similar to: How to calculate residual mean deviance in rpart

Hi Everyone, I'm working on building some classification trees, and up to this point I've been using rpart. However, I recently discovered the tree package, and found that it had some useful functions (in particular deviance(), which I would really like to use for my project). I can't seem to find an equivalent function for rpart. I've considered using tree() in place of

Running a binary logit model on the data df <- data.frame(y=sample(letters[1:3], 100, repl=T), x=rnorm(100)) reveals some residual deviance: summary(glm(y ~ ., data=df, family=binomial("logit"))) However, running a multinomial model on that data (multinom, nnet) reveals a residual deviance: summary(multinom(y ~ ., data=df)) On page 203, the MASS book says that "here the

A maybe trivial and stupid question: In the case of a lm or glm fit, it is quite informative (to me) to have a look to the null deviance and the residual deviance of a model. This is generally provided in the print method or the summary, eg: Null Deviance: 658.8 Residual Deviance: 507.3 and (a bit simpled minded) I like to think that the proportion of deviance 'explained' by the

The following code tr <- rpart(Y ~ ., dat, method="class") dev <- residuals(tr, "deviance") produces the following error Error in log(x) : Non-numeric argument to mathematical function > .Traceback [[1]] [1] "log(yhat)" # line 588 of rpart [[2]] [1] "switch(type, usual = as.integer(y != yhat), pearson = (1 - yhat)/yhat, " [2] "

Dear all, I'm trying to fit a statistical model to series of measurements. Unfortunately, my knowledge of statistics is rather limited, so I'm a bit at loss of what is going on with the model. First of all, I've prepared a histogram. Then, I've tried to fit a Poisson model to express the relation between the middle points of classes (mids) and the corresponding frequencies

I'm running a categorical data analysis with a two-way design of nominal by ordinal structure like the Political Ideology Example (Table 9.5) in Agresti's book Categorical Data Analysis. The nominal variable is Method while the ordinal variable is Quality (Bad, Moderate, Good, Excellent). I rank/quantify Quality with another variable QualityR (1, 2, 3, 4), and run the following:

I have used packages rpart, mvpart and tree for classification and regression trees. I want to calculate fraction of null deviance explained by each node and variable in the tree. For instance, at the first split, this would be (1 - (sum of residual deviance in each of the two leaves)/deviance at the root). In the subsequent splits, this formula is slightly different. There probably is a function

Hi, I am trying to use the rpart library with my own set of functions on a survival object. I get an immeadiate segmentation fault when i try calling rpart with my list of functions. I get the same problem with the logrank example from Therneau,s S-rpart library though their anova example works. Should I report this as a bug, as even if my functions are structured improperly, that should lead to

Dear all, I am running a logistic regression and this is the output: glm(formula = educationUniv ~ brncntr, family = binomial) Deviance Residuals: Min 1Q Median 3Q Max # ???? ????? ?? ???????? -0.8825 -0.7684 -0.7684 1.5044 1.6516 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.06869 0.01155 -92.487 <2e-16 *** brncntrNo

It seems that the standardized deviance residulas, that one gets on plots of a glm.object x with plot(x) are calculated as r <- residuals(x) s <- sqrt(deviance(x)/df.residual(x)) w <- weights(x) hii <- lm.influence(x)$hat r.w <- if (is.null(w)) r else (sqrt(w) * r) rs <- r.w/(s * sqrt(1 - hii)) This implies that, for example, for binomial B(ni,pi) data the devaince residials

dear List: glm(a~b+c,family=binomial,data=x)->fit deviance(fit) returns the same as the residual deviance. I don't not know much about logistic regression.Some book tells that: " Deviance (likelihood ratio statistic): Deviance = -2log( likelihoodof the currentmodel /likelihoodof thesaturated model) Note: (1). The current model is the model of interest. (2). The saturated model

I have been trying to write my own user defined function in Rpart.I imitated the anova splitting rule which is given as an example.In the work I am doing ,I am calculating the concentration index(ci) ,which is in between -1 and +1.So my deviance is given by abs(ci)*(1-abs(ci)).Now when I run rpart incorporating this user defined function i get the following error message: Error in

Hi all, I apologies in advance if I am missing something very simple here, but since I failed at resolving this myself, I'm sending this question to the list. I would appreciate any help in understanding how the rpart function is (exactly) computing the "improve" (which is given in fit$split), and how it differs when using the split='information' vs split='gini'

Full_Name: Bill Wheeler Version: 2.0.1 OS: Windows 2000 Submission from: (NULL) (67.130.36.229) The program to reproduce the error is below. I am calling rpart with a user-defined split function for a binary response variable and one continuous independent variable. The split function works for some datasets but not others. The error is: Error in "$<-.data.frame"(`*tmp*`,

I am comparing the Splus and R fits of a simple glm. In the following, foo is generated from rbinom with size = 20 p = 0.5. The coefficients (and SE's0 of the fitted models are the same, but the estimated deviances are quite different. Could someone please tell me why they are so different? I am using R version 1.3.1 and Splus 2000 release 3 on windows 2000. ++++++++++++++++++++++ foo

I have experimented with the cheese data example from McCullagh&Nelder, page 175. With a proportional odds model they obtain a residual deviance of 20.31. Estimating the same model with polr(MASS) gives a residual deviance of 762.11 !, while using ordglm(gnlm) gives a deviance of 523.94. Can anybody explain these differences? The data frame with the data are: > cheese Cheese N

Dear All, I encounter some discrepancies when comparing the deviance of a weighted and unweigthed model with the AIC values. A general example (from 'nls'): DNase1 <- subset(DNase, Run == 1) fm1DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal), DNase1) Now for a weighted fit: fm2DNase1 <- nls(density ~ SSlogis(log(conc), Asym, xmid, scal),

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

Dear R community, I am trying to write my own user defined split function for rpart. I read the example in the tests directory and I understand the general idea of the how to implement user defined splitting functions. However, I am having troubles with addressing the data frame used in calling rpart in my split functions. For example, in the evaluation function that is called once per node,

Dear all, I'm trying to manage with user defined split function in rpart (file rpart\tests\usersplits.R in http://cran.r-project.org/src/contrib/rpart_3.1-34.tar.gz - see bottom of the email). Suppose to have the following data.frame (note that x's values are already sorted) > D y x 1 7 0.428 2 3 0.876 3 1 1.467 4 6 1.492 5 3 1.703 6 4 2.406 7 8 2.628 8 6 2.879 9 5 3.025 10 3 3.494