similar to: Wishlist: Make screeplot() a generic (PR#9541)

Hi, I run the example for princomp for R211 I got the following error for biplot > ## The variances of the variables in the > ## USArrests data vary by orders of magnitude, so scaling is appropriate > (pc.cr <http://pc.cr> <- princomp(USArrests)) # inappropriate Erreur dans cov.wt(z) : 'x' must contain finite values only > princomp(USArrests, cor = TRUE) # =^=

I'm trying to make a screeplot including the Cumulative Proportion of the Variance, something that can easily be done in S-Plus with 'screeplot(pc.object,cumulative=T)'. How can I access the Proportion of Variance in an princomp object and how could I get the Cumulative Proportion of the Variance on the screeplot? Many thanks in advance, Jan:-) --

Hi, Suppose I've got a data set that I found the eigenvalues and eigenvectors. Then I want to draw a screeplot for the eigenvalues. However it returns: Error in matrix(w.m, nc = NC) : negative extents to matrix Then I tried a plot() function on the eigenvalues of the data set, and I can successfully draw it! And the output looks like what the screeplot would show to me. Is there any

My previous post about prcomp and princomp was done in some haste as I had long ago indicated to Kurt that I would try to have this ready for the June release, and it appeared that I would miss yet another release. I also need to get it out before it becomes hopelessly buried by other work. Brian Ripley kindly pointed out some errors, and also pointed out that I was suggesting replacing some

Dear all, I'm using R 2.8.1 under Vista. I programmed a Simulation with the code enclosed at the end of the eMail. After the simulation I want to analyse the columns of the single simulation-runs, i.e. e.g. Simulation[[1]][,1] sth. like that but I cannot address these columns... Can anybody please help? Best, Thomas ############################ CODE ############################

Dear all, perhaps I am using princomp.formula and prcomp.formula in a way that is not documented to work, but then the documentation just says: formula: a formula with no response variable. Thus, to avoid a lot of typing, it would be nice if one could use '.' and '-' in the formula, e.g. > library(DAAG) > res <- prcomp(~ . - case - site - Pop - sex, possum)

I do not understand, from a PCA point of view, the option center=F of prcomp() According to the help page, the calculation in prcomp() "is done by a singular value decomposition of the (centered and possibly scaled) data matrix, not by using eigen on the covariance matrix" (as it's done by princomp()) . "This is generally the preferred method for numerical accuracy"

Hello, I am using function princomp from the package psych. I have my principle component object mypc: mypc <- princomp(covmat=mycor) plot(mypc) # shows me a screeplot Question: how could I actually see the values displayed in the screeplot. I don't mean on the graph - I just want to know the actual value for each component (e.g., 10, 3.2, 1.8, etc.) I need to know how much variance,

According to R help: princomp() uses eigenvalues of covariance data. prcomp() uses the SVD method. yet when I run the (eg., USArrests) data example and compare with my own "hand-written" versions of PCA I get what looks like the opposite. Example: comparing the variances I see: Using prcomp(USArrests) ------------------------------------- Standard deviations: [1] 83.732400 14.212402

Hi , To my knowledge, there're two functions that can do principal component analysis, princomp and prcomp. I don't really know the difference; the only thing I know is that when the sample size < number of variable, only prcomp will work. Could someone tell me the difference or where I can find easy-to-read reference? To access the first PC using princomp:

Some webpage has described prcomp and princomp, but I am still not quite sure what the major difference between them is. Can they be used interchangeably? In help, it says 'princomp' only handles so-called R-mode PCA, that is feature extraction of variables. If a data matrix is supplied (possibly via a formula) it is required that there are at least as many units as

To my knowledge, this has not been reported previously, and doesn't seem to have been changed in R-devel or R-patched. If M is a matrix with coloumn names, and mod <- prcomp(M) # or princomp then predicting a single observation (row) with predict() gives the error Error in scale.default(newdata, object$center, object$scale) : length of 'center' must equal the number of

As of R 1.7.0, princomp no longer accept matrices with more coloumns than rows. I'm curious: Why was this decision made? I work a lot with data where more coloumns than rows is more of a rule than an exception (for instance spectroscopic data). To me, princomp have two advantages above prcomp: 1) It has a predict method, and 2) it has a biplot method. A biplot method shouldn't be too

In comparing the results of princomp and prcomp I find: 1. The reported standard deviations are similar but about 1% from each other, which seems well above round-off error. 2. princomp returns what I understand are variances and cumulative variances accounted for by each principal component which are all equal. "SS loadings" is always 1. 3. Same happens

Dear listserv, I am trying to perform a principal components analysis and create an output table of the eigenvalues for the dependent variables. What I want is to see which variables are driving each principal components axis, so I can make statements like, "PC1 mostly refers to seed size" or something like that. For instance, if I try the example from ?prcomp > prcomp(USArrests,

When I look through archives at https://stat.ethz.ch/pipermail/r-help/2003-October/040525.html I see this: Liaw, Andy wrote: >In the `Detail' section of ?princomp: > >princomp only handles so-called Q-mode PCA, that is feature extraction of >variables. If a data matrix is supplied (possibly via a formula) it is >required that there are at least as many units as variables. For

kalish at psy.uwa.edu.au wrote: > > I'm a newbie at R, and can't get libraries to really work. > I did this: > > library(help = mva) > cancor Canonical Correlations > cmdscale Classical (Metric) Multidimensional Scaling > dist Distance Matrix Computation > hclust Hierarchical Clustering

Hello, I have a large data set that has more columns than rows (sample data below). I am trying to perform a partitioning cluster analysis and then plot that using pca. I have tried using CLUSPLOT(), but that only allows for 'princomp' where I need 'prcomp' as I do not want to reduce my columns. Is there a way to edit the CLUSPLOT() code to use 'prcomp', please? #

After doing a PCA using princomp, how do you view how much each component contributes to variance in the dataset. I'm still quite new to the theory of PCA - I have a little idea about eigenvectors and eigenvalues (these determine the variance explained?). Are the eigenvalues related to loadings in R? Thanks, Paul -- View this message in context:

Hello folks, it may be regarded as a user error to scale() your data prior to prcomp() instead of using its 'scale.' argument. However, it is a user thing that may happen and sounds a legitimate thing to do, but in that case predict() with 'newdata' can give wrong results: x <- scale(USArrests) sol <- prcomp(x) all.equal(predict(sol), predict(sol, newdata=x)) ## [1]