similar to: A Tip: lm, glm, and retained cases

Hi again! Following up my previous posting below (to which no response as yet), I have located a report which situates this type of question in a longitudinal modelling context. http://www4.stat.ncsu.edu/~dzhang2/paper/glm.ps Generalized Linear Models with Longitudinal Covariates Daowen Zhang & Xihong Lin (This work seems to originally date from around 1999). They consider an outcome Y,

I have a sparse contingency table (most cells are 0): > xtabs(~.,data[,idx:(idx+4)]) , , x3 = 1, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 31 2 0 0 112 3 0 0 94 , , x3 = 2, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 3, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 1, x4

Hi Folks, Given a series of n independent Bernoulli trials with outcomes Yi (i=1...n) and Prob[Yi = 1] = Pi, I want P = Prob[sum(Yi) = r] (r = 0,1,...,n) I can certainly find a way to do it: Let p be the vector c(P1,P2,...,Pn). The cases r=0 and r=n are trivial (and also are exceptions for the following routine). For a given value of r in (1:(n-1)), library(combinat) Set <- (1:n)

He, I used the "coxph" function, with four covariates. Let's say something like that > model.1 <- coxph(Surv(Time,Event)~X1+X2+X3+X4,data=DATA) So I obtain the 4 coefficients B1,B2,B3,B4 such that h(t) = h0(t) exp(B1*X1+ B2*X2 + B3*X3 + B4*X4). When I use the function on the same data > predict.coxph(model.1,type="lp") how it works in making the prediction?

Hi All, I'm a little stumped by the following problem. I've got a dataset with the following structure: idxy ix iy country (other variables) 1 1 1 c1 x1 2 1 2 c1 x2 3 1 3 c1 x3 . . . . . 3739 55 67 c7 x3739 3740 55 68 c7 x3740 where ix and

AdvisoRs: Is the following a bug, feature, hinky error message, or dumb Bert? > mtest <- matrix(1:12,nr=4) > dftest <- data.frame(mtest) > ix <- cbind(1:2,2:3) > mtest[ix] <- NA > mtest [,1] [,2] [,3] [1,] 1 NA 9 [2,] 2 6 NA [3,] 3 7 11 [4,] 4 8 12 ## But ... > dftest[ix] <- NA Error in `[<-.data.frame`(`*tmp*`, ix, value

Hi Folks, I'm analysing some data which, in its simplest aspect, has 3 factors A, B, C each at 2 levels. If I do lm1 <- lm(y ~ A*B) say, and then summary(lm1, corr=T) I get the correlation matrix of the estimated coeffcients with numerical values for the correlations (3 coeffs in this case). Likewise with 'glm' instead of 'lm'. However, if I do lm2 <- lm(y ~

Good morning to all, I am encountering a blocking issue when using the function 'breackpoints' from package 'strucchange'. *Context:* I use a data frame, 248 observations of 5 variables, no NA. I compute a linear model, as y~x1+...+x4 x4 is a dummy variable (0 or 1). I want to check this model for structural changes. *Process & issues:* *First, I used function Fstats.* It

Greetings All. Once again, I am probably missing something fairly accessible, but since I can't find it I'd welcome advice! I have a dataframe derived from a text file of data in tabular format. For one of the variables, say X, I want to select the subsets which in which X equals a particular value. Such values are given in the text file like: 2.3978953, and each such value will occurr

Dear list, I am making a linear regression of the following format. lm(Y~X1+X2+log(X3)+log(X4)) Now I would like to check the linear regression above using generalized linear model. Please kindly if the following format is correct and thank you. (If it is wrong, please indicate why.) glm(Y~X1+X2+log(X3)+log(X4), family=poisson(link="log")) Also, please kindly suggest any references

Hello guys, i looked over the archive files and found nothing about this kind of error. I have a database of 33 elements described in 8 variables, i'm using the Leave-One-Out iterative process to take one of the elements to be the test element and make a regression with the other 32 and then I try to predict the clas of the element out. I'm using this call as a part of a Leave-One-Out

Hello I'm trying to do the following vector operation: given vector x = c(x1,x2,x3,x4...xn), produce vector y = c(x1,x1+x2,x1+x2+x3,...x1+...+xn). E.g., from x = c(1,3,2,2,5), produce y = c(1,4,6,8,13). The underlying problem is finding the cumulative distribution function given the empirical density distribution function. I have done some research on this but the only relevant

Hi.. i have an expression of the form: model1<-nls(y~beta1*(x1+(k1*x2)+(k1*k1*x3)+(k2*x4)+(k2*k1*x5)+(k2*k2*x6)+(k3*x7)+(k3*k4*x8)+(k3*k2*x9)+(k3*k3*x10)+ (k4*x11)+(k4*k1*x12)+(k4*k2*x13)+(k4*k3*x14)+(k4*k4*x15)+(k5*x16)+(k5*k1*x17)+(k5*k2*x18)+(k5*k3*x19)+

I have a couple quick questions about the use of cv.glm for cross-validation. 1. If we have a Poisson GLM with counts Y~Poisson(mu) and ln(mu)=beta0+beta1*x1+..., is the prediction error (delta) that is output from cv.glm provided in terms of the counts (y) or the (mu)? 2. Can cv.glm be used for negative binomial models fit using glm.nb? It appears to work, but since NB models aren't

Dette er en melding med flere deler i MIME-format. --=_alternative 004613C000257091_= Content-Type: text/plain; charset="US-ASCII" And some more informastion I forgot. R does not crash if I write out the formula: set.seed(123) x1 <- runif(1000) x2 <- runif(1000) x3 <- runif(1000) x4 <- runif(1000) x5 <- runif(1000) x6 <- runif(1000) x7 <- runif(1000) x8 <-

Hi, Using S=1000 and simdata <- replicate(S, generate(3000)) #If you want both "m1" and "m0" #here the missing values are 0 res1<-sapply(seq_len(ncol(simdata.psm1)),function(i) {x1<-merge(simdata.psm0[,i],simdata.psm1[,i],all=TRUE); x1[is.na(x1)]<-0; x1}) res1[,997:1000] #????? [,1]???????? [,2]???????? [,3]???????? [,4]??????? #x1??? Numeric,3000 Numeric,3000

Hi, A recent thread provided a (working) construct for lm: lm(as.matrix(freeny[ix]) ~., freeny[-ix]) Can someone explain what is meant by the formula in that expression, that is, what does "mymatrix~." do? I couldn't find any such example in the lm() or formula() help pages. thanks Carl

Hello, R experts, If I understand Ted's anwser correctly, then I can not contrast the mean yields between sections 1-8 and 9-11 under "Trt" but I can contrast mean yields for sections 1-3 and 6-11 because there exists significant interaction between two factors (Trt:section4, Trt:section5). Could I use the commands below to test the difference between sections 1-3 and 6-11 ?

Hello R users I like to extract z values for x1 and x2. I know how to extract coefficents using model$coef but I don't know how to extract z values for each of independent variable. I looked around using names(model) but I couldn't find how to extract z values. Any help would be appreciated. Thanks TM ######################################################### >summary(model) Call:

I am trying to write a function that will run a linear model and plot the regression coeficients with their corresponding means. I am having two problems. I can get the plot with the function below, but I am having trouble labeling the points. function(y,x1,x2,x3,x4){ outlm<-lm(y~x1+x2+x3+x4) imp<-as.data.frame(outlm$coef[-1]) meanvec<-c(mean(x1),mean(x2),mean(x3),mean(x4))