search for: glm2

Hello: I'd like to know how and if the GLM convergence problems are addressed in gamlss. For simplicity, let's focus on Normal and Negative Binomial with log link. The convergence issues of the glm() function were alleviated in 2011 when glm2 package was released. Package gamlss was released in 2012, so it might still use the glm-like solution or call glm() directly. Is that the case or the improvements from glm2 have been incorporated into gamlss also? Regards, Nik

...s of freedom Residual deviance: 1323.5 on 104 degrees of freedom (3 observations deleted due to missingness) AIC: 1585.2 Number of Fisher Scoring iterations: 5 #### the output for residual deviance and df does not change even when I use quasibinomial, is this ok? ##### library(MASS) > glm2<-glm(y~Pred,"quasibinomial") > summary(glm2) Call: glm(formula = y ~ Pred, family = "quasibinomial") Deviance Residuals: Min 1Q Median 3Q Max -9.1022 -2.7899 -0.4781 2.6058 8.4852 Coefficients: Estimate Std. Error t value Pr(>...

.../ Windows 2000 / P4/2.8 Ghz Z <- as.factor(rep(LETTERS[1:3],20)) Y <- rep(0:1, 30) X <- rnorm(60) glm1 <- glm(Y ~ Z + X + Z:X, family=binomial) summary(glm1)$coeff regTermTest( model=glm1 , test.terms=~Z:X) ZB <- ifelse(Z=="B",1,0) ZC <- ifelse(Z=="C",1,0) glm2 <- glm(Y ~ ZB + ZC + X + ZB:X + ZC:X, family=binomial) summary(glm2)$coeff ## Okay, same as glm1 regTermTest( model=glm2 , test.terms= c("ZB:X","ZC:X")) regTermTest( model=glm2 , test.terms= ~ ZB:X + ZC:X)

...> anova(glmm2$lme, glmm1$lme) Model df AIC BIC logLik Test L.Ratio p-value glmm2$lme 1 3 4340.060 4351.172 -2167.030 glmm1$lme 2 5 4292.517 4311.036 -2141.258 1 vs 2 51.54367 <.0001 > > # a linear model also works, though no p-value is reported > glm1 <- gam(y ~ x1 + x2) > glm2 <- gam(y ~ 1) > anova(glm1) Family: gaussian Link function: identity Formula: y ~ x1 + x2 Parametric Terms: df F p-value x1 1 45.58 7.69e-11 x2 1 13.96 0.000224 > anova(glm2, glm1) Analysis of Deviance Table Model 1: y ~ 1 Model 2: y ~ x1 + x2 Resid. Df Resid. Dev Df Deviance 1 299...

...I start digging further, can anyone see why dffits might be failing? Is there a problem with the data? Consider: # Load data dep <- read.table("http://www.sci.usq.edu.au/staff/dunn/Datasets/Books/Hand/Hand-R/factor1-R.dat", header=TRUE) attach(dep) dep # Fit Poisson glm dep.glm2 <- glm( Counts ~ factor(Depression) + factor(SLE) + factor(Children) + factor(Depression):factor(SLE), family=poisson( link=log) ) # Compute dffits dffits( dep.glm2 ) This produces the output: 1 2 3 4 5 6 1.4207746 -0.1513808...

...: 174.54 # Number of Fisher Scoring iterations: 5 # > sqrt(diag(vcovHC(glm1, type="HC0"))) # (Intercept) carrot0 # 0.1673655 0.1971117 # > ### Compare against ## http://www.ats.ucla.edu/stat/stata/faq/relative_risk.htm ## robust standard errors are: ## .1682086 .1981048 glm2 <- glm(lenses~carrot0 +gender1 +latitude, data=dat, family=poisson(link=log)) sqrt(diag(vcovHC(glm2, type="HC0"))) ### Result: # > summary(glm2) #Call: # glm(formula = lenses ~ carrot0 + gender1 + latitude, family = poisson(link = log), # data = dat) # Deviance Residuals:...

