similar to: generating an expression for a formula automatically

I try tu use mob() with my data.frame ('data.frame': 288 obs. of 81 variables; factors, numerics and ordered factors) My response is a binary variable and I should use for modelling a logistic regression (family=binomial). I read in the "MOB" Vignette that I could use a formula like this if I would like to have only partitioning variables apart from the response.

I require some help in debugging this code  library(neuralnet) ir<-read.table(file="iris_data.txt",header=TRUE,row.names=NULL) ir1 <- data.frame(ir[1:100,2:6]) ir2 <- data.frame(ifelse(ir1$Species=="setosa",1,ifelse(ir1$Species=="versicolor",0,""))) colnames(ir2)<-("Output") ir3 <- data.frame(rbind(ir1[1:4],ir2))

Hi list, One of my co-workers found this problem with 'cor' in his code and I confirm it too (see below). He's using R 2.2.1 under Win 2K and I'm using R 2.3.0 under Win XP. =========================================== > R.Version() $platform [1] "i386-pc-mingw32" $arch [1] "i386" $os [1] "mingw32" $system [1] "i386, mingw32" $status

How can I insert mathematical expressions for variable names in a lattice parallel plot? I tried to implement mathematical expressions in varnames, however, without success. For example, neither parallel(~iris[1:4] | Species, iris, varnames=c("P[Width]", "Petal[length]", "alpha[Width]", "Sepal[Length]")) nor parallel(~iris[1:4] | Species,

Hello, I'll use part of the iris dataset for an example of what I want to do. > data(iris) > iris<-iris[1:10,1:4] > iris Sepal.Length Sepal.Width Petal.Length Petal.Width 1 5.1 3.5 1.4 0.2 2 4.9 3.0 1.4 0.2 3 4.7 3.2 1.3 0.2 4 4.6 3.1 1.5

Greetings R experts, I'm having some difficulty recovering lda objects that I've saved within sublists using the save.image() function. I am running a script that exports a variety of different information as a list, included within that list is an lda object. I then take that list and create a list of that with all the different replications I've run. Unfortunately I've been

Hi! The kernlab function kpca() mentions that new observations can be transformed by using predict. Theres also an example in the documentation, but as you can see i am getting an error there (As i do with my own data). I'm not sure whats wrong at the moment. I haven't any predict functions written by myself in the workspace either. I've tested it with using the matrix version and the

Dear all Is there a simpler method to achieve the following: When I obtain an empty data.frame after subsetting, I need for it to contain one line of NAs. Here's a dummy example: > (.xb <- iris[ iris$Species=='zz', ]) [1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species <0 rows> (or 0-length row.names) > dim(.xb) [1] 0 5 > (.xa <-

I am making a battery of levelplots and wireframes for several fitted models. I wrote a function that takes the fitted model object as the sole argument and produces these plots. Various strange behavior ensued, but I have identified one very concrete issue (illustrated below): when my figure-drawing function includes the addition of points/lines to trellis plots, some of the

Hello, I am puzzled by the behavior of hist() when generating multiple plots per page on the pdf device. In the following example two pdf files are generated. The first results in 4 plots on one pdf page as expected. However, the second, which swaps one of the plot() calls for hist(), results in a 4 page pdf with one plot per page. How might I get the histogram with 3 other scatter

I tried to use ctree but am not sure about the meaning of the plot. My.data.ct<-ctree(Resp~., data=My.data) plot(My.data.ct) My data.frame contains 88 explanatory variables (continous,ordered/unordered multistate,count data) and one response with two groups. In the plot are only two variables shown (2 internal nodes) and 3 final nodes. Does it mean that only these two variables show a

Hi Dieter and R community: I tried both of these three versions with ylim as suggested, none work: I am getting only single (pch = 16) not overlayed (pch =3) everytime. *vs 1* require(lattice) xyplot(Sepal.Length ~ Sepal.Width | Species , data= iris, panel= function(x, y, subscripts) { panel.xyplot(x, y, pch=16, col = "green4", ylim = c(0, 10)) panel.lmline(x, y, lty=4, col =

Hi there: I have a question about generating mean value of a data.frame. Take iris data for example, if I have a data.frame looking like the following: --------------------- Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2

Hi there I want to pass arguments (i.e. the response variable and the subset argument) in a self-made function to glm. Here is one way I can do this: f.myglm <- function(y,subfact,subval) { glm(d.mydata[,y]~d.mydata[,'x1'],family=binomial,subset=d.mydata[,subfact]==subval) } > str(d.mydata) `data.frame': 15806 obs. of 3 variables: $ y : Factor w/ 2 levels

Dear R-help, First I apologize if my question is quite simple. I need add some of data in the first place my dataset, how can I do that. I have tried with rbind, but I did not succes. 0.1 3.6 0.4 0.9 rose 4.1 4.0 1.2 1.2 rose 4.4 3.2 1.9 0.5 rose 4.6 1.1 1.1 0.2 rose For example,

If I use latex(summary(X)) where X is a data frame with four variables I get something like Rainfall Education Popden Nonwhite Min. :10.00 Min. : 9.00 Min. :1441 Min. : 0.80 1st Qu.:32.75 1st Qu.:10.40 1st Qu.:3104 1st Qu.: 4.95 Median :38.00 Median :11.05 Median :3567 Median :10.40 Mean :37.37 Mean :10.97 Mean :3866

Dear all What is the easy way to drop a variable by using its name (and not its number)? Example: > data(iris) > head(iris) Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2 setosa 3 4.7 3.2 1.3 0.2 setosa 4 4.6 3.1

Hi, I tried several settings by using the "family=gaussian" in "gl1ce", but none of them works. For the case "glm" can work. Here is the error message I got: > glm(Petal.Width~Sepal.Length+Sepal.Width+Petal.Length ,data=iris,family=gaussian()) > gl1ce(Petal.Width~Sepal.Length+Sepal.Width+Petal.Length ,data=iris,family=gaussian()) Error in eval(expr, envir,

Hi all, I have to recode some values in a dataset. for example changing all zeros to "." or 999 would be also ok. does anybody know how to do this? thanks in advance. lars -- View this message in context: http://www.nabble.com/-R--Replacing-values-tf2841687.html#a7934402 Sent from the R help mailing list archive at Nabble.com.

Hi, I was wondering if it's possible to get the row numbers from a filtering. Here's an example: # give me the rows with sepal.length == 6.2 iris[(iris[,1]==6.2),] # output Sepal.Length Sepal.Width Petal.Length Petal.Width Species 69 6.2 2.2 4.5 1.5 versicolor 98 6.2 2.9 4.3 1.3 versicolor 127 6.2