search for: kyphosi

...that record does not have X then it should be reviewed(ie i need the row number/ID saved to a database). Generally i want to look at the misclassified records. This is a little hack i know, anyone got a better idea please let me know. Here is an example library(rpart) # grow tree fit <- rpart(Kyphosis ~ Age + Number + Start, method="class", data=kyphosis) #predict prediction<-predict(fit, kyphosis) #misclassification index function predict.function <- function(x){ for (i in 1:length(kyphosis$Kyphosis)) { #the idea is that if the record is "absent" but the predicti...

...ay), thus - turning for help. Thanks in advance for your help, -Saar The following code shows what the issue is; the first 3 trees are the same, but the following two (with uneven weights) turn out differently: ## playing with rpart weights require(rpart) dev.new() par(mfrow=c(2,3), xpd=NA) data(kyphosis) fitOriginal <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis, control=rpart.control(minsplit=15)) plot(fitOriginal) text(fitOriginal, use.n=TRUE) # this dataset is the original data repeated 3 times kyphosisRepeated <- rbind(kyphosis, kyphosis, kyphosis) fitRepeated <- rpart(Kyp...

...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) data(kyphosis) #fit1 from original unweighted data set fit1 <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) #modify loss matrix loss<-matrix(c(0,1,2,0),nrow=2,ncol=2) # true class? # [,1] [,2] #[1,] 0 2 #[2,] 1 0 predicted class? #modify priors prior=c(1/3,2/3) fit2<- r...

...ract the terms from an rpart object. Specifically, I would like to be able to know what is the response variable (so I could do some manipulation on it). But in general, such a method for rpart will also need to handle a "." case (see fit2) Here are two simple examples: fit1 <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) fit1$call fit2 <- rpart(Kyphosis ~ ., data=kyphosis) fit2$call Is there anything "prettier" then using string manipulation? Thanks. ----------------Contact Details:------------------------------------------------------- Contact me: Tal.G...

Hi, I am fairly new to GAM and started using package mgcv. I like the fact that optimal smoothing is automatically used (i.e. df are not determined a priori but calculated by the gam procedure). But the mgcv manual warns that p-level for the smooth can be underestimated when df are estimated by the model. Most of the time my p-levels are so small that even doubling them would not result

...parameter into the internal object name when tying to save it. Solving this one porblem will help me start blasting though 100's of models and saving their output for later use. I would call the script from command line as Rscript file_name.R argument for this example Rscript rpart_code.R Kyphosis #/bin/R args <- commandArgs(TRUE) # Read the argument require(rpart, quietly = TRUE) #for this example lets use the kyphosis data in the rpart package args <- "Kyphosis" X<-paste(args," ~.", sep ="") # create model formula call. X = "Kyphosis ~.&quot...

...with how rpart evaluates a formula and supporting arguments, in particular 'weights'. A simple contrived example is ----------------------------------------------------------------------------- library(rpart) ## using data from help(rpart), set up simple example myformula <- formula(Kyphosis ~ Age + Number + Start) mydata <- kyphosis myweight <- abs(rnorm(nrow(mydata))) goodFunction <- function(mydata, myformula, myweight) { hyp <- rpart(myformula, data=mydata, weights=myweight, method="class") prev <- hyp } goodFunction(mydata, myformula, myweight) cat(&...

Hi R-helpers. I have the following problem: I would like to apply my function gain(df,X,A) to a list of arguments. df is a data frame X,A are the varibales od data frame. When I do > gain(kyphosis,"Kyphosis",c("Start","Number")) [1] "Start" "Number" I get the following error... Error in unique.default(x) : unique() applies only to vectors I tried lapply and apply, it didn't seem to work... Can anybody tell me how to apply my function g...

I run the following code: library(rpart) data(kyphosis) fit <- rpart(Kyphosis ~ ., data=kyphosis) plot(fit) text(fit, use.n=TRUE) The text labels represent the count of each class at the leaf node. Unfortunately, the numbers are rounded and in scientific notation rather than the exact number of examples sorted by that node in each class. The plo...