...12 60 M 16 4 13 10 F 3 17 14 10 M 1 19 15 20 F 2 18 16 20 M 2 18 17 30 F 10 10 18 30 M 8 12 19 40 F 14 6 20 40 M 12 8 21 50 F 16 4 22 50 M 13 7 23 60 F 18 2 24 60 M 16 4 glm2<-glm(deadalive~Dose*Sex,family=binomial,data=rat.toxic) > anova(glm2,test="Chi") Analysis of Deviance Table Model: binomial, link: logit Response: deadalive Terms added sequentially (first to last) Df Deviance Resid. Df Resid. Dev P(>|Chi|) NULL...

...NOD NOD 1 2 NOD NOD F1 F1 NOD F1 1 3 NOD NOD F1 F1 NOD NOD 1 4 NOD F1 NOD NOD NOD NOD 1 5 NOD F1 F1 F1 NOD F1 1 If I want to use the information from names.select to fit a glm() on Cross, I can do something like: > one.glm2 <- glm(eval(substitute(Cross ~ x1 + x2 + x3 + x4, + list(x1 = as.name(names.select[1]), + x2 = as.name(names.select[2]), + x3 = as.name(names.select[3]), +...

...ing offset), and have resorted to re-estimating them by running a second polynomial (lm() this time) on the predicted values from predict() of the glm. This is clearly not a nice way of doing things. Could anyone please inform me of why this is happening and of a better way around this? Code: glm2<-glm(FEDSTATUS1~AGE+I(AGE^2), family=binomial(link="probit")) summary(glm2) ### first set of "wrong coefficients" nd1<-expand.grid(AGE=c(1:70)) Pred.Fed1<-predict(glm2,nd1,type="response") points(predict(glm2,nd1,type="response")~nd1$AGE, col=2) A...

..., 4, 9, 6, 9, 4, 7, 7, 6, 4, 8, 8, 9, 5, 8, 11, 13, 13, 3, 5, 7, 11, 11, 8, 6, 12, 11, 4, 13, 11, 6, 9, 12, 12, 7, 14, 4, 9, 11, 13, 11, 12, 16, 6, 13, 17, 17, 11, 8, 12, 16, 14) y <- cbind(successes, failures) data <- data.frame(y, x) glm1 <- glm(y ~ x, family= quasibinomial, data= data) glm2 <- glm(y ~ log(x), family=quasibinomial, data= data) # residual deviance is lower with log transformed x-value plot(x, successes) lines(x, predict(glm1, type= "response"), lwd=2) Firstly, because of the skewed distribution of the x variable I am not sure whether it should be lo...

...f samples simulated [1] 1000 >class(list.glm[[324]]) #any component of the list [1] "glm" "lm" >length(list.glm[[290]]$y) #sample size [1] 1000 Because length(list.glm) and the sample size are rather large, I've splitted the list into 10 sub-list, say: list.glm1, list.glm2,.... Now I'm using of course: out1<-lapply(list.glm1, myfun) out2<-lapply(list.glm2, myfun) .... However only the first works, for the second one it is: Error: cannot allocate vector of size 3 Kb In addition: Warning message: Reached total allocation of 255Mb: see help(memory.size) So I...

...far as I understand Baron and Li describe how to do this, but my question is now: how do I report this in an article? Can anyone recommend a particular article that shows a concrete example of how the results from te following simple modeling can be reported: glm1 = glm(DV ~ A, family = binomial) glm2 = glm(DV ~ A + B, family = binomial) anova(glm1, glm2, test = "Chisq") Any help on how this simple kind of modeling should be reported is appreciated. Greetings, Edwin Commandeur [[alternative HTML version deleted]]

...! i have run several glm analysises to estimate a mean rate of dung decay for independent trials. i would like to compare these results statistically but can't find any solution. the glm calls are: dung.glm1<-glm(STATE~DAYS, data=o_cov, family="binomial(link="logit")) dung.glm2<-glm(STATE~DAYS, data=o_cov_T12, family="binomial(link="logit")) as all the trials have different sample sizes (around 20 each), anova(dung.glm1, dung.glm2) is not applicable. has anyone an idea? thanks in advance! ps: my advisor urges me to use the z-test (the common test sta...

