similar to: R squared for lm prediction

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, I cannot find to explicitly get the R-squared or adjusted R-squared from summary(lm()) Thanks a lot! [[alternative HTML version deleted]]

Hi all, This could be very basic. I want to do exploratory factor analysis but I don't have the data, rather I have the correlation matrix. How do I do this with just the correlation matrix? I know for principal components, I can just find the eigen values. Best regards -- View this message in context: http://r.789695.n4.nabble.com/Factor-analysis-tp3040618p3040618.html Sent from the R

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

Dear R-devel list members, I've discovered an inconsistency in how lm() and similar functions handle logical predictors as opposed to factor or character predictors. An "lm" object for a model that includes factor or character predictors includes the levels of a factor or unique values of a character predictor in the $xlevels component of the object, but not the FALSE/TRUE values

I have a dataset and I wish to use two different models to predict. Both models are SVM. The reason for two different models is based on the sex of the observation. I wish to be able to make predictions and have the results be in the same order as my original dataset. To illustrate I will use iris: # Take Iris and create a dataframe of just two Species, setosa and versicolor, shuffle them

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

Dear Abby, > On Aug 30, 2019, at 8:20 PM, Abby Spurdle <spurdle.a at gmail.com> wrote: > >> I think that it would be better to handle factors, character predictors, and logical predictors consistently. > > "logical predictors" can be regarded as categorical or continuous (i.e. 0 or 1). > And the model matrix should be the same, either way. I think that

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',

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])

In using cor(data.frame), it is annoying that you have to explicitly filter out non-numeric columns, and when you don't, the error message is misleading: > cor(iris) Error in cor(iris) : missing observations in cov/cor In addition: Warning message: In cor(iris) : NAs introduced by coercion It would be nicer if stats:::cor() did the equivalent *itself* of the following for a data.frame:

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, May I ask how I can save the coefficients and the p values into a table? thanks. jiong The email message (and any attachments) is for the sole...{{dropped:11}}

Hi all, After trial and error by myself for a few hours, I decide to ask for your help. I have a training set which is a matrix of size 200 x 2, where the two columns denote each independent variable. I have 200 observations. ----------------- ss=data.frame(trainingSet); result=lm(trainingClass~ss$X1+ss$X2); ----------------- where trainingClass denotes the true classes of the training data.

Hi list, I want to write a general function so that it would take an lm object, extract its data element, then use the data at another R function (eg, glm). I searched R-help list, and found this would do the trick of the first part: a.lm$call$data this would return a name object but could not be recognized as a data.frameby glm. I also tried call(as.character(a.lm$call$data)) or

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

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

Hi, I'm trying to set up a prediction software, now i testing the performance of my method, so i need to calculate a ROC curve, specially auc, cut-off, sens and spec, i just looking at ROCH package, but it's a mass for me, i'm not a math guy and I'm getting lost Could any of you recommend me an easy-to-use package to do this task? i just have a list of positive/negative samples

hi all, Suppose I have four variables (X1,X2,Y1,Y2) I want to calculate conditional correlation of (X1,Y1) given (X2, Y2). How can I do it in R? Thanks Ming Chen

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