similar to: naive bayes help

Hello I have a rather simple code and for some reason it produces an error message. If someone can tell me why and how to fix it, I would be very greatful. Thank you in advance. ##### create data set.seed(10) n <- 200 # number of training points n.test <- 200 # number of test points p<-2 # dimension of input space z <-

Hi, I tried to use naivebayes in package 'e1071'. when I use following parameter, only one predictor, there is an error. > m<- naiveBayes(iris[,1], iris[,5]) > table(predict(m, iris[,1]), iris[,5]) Error in log(sapply(attribs, function(v) { : Non-numeric argument to mathematical function However, when I use two predictors, there is not error any more. > m<-

Hallo! I would need a code for 10-fold cross validation for the classifiers Naive Bayes and svm (e1071) package. Has there already been done something like that? I tried to do it myself by applying the tune function first: library(e1071) tune.control <- tune.control(random =F, nrepeat=1, repeat.aggregate=min.,sampling=c("cross"),sampling.aggregate=mean, cross=10, best.model=T,

Hey guys, I can't get my Naive Bayes model to predict. Forgive me if its simple... I've tried about everything and can't get it to work. Reproduceable code below. Thank you, Mike -- Michael Schumacher Manager Data & Analytics - ValueClick mike.schumacher@gmail.com * Functional Example Code from UCLA:

I am wondering if anybody here have a simple example in R for Naive Bayes. For example, I can do k-means clustering on the "iris" data - data(iris) cl <- kmeans(iris[,1:4], 3) cl$cluster cbind(1:150,iris$Species) =========== But how to do Naive Bayes classification in the same "iris" data? Many thanks! -- View this message in context:

hi there!! i am trying to implement the code of the e1071 package for naive bayes, but it doens't really work, any ideas?? i am very glad about any help!! i need a naive bayes with 10-fold cross validation: code: library(e1071) model <- naiveBayes(code ~ ., mydata) tune.control <- tune.control(random = FALSE, nrepeat = 1, repeat.aggregate = min, sampling = c("cross"),

Hi! I'm using the Naive Bayes classifier provided by the e1071 package ( http://cran.r-project.org/web/packages/e1071) and I've noticed that the predict function has a different behavior when the level set of the columns used for prediction is different from the ones used for fitting. From inspecting the predict.naiveBayes I came to the conclusion that this is due to the conversion of

This code gives unexpected result. library(data.table) library(lattice) set.seed(123) mydt <- data.table(date = seq.Date(as.IDate("2024-01-01"), by = 1, length.out = 50), xgroup = "A", x = runif(50, 0, 1)) mydt <- rbindlist(list(mydt, data.table(date = mydt$date, xgroup = "B", x = runif(50, 0, 3)))) mydt[, `:=`(xcumsum = cumsum(x)), by = .(xgroup)] mydt[,

I am trying to store regression objects in a data.table df <- data.frame(x = rnorm(20)) df[, "y"] <- with(df, x + 0.1 * x^2 + 0.2 * rnorm(20)) mydt <- data.table(mypower = c(1, 2), myreg = list(lm(y ~ x, data = df), lm(y ~ x + I(x^2), data = df))) mydt #?? mypower??? myreg #???? <num>?? <list> #1:?????? 1 <lm[12]> #2:?????? 2 <lm[12]> But mydt[1, 2]

I'm developing a naive bayes in R. I have the following data and am trying to predict on returned (class). dat = data.frame(home=c(0,1,1,0,0), gender=c("M","M","F","M","F"), returned=c(0,0,1,1,0)) str(dat) dat$home <- as.factor(dat$home) dat$returned <- as.factor(dat$returned) library(e1071) m <- naiveBayes(returned ~ ., dat) m

Hi everyone. I have these data: probClass<-seq(0,0.9,0.1) prob1<-c(0.0070,0.0911,0.1973,0.2949,0.3936,0.5030,0.5985,0.6869,0.7820,0.8822) prob2<-c(0.0066,0.0791,0.2358,0.3478,0.3714,0.3860,0.6667,0.6400,0.7000,1.0000) # which I'm plotting as follows: plot(probClass,prob1,xlim=c(0,1),ylim=c(0,1),xaxs='i',yaxs='i',type="n") lines(probClass,prob1)

Well, you may have good reasons to do things this way -- and you certainly do not have to explain them here. But you might wish to consider using R's poly() function and a basic nested list structure to do something quite similar that seems much simpler to me, anyway: x <- rnorm(20) df <- data.frame(x = x, y = x + .1*x^2 + rnorm(20, sd = .2)) result <- with(df,

Dear all, I would like to create a code for semiparametric Klein and Spady's estimator. For that I created a function that provides the log-likelihood function for each observation (so it is a function of betas and i, where i denotes the observation). Now, in order to maximize the log-likelihood function, I have to sum these log-likelihood functions for each i and so to get another function

Hi, I'm new to R and I've run into a problem that I'm not really sure how to express properly in the language. I've got a data table that I've read from a file containing some simple information about the performance of 4 algorithms. The columns are the name of the algorithm, the problem instance and the resulting score on that problem (if it wasn't solved I mark that

Thanks everyone for their responses. My data is organized in a data.table.? My goal is to perform analyses according to some groups.? The results of analysis are objects.? If these objects could be stored as elements of a data.table, this would help downstream summarizing of results. Let me try another example. carsdt <- setDT(copy(mtcars)) carsdt[, unique(cyl) |> length()] #[1] 3

How does one ideally handle and display multidimenstional contingency tables in R v. 2.6.2? E.g.: > prob1<- data.frame(victim=c(rep('white',4),rep('black',4)), + perp=c(rep('white',2),rep('black',2),rep('white',2),rep('black',2)), + death=rep(c('yes','no'),4), count=c(19,132,11,52,0,9,6,97)) > prob1 victim perp

I am calculating the mean of each column grouped by the variable 'id'. I do this using aggregate, data.table, and plyr. My aggregate results do not match the other two, and I am trying to figure out what is incorrect with my syntax. Any suggestions? Thanks. Here is the data. myData <- structure(list(var1 = c(31.59, 32.21, 31.78, 31.34, 31.61, 31.61, 30.59, 30.84, 30.98, 30.79, 30.79,

Hola, Bueno, puedes hacer el cálculo de una forma mucho más compacta y rápida. Esta forma es especialmente recomendable cuando tienes muchas columnas y muchas filas. > library(data.table) > myDT <- as.data.table(mtcars) > myDTlong <- melt(myDT, measure.vars=1:ncol(myDT)) > myDTlong[ , list(p_value = shapiro.test(value)$p.value, v_stat = shapiro.test(value)$statistic) , by

Sorry, I pasted the wrong file in earlier... this is the correct one: pValCalculator(b, n=20, m=20) { ind <- 1:min(c(n,m)) prob <- (1-pnorm(b,sd=std*sqrt(ind))) prob1 <- sum((n-ind+1)*(m-ind+1)*prob) prob1 } inputData <- scan("/users/lvssso/projects/LAMA/output/pValLamaScore.tmp", list(block1 = "",block2 = "",width1 = 0,width2 = 0,alignment = 0,score

Hi, I have a small doubt regarding naive Bayes. I am able to classify the data's properly but i am just stuck up with how to get the probability values for naive bayes. In the case of SVM we have "attr" function that helps in displaying the probability values. Is there any function similar to "attr" in naive Bayes that can be used for displaying the attribute values. my