search for: minsplit

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 obser...

...r, > arbol.bsvg.02 n= 100000 node), split, n, loss, yval, (yprob) * denotes terminal node 1) root 100000 6657 0 (0.9334300 0.0665700) * And I get an "empty" tree. But there were branches in the original tree with more than 1000 observations. Something similar happens if I set minsplit (or both minbucket and minsplit) to a similar value: I end up with the same root, branch-less tree. Am I misreading something? Can anybody cast a light on the correct usage of the minbucket (and/or minsplit) for me? Sincerely, Carlos J. Gil Bellosta http://www.datanalytics.com

...un it n=50 times and from each output pick the appropriate tree size and post it to a datafile where I can then look at the frequency distribution of tree sizes. Here is the code and output from a single run > fit1 <- rpart(CHAB~.,data=chabun, method="anova", control=rpart.control(minsplit=10, cp=0.01, xval=10)) > printcp(fit1) Regression tree: rpart(formula = CHAB ~ ., data = chabun, method = "anova", control = rpart.control(minsplit = 10, cp = 0.01, xval = 10)) Variables actually used in tree construction: [1] EXP LAT POC RUG Root node error: 35904/33 = 1088 n= 33...

...um_genes","position", "acid_per", "base_per", "charge_per", "hydrophob_per", "polar_per", "out") train<-train[myvar] # update training set valid<-valid[myvar] control<-rpart.control(xval=10, cp=0.01, minsplit=5, minbucket=5) #control the size of the initial tree tree.fit <- rpart(out ~ ., method="class", data=train, control=control) # model fitting p.tree<- prune(tree.fit, cp=tree.fit$cptable[which.min(tree.fit$cptable[,"xerror"]),"CP"]) # prune the tree...

...ultiple diet taxa abundances (prey.data) Neither rpart or mvpart seem to allow me to do multiple responses. (Or if they can, I'm not using the functions properly.) > library(rpart) > turtle.rtree<-rpart(prey.data~., data=turtle.data$Clength, method="anova", maxsurrogate=0, minsplit=8, minbucket=4, xval=10); plot(turtle.rtree); text(turtle.rtree) Error in terms.formula(formula, data = data) : '.' in formula and no 'data' argument When I switch response for predictor, it works. But this is the opposite of what I wanted to test and gives me splits at a...

...ear 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 this colnamesUsed<-unique(rownames(rpartModel$splits)); When later I apply the rpart model to predict the new data I strip the input data from unneccessary columns and only use X columns that exist in c...

Still about the mvpart. Is there any way I can control for the number of elements in each node in the function mvpart? Specifically, how can I ask partition to ignore node with elements less than 10? Thanks! -Shu

Hi, I am using ada to predict a data set with 36 variables ada(x~.,data=train,iter=Iter, control=rpart.control(maxdepth=4,cp=-1,minsplit=0,xval=0)) can any one tell me in in laymans terms maxdepth- how do you set this, how do you change this to improve predictions success cp- same question for this also minsplit- same question for this also how do I change all this parameters to my advantage/ Greatly appreciated the help pc...

...in advance! And thanks for Eric who helped with my previous question about starting "rpart". Olga > fit <- rpart(Retention ~ Info_G+AOPD+Mail+Xref_Umbr+Ins_Age+Discount+Xref_A + Con6 + + Con5 + Con4 + + Con3 + Con2 + + Con1 , data=Home,control=rpart.control(minsplit=5)) > > fit n= 48407 node), split, n, deviance, yval * denotes terminal node 1) root 48407 4730.642 0.8902225 2) Info_G< 0.5 14280 1999.293 0.8316527 * 3) Info_G>=0.5 34127 2661.865 0.9147303 * > [[alternative HTML version deleted]]

...de for selecting and plotting the frequency histogram of minimum xerror. Here is the output that is being referenced in the code below Regression tree: rpart(formula = chbiomsq ~ HC + BC + POC + RUG + Depth + Exp + DFP + FI + LAT, data = ch, method = "anova", control = rpart.control(minsplit = 10, cp = 0.01, xval = 10)) Variables actually used in tree construction: [1] BC Depth DFP Exp Root node error: 47456/99 = 479.35 n= 99 CP nsplit rel error xerror xstd 1 0.344626 0 1.00000 1.02074 0.139585 2 0.179054 1 0.65537 0.76522 0.107470 3 0.072037...

