Perhaps your approach reflects a method of producing a prediction dataframe that is just unfamiliar to me, but it looks to me like you have created two predictor variables based on the names of the levels of the original predictor (predictors.train1, predictors.train2). I don't know how the glm function would know that predictors.train1 and predictors.train2 are two subs for predictors.train. Maybe try just using one prediction variable, and give it the original variable name (predictors.train). If this works, just repeat for your second set of values.
> Mark Fowler > Marine Fish Division > Bedford Inst of Oceanography > Dept Fisheries & Oceans > Dartmouth NS Canada > [EMAIL PROTECTED] > -----Original Message----- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: September 22, 2004 2:53 PM To: [EMAIL PROTECTED] Subject: [R] Issue with predict() for glm models Hello everyone, I am having a problem using the predict (or the predict.glm) function in R. Basically, I run the glm model on a "training" data set and try to obtain predictions for a set of new predictors from a "test" data set (i.e., not the predictors that were utilized to obtain the glm parameter estimates). Unfortunately, every time that I attempt this, I obtain the predictions for the predictors that were used to fit the glm model. I have looked at the R mailing list archives and don't believe I am making the same mistakes that have been made in the past and also have tried to closely follow the predict.glm example in the help file. Here is an example of what I am trying to do: ######################################################## set.seed(545345) ################ # Necessary Variables # ################ p <- 2 train.n <- 20 test.n <- 25 mean.vec.1 <- c(1,1) mean.vec.2 <- c(0,0) Sigma.1 <- matrix(c(1,.5,.5,1),p,p) Sigma.2 <- matrix(c(1,.5,.5,1),p,p) ############### # Load MASS Library # ############### library(MASS) ################################### # Data to Parameters for Logistic Regression Model # ################################### train.data.1 <- mvrnorm(train.n,mu=mean.vec.1,Sigma=Sigma.1) train.data.2 <- mvrnorm(train.n,mu=mean.vec.2,Sigma=Sigma.2) train.class.var <- as.factor(c(rep(1,train.n),rep(2,train.n))) predictors.train <- rbind(train.data.1,train.data.2) ############################################## # Test Data Where Predictions for Probabilities Using Logistic Reg. # # From Training Data are of Interest # ############################################## test.data.1 <- mvrnorm(test.n,mu=mean.vec.1,Sigma=Sigma.1) test.data.2 <- mvrnorm(test.n,mu=mean.vec.2,Sigma=Sigma.2) predictors.test <- rbind(test.data.1,test.data.2) ############################## # Run Logistic Regression on Training Data # ############################## log.reg <- glm(train.class.var~predictors.train, family=binomial(link="logit")) log.reg #> log.reg #Call: glm(formula = train.class.var ~ predictors.train, family = #binomial(link = "logit")) # #Coefficients: # (Intercept) predictors.train1 predictors.train2 # 0.5105 -0.2945 -1.0811 # #Degrees of Freedom: 39 Total (i.e. Null); 37 Residual #Null Deviance: 55.45 #Residual Deviance: 41.67 AIC: 47.67 ########################### # Predicted Probabilities for Test Data # ########################### New.Data <- data.frame(predictors.train1=predictors.test[,1], predictors.train2=predictors.test[,2]) logreg.pred.prob.test <- predict.glm(log.reg,New.Data,type="response") logreg.pred.prob.test #logreg.pred.prob.test # [1] 0.51106406 0.15597423 0.04948404 0.03863875 0.35587589 0.71331091 # [7] 0.17320087 0.14176632 0.30966718 0.61878952 0.12525988 0.21271139 #[13] 0.70068113 0.18340723 0.10295501 0.44591568 0.72285161 0.31499339 #[19] 0.65789420 0.42750139 0.14435889 0.93008117 0.70798465 0.80109005 #[25] 0.89161472 0.47480625 0.56520952 0.63981834 0.57595189 0.60075882 #[31] 0.96493393 0.77015507 0.87643986 0.62973986 0.63043351 0.45398955 #[37] 0.80855782 0.90835588 0.54809117 0.11568637 ######################################################## Of course, notice that the vector for the predicted probabilities has only 40 elements, while the "New.Data" has 50 elements (since n.test has 25 per group for 2 groups) and thus should have 50 predicted probabilities. As it turns out, the output is for the training data predictors and not for the "New.Data" as I would like it to be. I should also note that I have made sure that the names for the predictors in the "New.Data" are the same as the names for the predictors within the glm object (i.e., within "log.reg") as this is what is done in the example for predict.glm() within the help files. Could some one help me understand either what I am doing incorrectly or what problems there might be within the predict() function? I should mention that I tried the same program using predict.glm() and obtained the same problematic results. Thanks and take care, Joe Joe Rausch, M.A. Psychology Liaison Lab for Social Research 917 Flanner Hall University of Notre Dame Notre Dame, IN 46556 (574) 631-3910 "If we knew what it was we were doing, it would not be called research, would it?" - Albert Einstein ______________________________________________ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html ______________________________________________ [EMAIL PROTECTED] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
