similar to: Problem: Update of glm-object cannot find where the data object is located

Hello - I am using glmnet to generate a model for multiple cohorts i. For each i, I run 5 separate models, each with a different x variable. I want to compare the fit statistic for each i and x combination. When I use auc, the output is in some cases is < .5 (.49). In addition, if I compare mean MSE (with upper and lower bounds) ... there is no difference across my various x variables, but

Hi everyone, I have a data frame (d), wich has the results of mosquitoes trapping in three different places. I suspect that one of these places (Local=='Palm') is biased by low numbers and will yield slower slopes in the variance-mean regression over the areas. I wonder if these slopes are diferents. I've looked trought the support list for methods for comparing slopes and found 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

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 John, Brilliant solution and the best sort - when you finally solve your problem by yourself. Jim On Thu, Oct 1, 2020 at 2:52 AM array chip <arrayprofile at yahoo.com> wrote: > > Hi Jim, > > I found out why clip() does not work with lines(survfit.object)! > > If you look at code of function survival:::lines.survfit, in th middle of the code: > > do.clip <-

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)

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))

Should I repost the question with reply-all? On Tue, Dec 19, 2017 at 6:13 PM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > 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

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

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)) #

Dear R helpers, I would like to ask why polr occasionally generates results that look very odd. I have been trying to compare the power of proportional odds logistic regression with the Wilcoxon test. I generated random samples, applied both tests and extracted and compared the p-values, thus:- library(MASS) c1=rep(NA,100); c2=c1 for (run in 1:100) { dat=c(rbinom(20,12,0.65),rbinom(20,12,0.35))

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 +

Hi, I am working with R version 2.1.0, and I seem to have run into what looks like a bug. I get the same error message when I run R on Windows as well as when I run it on Linux. When I call anova to do a LR test from inside a function, I get an error. The same call works outside of a function. It appears to not find the right environment when called from inside a function. I have provided

On Wed, December 22, 2004 12:11 am, r.ghezzo at staff.mcgill.ca said: > ----- Forwarded message from r.ghezzo at staff.mcgill.ca ----- > Date: Mon, 20 Dec 2004 15:45:11 -0500 > From: r.ghezzo at staff.mcgill.ca > Reply-To: r.ghezzo at staff.mcgill.ca > Subject: [R] problems with limma > To: r-help at stat.math.ethz.ch > > I try to send this message To Gordon

Dear all, I have 3 models (from simple to complex) and I want to compare them in order to see if they fit equally well or not. From the R prompt I am not able to see where I can get this information. Let´s do an example: fit1<- lm(response ~ stimulus + condition + stimulus:condition, data=scrd) #EQUIVALE A lm(response ~ stimulus*condition, data=scrd) fit2<- lm(response ~ stimulus +

Hello, I am trying to do a fit a loglikelihood function with Multivariate distribution via copulas with fitMdvc. The problem is that it doesn't recognize that my beta is a vector of km parameter and when I try to run it it say that the length of my initial values is not the same as the parameter. Can somebody guide me where my mistake is. Thanks, Elisa. #################################

Hey, I have stacked a couple of garchFit objects in a list with names $fit1, $fit2, ..., $fiti assigning objects names using a loop, i.e. after running the loop modelStack = list($fit1, $fit2,...,$fiti). Thus the following apply; a = modelStack$fit2, then a is the second garchFit object of formal class 'fGarch' with 11 slots, @call, @formula... etc. I then want to extract information in

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

Dear R-Community! For example I have a study with 4 treatment groups (10 subjects per group) and 4 visits. Additionally, the gender is taken into account. I think - and hope this is a goog idea (!) - this data can be analysed using lme as below. In a balanced design everything is fine, but in an unbalanced design there are differences depending on fitting y~visit*treat*gender or

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 <-