similar to: Anova Type II and Contrasts

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

First, here is your message as it appears on R-help. On 10/14/2012 05:00 AM, r-help-request@r-project.org wrote: > 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 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),

Homework questions are not answered on this list. Best, Uwe Ligges On 04.10.2024 10:32, Steven Yen wrote: > The following line calculates standard deviations of a column vector: > > se<-apply(dd,1,sd) > > How can I calculate the covariance matrix using apply? Thanks. > > ______________________________________________ > R-help at r-project.org mailing list -- To

I have a fairly large model: > length(Y) [1] 3051 > dim(covariates) [1] 3051 211 All of these 211 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

Hello, I am a new R user and trying to learn how to implement the mahalanobis function to measure the distance between to 2 population centroids. I have used STATISTICA to calculate these differences, but was hoping to learn to do the analysis in R. I have implemented the code as below, but my results are very different from that of STATISTICA, and I believe I may not have interpreted the help

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 =

On 10/4/2024 5:13 PM, Steven Yen wrote: > Pardon me!!! > > What makes you think this is a homework question? You are not > obligated to respond if the question is not intelligent enough for you. > > I did the following: two ways to calculate a covariance matrix but > wonder how I might replicate the results with "apply". I am not too > comfortable with the

Pardon me!!! What makes you think this is a homework question? You are not obligated to respond if the question is not intelligent enough for you. I did the following: two ways to calculate a covariance matrix but wonder how I might replicate the results with "apply". I am not too comfortable with the online do of apply. > set.seed(122345671) > n<-3 > x<-rnorm(n); x

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

Even if this is not a homework question, it smells like one. If you read the Posting Guide it warns you that homework is off-topic, so when you impose an arbitrary constraint like "must use specific unrelated function" we feel like you are either cheating or wasting our time, and it is up to you to explain why we should follow you down this rabbit hole, keeping in mind that statistics

OK. Thanks to all. Suppose I have two vectors, x and y. Is there a way to do the covariance matrix with ?apply?. The matrix I need really contains the deviation products divided by the degrees of freedom (n-1). That is, the elements (1,1), (1,2),...,(1,n) (2,1), (2,2),...., (2,n) .... (n,1),(n,2),...,(n,n). > Hello, > > This doesn't make sense, if you have only one vector you

Hello, If you have a numeric matrix or data.frame, try something like cov(mtcars) Hope this helps, Rui Barradas ?s 10:15 de 04/10/2024, Steven Yen escreveu: > On 10/4/2024 5:13 PM, Steven Yen wrote: > >> Pardon me!!! >> >> What makes you think this is a homework question? You are not >> obligated to respond if the question is not intelligent enough for you.

Hello, I have a question regarding competing risk regression using cmprsk package (function crr()). I am using R2.9.1. How can I do to assess the effect of qualitative predictor (gg) with more than two categories (a,b,c) categorie c is the reference category. See above results, gg is considered like a ordered predictor ! Thank you for your help Jan > # simulated data to test > set.seed(10)

Hello I have a vector: set.seed(123) > n<-3 > x<-rnorm(n); x [1] -0.56047565 -0.23017749 1.55870831 I like to create a matrix with elements containing variances and covariances of x. That is var(x[1]) cov(x[1],x[2]) cov(x[1],x[3]) cov(x[2],x[1]) var(x[2]) cov(x[2],x[3]) cov(x[3],x[1]) cov(x[3],x[2]) var(x[3]) And I like to do it with "apply". Thanks. On 10/4/2024 6:35

Hello, This doesn't make sense, if you have only one vector you can estimate its variance with var(x) but there is no covariance, the joint variance of two rv's. "co" or joint with what if you have only x? Note that the variance of x[1] or any other vector element is zero, it's only one value therefore it does not vary. A similar reasonong can be applied to cov(x[1],

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

It's still hard to figure out what you want. If you have two vectors you can compute their (2x2) covariance matrix using cov(cbind(x,y)). If you want to compute all pairwise squared differences between elements of x and y you could use outer(x, y, "-")^2. Can you explain a little bit more about (1) the context for your question and (2) why you want/need to use apply() ? On

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, Alan Really very excited to receive your email and sorry to be slow replying, it has been exceptionally busy over the last few days ;( Your explanation made the problem clear to me. So gcov branch coverage should be called condition coverage and clang region coverage is branch coverage in fact(also known as *decision/C1*), right? And llvm/clang will support all the following coverage