similar to: Logistic Regression Fitting with EM-Algorithm

Hi, I am trying to fit a logistic regression using glm, but my explanatory variables are of mixed type: some are numeric, some are ordinal, some are categorical, say If x1 is numeric, x2 is ordinal, x3 is categorical, is the following formula OK? *model <- glm(y~x1+x2+x3, family=binomial(link="logit"), na.action=na.pass)* * * *Thanks,* * * *-Jack* [[alternative HTML version

How do I construct a figure showing predicted value plots for the dependent variable as a function of each explanatory variable (separately) using the results of a logistic regression? It would also be helpful to know how to show uncertainty in the prediction (95% CI or SE). Thanks- This email has been processed by SmoothZap - www.smoothwall.net

Hi R-experts, Does anyone have experience using R for handling large scale data (millions of rows, hundreds or thousands of features)? What is the largest size of data that anyone has used with glm? Also, is there a library to read data in sparse data format (like SVMlight format)? Thanks Pradheep [[alternative HTML version deleted]]

Hello Folks, I hope that you are well. I have installed CentOS 4.1 and I have an MS mouse. The mouse pointer can be moved around BUT I cannot click on anything menus and such, even during the install the mouse isn't operational for clickage. Any thoughts. Any help would be most welcome. I am using an Avocent SwitchView MP, but its made no difference in installing Core 3 or any other OS.

Hi, I have a dichotomous variable (Q1) whose answers are Yes or No. Also I have 2 categorical explanatory variables (V1 and V2) with two levels each. I used logistic regression to determine whether there is an effect of V1, V2 or an interaction between them. I used the R and SAS, just for the conference. It happens that there is disagreement about the effect of the explanatory variables

Dear R-list, *When comparing two logistic regression models with the anova CHi test, I obtain the following error: (there are no NA's in the time series). How can this be solved such that I can compare two models on the same dataset were different explanatory variables are used? l.KBDI <- glm(zna.arson2 ~ zna.KBDI,family = binomial) l.NDWI <- glm(zna.arson2 ~ zna.NDWI,family

Dear R-helpers, I tried to create a 3D surface showing the interaction between two continuous explanatory variables; the response variable is binary (0/1). The model is: model<-glm(incidence~sun*trees,binomial) then I used "wireframe" to create a 3D plot: xyz<-expand.grid(sun=seq(30,180,1),trees=seq(0,4000,10)) xyz$incidence<-as.vector(predict(model,xyz))

Hello all, I've been having trouble with assessing a breakpoint in a logistic GLM with two explanatory variables. For this analysis I've been using the 'segmented' package version 0.2-9.1. But I keep getting an error and I don't see where I would be going awry. The situation is the following: Two explanatory variables: bedekking - a variable with possible values between 0 and

Dear R list, I have one very elementary question regrading correlation between two variables. x = c(44,46,46,47,45,43,45,44) y = c(44,43,41,41,46,48,44,43) > cov(x, y) [1] -2.428571 However, if I try to calculate the covariance using the formula as covariance = sum((x-mean(x))*(y-mean(y)))/8       # no of of paired obs. = 8 or     covariance = sum(x*y)/8-(mean(x)*mean(y)) gives

Dear Masters, first I'd like to wish u all a great 2011 and happy holydays by now, second (here it come the boring stuff) I have a question to which I hope u would answer: I run a logistic regression by glm(), on the following data type (y1=1,x1=x1); (y2=0,x2=x2);......(yn=0,xn=xn), where the response (y) is abinary outcome on 0,1 amd x is any explanatory variable (continuous or not)

Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c("A1","B1") L[2] = c("A2","B2") L[3] = c("A3","B3")

Hello, I have a problem concerning logistic regressions. When I add a quadratic term to my linear model, I cannot draw the line through my scatterplot anymore, which is no problem without the quadratic term. In this example my binary response variable is "incidence", the explanatory variable is "sun": > model0<-glm(incidence~1,binomial) >

Greetings All. My apologies for a question whose answer is probably readily available somewhere (for some interpetation of "somewhere") ... Say I have just typed (from a sheet of paper) several lines into the R command-line, and what I see is: > chisq.test(matrix(c(3,6,3,4,4, + 4,1,4,6,5, + 2,7,4,2,5, + 8,2,4,4,2, +

Hi Folks, According to the PDF file blogit.pdf in the Zelig documentation: "Use the bivariate logistic regression model ["blogit"] if you have two binary dependent variables (Y1,Y2), and and wish to model them jointly as a function of some explanatory variables. Each pair of dependent variables (Yi1,Yi2) has four potential outcomes, (Yi1=1,Yi2=1), (Yi1=1,Yi2=0),

many languages have shorthands for that operation like: variable += 1 or ++variable is there something like that in R ? -- View this message in context: http://r.789695.n4.nabble.com/increase-or-decrease-variable-by-1-tp3076390p3076390.html Sent from the R help mailing list archive at Nabble.com.

Hello all, Could someone tell me what I am doing wrong here? I am trying to fit a binary logistic model using the lrm function in Design. The dataset I am using has a dichotomous response variable, 'covered' (1-yes, 0-no) with explanatory variables, 'nepall', 'title', 'abstract', 'series', and 'author1.' I am running the following script and

Dear R-help members, I'd like to run a binomial logistic stepwise regression with ten explanatory variables and as many interaction terms as R can handle. I'll come up with the right R command sooner or later, but my real question is whether and how the criterion for the evaluation of the different models can be set to be the probability of the residual deviance in the Chi-Square

Hi, I am seeking help on designing an algorithm to identify the locations of stretches of 1s in a vector of 0s and 1s. Below is an simple example: > dat<-as.data.frame(cbind(a=c(F,F,T,T,T,T,F,F,T,T,F,T,T,T,T,F,F,F,F,T) ,b=c(4,12,13,16,18,20,28,30,34,46,47,49,61,73,77,84,87,90,95,97))) > dat a b 1 0 4 2 0 12 3 1 13 4 1 16 5 1 18 6 1 20 7 0 28 8 0 30 9 1 34 10 1 46 11 0

Dear R-list. I have done a logistic glm using Age as explanatory variable for some allergic event. #the model model2d<-glm(formula=AEorSAEInfecBac~Age,family=binomial("logit"),data=emrisk) #predictions for age 30 and 60 preds<-predict(model2d,data.frame(Age=c(30,60)),se.fit=TRUE) # prediction interval

Dear All, I am learning the ropes about logistic regression in R. I found some interesting examples http://bit.ly/Vq4GgX http://bit.ly/W9fUTg http://bit.ly/UfK73e but I am a bit lost. I have several questions. 1) For instance, what is the difference between glm.out = glm(response ~ poverty + gender, family=binomial(logit), data=mydata) and glm.out = glm(response ~ poverty * gender,