similar to: Classification trees problem.

Hello everyone, I want make a variable selection  from a dataframe, but when I build this new object (using cbind) the new table (which is a matrix) I lost all the original factor names in the variables. I get 1,2,3.... Someone would be so kind and tell me if there's is a ny way to get variables in a dataframe, but not changing these factor to numbers? Thank you all. José [[alternative

 Hello everyone, I'm trying to change the name variables of a big dataset. Here's more than 300 variables. The point is that I have to match it with another dataset that have same variables, but in lowercase, the I can use rbind and do my work. Is there any function for changing uppercase or lowercase variables? Thank you in advance! José [[alternative HTML version deleted]]

Hola everyone, I'm doing an analisys about abundance of a planctonic specie in Robinson Crusoe Islan and I am having a hard time just adding the superscripts and the 100 (one hundred) in the measure of abundance. Perhaps this is so basic to you, but I need to put the real unit of measure. Individuals / 100 cubic meters. I have been looking in websites and R-list and there is not any

Dear R users.   I'm trying to know how to create a function. I have developed a big code that use a file and makes many process and then creates a table that is exported as txt. The point is to make something like a function.  I need to make it easy for others, so they can run it. This has a particular name? I will be very gratefull and placed if any of you can provide some lights.

Hello all, I wish to extract 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 <-

Greetings All, I am looking to write a parametrized Rscript that will accept a variable name(that also is the name of the flat file), transform the data into a data frame and preform various modeling on the structure and save the output and plot of the model. In this example i am using a rpart decision tree. The only problem i am having is integrating the parameter into the internal object name

Estimados, A todos quienes estan insterados y estan en Stgo de Chile les tengo una muy buena noticia. Desde hace ya algunas semanas, nos hemos sentado a la mesa con Alex (Epidemiologo) a discutir la necesidad de desarrollar un curso de R para principiantes. No solo porque es necesario unirse en torno a un projecto como es el R, sino que es una oportunidad para muchos cientificos (como

Greetings R world, I know some version of the this question has been asked before, but i need to save the output of a loop into a data frame to eventually be written to a postgres data base with dbWriteTable. Some background. I have developed classifications models to help identify problem accounts. The logic is this, if the model classifies the record as including variable X and it turns out

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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 absent (1.0000000

Dear R-users I am using RPART package to get regression trees. However having trouble getting the text function to put the full splitting rule number on the plot, instead to puts it in scientific notation. When a covariate has 1e4 or greater number of digits then the splitting rule number displayed on the plot is in scientific notation. But print.rpart displays the splitting rules in full.

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 plot is supposed to look like

Philipp Benner reported a Debian bug report against r-cran-rpart aka rpart. In short, the issue has to do 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 <-

Hello, Due to exploration of the JIT capabilities offered through the {compiler} package, I came by the fact that using enableJIT(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

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

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

I ran this code (several times) from the Quick-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,

I am using rpart to build a model for later predictions. To save the prediction across restarts and share the data across nodes I have been using "save" to persist the result of rpart to a file and "load" it later. But the saved size was becoming unusually large (even with binary, compressed mode). The size was also proportional to the amount of data that was used to create the

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

? 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