similar to: Extract estimate of error variance from glm() object

... but do note: glm(lot1 ~ log(u), data = clotting, family = gaussian) is a plain old *linear model*, which is of course a specific type of glm, but not one that requires the machinery of glm() to fit. That is, the above is exactly the same as: lm(lot1 ~ log(u), data = clotting) and gives exactly the same sigma() ! (and I would therefore hazard the guess that the poster may misunderstand

I think vcov() gives estimates of VCV for coefficients. I want estimate of SD for residuals On Tue, Dec 24, 2024 at 7:24?PM Ben Bolker <bbolker at gmail.com> wrote: > > vcov(). ? > > > On Tue, Dec 24, 2024, 8:45 AM Christofer Bogaso <bogaso.christofer at gmail.com> wrote: >> >> Hi, >> >> I have below GLM fit >> >> clotting <-

?deviance ?anova Bert On Tue, Dec 24, 2024 at 6:22?AM Christofer Bogaso <bogaso.christofer at gmail.com> wrote: > > I think vcov() gives estimates of VCV for coefficients. > > I want estimate of SD for residuals > > On Tue, Dec 24, 2024 at 7:24?PM Ben Bolker <bbolker at gmail.com> wrote: > > > > vcov(). ? > > > > > > On Tue, Dec 24,

vcov(). ? On Tue, Dec 24, 2024, 8:45 AM Christofer Bogaso <bogaso.christofer at gmail.com> wrote: > Hi, > > I have below GLM fit > > 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)) > summary(glm(lot1 ~ log(u), data = clotting, family = gaussian)) > >

Dear all, I'm trying to use the "optim" function to replicate the results from the "glm" using an example from the help page of "glm", but I could not get the "optim" function to work. Would you please point out where I did wrong? Thanks a lot. The following is the code: # Step 1: fit the glm clotting <- data.frame( u =

Hello, silly question I suppose, but somehow I can't manage to extract the probabilities from a glm.predict() result: > str(res) Named num [1:9] 0.00814 0.01877 0.025 0.02941 0.03563 ... - attr(*, "names")= chr [1:9] "1" "2" "3" "4" ... I got from: # A Gamma example, from McCullagh & Nelder (1989, pp. 300-2) clotting <-

Hello, I am trying to figure out how to plot the profile likelihood curve of a GLM parameter with 95% pCI's on the same plot. The example I have been trying with is below. The plots I am getting are not the likelihood curves that I was expecting. The y-axis of the plots is tau and I would like that axis to be the likelihood so that I have a curve that maxes at the parameter estimate. I am

> d = data.frame(gender=rep(c('f','m'), 5), pos=rep(c('worker', 'manager', 'speaker', 'sales', 'investor'), 2), lot1=rnorm(10), lot2=rnorm(10)) > d gender pos lot1 lot2 1 f worker 1.1035316 0.8710510 2 m manager -0.4824027 -0.2595865 3 f speaker 0.8933589 -0.5966119 4 m sales

Hi R users, I want to use dglm Package. I run the examples and it give me an error: Error en dglm(lot1 ~ log(u), ~1, data = clotting, family = Gamma) : no se pudo encontrar la funci?n "as.family" dglm can't find "as.family" function why ? Thank you for your help

Hi, I have a data set (weight) that does not follow the Gaussian (Normal) distribution. However, I have to transform the data before applying the Gaussian distribution. I used this syntax and used log(weight) as: posJy.model<-glm(log(weight) ~ factor(pos), family=gaussian(link='identity'), subset=Soil=="Jy"). This syntax COULD NOT transform the data. But if I transform the

Below is my full implementation (tried to make it simple as for demonstration) Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Lapply_me(as.list(1:4), function(xx) { if (xx ==

Hi all, please see my code: > library(zoo) > a <- as.yearmon("March-2010", "%B-%Y") > b <- as.yearmon("May-2010", "%B-%Y") > > nn <- (b-a)*12 # number of months in between them > nn [1] 2 > as.integer(nn) [1] 1 What is the correct way to find the number of months between "a" and "b", still

The reason that it works for Apply_MC=TRUE is that in that case you call mclapply(X,FUN,...) and the mclapply() function strips off the mc.cores argument from the "..." list before calling FUN, so FUN is being called with zero arguments, exactly as it is declared. A quick workaround is to change the line Lapply_me(as.list(1:4), function(xx) { to Lapply_me(as.list(1:4),

My modified function looks below : Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Here, I am not passing ... anymore rather passing myList On Sun, Mar 4, 2018 at 10:37 PM,

@Eric - with this approach I am getting below error : Error in FUN(X[[i]], ...) : unused argument (list()) On Sun, Mar 4, 2018 at 10:18 PM, Eric Berger <ericjberger at gmail.com> wrote: > Hi Christofer, > You cannot assign to list(...). You can do the following > > myList <- list(...)[!names(list(...)) %in% 'mc.cores'] > > HTH, > Eric > > On Sun, Mar

Dear all, let say I have following list: Dat <- vector("list", length = 26) names(Dat) <- LETTERS My_Function <- function(x) return(rnorm(5)) Dat1 <- lapply(Dat, My_Function) However I want to apply my function 'My_Function' for all elements of 'Dat' except the elements having 'names(Dat) == "P"'. Here I have specified the name

Hi again, I am struggling to extract the number part from below string : "\"cm_ffm\":\"563.77\"" Basically, I need to extract 563.77 from above. The underlying number can be a whole number, and there could be comma separator as well. So far I tried below : > library(stringr) > str_extract("\"cm_ffm\":\"563.77\"",

Dear all, when I start R, I want that the console window should be in the Maximized size automatically. Can somebody help me how to achieve that? Thanks and regards,

Hello again, I want to remove "$" sign and replace with nothing in my text. Therefore I used following code: > gsub("$|,", "", "$232,685.35436") [1] "$232685.35436" However I could not remove '$' sign. Can somebody help me why is it so? Thanks and regards

Dear all. Let say I have a group of codes which will be used in many places in my overall R-code files. These group of codes will be used within a for-loop (with a big length, like 10000 times) and also many other places outside of that for loop. As this group of codes are being used in many places, I thought to put them within a user-defined function. Here my question is, is there any speed