similar to: 3-parametric Weibull regression

hi all i have a function that uses two inputs, say xdata and ydata. An example is the following, simple1<-function(xdata,ydata) { ofit<-lm(ydata~xdata) list(ofit) } say i use arbitray number for xdata and ydata such that D = x1 x2 y 1 1 10 2 6 6 3 10 7 x<-D[,1:2] and y<-D[,3] if one uses these inputs and rund the program we get the following: >simple(xdata=x,ydata=y)

Hello I was trying to estimate the weibull model using nls after putting OLS values as the initial inputs to NLS. I tried multiple times but still i m getting the same error of Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial parameter estimates. The Program is as below > vel <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14) > df <- data.frame(conc, vel) >

I am using the lm function in R to fit several linear models to a fair-sized dataset (~160 collections of ~1000 data points each). My data have intrinsic, systematic uncertainty much greater than the measurement errors on any individual point. My thought is to use the residuals of my linear fits to quantify this intrinsic uncertainty, but I am puzzled over the correct interpretation of R's

Dear R Helpers, I have noticed that when I use lmer to analyse data, the summary function gives different values for the AIC, BIC and log-likelihood compared with the anova function. Here is a sample program #make some data set.seed(1); datx=data.frame(array(runif(720),c(240,3),dimnames=list(NULL,c('x1','x2','y' )))) id=rep(1:120,2); datx=cbind(id,datx) #give x1 a

Hello dear R help members. I am trying to understand the anova.rq, and I am finding something which I can not explain (is it a bug?!): The example is for when we have 3 nested models. I run the anova once on the two models, and again on the three models. I expect that the p.value for the comparison of model 1 and model 2 would remain the same, whether or not I add a third model to be compared

R 2.8.1 Windows XP I am trying to plot the results of a coxph using plot(survfit()). The plot should, I believe, show two lines one for survival in each of two treatment (Drug) groups, however my plot shows only one line. What am I doing wrong? My code is reproduced below, my figure is attached to this EMail message. John > #Create simple survival object >

Dear R developers, I noticed a bug in the stats4 package, specifically in the confint method applied to ?mle? objects. In particular, when some ?fixed? parameters define the log likelihood, these parameters are stored within the mle object but they are not used by the ?confint" method, which retrieves their value from the global environment (whenever they still exist). Sample code: >

I am trying to find methods for testing and visualizing calibration to Cox models with time-depended coefficients. I have read this nice article <http://journals.sagepub.com/doi/10.1177/0962280213497434>. In this paper, we can fit three models: fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) p <- log(predict(fit0, newdata = data1, type = "expected")) lp

Hi , I want to apply the following code fo my data with 400 predictors. I was wondering if there ia an alternative way instead of typing 400 predictors for the following code. I really appreciate your help. fit0<-lm(Y~1, data= mydata) fit.final<- lm(Y~X1+X2+X3+.....+X400, data=mydata) ??? step(fit0, scope=list(lower=fit0, upper=fit.final), data=mydata, direction="forward")

First, as others have said please obey the mailing list rules and turn of First, as others have said please obey the mailing list rules and turn off html, not everyone uses an html email client. Here is your code, formatted and with line numbers added. I also fixed one error: "y" should be "status". 1. fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) 2. p

windows xp R 2.8.1 I am trying to plot the -log(log(survival)) to visually test the proportional hazards assumption of a Cox regression. The plot, which should give two lines (one for each treatment) gives only one line and a warning message. I would appreciate help getting two lines, and an explanation of the warning message. My problem may the that I have very few events in one of my strata,

Here is an example, modified from the help page to use test="Cp": -------------------------------------------------------------------------------- > fit0 <- lm(sr ~ 1, data = LifeCycleSavings) > fit1 <- update(fit0, . ~ . + pop15) > fit2 <- update(fit1, . ~ . + pop75) > anova(fit0, fit1, fit2, test="Cp") Error in `[.data.frame`(table, , "Resid.

Dear List, After including cluster() option the coxreg (from eha package) produces results slightly different than that of coxph (from survival) in the following time-dependent treatment effect calculation (example is used just to make the point). Will appreciate any explaination / comment. cheers, Ehsan ############################ require(survival) require(eha) data(heart) # create weights

Hello, All: ????? The default "simulate" method for lm and glm seems to ignore the sampling variance of the parameter estimates;? see the trivial lm and glm examples below.? Both these examples estimate a mean with formula = x~1.? In both cases, the variance of the estimated mean is 1. ??? ??????? * In the lm example with x0 = c(-1, 1), var(x0) = 2, and

Hi everybody, I'm using the method described here to make a linear regression: http://www.apsnet.org/education/advancedplantpath/topics/Rmodules/Doc1/05_Nonlinear_regression.html > ## Input the data that include the variables time, plant ID, and severity > time <- c(seq(0,10),seq(0,10),seq(0,10)) > plant <- c(rep(1,11),rep(2,11),rep(3,11)) > > ## Severity

Dear r-helpers, Spencer Graves and Manual Morales proposed the following methods to simulate p-values in lme4: ************preliminary************ require(lme4) require(MASS) summary(glm(y ~ lbase*trt + lage + V4, family = poisson, data = epil), cor = FALSE) epil2 <- epil[epil$period == 1, ] epil2["period"] <- rep(0, 59); epil2["y"] <- epil2["base"]

How can I extract K (number of parameters) from an AIC calculation, both to report K itself and to calculate AICc? I'm aware of the conversion from AIC -> AICc, where AICc = AIC + 2K(K+1)/(n-K-1), but not sure of how K is calculated or how to extract that value from either an AIC or logLik calculation. This is probably more of a basic statistics question than an R question, but I thank

Dear all, I was wondering if there is any way i could do a "Grid Search" on a parameter space using R (as SAS 6.12 and higher can do it) to start the Newton-Gauss Linearization least squares method when i have NO prior information about the parameter. W. N. Venables and B. D. Ripley (2002) "Modern Applied Statistics with S", 4 th ed., page 216-7 has a topic

Folks: I'm having trouble doing a forward variable selection using step() First, I fit an initial model: fit0 <- glm ( est~1 , data=all, subset=c(n>=25) ) then I invoke step(): fit1 <- step( fit0 , scope=list(upper=est~ pcped + pchosp + pfarm ,lower=est~1)) I get the error message: Error in eval(expr, envir, enclos) : invalid 'envir' argument I looked at the

Dear WizaRds, I am sorry to bother you with a newbie question, but although I tried to solve my problem using the various .pdf files (Introduction, help pages etc.), I have come to a complete stop. Please be so kind as to guide me a little bit along my way of exploring multivariate analysis in R. I want to test wether the means-vector mu1 of X, consisting of the means per column of that matrix