similar to: predict.rpart help

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

Hi, I am trying to get an error estimation for a classification done using lda. The examples work fine, however I don't get my own code to work. The data is in object d > d class hydrophobicity charge geometry 1 2 6490.0400 1434.9700 610.99902 2 2 1602.0601 400.6030 -5824.00000 3 2 969.0060 260.1360 -415.00000 4 1

Hi, I am using rpart decision trees to analyze customer churn. I am finding that the decision trees created are not effective because they are not able to recognize factors that influence churn. I have created an example situation below. What do I need to do to for rpart to build a tree with the variable experience? My guess is that this would happen if rpart used the loss matrix while creating

Hi, When I look at the summary of an rpart object run on my data, I get 7 nodes but when I plot the rpart object, I get only 3 nodes. Should the number of nodes not match in the results of the 2 functions (summary and plot) or it is not always the same? Look forward to your reply, Carol -------------------------------------------- ?summary(rpart.res) Call: rpart(formula = mydata$class ~ ., data

I am performing a binary classification using a classification tree. Ironically, the data themselves are 2483 tree (real biological ones) locations as described by a suite of environmental variables (slope, soil moisture, radiation load, etc). I want to separate them from an equal number of random points. Doing eda on the data shows that there is substantial difference between the tree and random

? data=read.table("http://statcourse.com/research/boston.csv", , sep=",", header = TRUE) ? library(rpart) ? fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT) predict(fit,data[4,]) plot only reveals part of the tree in contrast to the results on obtains with CART or C5 -------- Original Message -------- Subject: Re: [R] help with rpart From: Sarah

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

Dear useRs: Is there a way I could predict the terminal node associated with a new data entry in an rpart environment? In the example below, if I had a new data entry with an AM of 5, I would like to link it to the terminal node 2. My searches led to http://tolstoy.newcastle.edu.au/R/e4/help/08/07/17702.html but I do not seem to be able to operationalize Professor Ripley's suggestions. Many

I am doing > library(rpart) > m <- rpart("y ~ x", D[insample,]) > D[outsample,] y x 8 0.78391922 0.579025591 9 0.06629211 NA 10 NA 0.001593063 > p <- predict(m, newdata=D[9,]) Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid result from na.action How do I persuade him to give me NA

Dear All, I have a question regarding predict.rpart. I use rpart to build classification and regression trees and I deal with data with relatively large number of input variables (predictors). For example, I build an rpart model like this rpartModel <- rpart(Y ~ X, method="class", minsplit =1, minbucket=nMinBucket,cp=nCp); and get predictors used in building the model like

Hi, I am using rpart to do leave one out cross validation, but met some problem, Data is a data frame, the first column is the subject id, the second column is the group id, and the rest columns are numerical variables, > Data[1:5,1:10] sub.id group.id X3262.345 X3277.402 X3369.036 X3439.895 X3886.935 X3939.054 X3953.777 X3970.352 1 32613 HAM_TSP 417.7082 430.4895 619.4776 720.8246

Hi folks, I'm getting the following error message when using predict.rpart on a large dataset (say 100,000 records) within Splus: Terminating S Session: Signal: bad address signal Has anyone run into this error, discovered a work-around, etc.? More generally, I'm trying to use predict.rpart to generate predictions from very large, geographic datasets (we're talking all of

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

We are trying to make a decision tree using rpart and we are continually running into the following error: > fit_rpart=rpart(ENROLL_YN~MINORITY,method="class") > summary(fit_rpart) Call: rpart(formula = ENROLL_YN ~ MINORITY, method = "class") n= 5725 CP nsplit rel error 1 0 0 1 Error in yval[, 1] : incorrect number of dimensions ENROLL_YN is a

I am using an algorigm to split my data set into two random sections repeatedly and constuct a model using rpart() on one, test on the other and average out the results. One of my variables is a factor(crop) where each crop type has a code. Some crop types occur infrequently or singly. when the data set is randomly split, it may be that the first data set has a crop type which is not present in

Hi, the predict.rpart() function from the rpart library allows for calculating the class probabilities for a given test case instead of a discrete class label. How are these class probabilities derived? Is it simply the proportion of the majority class to all cases in a leaf node? Thanks in advance, Chris

Hello All, Am I misreading the documentation? The text.rpart documentation says: "label a column name of x$frame; values of this will label the nodes. For the "class" method, label="yval" results in the factor levels being used, "yprob" results in the probability of the winning factor level being used, and ?specific yval level? results in the probability of

Hi there. I was wondering if somebody knows how to perform a bagging procedure on a classification tree without running the classifier with weights. Let me first explain why I need this and then give some details of what I have found out so far. I am thinking about implementing the bagging procedure in Matlab. Matlab has a simple classification tree function (in their Statistics toolbox) but

Hi, I think I'm missing something very obvious, but I am missing it, so I would be very grateful for help. I'm using rpart to analyse data on skull base morphology, essentially predicting sex from one or several skull base measurements. The sex of the people whose skulls are being studied is known, and lives as a factor (M,F) in the data. I want to get back predictions of gender, 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,