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...d predictor variables) for each country of the world; I started by fitting standard GLM and tested for spatial correlation using variogram models (geoR) fitted to the residuals of the GLM. Spatial autocorrelation is significant. Therefore, I think about using general estimation equations (geeglm or geese in geepack) allowing for residual spatial correlation . Several questions: 1. Are GEE’s the right choice? 2. How can I specify a user-defined spatial (exponential) correlation structure in geeglm or geese? I did not really get this from the help-sites. I guess I need to set corstr = "user...

Hello, Is it possible to derive Pan's QIC (2001 Biometrics 57:120) from either a fitted gee() object in the gee package or from a geese() fit in the geepack package? If so, would anyone be kind enough to provide me with code to do so? I realize that QIC is part of the output from yags() but I would like to use one of the other functions. Thanks. Richard

We are currently analyzing data on children clustered in day care-centers (DCC). We have tried to use geepack and gee libraries to estimate an overall incidence rate for absences (=number of absences/risk time) by specifying geese(number.absences ~ offset(log(risktime)), id=day.care.id, family=poisson("log"), data=dcc, corstr="exch", sca.link="log", cor.link="fisherz") gee(number.absences ~ offset(log(risktime)), id=day.care.id,...

Dear R users, I fitted a GEE model using the function 'geese' (or 'geeglm') with user defined correlation matrix. I want to get the var-cov matrix of the regression coefficients. But the output provides only limited information. I would be very much thankful if you could kindly let me know how to get it..since I am struggling lot getting thi...

Good evening, In my research I am studying the marginal models, where the main goal is on the structure of the association. My practical example has cluster with up to 600 observations and with this database, the function geese() return me the following message: This application has requested the runtime to terminate it is on unusual way. Please contact the application’s support team for more information why this happens? In other hand, when I use the ALR function ordgee(), after a few hours, models converge....

Hi, In my research I am studying the marginal models, where the main goal is on the structure of the association. My practical example has cluster with up to 600 observations and with this database, the function geese() return me the following message: This application has requested the runtime to terminate it is on unusual way. Please contact the application’s support team for more information why this happens? In other hand, when I use the ALR function ordgee(), after a few hours, models converge....

...geable", std.err = "san.se") mm3 <- yags(breaks ~ tension, id=wool, data=warpbreaks, corstr="exchangeable", alphainit=0.) # S.E. # gee: sqrt(diag(mm$robust.variance)) #(Intercept) tensionM tensionH # 5.77 7.46 3.73 # geeglm: sqrt(diag(mm2$geese$vbeta)) # [1] 8.17 10.56 5.28 # yags: sqrt(diag(slot(mm3, "robust.parmvar"))) # [1] 5.77 7.46 3.73 Any explanation of this behavior is welcome. Best wishes Torleif Markussen Lunde PhD candidate Centre for International Health/ Bjerkens Centre for Climate Research University of Bergen...

I am working with geese from geepack, does anyone know how to get the predicted values from this program. I tried the predict function and got the error Error in predict(results.e, newdat = prog) : no applicable method for "predict" Thank you for any help you can send, Suzette ========================...

...ual = a code indicating the id of the specimen. I have up to 12 replicate observations for individual. Some observation are missing and I organized the data frame to have exactly 12 rows for each specimen, with NAs where there is a missing observation. The following model worked fine: gee1<-geese(Activity~Water, id=Individual, data=dataF, family=gaussian) but when I use other variables e.g gee2<-geese(Activity~Hgt+Size+Water, id=Individual, data=dataF, family=gaussian) I get the error message "Error in geese.fit(x, y, id, offset, soffset, w, waves, zsca, zcor, corp, : nrow(...

...em (I get the same results using 'binomial(link="logit")' or 'binomial(link="corlogit")'). But with library(geepack) (Version: 0.2-10) I get different results for logit and test.logit as link functions. First there was an error, so that I had to modify 'geese.fit' (I only added test.logit where I found logit). With library(gee) (Version: 4.13-10) R exits without error when fitting a GEE with binomial(link="corlogit"). Because of the error 'unkown link' I modified 'gee' the same way as for geepack(). I never changed t...

....gee<-gee(score ~ chem + time, id=id, family=gaussian,corstr="exchangeable",data=dat) and library(yags) fit.yags <- yags(score ~ chem + time, id=id, family=gaussian,corstr="exchangeable",data=dat,alphainit=0.05) However, I am making a mistake with: library(geepack) fit.geese <- geese(score ~ chem + time, id=id, family=gaussian,corstr="exch",data=dat) I obtain the following error: Error in geese.fit(x, y, id, offset, soffset, w, waves, zsca, zcor, corp, : nrow(zsca) and length(y) not match Could someone tell me what I have done incorrectly. Thanks for...