...ays the splitting rules in full. I have tried using digits option in text function but this only alters digits of the mean displayed at terminal nodes. How do I get the full splitting rule number displayed on the tree plot? Below is an example. library(rpart) temp <- as.data.frame(cbind(kyphosis, Start2=kyphosis$Start)) temp$Start2 <- temp$Start2+10000 fit1 <- rpart(Age~ Kyphosis +Number + Start2, data=temp) plot(fit1) text(fit1, use.n=TRUE, digits=5) Thanks Dr Rebecca O'Leary, PhD Biostatistician Senior Research Officer UWA Centre for Child Health Research Te...

I am using the rpart package and seem to have trouble with data sets that have columns with no data. I look at the column data in R and all values are NA. When this occurs, I get nothing back from the rpart function. Is there a way to get the rpart package to ignore these columns, without knowing what columns are empty? I have tried the na.action=na.omit and na.action=na.exclude, but neither one

Hi, Suppose i have generated an object using the following : fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) And when i print fit, i get the following : n= 81 node), split, n, loss, yval, (yprob) * denotes terminal node 1) root 81 17 absent (0.7901235 0.2098765) 2) Start>=8.5 62 6 absent (0.9032258 0.0967742) 4) Start>=14.5 29 0 abse...

I am working with a non-parametic smoothing operation using a Generalized Additive Model. It is a bivariate data set. I know how to do the smooth, and out comes a nice smooth curve. Now I want to find the value of the smoothed curve for several values of x (the abscissa). This can be done (please correct me if I am wrong) by using the predict.gam function. You feed the predict.gam function a

...nableJIT(2) can *slow* the rpart function (from the {rpart} package) by a magnitude of about 10 times. Here is an example code to run: library(rpart) require(compiler) enableJIT(0) # just making sure that JIT is off # We could also use enableJIT(1) and it would be fine fo <- function() {rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)} system.time(fo()) # user system elapsed # 0 0 0 # this can also be 0.01 sometimes. enableJIT(2) # also happens for enableJIT(3) system.time(fo()) # user system elapsed # 0.12 0.00 0.12 Which brings me to my *questions*: 1...

...of the Harrell (2001) book Googling located a thread in R-help back in 2003, which seemed dated. Many thanks in advance for the help, Yuelin. http://idecide.mskcc.org ------------------------------- > library(Hmisc) Loading required package: survival Loading required package: splines > data(kyphosis, package = "rpart") > kp <- lapply(kyphosis, function(x) + { is.na(x) <- sample(1:length(x), size = 10); x }) > kp <- data.frame(kp) > kp$kyp <- kp$Kyphosis == "present" > set.seed(7) > imp <- aregImpute( ~ kyp + Age + Start + Number, dat =...

...k-R web page ( http://www.statmethods.net/advstats/cart.html) but my cross-validation errors increase instead of decrease (same thing happens with an unrelated data set). Why does this happen? Am I doing something wrong? # Classification Tree with rpart library(rpart) # grow tree fit <- rpart(Kyphosis ~ Age + Number + Start, method="class", data=kyphosis) printcp(fit) # display the results plotcp(fit) # visualize cross-validation results Thank you, Claudia [[alternative HTML version deleted]]

...ssification trees with categorical explanatory variables using library rpart and I would like to do a prediction for some data imputs. I don't know where's a function or how can I do it?. Is there someone can help ?? ¿. Here's the code that I'm using. library(rpart) fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) plot(fit) text(fit, use.n=TRUE) #Someone can help me?? I hope someone can help! Respect, José Bustos [[alternative HTML version deleted]]

Hi, I am using rpart package to fit classification trees. library(rpart) fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) plot(fit,uniform=T) text(fit, use.n=TRUE) But I am unable to label the branches (not the nodes) of the tree. Can somebody help me out in this? Thank you, Regards Utkarsh Singhal | Amba Research Ph +91 80 3980 8017 | Mob +91 99 0295 8815 Bangalor...

Hi, is it possible to put a summary of an rpart-Object into a tktext-window? Here is what I'm trying to do: fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis) tt <- tktoplevel() tex <- tktext(tt) tkpack(tex) tkinsert(tex, "end", summary(fit)) But since the summary of an object is a list, I always get back the following error-message: cannot handle object of mode 'list' So, is there a di...

Simple example: # Classification Tree with rpart library(rpart) # grow tree fit <- rpart(Kyphosis ~ Age + Number + Start, method="class", data=kyphosis) Now I would like to know how can I measure the "importance" of each of my three explanatory variables (Age, Number, Start) in the model? If this was a regression model, I could have looked at p values from the &quot...