Re: [R] rpart package: why does predict.rpart require values for unused predictors?
Jason, In the help file for predict.rpart it says, The predictors referred to in the right side of formula(object) must be present by name in newdata. ?predict.rpart So, that's just the way it is. There are a couple ways to work around this, if you wish. You could create a data frame with all NAs for the unused predictor(s). For example, newdata2 - data.frame(Disp.=car.test.frame$Disp., Weight=car.test.frame$Weight, HP=as.numeric(rep(NA, dim(car.test.frame)[1]))) predict(model, newdata=newdata2) Or, you could refit the model using only the important factors. For example, model2 - rpart(Mileage ~ Weight + Disp., car.test.frame) predict(model2, newdata=newdata) Jean Jason Roberts jason.robe...@duke.edu wrote on 08/01/2012 05:17:38 PM: After fitting and pruning an rpart model, it is often the case that one or more of the original predictors is not used by any of the splits of the final tree. It seems logical, therefore, that values for these unused predictors would not be needed for prediction. But when predict() is called on such models, all predictors seem to be required. Why is that, and can it be easily circumvented? Consider this example: model - rpart(Mileage ~ Weight + Disp. + HP, car.test.frame) model n= 60 node), split, n, deviance, yval * denotes terminal node 1) root 60 1354.58300 24.58333 2) Disp.=134 35 154.4 21.4 4) Weight=3087.5 22 61.31818 20.40909 * 5) Weight 3087.5 13 34.92308 23.07692 * 3) Disp. 134 25 348.96000 29.04000 6) Disp.=97.5 16 101.75000 27.12500 * 7) Disp. 97.5 9 84.2 32.4 * newdata - data.frame(Disp.=car.test.frame$Disp., Weight=car.test.frame$Weight) predict(model, newdata=newdata) Error in eval(expr, envir, enclos) : object 'HP' not found In this model, Disp. and Weight were used in splits, but HP was not. Thus I expected to be able to perform predictions by providing values for just Disp. and Weight, but predict() failed when I tried that, complaining that HP was not also provided. Thanks for any help you can provide. My apologies if I simply do not understand how this works. Best regards, Jason [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] rpart package: why does predict.rpart require values for unused predictors?
Jean, Thanks for your quick reply and suggestions! In the help file for predict.rpart it says, The predictors referred to in the right side of formula(object) must be present by name in newdata. I was aware of that statement from the help file. I wondered about the reason for that requirement. It would be convenient for the caller to not have to provide values for unused predictors. I wondered whether the requirement to provide them all was related to something I did not understand, such as surrogate splits, or whether imposing it simply made rpart itself easier to implement. (No offence intended to the authors for taking a shortcut, if indeed they did.) Are you pretty confident that your suggested workarounds will result in a model that produces identical predictions? I only ask because I'm aware that rpart has the ability to use surrogate variables in place of predictors that are missing. But I do not fully understand how that capability works. I do not know whether it is only used during fitting and not prediction. Continuing my example, I can see that printcp produces some output Variables actually used in tree construction: printcp(model) Regression tree: rpart(formula = Mileage ~ Weight + Disp. + HP, data = car.test.frame) Variables actually used in tree construction: [1] Disp. Weight ... I can see in the source for printcp how those variables were obtained. But when doing predictions, is it really safe to only provide them and not HP, if I expect that there could be missing values for them? When I call summary, I can see surrogate splits that reference the HP variable: summary(model) Call: rpart(formula = Mileage ~ Weight + Disp. + HP, data = car.test.frame) n= 60 CP nsplit rel errorxerror xstd 1 0.62840234 0 1.000 1.0326274 0.17828576 2 0.12032318 1 0.3715977 0.5271278 0.08627909 3 0.04293478 2 0.2512745 0.4092689 0.07260291 4 0.0100 3 0.2083397 0.3629544 0.06865150 Node number 1: 60 observations,complexity param=0.6284023 mean=24.58333, MSE=22.57639 left son=2 (35 obs) right son=3 (25 obs) Primary splits: Disp. 134to the right, improve=0.6284023, (0 missing) Weight 2567.5 to the right, improve=0.5953491, (0 missing) HP 104.5 to the right, improve=0.4085043, (0 missing) Surrogate splits: Weight 2747.5 to the right, agree=0.900, adj=0.76, (0 split) HP 104.5 to the right, agree=0.817, adj=0.56, (0 split) ... Assuming that the answer is: 1. The best predictions will be obtained by providing values for the variables actually used in tree construction plus those used as surrogates, and: 2. If a variable is neither actually used in tree construction nor as a surrogate, it can be safely set to NA for the prediction. Do you know of a way to easily identify the variables used as surrogates? Thanks again for your help, and sorry to write a book in response, Jason __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] rpart package: why does predict.rpart require values for unused predictors?
After fitting and pruning an rpart model, it is often the case that one or more of the original predictors is not used by any of the splits of the final tree. It seems logical, therefore, that values for these unused predictors would not be needed for prediction. But when predict() is called on such models, all predictors seem to be required. Why is that, and can it be easily circumvented? Consider this example: model - rpart(Mileage ~ Weight + Disp. + HP, car.test.frame) model n= 60 node), split, n, deviance, yval * denotes terminal node 1) root 60 1354.58300 24.58333 2) Disp.=134 35 154.4 21.4 4) Weight=3087.5 22 61.31818 20.40909 * 5) Weight 3087.5 13 34.92308 23.07692 * 3) Disp. 134 25 348.96000 29.04000 6) Disp.=97.5 16 101.75000 27.12500 * 7) Disp. 97.5 9 84.2 32.4 * newdata - data.frame(Disp.=car.test.frame$Disp., Weight=car.test.frame$Weight) predict(model, newdata=newdata) Error in eval(expr, envir, enclos) : object 'HP' not found In this model, Disp. and Weight were used in splits, but HP was not. Thus I expected to be able to perform predictions by providing values for just Disp. and Weight, but predict() failed when I tried that, complaining that HP was not also provided. Thanks for any help you can provide. My apologies if I simply do not understand how this works. Best regards, Jason __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