Hi R-users, I have a problem for updating the estimates of correlation coefficient in simulation loop. I want to get the matrix of correlation coefficients (matrix, name: est) from geese by using loop(500 times) . I used following code to update, nsim<-500 est<-matrix(ncol=2, nrow=nsim) for(i in 1:nsim){ fit <- geese(x ~ trt, id=subject, data=data_gee, family=binomial, corstr="exch", scale.fix=TRUE) ............. corr_gee<-summary(fit)$correlation[1] se_corr...

...f order 1. Surprisingly for me, I found very different results. Coefficients are slightly different in value but sometimes opposite in sign. Moreover, the estimate of rho (correlation coefficient) given by gee is 0.5759 (see element 1.2 in the working correlation matrix) while the estimate given by geese is 0.4519. Could somebody explain me what happened? Data are in dati22, which is only a part of the data for the study I intend to perform. Here what I did ,using first gee and then geese: > str(dati22) `data.frame': 47 obs. of 7 variables: $ TR : Factor w/ 4 levels "7",...

I want to run a GEE for a time series of counts. The data are daily respiratory mortality counts and so there aren't any 'clusters' in the longitudinal sense. Neither the gee or geese packages work. The gee one wont run at all and the geese one produces NaNs or just runs indefinitely depending on how long the time series is. Any ideas how to make these work of any other packages that might do the trick? Duncan Lee d.p.lee@maths.bath.ac.uk mapdpl@bath.ac.uk [[alternative...

Hello, I have the following r-codes for solving a quasilikelihood estimating equation: >library(geepack) >fit<-geese(y~x1+x2+x3,jack=TRUE,id=id,scale.fix=TRUE,data=dat,mean.link = "logit", corstr="independence") Now my question is how can I calculate the confidence interval of the parameters of the above model "fit"? [[alternative HTML version deleted]]

...ut without any adjustments for covariates. Q: It looks like it works but have I specified it correctly to get the statistics that I want? geeout <- geeglm(outcome ~ 1, id=familyID, family=gaussian, data=dat, corstr="unstructured") # Then the adjusted mean would be in: summary(geeout)$geese$mean # and the adjusted variance would be the estimate in: summary(geeout)$geese$scale If you know of a better way to analyse this data, I would love to know this as well. Thank you very much for your time and any help you can offer. Jillian Dwyer [[alternative HTML version deleted]]

...a normal category . This only gives me a obese not obese 14, but when I did with the wide format hoping to get a obese14,normal14,overweight 14 Vs hibp 21, i could not complete any of the models. This time I classified obese=1 & overweight=1 as obese itself. Can you tell me if I can use the geese or geeglm function with this data eg: : HIBP~ time* Age Here age is a factor with 3 levels, time: 2 levels, HIBP = yes/no. It says geese/geeglm: contrast can be applied only with factor with 2 or more levels. What is the way to overcome this. Can I manipulate the data to make it work. I need to k...

...ry(pseudo) pseudo <- pseudoci(time=a$TIME,event=a$icr,tmax=cutoffs) #rearranging data b <- NULL for(it in 1:length(pseudo$time)){ b <- rbind(b,cbind(a,pseudo = pseudo$pseudo[[1]][,it], tpseudo = pseudo$time[it],id=1:nrow(a))) } b <- b[order(b$id),] library(geepack) fit3 <- geese(pseudo ~ as.factor(AGE) + as.factor(WEIGHT) + as.factor(HEIGHT) , data =b, id=id, jack = TRUE, scale.fix=TRUE, mean.link = "logit", corstr="unstructured") #The results using the AJ variance estimate cbind(mean = round(fit3$beta,4), SD = round(sqrt(diag(fit3$vbeta.ajs)),4), Z =...

...0.358 169 -4.49 0.0000 --- glmmML/glmmML/Brostr?m coef se(coef) z Pr(>|z|) (Intercept) 3.579 0.701 5.10 3.3e-07 trtdrug -1.369 0.694 -1.97 4.8e-02 trtdrug+ -0.789 0.700 -1.13 2.6e-01 I(week > 2)TRUE -1.627 0.482 -3.38 7.3e-04 --- geese/geepack/Jun Yan estimate san.se wald p (Intercept) 2.844 0.529 28.92 7.56e-08 trtdrug -1.113 0.586 3.61 5.76e-02 trtdrug+ -0.634 0.544 1.36 2.44e-01 I(week > 2)TRUE -1.325 0.368 12.96 3.18e-04 --- glmm/repeated/J.K.Lindsey Es...