similar to: User defined function and nonlinear least-squares fit

Hi all, I met with a problem about the weighted least square regression. 1. I simulated a Normal vector (sim1) with mean 425906 and standard deviation 40000. 2. I simulated a second Normal vector with conditional mean b1*sim1, where b1 is just a number I specified, and variance proportional to sim1. Precisely, the standard deviation is sqrt(sim1)*50. 3. Then I run a WLS regression without the

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.

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)

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

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. I am having a problem setting up a self-starting function for use in nonlinear regression (and eventually in the mixed model version). The function is a non-rectangular hyperbola - called "NRhyperbola" - which is used for fitting leaf photosynthetic rate to light intensity. It has one independent variable (Irr) and four parameters (theta, Am, alpha and Rd). I have created this

> From: "Alan T. Arnholt" <arnholt at math.appstate.edu> > To: Bill Venables <William.Venables at cmis.CSIRO.AU> > Cc: r-help at stat.math.ethz.ch, arnholt at math.appstate.edu > Subject: Re: [R] Lack of Fit test > Date: Wed, 23 Feb 2000 09:40:21 -0500 (EST) > X-Authentication: none > > > I guess my question was not adequately stated when I sent

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

Dear list, I am getting weired with fitting data with a 1/x-polynomial. Suggest I have the following data: x <- c(1,2,3,4,5,6,7) y <- c(100,20,4,2,1,.3,.1) I may fit this with a linear model fit1 = lm(y ~ I(x)) Getting plot out of this model I applied library(polynom) pol1 = polynomial(fit1$coefficients) f1 = as.function(pol1) plot(x,y) lines(x, f1(x), col = 2) Clearly, this model

I am trying to extract coefficients from lars fit and can't find how to get intercept. E.g. y = rnorm(10) x = matrix(runif(50),nrow=10) X = data.frame(y,x) fit1 = lars(as.matrix(X[,2:6]),as.matrix(X[,1])) fit2 = lm(y~.,data=X) Then, if I do: > predict(fit1,s=1,mode='fraction',type='coefficients')$coef X1 X2 X3 X4 X5 0.3447570

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 <-

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 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)) #

Hi everybody, I want to ask your help to explain what is going on with my following code: > mydata <- data.frame(y=rbinom(100, 1, 0.5), x1=rnorm(100), x2=rnorm(100)) > glm.fit.method <- function(model,data,...){glm(formula=model,data=data,family="binomial",. ..)} > fit1 <- glm(y ~ x1 + x2, data=mydata, family=binomial()) > update(fit1, .~1) Call: glm(formula =

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 +