search for: cov1

I?m trying to set up proportional hazard model that is stratified with respect to covariate 1 and has an interaction between covariate 1 and another variable, covariate 2. Both variables are categorical. In the following, I try to illustrate the two problems that I?ve encountered, using the lung dataset. The first problem is the warning: To me, it seems that there are too many dummies

I can use stepAIC on an lme object in 1.6.2, but I get the following error if I try to do the same in 1.7.0: Error in lme(fixed = resp ~ cov1 + cov2, data = a, random = structure(list( : unused argument(s) (formula ...) Does anybody know why? Here's an example: library(nlme) library(MASS) a <- data.frame( resp=rnorm(250), cov1=rnorm(250), cov2=rnorm(250), group=rep(letters[1:10],25) ) mod1 <- lme(re...

...Please try to fix this in the future (Nabble issue?) As to the problems: handling strata by covariate interactions turns out to be a bit of a pain in the posteriorin the survival code. It would have worked, however, if you had done the following: fit <- coxph(Surv(time, status) ~ strata(cov1) * cov2, data=...) or ~ strata(cov1):cov2 or ~ strata(cov1):cov2 + cov2 But by using ~ strata(cov1) + cov1:cov2 you fooled the program into thinking that there was no strata by covariate interaction, and so it did not follow the special logic necessary for that case. Second is...

Hi all I do not understand why I am getting the following error message. Can anybody help me with this? Thanks in advance. install.packages("cmprsk") library(cmprsk) result1 <-crr(ftime, fstatus, cov1, failcode=1, cencode=0 ) one.pout1 = predict(result1,cov1,X=cbind(1,one.z1,one.z2)) predict.crr(result1,cov1,X=cbind(1,one.z1,one.z2)) Error: could not find function "predict.crr" [[alternative HTML version deleted]]

...drastically slower under current R-devel vs. current R-patched - one example is below using a ttest. I have the following snippet of code that demonstrates the problem while avoiding "real" code that takes an extremely long time to finish on R-devel: library(genefilter) data(eset) eset$cov1 z <- ttest(eset$cov1, p=0.01) gf <- filterfun(z) system.time(genefilter(eset, gf)) (this requires the genefilter & Biobase packages from Bioconductor) Using R-1.7.0p I had the following output: > data(eset) > eset$cov1 [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 >...

...better than m2. So far so good. Now I have one (1) measure from each Trial, of soil factors weather and such, that I want to evaluate. Remember: only one value of the covariate for each Trial. The suggestion I have got from my local guru is to base this in m1 like: m3 <- lme(Yield ~ Treat + Cov1 + Treat:Cov1, data=data, random =~1| Trial/Block) thus giving a model where the major random factor (Trial) is represented both as a (1) measure of Cov1 in the fixed part and by itself in the random part. Trying the simpler call: m4 <- lm(Yield ~ Treat + Cov1 + Treat:Cov1, data=...

the study design of the data I have to analyse is simple. There is 1 control group (CTRL) and 2 different treatment groups (TREAT_1 and TREAT_2). The data also includes 2 covariates COV1 and COV2. I have been asked to check if there is a linear or quadratic treatment effect in the data. I created a dummy data set to explain my situation: df1 <- data.frame( Observation = c(rep("CTRL",15), rep("TREAT_1",13), rep("TREAT_2", 12)), COV1 = c(rep(&...

...n 1:length(ed$ID)) { ed$base[k]<-100*exp(rnorm(1,0,0.05)) ed$drop[k]<-0.2*exp(rnorm(1,0,0.01)) ed$frac[k]<-0.5*exp(rnorm(1,0,0.1)) } comb1<-merge(data1[, c("ID","TIME")], ed) comb2<-comb1 comb2$score<-comb2$base*exp(-comb2$drop*comb2$TIME) func1<-function(t,cov1,beta1, change,other) { ifelse(t==0,cov1, cov1*exp(beta1*change+other)) } comb3<-comb2 comb3$cmhz=0 comb3<-comb3[order(comb3$ID, comb3$TIME), ] for (q in 1:length(comb3$ID)) { comb3$cmhz[q]<-integrate(func1, lower=0, upper=comb3$TIME[q], cov1=0.001,beta1=0.02, change=comb...

