similar to: R squared for lm prediction

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, 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 # ---

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 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

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:

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

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

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

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}}

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

Muy buenas, quiero calcular correlaciones de Pearson entre dos variables (a,b) teniendo en cuenta una tercera (c). Para ello estoy usando una función llamada "pcor.test" (http://www.yilab.gatech.edu/pcor.html), que en realidad no está en ningún paquete de R, que yo sepa. ¿Alguien conoce una función similar en alguna librería de R? Por otro lado, para ver si me cuadraban los resultados,

Dear Sir/Madam, I am trying to do PCA analysis with "iris" dataset and trying to interpret the result. Dataset contains 150 obs of 5 variables 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

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

Dear list, I have the following how-to-do it in R, questions. Suppose I have ten independent variables, and one dependent variable. I want to find the Pearson correlation of all the IVs with the DV, but not the correlation between the IVs. What I know so far, about R, that I have to type the cor () function ten times, each time requesting for a correlation between one IV and the DV. I was

Could somebody program this kind of plot type to R, if none exists, based on mds or correlation tables or some more suitable method? What do you think about idea? Does it work? None similar or better exists? http://weightedassociationmap.blogspot.com/ Atte Tenkanen University of Turku, Finland

Hi For first question, maybe I am completely wrong but cor(swiss[,-1], swiss[,1]) should give you what you want in one step. Second question Without an example it is hard to say but maybe aggregate is the way forward. > aggregate(iris[,1:4], list(iris$Species), function (x) c(mean=mean(x), sd=sd(x))) Group.1 Sepal.Length.mean Sepal.Length.sd Sepal.Width.mean Sepal.Width.sd 1