...or rror in if (nrow(object$splits) > 0) { : argument is of length zero when I am running the following codes. train <- c(sample(1:27,18), sample(28:54, 18), sample(55:81, 8)) a2011.adaboost <- boosting(median_kod ~ ., data = b[train, ], boos=TRUE, mfinal = 10, control = rpart.control(minsplit = 0)) Regards, Greg [[alternative HTML version deleted]]

...tors, if an observation contains a level not used to grow the tree, it is left at the deepest possible node and frame$yval at the node is the prediction. " Many thanks. > mod <- rpart(y~., data=data.frame(y=y,x=x), method="anova", + cp=0.05, minsplit=100, minbucket=50, maxdepth=5) > predictLost <- predict(mod, newdata=data.frame(y=yLost, x=xLost), type="vector") Error in model.frame.default(Terms, newdata, na.action = act, xlev = attr(object, : factor 'x.Incoterm' has new level(s) MTD ---- Ch...

...ult of R command is returned in REXP and we use geterrMessage() to retrieve the error. When we execute the following command, cnr_model<-rpart(as.factor(Species)~Sepal Length+Sepal Width+Petal Length, method="class", parms=list(split="gini",prior=c()), control=rpart.control(minsplit=2, na.action=na.pass,cp=0.001,usesurrogate=1,maxsurrogate=2,surrogatestyle=0,maxdepth=20,xval=10)) we get an error message* "Error: unexpected symbol in "a<-cnr_model<-rpart(as.factor(Species)~Sepal Length""* The REXP returned in not null and the geterrMessage() call doe...

...Criterion <- factor(paste("Leaf", 1:5)) Node <- factor(1:5) assign("tree.df", data.frame(Criterion = Criterion, Node = Node)) nobs <- dim(tree.df)[[1]] u.tree <- rpart(Node ~ Criterion, data = tree.df, all = F, control = list(minsplit = 2, minbucket = 1, cp = 9.999999999999998e-008)) plot(u.tree, uniform=T) text(u.tree) --please do not edit the information below-- Version: platform = i386-pc-mingw32 arch = x86 os = Win32 system = x86, Win32 status = major = 1 minor = 3.0...

Hi, I am currently studying Decision Trees by using rpart package in R. I artificially created a data set which includes the dependant variable (y) and a few independent variables (x1, x2...). The dependant variable y only comprises 0 and 1. 90% of y are 1 and 10% of y are 0. When I apply rpart to it, there is no splitting at all. I am wondering whether this is because of the

Hi, Being a novice this is my first usage of R. I am trying to use rpart for building a decision tree in R. And I have the following dataframe Outlook Temp Humidity Windy Class Sunny 75 70 Yes Play Sunny 80 90 Yes Don't Play Sunny 85 85 No Don't Play Sunny 72 95 No Don't Play Sunny 69 70 No Play Overcast 72 90 Yes Play Overcast 83 78 No Play Overcast 64 65 Yes Play Overcast 81 75

Dear all, I know in rpart(), one can control the tree size (i.e. number of terminal nodes) through rpart.control(), e.g. minsplit, minbucket, maxdepth etc. But is there any more direct way to specify the number of terminal nodes when rpart() does the recursive partitioning? Your help is highly appreciated! Regards, -Ji -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list --...

...formatg(yval,digits),"\nn=", n,sep="")} else{paste(formatg(yval,digits))} }) } tst.lst<-list(eval=tst.eval, split=tst.split, init=tst.init) data(lung) fit1 <- rpart(Surv(time, status) ~ age + ph.karno + meal.cal,data=lung,control=rpart.control(minsplit=30, xval=0, cp=.011),method=tst.lst) -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not th...

...t, loss = loss); model.blackboost <- blackboost(tr[,1:DIM], tr.y, family=CSAdaExp, weights=NULL, control=boost_control(mstop=MSTOP, nu=0.1,savedata=TRUE,save_ensembless=TRUE,trace=TRUE), tree_controls=ctree_control(teststat = "max",testtype = "Teststatistic",mincriterion = 0,minsplit = 2000, minbucket = 700,maxdepth = TREEDEPTH)); -------------------------------- regards, Yuchun Tang, Ph.D. Principal Engineer, Lead McAfee, Inc. 4501 North Point Parkway Suite 300 Alpharetta, GA 30022 Main: 770.776.2685 www.mcafee.com www.trustedsource.org www.linkedin.com/in/yuchuntang

....seed(100) x <- matrix(rnorm(n*p), ncol=p) y <- as.factor(c(-1,1)[as.numeric(apply(x^2, 1, sum) > 9.34) + 1]) indtrain <- sample(1:n, 2000, FALSE) train <- data.frame(y=y[indtrain], x[indtrain,]) test <- data.frame(y=y[-indtrain], x[-indtrain,]) control <- rpart.control(cp = -1,minsplit = 0,xval = 0,maxdepth = 1) gdis <- ada(y~., data = train, iter = 400, bag.frac = 1, nu = 1, control = control, test.x = test[,-1], test.y = test[,1]) gdis plot(gdis, TRUE, TRUE) summary(gdis, n.iter = 398) My problem is that my confusion matrix, testing results and diagnostic plots differ from...