similar to: How R calculates SE of prediction for Logistic regression?

?s 12:13 de 13/07/2024, Christofer Bogaso escreveu: > Hi, > > I ran below code > > Dat = read.csv('https://raw.githubusercontent.com/sam16tyagi/Machine-Learning-techniques-in-python/master/logistic%20regression%20dataset-Social_Network_Ads.csv') > head(Dat) > Model = glm(Purchased ~ Gender, data = Dat, family = binomial()) > head(predict(Model,

Hi, I ran below code Dat = read.csv('https://raw.githubusercontent.com/sam16tyagi/Machine-Learning-techniques-in-python/master/logistic%20regression%20dataset-Social_Network_Ads.csv') head(Dat) Model = glm(Purchased ~ Gender, data = Dat, family = binomial()) head(predict(Model, type="response")) My_Predict = 1/(1+exp(-1 * (as.vector(coef(Model))[1] * as.vector(coef(Model))[2] *

In the 'doBy' package there is an esticon() function for calculating linear contrasts for various model types. I have defined an S3-method 'esticon.mer()' for 'mer' objects from the lme4 package. Building the package and invoking the method gives: > esticon(fm1, c(1,1)) Confidence interval ( WALD ) level = 0.95 Error in as.integer(x) : cannot coerce type 'S4'

Hi, I am trying to replicate the R's result for VCV matrix of estimated coefficients from linear model as below data(mtcars) model <- lm(mpg~disp+hp, data=mtcars) model_summ <-summary(model) MSE = mean(model_summ$residuals^2) vcov(model) Now I want to calculate the same thing manually, library(dplyr) X = as.matrix(mtcars[, c('disp', 'hp')] %>% mutate(Intercept =

sigma(model)^2 will give the correct MSE. Also note that your model matrix has intercept at the end whereas vcov will have it at the beginning so you will need to permute the rows and columns to get them to be the same/ On Wed, Sep 4, 2024 at 3:34?PM Daniel Lobo <danielobo9976 at gmail.com> wrote: > > Hi, > > I am trying to replicate the R's result for VCV matrix of

Dear List, I'm just teaching myself semi-parametric techniques. Apologies in advance for the long post. I've got observational data and a longitudinal, semi-parametric model that I want to fit in GAM (or potentially something equivalent), and I'm not sure how to do it. I'm posting this to ask whether it is possible to do what I want to do using "canned" commands

Hi all, Can anyone please guide me how to draw a Concentration ellipsoid for a bivariate system with a bivariate normal dist. having a VCV matrix : Sigma <- matrix(c(1,2,2,5), 2, 2) I would like to draw in using GGPLOT. Your help will be highly appreciated. Thanks, -- View this message in context: http://www.nabble.com/Concentration-ellipsoid-tp25315705p25315705.html Sent from the R help

Hello, I'm doing a comparative analysis of mammal brain and body size data. I'm following Charlie Nunn and Natalie Cooper's instructions for "Running PGLS in R using caper". I run into the following error when I create my comparative dataset, combining my phylogenetic tree (mammaltree) and taxon measures (mammaldata): "Error in phy$node.label[which(newNb > 0) -

I am fitting a logistic binomial model of the form glm(y ~ a*x,family=binomial) where a is a factor (with 5 levels) and x is a continuous predictor. To assess how much ``impact'' x has, I want to compare the fitted success probability when x = its maximum value with the fitted probability when x = its mean value. (The mean and the max are to be taken by level of the factor

## I clicked the send-button too quickly, before changing the title of the message etc... Sorry.## I am running following phylogenetic analyses with the caper package: data=read.table(file="data.txt",header=T,sep="\t") tree = read.nexus("Tree.nex") primate = comparative.data(phy=tree, data=data,              names.col=Species, vcv=TRUE, na.omit=FALSE,

Hello, I am trying to carry out a phylogenetic autoregression to test whether my data show a phylogenetic signal, but I keep calculating bizzare R-squared values. My script is: > library(ape) > x <-

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 ==

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,

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

Hello. I want to use conditional logistic regression to calculate the probability of winning one of three players in golf. I was able to calculate these probabilities in Stata10, and I now want to transfer the code in the R Project, because it is can get data directly form MySQL. Unfortunately, I'm novice in R and I can't calculate the probability using the predict function when trying to

That's fine. The issue is how you called Lapply_me(). What did you pass as the argument to FUN? And if you did not pass anything that how is FUN declared? You have not shown that in your email. On Sun, Mar 4, 2018 at 7:11 PM, Christofer Bogaso < bogaso.christofer at gmail.com> wrote: > My modified function looks below : > > Lapply_me = function(X = X, FUN = FUN, Apply_MC =

@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

Hi Christofer, Before you made the change that I suggested, your program was stopping at the statement: list(...) = list(..) .etc This means that it never tried to execute the statement: return(lapply(X,FUN,...)) Now that you have made the change, it gets past the first statement and tries to execute the statement: return(lapply(X,FUN,...)). That attempt is generating the error message because

Bert, Posit (formerly RStudio) has moved from RMarkdown to Quarto. They still support RMarkdown but major new features will be in Quarto. For new users a better choice would be Quarto. See https://quarto.org/docs/faq/rmarkdown.html Secondly, the OP stated he was using the VS-Code IDE, so there is no reason to point him to the Posit/RStudio IDE for this functionality which VS-Code supports very