similar to: SoS! How to predict new values using linear regression models?

Dear list, I'm quite confused when interpreting results from a mixed linear model. For example, working on Iris data frame, I want to know the effect of species on slope of the model "Petal.Length~Sepal.Length" I write this : data(iris) reg01 <- lm(Petal.Length~Sepal.Length + Sepal.Length:Species, data=iris) summary(reg01) It gives me a summary table with lm for the first

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

Hi, First of all, I'm a complete rookie to R (~2 weeks). But anyway, I'm trying to use the RWeka interface for C4.5 (J48) classification. As a proof of concept I'm using the Iris data set to create a training set of 30 instances (10 per species) and use the remaining 120 instances as my test set. This is what I do: trainingIndices <- rep(1:10, 3) + rep(0:2, each=10) * 50

Hi All, I created a two variable lm() model slm<-lm(y[1:3000,8]~y[1:3000,12]+y[1:3000,15]) I made two predictions predict(slm,newdata=y[201:3200,]) predict(slm,newdata=y[601:3600,]) there is no error message for either of these. the results are identical, and identical to slm$fitted as well. if this is not the right way to apply the model coefficients to a new set of inputs, what is

I would like to ask a general question about the randomForest predict function and how it handles No Data values. I understand that you can omit No Data values while developing the randomForest object, but how does it handle No Data in the prediction phase? I would like the output to be NA if any (not just all) of the input data have an NA value. It is not clear to me if this is the default or

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<-

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

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

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

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

Dear all, I'd like to get a percentage variable based on a group, but without creating a new data frame. For example: data(iris) iris$percent <-unlist(tapply(iris$Sepal.Length,iris$Species,function(x) x/sum(x, na.rm=TRUE))) This does not work, I should have only three standard values, respectively for setosa, versicolor, and virginica. How can I do this? MANY THANKS, Karine

Por si alguno pudiera ayudarme. Al realizar el t.test para una muestra, junto con el valor de t y el p-valor, la función proporciona la estimación de la media y su INTERVALO DE CONFIANZA. Desde el punto de vista de la estadística de rangos esto se puede hacer mediante: > iris$MEDIANA <- with(iris, 2.95) > median(iris$Sepal.Width - iris$MEDIANA, na.rm=TRUE) # median difference [1]

I have a data set where I want the correlations between 2 variables conditional on a students grade level. This code works just fine. by(tmp[,c('mtsc07', 'DCBASmathscoreSPRING')], tmp$Grade, cor, use='complete', method='pearson') However, this generates an error by(tmp[,c('mtsc07', 'DCBASmathscoreSPRING')], tmp$Grade, cor, use='complete',

I would like to create an xyplot with varying y-axis limits and horizontal labels at the y-axis tickmarks. The following does not seem to work, although I think it should, going by the documentation for par. R version 2.5.1, Windows XP Prof. Thanks for a clue. Andreas Krause library(lattice) # axis labels for y-axis are horizontal xyplot(Sepal.Length ~ Sepal.Width | Species, data=iris) #

Dear all When using Sweave, I'm always hitting the same bump: I want to group repetitive calls in a function, but I want both the results and the function calls in the printed output. Let me explain myself. Consider the following computation in an Sweave document: summary(iris[,1:2]) cor(iris[,1:2]) When using these two calls directly, I obtain the following output: > summary(iris[,1:2])

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, The below very strange: # a) aov function av <- aov(Sepal.Length ~ Species, data=iris) # Error in parse(text = x) : # unexpected symbol in "Sepal(Sepal.Length+Species)Length" av <- aov(iris[, 1] ~ iris[, 5]) # summary(av) # Df Sum Sq Mean Sq F value Pr(>F) # iris[, 5] 2 63.2 31.6 119 <2e-16 *** # Residuals 147 39.0 0.3 # ---

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 <-

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

Hallo, I'd like to add grid lines to a lattice graph having 2 series of Y data. See these 2 examples: data(iris) [1] xyplot(Sepal.Length + Sepal.Width ~ Petal.Length , data = iris, allow.multiple = TRUE, scales = "same",type="l", ) [2] xyplot(Sepal.Length + Sepal.Width ~ Petal.Length , data = iris, allow.multiple = TRUE, scales =