similar to: update forgets about offset() (PR#6656)

Dear all, the help file for predict.gbm states that "The predictions from gbm do not include the offset term. The user may add the value of the offset to the predicted value if desired." I am just not sure how exactly, especially for a Poisson model, where I believe the offset is multiplicative ? For example: library(MASS) fit1 <- glm(Claims ~ District + Group + Age +

Dear R users, Could somebody please help me to find a way of comparing nonlinear, non-nested models in R, where the number of parameters is not necessarily different? Here is a sample (growth rates, y, as a function of internal substrate concentration, x): x <- c(0.52, 1.21, 1.45, 1.64, 1.89, 2.14, 2.47, 3.20, 4.47, 5.31, 6.48) y <- c(0.00, 0.35, 0.41, 0.49, 0.58, 0.61, 0.71, 0.83, 0.98,

Hello, Can anyone suggest a simple way to generate a Kaplan-Meier plot with 2 survfit objects, just like this one:? https://drive.google.com/file/d/1fEcpdIdE2xYtA6LBQN9ck3JkL6-goabX/view?usp=sharing Suppose I have 2 survfit objects: fit1 is for the curve on the left (survtime has been truncated to the cutoff line: year 5), fit2 is for the curve on the right (minimum survival time is at the

Hi, I am trying to do logistic regression for data of 104 patients, which have one outcome (yes or no) and 15 variables (9 categorical factors [yes or no] and 6 continuous variables). Number of yes outcome is 25. Twenty-five events and 15 variables mean events per variable is much less than 10. Therefore, I tried to analyze the data with penalized regression method. I would like please some of the

Hello all, I wish to extract the terms from an rpart object. Specifically, I would like to be able to know what is the response variable (so I could do some manipulation on it). But in general, such a method for rpart will also need to handle a "." case (see fit2) Here are two simple examples: fit1 <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) fit1$call fit2 <-

You also need to reply-all so the mailing list stays in the loop. -- Sent from my phone. Please excuse my brevity. On December 19, 2017 4:00:29 PM PST, Timothy Axberg <axbergtimothy at gmail.com> wrote: >Sorry about that. Here is the code typed directly on the email. > >qe = (Qmax * Kl * ce) / (1 + Kl * ce) > >##The data >ce <- c(15.17, 42.15, 69.12, 237.7, 419.77)

Hello R users, I've used the following help two compare two regression line slopes. Wanted to test if they differ significantly: Hi, I've made a research about how to compare two regression line slopes (of y versus x for 2 groups, "group" being a factor ) using R. I knew the method based on the following statement : t = (b1 - b2) / sb1,b2 where b1 and b2 are the two slope

I try to send this message To Gordon Smyth at smyth at vehi,edu.au but it bounced back, so here it is to r-help I am trying to use limma, just downloaded it from CRAN. I use R 2.0.1 on Win XP see the following: > library(RODBC) > chan1 <- odbcConnectExcel("D:/Data/mgc/Chips/Chips4.xls") > dd <- sqlFetch(chan1,"Raw") # all data 12000 > # > nzw <-

I have the following results for an ANOVA comparing two nested models. I wasn't sure how I am supposed to report this result in the area of psychology. Specifically, am I supposed to report the DF's or just the F ratio? I could manually calculate the degrees of freedoms, but there must be a reason why R does not give this information, i.e. those are not conventionally used in the

Hello, Is there a function in the survival package that will allow me to test a subset of independent variables for joint significance? I am thinking along the lines of a Wald, likelihood ratio, or F-test. I am using the survreg procedure to estimate my parameters. Thank you. Geoff Geoffrey Smith Visiting Assistant Professor Department of Finance University of Illinois at Urbana-Champaign

Hi, I want to do a global likelihood ratio test for the proportional odds logistic regression model and am unsure how to go about it. I am using the polr() function in library(MASS). 1. Is the p-value from the likelihood ratio test obtained by anova(fit1,fit2), where fit1 is the polr model with only the intercept and fit2 is the full polr model (refer to example below)? So in the case of the

Hi, I am computing the Nelson-Aalen (NA) estimate of baseline cumulative hazard in two different ways using the "survival" package. I am expecting that they should be identical. However, they are not. Their difference is a monotonically increasing with time. This difference is probably not large to make any impact in the application, but is annoyingly non-trivial for me to just

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

Hi, I just encountered what I thought was strange behavior in MDS. However, it turned out that the mistake was mine. The lesson learned from my mistake is that one should plot on a square pane when plotting results of an MDS. Not doing so can be very misleading. Follow the example of an equilateral triangle below to see what I mean. I hope this helps others to avoid this kind of headache.

I stumbled across a mild glitch when trying to compare the result of gam() fitting with the result of lm() fitting. The following code demonstrates the problem: library(gam) x <- rep(1:10,10) set.seed(42) y <- rnorm(100) fit1 <- lm(y~x) fit2 <- gam(y~lo(x)) fit3 <- lm(y~factor(x)) print(anova(fit1,fit2)) # No worries. print(anova(fit1,fit3)) # Likewise. print(anova(fit2,fit3)) #

Hi! I have fitted two glms assuming a poisson distribution which are: fit1 <- glm(Aids ~ Year, data=aids, family=poisson()) fit2 <- glm(Aids ~ Year+I(Year^2), data=aids, family=poisson()) I am trying to work out how to represent the fitted regression curves of fit1 and fit2 on the one graph. I have tried: graphics.off() plot(Aids ~ Year, data = aids) line(glm(Aids ~ Year,

Dear List, I used rlm to calculate two regression models for two data sets (rlm due to two outlying values in one of the data sets). Now I want to compare the two regression slopes. I came across some R-code of Spencer Graves in reply to a similar problem: http://www.mail-archive.com/r-help at stat.math.ethz.ch/msg06666.html The code was: > df1 <- data.frame(x=1:10, y=1:10+rnorm(10))

Hello All, Does anyone know why length(fit1$time) < length(fit2$n) in survfit.coxph output? Why is the predicted time length is not the same as the number of samples (n)? I tried: example(survfit.coxph). Thanks, parmee > fit2$n [1] 241 > fit2$time [1] 0 31 32 60 61 152 153 174 273 277 362 365 499 517 518 547 [17] 566 638 700 760 791

Dear R-users, I have noticed small discrepencies in the reported estimate of the variance of the frailty by the print method for survreg() and the 'theta' component included in the object fit: # Examples in R-2.6.2 for Windows library(survival) # version 2.34-1 (2008-03-31) # discrepancy fit1 <- survreg(Surv(time, status) ~ rx + frailty(litter), rats) fit1 fit1$history[[1]]$theta

Hi all, I was trying to fit a Gamma hierarchical model using "glmer", but got weird error message that I could not understand. On the other hand, a similar call to the glmmPQL leads to results that are close to what I expect. I also tried to change tha "nAGQ" argument in "glmer", but it did not solve the problem. The model I was fitting has a simple structure - one