yb[i]))) log.res2[i,t]=log(res2[i,t]) res...

...------------------------------------ s <- matrix(runif(1600,0,1000), nrow=20) a.res1 <- pvclust(t(s), method.hclust="complete", method.dist="euclidian", nboot=500) a.res2 <- pvclust(t(s), method.hclust="complete", method.dist="maximum", nboot=500) a.res3 <- pvclust(t(s), method.hclust="complete", method.dist="canberra", nboot=500) a.res4 <- pvclust(t(s), method.hclust="ward", method.dist="euclidian", nboot=500) a.res5 <- pvclust(t(s), method.hclust="ward", method.dist="maximum", n...

...y = 0.1), ylim = c(0,1)) curve(DichteGam, 0,4, add = TRUE, col = 2) hist(res2, freq = FALSE, xlim = c(0,4), breaks = seq(0,100,by = 0.1), ylim = c(0,1)) curve(DichteGam, 0,4, add = TRUE, col = 2) ## If we repeat the procedure, using the fix intial value 1, ## the situation is even worse res3 <- NULL for(i in 1:1000) res3[i] <- arms(1, logDichteGam, function(x) (x>0)&(x<100), 1) ## Generating a sample of thousand observations with 1 call of arms res4 <- arms(1, logDichteGam, function(x) (x>0)&(x<100), 1000) ## Plot of the samples par(mfrow = c(2,2)) plo...

...run my model 3 times with different random number seeds and differently dispersed multivariate normal priors. For example: res1 = MCMClogit(y~x,b0=0,B0=0.001,data=mydat, burnin=500, mcmc=5500, seed=1234, thin=5) res2 = MCMClogit(y~x,b0=1,B0=0.01,data=mydat, burnin=500, mcmc=5500, seed=5678, thin=5) res3 = MCMClogit(y~x,b0=5,B0=0.0001,data=mydat, burnin=500, mcmc=5500, seed=91011, thin=5) Each result produces an object of class mcmc. In order to use the Gelman-Rubin diagnostic test via coda, I need to "combine" these 3 mcmc objects appropriately. I thought that this would be possible us...

Hi everyone- I have a dataset with multiple "pre" and "post" variables I want to compare. The variables are named "apple_pre" or "pre_banana" with the corresponding post variables named "apple_post" or "post_banana". The variables are in no particular order. apple_pre orange_pre orange_post pre_banana apple_post post_banana person_1

...se t-test (for location, and a simple t-test for job) appropriate in this case? data = read.table("data.txt", header=T, nrows=90) attach(data) res1 = pairwise.t.test(all, location, p.adj="bonf") print(res1) res2 = pairwise.t.test(M, location, p.adj="bonf") print(res2) res3 = pairwise.t.test(OA, location, p.adj="bonf") print(res3) res4 = pairwise.t.test(UE, location, p.adj="bonf") print(res4) res1 = t.test(all~job) print(res1) res2 = t.test(M~job) print(res2) res3 = t.test(OA~job) print(res3) res4 = t.test(UE~job) print(res4) I'd also like to...

...) 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") ?identical(resNew,res2) #[1] TRUE A.K. So I'm having a prob...

...simpleRDA2(Y,X[,j],SS.Y) if(toto$Rsquare > R2cum) { R2cum <- toto$Rsquare no.sup <- j } } mm <- 1 FP <- FPval(R2cum,R2prev,n,mm,p) if(FP$pval <= alpha) { adjRsq <- RsquareAdj(R2cum,n,mm) res1 <- var.names[no.sup] res2 <- no.sup res3 <- R2cum res4 <- R2cum res5 <- adjRsq res6 <- FP$Fstat res7 <- FP$pval X.out[no.sup] <- 0 delta <- R2cum } else { stop("Procedure stopped (alpha criterion): pvalue for variable ",no.sup," is ",FP$pval) } ## Add variables X...

...ry))) { current_try<-ltry[j] for (k in (1:length(lalgo))) { current_algo<-lalgo[k] res2<-res[res$instance==current_instance & res$try==current_try & res$idalgo==current_algo,] # res2 contains for a given instance, a given try, and a given algo, all results res3<-res2[res2$best==min(res2$best),] res4<-res3[res3$time==min(res3$time),] if (nrow(res4)>1) { res4<-res4[1,] } if (nrow(res4)==1) { res4$best<-(res4$best*100/optimal_values[optimal_values $instance==linstance[i],]$optimum)-100 print(res4) bestal...