...covariates need to be nested hierarchically within a grouping "class", of which there are 8. I have an accessory vector, " cov2class" that specifies the mapping between covariates and the 8 classes. Now, I understand I can break all this information up into individual vectors (cov1, cov2, ..., cov211, class1, class2, ..., class8), and do something like this: model <- lmer(Y ~ 1 + cov1 + cov2 + ... + cov211 + (cov1 + cov2 + ... | class1) + (...) + (... + cov210 + cov211 | class8) But I'd like keep things syntactica...

Hello. Im hoping someone can help with a grouping question related to the "random=" statement within the nlme function. How do you specify that some grouping levels are inner to others? I tried several things, given below. Lets say I have a data frame with five variables, resp, cov1, ran1, ran2, group1, and group 2. The formula is resp~cov1 + ran1 + ran2, where the ran are random variables. The data is of length 80, and there are 4 unique factors in group1 and 20 unique factors in group2. These are factors related to ran1 and ran2, respectively. The difficult part is tha...

...matrix (the third argument supplied to the mahalanobis funtion). Any help or guidance would be greatly appreciated. Thank you! RL CODE fit<-lda(pop~v1 + v2 + v3 +...+vn, data=my.data) x1<-subset(my.data, pop==1) x2<-subset(my.data, pop==2) #Save Covariance Matices for each group cov1<-cov(x1) cov2<-cov(x2) #Determine number of rows in each matrix n1<-nrow(x1); n2<-nrow(x2); n.rows<-c(n1,n2) #store mean vectors from lda object mu1<-fit$means[1,] mu2<-fit$means[2,] #Calculate the common Covariance Matrix S<-(((n.rows[1]-1)*cov1)+((n.rows[2]-1)*cov2...

Dear r-helpers, I have a question about multiple testing. Here an example that puzzles me: All matrixes and contrast vectors are presented in treatment contrasts. 1. example: library(multcomp) n<-60; sigma<-20 # n = sample size per group # sigma standard deviation of the residuals cov1 <- matrix(c(3/4,-1/2,-1/2,-1/2,1,0,-1/2,0,1), nrow = 3, ncol=3, byrow=TRUE, dimnames = list(c("A", "B", "C"), c("C.1", "C.2", "C.3"))) # cov1 = variance covariance matrix of the beta coefficients of a # 2x2 factorial design (see Pian...

...of them are, of course, visible even if they are drawn on a non visible "face". Any suggestion to avoid this problem ? Thank you Etienne Example : trans3d <- function(x,y,z, pmat) { tr <- cbind(x,y,z,1) %*% pmat list(x = tr[,1]/tr[,4], y= tr[,2]/tr[,4]) } mean1 = c(2,4) cov1 = matrix(c(0.7, 0.2,0.2,0.7), ncol=2) x = seq(0, 5, by = 0.1) y = seq(0,10, by= 0.1) z = matrix(nrow = length(x), ncol=length(y)) for(i in 1:length(x)) for(j in 1:length(y)) { z[i,j]=dmvnorm(c(x[i],y[j]),mean1,cov1) } pmat1 = persp(x, y, z, col= "red",shade = 0.25, border=NA)...

...s="blue", col.grid="lightblue",angle=120, pch=20) em2mn<- function(y) { n<-length(y[,1]) p<-matrix(0,n,1) f1<-matrix(0,n,1) f2<-matrix(0,n,1) tau<-matrix(0,n,2) eps<-0.0001 mu01<-c(0,0) mu02<-c(0,0) sd01<-matrix(0,2,2) sd02<-matrix(0,2,2) cov1<-matrix(0,2,2) cov2<-matrix(0,2,2) # 1 inizializzare i valori alpha0= runif(1,0,1) for (j in 1:2) { mu01[j] <- runif(1,min=quantile(y[,j], probs =0.25), max=quantile(y[,j], probs =0.75)) mu02[j] <- runif(1,min=quantile(y[,j], probs =0.25), max=quantile(y[,j], probs =0.75)) } sd01<-...

