search for: model3

...ata x = c(0,0,0,1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7); y0 = c(.1,.1,.1, .5,.5,.5, 1,1,1, 2,2,2, 4,4,4, 7,7,7, 9,9,9, 10,10,10); yp = c(0,.03,.05, 0,.05,.01, 0,.07,.03, .1,0,.2, .2,.1,0, .3,0,.1, 0,.3,.4, .3,.5,0); y = y0 + 4*yp; data = data.frame(x=x,y=y); #Run model library(nlme) model3 = try(gnls(y ~ SSfpl(x,A,B,xmid,scal),data=data,weights=varConstPower(const=1,power=0,form= ~fitted(.)))); #Examine results model3; #const = .6551298, power = .8913665 summary(model3); #const = .6551298, power = .8913665...

Hi, I'm trying to compare models, one of which has all parameters fixed using offsets. The log-likelihoods seem reasonble in all cases except the model in which there are no free parameters (model3 in the toy example below). Any help would be appreciated. Cheers, Jarrod x<-rnorm(100) y<-rnorm(100, 1+x) model1<-lm(y~x) logLik(model1) sum(dnorm(y, predict(model1), summary(model1)$sigma,log=TRUE)) # no offset - in agreement model2<-lm(y~offset(rep(1,100))+x-1) logLik(model2)...

...v:group) model1.summ <- summary(model1) model1.anv <- anova(model1) # separate regression lines for each group, but with the same slope model2 <- lm(lgVar ~ lgCov + group) model2.summ <- summary(model2) model2.anv <- anova(model2) # same regression line for all groups model3 <- lm(lgVar ~ lgCov) model3.summ <- summary(model3) model3.anv <- anova(model3) compare <- anova(model1, model2, model3) # compare all models # plots par(mfcol=c(1,2)) boxplot(lgVar ~ group, col="darkgoldenrod1") # plot the variate and covariate by group plot(lgVar...

...1,2,3),5))) # factor mid<-c(rep(1:5,4)) # covariate df<-data.frame(id, n.1,n.2, g, var.g,mod, mid) # Examples # Random Effects model1<-rma(g,var.g,mods=~mid,method="REML") # covariate model model2<-rma(g,var.g,mods=~mod,method="REML") # factor model model3<-rma(g,var.g,mods=~mid+mod,method="REML") # multiple # example matrix for predicting against model3 newdat<-expand.grid(c(1,2,3,4,5),c(1,2,3)) predict(model1,newmods=c(1,2,3,4,5)) predict(model2,newmods=c(1,2,3)) predict(model3,newmods=newdat) -- Andr...

...spencer.graves at pdf.com] Sent: Sunday, January 15, 2006 6:41 PM To: Rand, Hugh Cc: 'r-help at lists.R-project.org' Subject: Re: [R] trouble with extraction/interpretation of variance structure para meters from a model built using gnls and varConstPower How about this: > exp(coef(model3$modelStruct$varStruct)["const"]) const 0.6551298 Does that answer the question about not understanding the connection between summary(model3) and coef(model3$modelStruct$varStruct)["const"]? Regarding the question about R not being able to find 'coef.varFunc...

...1 7 9.702400 0.0170 adh31 1 2 0.015683 0.9118 awpm 1 2 2.824076 0.2349 apsm 1 2 1.520431 0.3428 orien 1 2 6.167557 0.1310 > > > model2<-lme(awsm~adh31+awpm+apsm+orien,random=~1|viney,method="ML") > model3<-lme(awsm~adh31+awpm+orien,random=~1|viney,method="ML") > anova(model2,model3) Model df AIC BIC logLik Test L.Ratio p-value model2 1 7 42.44324 46.91664 -14.22162 model3 2 6 41.47847 45.31281 -14.73924 1 vs 2 1.035230 0.3089 > > model3.1...

...+ alpha12 * x1[i] * x2[i] + b[i] } alpha0 ~ dnorm(0, 1.0E-6) alpha1 ~ dnorm(0, 1.0E-6) alpha2 ~ dnorm(0, 1.0E-6) alpha12 ~ dnorm(0, 1.0E-6) tau ~ dgamma(0.001, 0.001) sigma <- 1 / sqrt(tau) } " print(modelString) writeLines(modelString,con="model3.txt") modelCheck( "model3.txt" ) Which (usually) produces: > modelCheck( "model3.txt" ) Error in handleRes(res[[3]]) : An OpenBUGS module or procedure was called that did not exist. I've copied at the end of this message an example from a single R session that s...

Dear R users, Assume I have three models with the following AIC values: model AIC df model1 -10 2 model2 -12 5 model3 -11 2 Obviously, model2 would be preferred, but it "wastes" 5 df compared to the other models. Would it be allowed to select model3 instead, simply because it uses up less df and the delta-AIC between model2 and model3 is just 1? Many thanks for any suggestions/comments! Best wishes C...

...artment of Commerce, Sydney. I am using R 2.9.1 on Windows XP. This has reference to the package “MCMCpack”. My objective is to select a better model using various alternatives. I have provided here an example code from MCMCpack.pdf. The matrix of Bayes Factors is: model1 model2 model3 model1 1.000 14.08 7.289 model2 0.071 1.00 0.518 model3 0.137 1.93 1.000 > PostProbMod(BF) model1 model2 model3 0.82766865 0.05878317 0.11354819 I need assistance to interpret the results. If I am wrong, please correct me. 1. By looking at the matrix...

