similar to: Parametrized object name in Save statement

Displaying 20 results from an estimated 5000 matches similar to: "Parametrized object name in Save statement"

2011 Jan 26
2
Extracting the terms from an rpart object
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 <-
2011 May 12
1
Saving misclassified records into dataframe within a loop
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
2010 Dec 13
2
rpart.object help
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
2009 Dec 14
1
RPART - printing full splitting rule number on tree plot
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.
2012 Mar 04
1
rpart package, text function, and round of class counts
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
2007 Jun 15
2
model.frame: how does one use it?
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 <-
2012 Apr 12
2
enableJIT(2) causes major slow-up in rpart
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
2010 Mar 07
1
Is there an equivalence of lm's “anova” for an rpart object ?
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
2009 Feb 25
1
how to label the branches of a tree
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
2011 Aug 08
1
Classification trees problem.
Hello Everyone, I'm doing a Classification 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)
2010 May 03
1
rpart, cross-validation errors question
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,
2009 Feb 03
5
Large file size while persisting rpart model to disk
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
2009 Sep 14
1
summary of rpart-Object in tktext window?
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
2011 Jul 29
1
help with predict.rpart
? 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
2011 Mar 31
2
fit.mult.impute() in Hmisc
I tried multiple imputation with aregImpute() and fit.mult.impute() in Hmisc 3.8-3 (June 2010) and R-2.12.1. The warning message below suggests that summary(f) of fit.mult.impute() would only use the last imputed data set. Thus, the whole imputation process is ignored. "Not using a Design fitting function; summary(fit) will use standard errors, t, P from last imputation only. Use
2002 Apr 29
2
RPart
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
2011 Sep 08
1
"rpart" or "tree" function issue
I am trying to create a classification tree using either tree or rpart functions but when it comes to plotting the results the formatting I get is different than what I see in all the tutorials (like http://www.youtube.com/watch?v=9XNhqO1bu0A or http://www.youtube.com/watch?v=m3mLNpeke0I&feature=related or http://www.statmethods.net/advstats/cart.html "tree for kyphosis"). I am
2008 Mar 06
1
Rpart and bagging - how is it done?
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
2011 Jul 28
2
help with rpart
1. How can I plot the entire tree produced by rpart? 2. How can I submit a vector of values to a tree produced by rpart and have it make an assignment? Mark
2011 Jan 26
1
Inconsistencies in the rpart.object help file?
Hello all, I'm was going through the help for ?rpart.object And noticed some inconsistencies, Some might be a mistake in the help file and some might be my misunderstanding. The help in the section: value -> frame (first paragraph), states that: > yval, the fitted value of the response at each node, *and splits, a two > column matrix of left and right split labels for each node. *