...to assign colnames within the loop, so I won't have to type the same thing 20 times over. I have concocted this really goofy example which constructs two datasets: ----- male <- rep(0:1, each=5) age <- factor(c(10:14,10:14)) DF <- data.frame(male, age, res1=rnorm(10), res2=rnorm(10), res3=rnorm(10)) for(n in 0:1) { assign(paste("test",n, sep="."), as.data.frame(t(subset(DF, male==n, select=c(res1, res2, res3))))) colnames(get(paste("test",n, sep="."))) <- paste("age",levels(age), "m", n, sep="") # This line...

...598 -0.07347383 0.7608840 [2,] -0.2319540 0.79957999 0.3438537 [3,] 0.2745904 0.41276093 -0.4028185 [4,] 0.5186794 0.23838204 0.1543444 [5,] -0.2170828 -0.32631616 0.3236952 [6,] -0.6661196 0.14694766 0.1093469 Here, the values in $d are not the squares of the standard deviations. > res3=test %*% res2$v > res3 [,1] [,2] [,3] [1,] -1.434507 3.1212479 -7.494005e-16 [2,] -2.685074 -2.5476217 3.996803e-15 [3,] 4.119582 -0.5736263 -2.053913e-15 > apply(res3,2,sd) [1] 3.622043e+00 2.877639e+00 3.184320e-15 As shown above, taking the standard deviation o...

...I len +.res1 db 0 ; Reserved +.res2 db 0 ; Reserved +.bustype equ ''PCI '' ; Host bus type (4 bytes, space padded) +.inttype equ ''ATA '' ; Interface type (8 bytes, spc. padded) +.intpath dd 0, 0 ; Interface path +.devpath dd 0, 0, 0, 0 ; Device path +.res3 db 0 ; Reserved +.chksum db 0 ; DPI checksum (not implemented) =20 %endif =20 diff --git a/memdisk/setup.c b/memdisk/setup.c index 98c4b69..b8efd8a 100644 --- a/memdisk/setup.c +++ b/memdisk/setup.c @@ -80,6 +80,16 @@ struct edd_dpt { uint64_t sectors; /* Total sectors */ uint16_t...

Dear all! I have a data frame composed by 13 columns (year, and 12 months). I want to transform this data base in another like this year month values 1901 1 1901 2 1901 3 ..... 1901 12 1902 1 1902 2 .... 1902 12 Is there a possibility to succeed that in R? Thank you! best regards! CR -- --- Catalin-Constantin ROIBU Lecturer PhD, Forestry engineer Forestry Faculty of Suceava Str.

Hi, Suppose, if I create 15 files in my working directory. set.seed(48) lapply(1:15,function(i) {m1 <- matrix(sample(1:20,1686*2,replace=TRUE),nrow=1686,ncol=2); write.table(m1,paste0("file_",i,".txt"),row.names=FALSE,quote=FALSE)}) ?D <-dir() D1 <- D[order(as.numeric(gsub("\\D+","",D)))] D1 ?res <- t(sapply(D1,function(x) {x1<-

...-- any suggestions greatly appreciated. Cheers, Berkan Here's the C code: // gtest.c // gcc -Wall -g gtest.c gtest -I/usr/lib/R/include -lRmath -lm #include <stdio.h> #define MATHLIB_STANDALONE #include <Rmath.h> int main (int argc, char *argv[]) { double res1, res2, res3; res1 = pgamma(98, 100, 1, 1, 0); res2 = pgamma(21, 100, 5, 1, 0); res3 = pgamma(2, 100, 69, 1, 0); printf ("pgamma(98, 100, 1) = %f\n\ pgamma(21, 100, 5) = %f\n\ pgamma(2, 100, 69) = %f\n", res1, res2, res3); exit(0); } -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-....

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

...e? I'm unsure; results are slightly different. # I understand that there is no predict for lmer models. anova does not give me the significance for lmer modes, so I guess I should enter the variables by steps and compare models with anova(m1, m2, ...) # and now the binary data. Is this correct? res3 <- lmer(pr01 ~ rain + year + site + (1|tree), data=mydat, family=binomial); anova(res3) res3 Many thanks for any comments Juli -- http://www.ceam.es/pausas