predict,bbl-method {bbl} | R Documentation |
bbl
ModelMake prediction of response group identity based on trained model
## S4 method for signature 'bbl' predict(object, newdata = NULL, logit = TRUE, useC = TRUE, verbose = 1, naive = FALSE, progress.bar = FALSE)
object |
Object of class |
newdata |
Data frame of new data for which prediction is to
be made. Columns must contain all of those in |
logit |
Return predictors whose logistic function gives probability; otherwise return probability itself. |
useC |
Use |
verbose |
Verbosity level |
naive |
Naive Bayes. Skip all interaction terms. |
progress.bar |
Display progress of response group probability. Useful for large samples. |
Will use new data set for predictors and trained bbl
model
parameters and compute posterior probabilities of response group
identity.
Matrix of predictors/posterior proabilities with samples in rows and response groups in columns.
set.seed(154) m <- 5 L <- 3 n <- 1000 predictors <- list() for(i in 1:m) predictors[[i]] <- seq(0,L-1) par0 <- randompar(predictors=predictors, h0=0, J0=0, dJ=0.5) xi0 <- sample_xi(nsample=n, predictors=predictors, h=par0$h, J=par0$J) par1 <- randompar(predictors=predictors, h0=0.1, J0=0.1, dJ=0.5) xi1 <- sample_xi(nsample=n, predictors=predictors, h=par1$h, J=par1$J) xi <- rbind(xi0,xi1) y <- c(rep(0,n),rep(1,n)) dat <- cbind(data.frame(y=y),xi) dat <- dat[sample(2*n),] dtrain <- dat[seq(n),] dtest <- dat[seq(n+1,2*n),] ytest <- dtest[,'y'] model <- bbl(data=dtrain) model <- train(model) pred <- predict(object=model, newdata=dtest) yhat <- apply(pred,1,which.max)-1 score <- mean(ytest==yhat) score auc <- pROC::roc(response=ytest, predictor=pred[,2], direction='<')$auc auc