similar to: QUestion on prediction of class from rpart

Hi. I uses rpart to build a regression tree. Y is continuous. Now, I try to predict on a new set of data. In the new set of data, one of my x (call Incoterm, a factor) has a new level. I wonder why the error below appears as the guide says "For factor predictors, if an observation contains a level not used to grow the tree, it is left at the deepest possible node and

I'm in a situation where I say: > predict(m.rpart, newdata=D[N1+t,]) 0 1 173 0.8 0.2 which I interpret as meaning: an 80% chance of "0" and a 20% chance of "1". Okay. This is consistent with: > predict(m.rpart, newdata=D[N1+t,], type="class") [1] 0 Levels: 0 1 But I'm puzzled at the following. If I say: > predict(m.rpart,

Buenos días, hago un mapa con ggplot: world<-map_data('world') windows();ggplot(legend=FALSE) + ... geom_point(data=Data,aes(x=lon,y=lat,color=Clst),size=1.25) + scale_color_manual(values=c("grey45","navy","skyblue","gold","green3","darkgreen")) + geom_path( data=world, aes(x=long, y=lat,group=group)) + labs(title =

Gracias Emilio y Jorge. Tengo que explicarlo mejor. Mostrando a una audiencia cómo hacer un tipo de análisis, se hace un loop (abajo) que analiza un mapa por regiones longitudinales. Tal y como está el script, print(i) te indica la longitud por la que va (de 10º en 10º) pero me gustaría que en vez de eso te fuese representando una línea vertical sobre el mapa, que he representado previamente con

Dear r-help mailing list, Is there a way to incorporate weights into the minsplit criteria in rpart, when the weights are uneven? I could not find a way for the minsplit threshold to take the weights into account, and when the weights are uneven it becomes an issue, as the following example shows. My current workaround is to expand the data into one in which each row is an observation, but that

Dear R users, I'm working with the rpart package and want to evaluate the performance of user defined split functions. I have some problems in understanding the meaning of the xval argument in the two functions rpart.control and xpred.rpart. In the former it is defined as the number of cross-validations while in the latter it is defined as the number of cross-validation groups. If I am

Dear list For my work, it would be helpful if rpart worked seamlessly with an empty model: library(rpart); rpart(formula=y~0, data=data.frame(y=factor(1:10))) Currently, an unrelated error (originating from na.rpart) is thrown. At some point in the near future, I'd like to release a package to CRAN which uses rpart and relies on that functionality. I have prepared a patch (minor

There have been various messages about packages tree and rpart whilst I have been travelling, and I have now prepared updates. tree ==== Tree is one of the oldest packages on CRAN (Feb 2000 apart from adding the maintainer field), and I had been hoping that it would fade away in favour of rpart. 1) sys.parent needed to be replaced by parent.frame in all but the most recent R (post 1.3.0).

Hello, I just upgraded my Ubuntu Xenial system to R 3.5.1 (from 3.4.?) by changing the sources.list entry and doing an "apt-get dist-upgrade". Everything works except loading the rpart package in R: > library(rpart) Error: package or namespace load failed for ?rpart?: package ?rpart? was installed by an R version with different internals; it needs to be reinstalled for use with

On 08/14/2014 05:00 AM, r-devel-request at r-project.org wrote: > Dear list > > > For my work, it would be helpful if rpart worked seamlessly with an > empty model: > > library(rpart); rpart(formula=y~0, data=data.frame(y=factor(1:10))) > > Currently, an unrelated error (originating from na.rpart) is thrown. > > At some point in the near future, I'd like to

I've written a small adaptation of the text.rpart function of the rpart package to better suite my tree presentation needs (I basically left the code untouched, only changing some numbers relating to label positioning). When I changed to version 1.7.0 this function stopped working printing the error : couldn't find function "rpartco" As far as I understand this has to do

Dear R users, I know cross-validation does not work in rpart with user defined split functions. As Terry Therneau suggested, one can use the xpred.rpart function and then summarize the matrix of the predicted values into a single "goodness" value. I need only a confirmation: set for example xval=10, if I correctly understood a single column of the matrix obatined by xpred.rpart gives

I tried using rpart, as below, and got this error message "rpart Error in yval[, 1] : incorrect number of dimensions". Thinking it might somehow be related to the large number of missing values, I tried using complete data, but with the same result. Does anyone know what may be going on, and how to fix it? I have traced two similar error messages in the Archive, but following the

A few weeks back I posted that the subset feature of rpart was not working when predicting a categorical variable. I was able to figure out a simple solution to the problem that I hope can be included in future editions of rpart. I also include a fix for another related problem. The basic problem is that when predicting a categorical using a subset, the subset may not have all the categories

R-users E-mail: r-help@r-project.org Hi! R-users. http://finzi.psych.upenn.edu/R/library/mvpart/html/xpred.rpart.html says: data(car.test.frame) fit <- rpart(Mileage ~ Weight, car.test.frame) xmat <- xpred.rpart(fit) xerr <- (xmat - car.test.frame$Mileage)^2 apply(xerr, 2, sum) # cross-validated error estimate # approx same result as rel. error from printcp(fit) apply(xerr, 2,

Hi, I have a technical question about rpart: according to Breiman et al. 1984, different costs for misclassification in CART can be modelled either by means of modifying the loss matrix or by means of using different prior probabilities for the classes, which again should have the same effect as using different weights for the response classes. What I tried was this: library(rpart)

Hi, I am new to R and need help with rpart. I am trying to create a classification tree using rpart. In order to plot the reults I use the plot function and the text function to label the plot of the tree dendrogram with text. The documentation of text.rpart says : "For the "class" method, label="yval" results in the factor levels being used, "yprob" results

(Apparently, I posted this to the wrong place. I am hopefully posting this is the correct place now. If not, please advise.) A few weeks back I posted that the subset feature of rpart was not working when predicting a categorical variable. I was able to figure out a simple solution to the problem that I hope can be included in future editions of rpart. I also include a fix for another related

(I'm sending to r-bugs because rpart is one of the recommended packages and is always installed. I'm also sending it directly to Dr. Ripley, as the maintainer.) rpart working as a classifier does not work (produces no splits) when the class indicator has no instances of one of the factor levels, as long as the factor level is not the final level. I have at least a partial fix, which I

Hello there. I have fitted a rpart model. > rpartModel <- rpart(y~., data=data.frame(y=y,x=x),method="class", ....) and can use rpart$where to find out the terminal nodes that each observations belongs. Now, I have a set of new data and used predict.rpart which seems to give only the predicted value with no information similar to rpart$where. May I know how