similar to: what missed ----- CART

So I have 2 sets of data - a training data set and a test data set. I've been doing the analysis on the training data set and then using predict and feeding the test data through that. There are 114 rows in the training data and 117 in the test data and 1024 columns in both. It's actually the same set of data split into two. The rows are made of 5 different numbers. They do represent

> From: Peter Dalgaard BSA <p.dalgaard@biostat.ku.dk> > > Martin Maechler <maechler@stat.math.ethz.ch> writes: > > > Reading the long output of ``cc -flags'', > > I see that more than average optimization is done using > > cc -xO[1-4] > > # And -O is -xO2. If you do want higher speed, you need to use other flags too, and -fast is a

Dear R community I have a question regarding the value of cost complexity parameter "k" used in "tree" package for pruning purpose. Any help in finding the optimum value of "k" is requested. Please give some suggestion in this regard. In the example below i used k=0 but i don't know why? But if i use k=NULL, then it will not plot the resultant tree.

Using the data set fgl in MASS the following code layout(matrix(1:9,3,3)) for(i in 1:9){ boxplot(fgl[,i] ~ type, data = fgl,main=dimnames(fgl)[[2]][i])} produces a 3 by 3 array of plots, each one of which consists of six boxplots. Is it possible to do this in lattice? Steve "R version 2.7.2 (2008-08-25)" on Ubuntu 6.06

HI Is there any function in R that tells us error rate(misclassification rate) for logistic regression type classification? i also want to know the function to determine type I and type II error. I have found a link where "misclass" and "confusion" are used. But I dont know the package name. http://alumni.media.mit.edu/~tpminka/courses/36-350.2001/lectures/day32/

> On Jun 18, 2017, at 6:24 AM, Manoranjan Muthusamy <ranjanmano167 at gmail.com> wrote: > > I am using nlsLM {minpack.lm} to find the values of parameters a and b of > function myfun which give the best fit for the data set, mydata. > > mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45)) > > myfun=function(a,b,r,t){ > prd=a*b*(1-exp(-b*r*t)) >

Dear sir or ma'am, I have a question about the dataset "fgl." The dataset seems to be in the "VR" package, so I tried to download it from CRAN. However, after downloading, when I tried to load the package, it was not in my package list. I am wondering what is wrong. Any advice on how to access the fgl dataset would be appreciated. Thanks. Seungho Huh

I ran the following script. I satisfied the constraint by making a*b a single parameter, which isn't always possible. I also ran nlxb() from nlsr package, and this gives singular values of the Jacobian. In the unconstrained case, the svs are pretty awful, and I wouldn't trust the results as a model, though the minimum is probably OK. The constrained result has a much larger sum of squares.

https://cran.r-project.org/web/views/Optimization.html (Cran's optimization task view -- as always, you should search before posting) In general, nonlinear optimization with nonlinear constraints is hard, and the strategy used here (multiplying by a*b < 1000) may not work -- it introduces a discontinuity into the objective function, so gradient based methods may in particular be

I'm drawing a fitted normal distribution over a histogram. The use case is trivial (fitting normal distributions on densities) but I want to extend it to other fitting scenarios. What has stumped me so far is how to take the list that is returned by nls() and use it for curve(). I realize that I can easily do all of this with a few intermediate steps for any specific case. But I had expected

I've seen a number of problems like this over the years. The fact that the singular values of the Jacobian have a ration larger than the usual convergence tolerances can mean the codes stop well before the best fit. That is the "numerical analyst" view. David and Jeff have given geometric and statistical arguments. All views are useful, but it takes some time to sort them all out and

I haven't followed your example closely, but can't you use the predict() method for this? To draw a curve, the function that will be used in curve() sets up a newdata dataframe and passes it to predict(fit, newdata= ...) to get predictions at those locations. Duncan Murdoch On 17/10/2020 5:27 a.m., Boris Steipe wrote: > I'm drawing a fitted normal distribution over a

I am not as expert as John, but I thought it worth pointing out that the variable substitution technique gives up one set of constraints for another (b=0 in this case). I also find that plots help me see what is going on, so here is my reproducible example (note inclusion of library calls for completeness). Note that NONE of the optimizers mentioned so far appear to be finding the true best

Hi, I have problem getting the standard deviation from the manova output. I have used the manova function: myfit <- manova(cbind(y1, y2) ~ x1 + x2 + x3, data=mydata) . I tried to get the predicted values and their standard deviation by using: predict(myfit, type="response", se.fit=TRUE) But the problem is that I don't get the standard deviation values, I only

Dear List, I think I'm going crazy here...can anyone explain why do I get the same predictions in train and test data sets below when the second has a missing input? y <- rnorm(1000) x1 <- rnorm(1000) x2 <- rnorm(1000) train <- data.frame(y,x1,x2) test <- data.frame(x1) myfit <- glm(y ~ x1 + x2, data=train) summary(myfit) all(predict(myfit, test) == predict(myfit, train))

I use knncat to make a predictive model and get misclass rate > knncat.m<-knncat(training.new,k=c(10,20),classcol=5) > knncat.m Training set misclass rate: 36.88% then I try to calculate prediction accuracy by the following: > pr.knncat.train <- predict (knncat.m,training.new,training.new,train.classcol=5,newdata.classcol=5) > tb.knncat.train <-table (pr.knncat.train,

Hi R-Helpers: I?m trying to fit the same linear model to a bunch of variables in a data frame, so I was trying to adapt the codes John Fox, Spencer Graves and Peter Dalgaard proposed and discused yesterday on this e-mail list: for (y in df[, 3:5]) { mod = lm(y ~ Trt*Dose, data = x, contrasts = list(Trt = contr.sum, Dose = contr.sum)) Anova(mod, type = "III") } ## by John Fox or for

Hi, I am trying to estimate the effects of covariates on the hazard function, rather than on the survival. I know this is actually the same thing. For example, using the survival package, and doing: > myfit <- survreg( Surv(time, event) ~ mymodel ) all I have to do to get the quantities of my interest is > -myfit$coefficients/myfit$scale The standard erros are easily worked out, as

Hello R users, A beginners question which I could not find the answer to in earler posts. My thought process: Here "z" is a 119 x 15 data matrix Step 1: start at column one, bind every column with column 1 Step2: use the new matrix, "test", in the fitCopula package Step3: store each result in myfit, bind each result to "answer" Step4: return "answer"

I'm trying to get the following section of code to work, I think the problem is being caused by the assignment of data to the lm function not evaluating to "train" in the parent environment but I can't seem to figure out how to do this. fitmodel <- function(trial,data) { wrap.lm <- function(formula,data,...) { cat("in wrap lm",NROW(data),"\n");