...the argument "coeff". The two factors are Y (Ya, Yb,..., Ye) and Auto (Auto1, Auto2,...., Auto10) The responses are binomial (counts in the matrice "mat") and gaussian (variable "tot"). I use the glm procedure : glm1 <- glm (mat ~ y*auto, data, family=binomial) glm2 <- glm (tot ~ y*auto, data, family=gaussian) Subsequently I use the Type III anova : Anova (glm1, type="III") Anova (glm2, type="III") Using the contrast.treatment for both factors Y and Auto, R gives coherent results, but I really not sure that this contrasts are appropr...

...R (3rd Ed. Ch7) and Chambers & Hastie (1993) were very helpful but I wasn't sure I got all the answers. In a simplistic example suppose I want to explore how disability (3 levels, profound, severe, and mild) affects the dichotomized outcome. The glm1 model (see below) is the same as glm2 (2 dummy variables coding sever and mild). In both models the reference level is profound disability. My questions are: 1. How do I interpret the coefficients in the third model (the intercept is removed)? 2. Does glm3 make sense? Does anyone ever want to construct a model like glm3? If...

Dear Listers, If I have several glm objects with names glm1, glm2.... and want to apply new data to these objects. Instead of typing "predict(glm1, newdata)..." 100 times, is there way I could do so in a loop? Thank you so much! wensui [[alternative HTML version deleted]]

...Any assistance would be greatly appreciated. The function is as follows: > colonization <- function(x,y){ library(segmented) col <- subset(final,final[,x] == 2 | final[,x] == 4) col[,x] <- ifelse(col[,x] == 2,1,0) glm1 <- glm(col[,x]~col[,"PLAND"],family=binomial) glm2 <- segmented.glm(obj = glm1, seg.Z = ~PLAND, psi=y, it.max = 100) } > colonization("PIWO",48) Error in eval(expr, envir, enclos) : object "x" not found Thank you.

...cremo.html This package addresses the problem of assembling and rendering results of multiple linear model fits in formats suitable for presentation in applications contexts. An extract from the output of 'example(cremo)': cremo> data(stackloss) cremo> attach(stackloss) cremo> glm2 <- glm(stack.loss ~ Air.Flow + Water.Temp + Acid.Conc.) cremo> glm3 <- glm(stack.loss ~ Air.Flow + Water.Temp) cremo> glm4 <- glm(stack.loss ~ Air.Flow * Water.Temp) cremo> try(print(cremo(list(glm2, glm3, glm4), adeqel = c("deviance", "aic")), quote =...

...hts = rep(1, 10)) add1(lm0, scope = ~ x) ## works ok add1(lm1, scope = ~ x) ## error lm2 <- lm(y ~ 1, offset = 1:10) add1(lm0, scope = ~ z) ## works ok add1(lm2, scope = ~ z) ## gives incorrect results (ignores the offset) glm0 <- glm(y ~ 1) glm1 <- glm(y ~ 1, weights = rep(1, 10)) glm2 <- glm(y ~ 1, offset = rep(0, 10)) add1(glm0, scope = ~ x) ## error add1(glm1, scope = ~ x) ## error add1(glm2, scope = ~ x) ## error As I see it, the two problems are: 1. add1.lm ignores any offset present in its "object" argument. 2. add1.lm and add1.glm both take weights...

...interpreting the theory behind this. Thanks for any help you can give. Max clotting <- data.frame( + u = c(5,10,15,20,30,40,60,80,100), + lot1 = c(118,58,42,35,27,25,21,19,18), + lot2 = c(69,35,26,21,18,16,13,12,12)) glm(lot2 ~ log(u), data=clotting, family=Gamma) prof<-profile(glm2) plot(prof) [[alternative HTML version deleted]]