...s="blue", col.grid="lightblue",angle=120, pch=20) em2mn<- function(y) { n<-length(y[,1]) p<-matrix(0,n,1) f1<-matrix(0,n,1) f2<-matrix(0,n,1) tau<-matrix(0,n,2) eps<-0.0001 mu01<-c(0,0) mu02<-c(0,0) sd01<-matrix(0,2,2) sd02<-matrix(0,2,2) cov1<-matrix(0,2,2) cov2<-matrix(0,2,2) # 1 inizializzare i valori alpha0= runif(1,0,1) for (j in 1:2) { mu01[j] <- runif(1,min=quantile(y[,j], probs =0.25), max=quantile(y[,j], probs =0.75)) mu02[j] <- runif(1,min=quantile(y[,j], probs =0.25), max=quantile(y[,j], probs =0.75)) } sd01<-...

Hey, I am confused about how to generate the sample data from a mixture of Multivariate Gaussian ditribution. For example, there are 2 component Gaussian with prior probability of 0.4 and 0.6, the means and variances are u1=[1 1]', Cov1=[1 0;0 1] and u2=[-1 -1]', Cov2=[1 0;0 1] repectively. So how can I generate a sample of 500 data from the above mixture distribution? Thanks. Fred

Hi, what I would need are 2 vector pairs (x,y) and (x1,y1). x and x1 must have the same sd. y and y1 should also exhibit the same sd's but different ones as x and x1. Plotting x,y and x1,y1 should produce a plot with 2 vectors having a different slope. Plotting both vector pairs in one plot with fixed axes should reveal the different slope. many thanks syrvn -- View this message in

...64824,10465,58176,36662,19769,9080,45041,16080,49648,59638 ,23339,40697,47107,8536,57035,55090,44414,1321,12861,21108,32654,27068,38365,22255,31550,11789,45404,53969 ,13509,36350) Dist <- sqrt(outer(x,x, "-")^2 + outer(y,y, "-")^2) Dist <- Dist/max(Dist) library(spatial) Cov1 <- sphercov(Dist, 0.8, alpha=0, se=sqrt(2)) Cov2 <- sphercov(Dist, 0.6, alpha=0, se=sqrt(0.55)) library(geoR) Cov1b <- cov.spatial(Dist, cov.model= "spherical", cov.pars=c(2, 0.8)) Cov2b <- cov.spatial(Dist, cov.model= "spherical", cov.pars=c(0.55, 0.6)) library(mvt...

...ence: TRUE coefficients: x1 x2 x3 gg 0.2624 0.6515 -0.8745 -0.1144 standard errors: [1] 0.3839 0.3964 0.4559 0.1452 two-sided p-values: x1 x2 x3 gg 0.490 0.100 0.055 0.430 > summary(z) Competing Risks Regression Call: crr(ftime = ftime, fstatus = fstatus, cov1 = cov2) coef exp(coef) se(coef) z p-value x1 0.262 1.300 0.384 0.683 0.490 x2 0.652 1.918 0.396 1.643 0.100 x3 -0.874 0.417 0.456 -1.918 0.055 gg -0.114 0.892 0.145 -0.788 0.430 [[alternative HTML version deleted]]

...after 3 months. ----------------------------------------------- ERROR Errors in running code in vignettes: when running code in ?compete.Rnw? ... > temp$fstat <- as.numeric(event) > temp$msex <- with(temp, 1 * (sex == "M")) > fgfit1 <- with(temp, crr(etime, fstat, cov1 = cbind(age, + msex, mspike), failcode = 2, cencode = 1, variance = TRUE)) When sourcing ?compete.R?: Error: could not find function "crr" Execution halted * checking re-building of vignette outputs ... NOTE Error in re-building vignettes: ... Warning in coxph(Surv(futime, deat...