...way to analyze a mixed model with the MatingPair as the fixed effect, DrugPair as the random effect and the interaction between these two as the random effect as well. Please confirm if that seems correct. 2. Assuming the above code is correct, I have model2 in which I remove the interaction term, model3 in which I remove the DrugPair term and model4 in which I only keep the fixed effect of MatingPair. 3. I want to perform the log likelihood ratio test to compare these models and that's why I have REML=F. However the code anova(model1, model2, model3, model4) gives me a chisq estimate and a p-...

...ial", na.action = na.exclude) A second nested model could be: model2 <-glm(formula = exitus ~ age+gender, family = "binomial", na.action = na.exclude) To compare these two GLM models the instruction is: anova(model1,model2, test="F") Similarly for GAM models model3 <-gam(formula = exitus ~ s(age)+gender, family = "binomial", na.action = na.exclude) "R" allows to compare these two models (GLM vs. GAM) anova(model2,model3, test="F") This instruction returns a p-value with no error or warning, but this test is based on max...

Hi, I ran the following model using nlme: model2<-lme(log(malrat1)~I(year-1982),random=~1|Continent/Country,data=wbmal10) I'm trying to run a Poisson GlMM to avoid the above transformation but I don't know how to specify the model using lmer in the lme4 library: model3<-lmer((malrat1)~I(year-1982) + ??,data=wbmal10,family=poisson) How can I introduce a random factor of the form= ~1|Continent/Country? Thanks; -- Luis Fernando Chaves

Is there any way to put an argument into an object name. For example, say I have 5 objects, model1, model2, model3, model4 and model5. I would like to make a vector of the r.squares from each model by code such as this: rsq <- summary(model1)$r.squared for(i in 2:5){ rsq <- c(rsq, summary(model%i%)$r.squared) } So I assign the first value to rsq then cycle through models 2 through 5 gathering th...

If I have a model line = lm(y~x1) and I want to use a for loop to change the number of explanatory variables, how would I do this? So for example I want to store the model objects in a list. model1 = lm(y~x1) model2 = lm(y~x1+x2) model3 = lm(y~x1+x2+x3) model4 = lm(y~x1+x2+x3+x4) model5 = lm(y~x1+x2+x3+x4+x5)... model10. model_function = function(x){ for(i in 1:x) { } If x =1, then the list will only add model1. If x =2, then the list will add both model1 and model2. If x=3, then the list will add model1 model 2 and model3 and s...

...with the example shown below: * model1<-gam(nsdall ~ s(jdaylitr2), data=datansd) newd1 <- data.frame(jdaylitr2=(244:304)) pred1 <- predict.gam(model1,newd1,type="response")* The problem I am encountering now is that I cannot seem to get it done for the following type of model: *model3<-gam(y_no~s(day,by=mapID),family=binomial, data=mergeday)* My mapID consists of 8 levels of which I get individual plots with * plot(model3)*. When I do predict with a newdata in it just like my first model I need all columns to have the same amount of rows or else R will not except it ofcourse...

...ata.frame( response=rpois(40,1:40), explanatory=1:40) attach(poissondata) However, I have run into a problem because it looks like the lm model (with sqrt-transformation) fits the data best: ## model1=lm(response~explanatory,poissondata) model2=lm(sqrt(response+0.5)~explanatory,poissondata) model3=lm(log(response+1)~explanatory,poissondata) model4=glm(response~explanatory,poissondata,family=poisson) model5=glm(response~explanatory,poissondata,family=quasipoisson) model6=glm.nb(response~explanatory,poissondata) model7=glm(response~explanatory,quasi(variance="mu",link="identity&...

...sobel.lme<-function (pred, med, out, grpid) { NEWDAT <- data.frame(pred = pred, med = med, out = out, grpid=grpid) NEWDAT <- na.exclude(NEWDAT) model1 <- lme(out ~ pred, random=~1|grpid,data = NEWDAT) model2 <- lme(out ~ pred + med, random=~1|grpid, data = NEWDAT) model3 <- lme(med ~ pred, random=~1|grpid, data = NEWDAT) mod1.out <- summary(model1)$tTable mod2.out <- summary(model2)$tTable mod3.out <- summary(model3)$tTable indir <- mod3.out[2, 1] * mod2.out[3, 1] effvar <- (mod3.out[2, 1])^2 * (mod2.out[3, 2])^2 + (mod2.out[3,...

.... #group3 -2.71500 0.61412 -4.421 4.67e-05 *** #group4 -1.25833 0.61412 -2.049 0.045245 * #group5 0.08667 0.61412 0.141 0.888288 # Use sum contrasts to compare each group against grand mean. options(contrasts = c("contr.sum","contr.poly")) model3 <- lm(dv ~ group) summary(model3) # Again, this is as expected. Intercept is grand mean and others are deviatoions from that. #Coefficients: # Estimate Std. Error t value Pr(>|t|) # (Intercept) 0.7997 0.1942 4.118 0.000130 *** # group1 1.0028...

The Rows function which is called from plot.compareFits in the nlme package is not found. > plot(compareFits(coef(bp.model3),coef(bp.model3M))) Error in plot.compareFits(compareFits(coef(bp.model3), coef(bp.model3M))) : couldn't find function "Rows" > Can I find it elswhere? Have I missed a required package? Thanks Ross Darnell > library(help=nlme) nlme Linear and nonlinear mixed e...

Hi all, I'm trying to fit a nonlinear regression by "nlrob": model3=nlrob(y~a1*x^a2,data=transient,psi=psi.bisquare, start=list(a1=0.02,a2=0.7),maxit=1000) However an error message keeps popping up saying that the function psi.bisquare doesn't exist. I also tried psi.huber, which is supposed to be the default for nlrob: model3=nlrob(y~a1*x^a2,data=transient,...