[R] LDA decission boundaries

Hello, I would like to plot the results of a LDA analysis plotting the discriminant scores with the decission boundaries on it with rggobi. I have GGobi already installed on my computer. I have three classes, so the plot would be LD1xLD2 plus the decission boundaries. Here there is the code I use

Re: [R] LDA decission boundaries

2007/8/20, Dani Valverde [EMAIL PROTECTED]: Hello, I would like to plot the results of a LDA analysis plotting the discriminant scores with the decission boundaries on it with rggobi. I have GGobi already installed on my computer. I have three classes, so the plot would be LD1xLD2 plus the

[R] LDA and RDA: different training errors

Hello I try to fit a LDA and RDA model to the same data, which has two classes. The problem now is that the training errors of the LDA model and the training error of the RDA model with alpha=0 are not the same. In my understanding this should be the case. Am I wrong? Can someone explain what the

Re: [R] LDA and RDA: different training errors

dominic senn wrote: Hello I try to fit a LDA and RDA model to the same data, which has two classes. The problem now is that the training errors of the LDA model and the training error of the RDA model with alpha=0 are not the same. In my understanding this should be the case. Am I wrong?

[R] lda and maximum likelihood

Dear all, I am trying to compare several methods for classify data into groups. In that purpose I 'd like to developp model comparison and selection using AIC. In the lda function of the MASS library, the maximum likelihood of the function is not given in the output and the script is not

Re: [R] lda and maximum likelihood

On Mon, 6 Aug 2007, [EMAIL PROTECTED] wrote: I am trying to compare several methods for classify data into groups. In that purpose I 'd like to developp model comparison and selection using AIC. In the lda function of the MASS library, the maximum likelihood of the function is not given in

[R] lda() function and Fisher's LDA

Hi, I am using the function lda() from MASS for finding reduced-dimensional representations of a datset. In reading various texts to compare LDA with Fisher's LDA approach (including Ripley's Modern Applied Statistics with S-Plus), it is still unclear to me whether or not they produce the

Re: [R] LDA newbie question

I am using lda for the first time. I am using version 2.3.1 of R. When I ran the lda I did not get Proportion of trace in the output. Is there another way to get this or is there a bug in my version? Sarah Hodgson __ R-help@stat.math.ethz.ch

Re: [R] lda plotting: labeling x axis and changing y-axis scale

--- Wade Wall [EMAIL PROTECTED] wrote: Hi all, I have performed an lda on two groups and have plotted using plot(x.lda), with x.lda being my lda results. I have forgotten how to change the labels of the of the x-axes (they are currently listed as Group1 and Group 13), and to rescale

Re: [R] lda plotting: labeling x axis and changing y-axis scale

Sorry I wasn't clearer. I believe that it was a specialized function, but it may have been plot(). What I am basically trying to do is alter the y-axis to represent frequency and change the labels on the plotting of the linear discriminant analysis results. I can't seem to do this with plot(),

[R] lda plotting: labeling x axis and changing y-axis scale

Hi all, I have performed an lda on two groups and have plotted using plot(x.lda), with x.lda being my lda results. I have forgotten how to change the labels of the of the x-axes (they are currently listed as Group1 and Group 13), and to rescale the y-axis to reflect frequency. If anyone knows

[R] lda

hi, i am wondering if i could use lda$scaling (i.e. coeff) to evaluate variables' importance if all the x's are normalized before put into model? thanks. -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. Did you always know? No, I did not. But I believed... ---Matrix III

Re: [R] lda

Pieter Vermeesch wrote: I'm trying to do a linear discriminant analysis on a dataset of three classes (Affinities), using the MASS library: data.frame2 - na.omit(data.frame1) data.ld = lda(AFFINITY ~ ., data.frame2, prior = c(1,1,1)/3) Error in var(x - group.means[g, ]) : missing

Re: [R] lda

Pieter == Pieter Vermeesch [EMAIL PROTECTED] on Mon, 16 Oct 2006 19:15:59 +0200 writes: Pieter I'm trying to do a linear discriminant analysis on a Pieter dataset of three classes (Affinities), using the Pieter MASS library: ^^^ No, no!MASS *package*

Re: [R] lda

Dear Martin and Uwe, I did indeed have a few -Inf values in my data frame. Few enough that I didn't notice them when I inspected my data. Thanks a lot for helping me better understand the MASS *package* :-) Pieter On 10/17/06, Martin Maechler [EMAIL PROTECTED] wrote: Pieter == Pieter

[R] lda

I'm trying to do a linear discriminant analysis on a dataset of three classes (Affinities), using the MASS library: data.frame2 - na.omit(data.frame1) data.ld = lda(AFFINITY ~ ., data.frame2, prior = c(1,1,1)/3) Error in var(x - group.means[g, ]) : missing observations in cov/cor What does

[R] LDA

Dears, Using the LDA function of MASS package I can project my data into one dimension. It returns the coefficients that leads to the dimension of maximum separation (e.g. model$scaling), so far so good. Question: how can I get two or four dimensions? That is, instead of project my data using

[R] lda discriminant functions

Hi list, I'm looking about lda function. I'd like to know how calcolate the value of the discriminant functions for the original datas. I see that in the result object lda there is $scaling a matrix which transforms observations to discriminant functions, normalized so that within groups

Re: [R] lda discriminant functions

--- Leonardo Lami [EMAIL PROTECTED] wrote: Hi list, I'm looking about lda function. I'd like to know how calcolate the value of the discriminant functions for the original datas. I see that in the result object lda there is $scaling a matrix which transforms observations to

[R] lda function

Hello, I'm tring to make a linear discriminant analysis eith lda (MASS). In the value of the resulting object is there information about the value of the centroids of the discriminated group for the discriminant functions? Thank you in advance for your help! Leonardo -- Leonardo Lami [EMAIL

[R] lda: scaling to 'disctiminant function'

Friends Briefly... In the documentation for lda in MASS it describes the value 'scaling' as 'a matrix which transforms observations into discriminint functions...'. How? Verbosely... I have a matrix of data. 9 independent variables and describing 3-classes. About 100 observations in total.

[R] lda

Dear R-helpers, if I am right a discriminant analysis can be done with lda. My questions are: 1. What method to discriminate the groups is used by lda (Fisher's linar discriminant function, diagonal linear discriminant analysis, likelihood ratio discriminant rule, ...)? 2. How can I see, which

Re: [R] lda

On Sat, 14 May 2005, K. Steinmann wrote: if I am right a discriminant analysis can be done with lda. My questions are: 1. What method to discriminate the groups is used by lda (Fisher's linar discriminant function, diagonal linear discriminant analysis, likelihood ratio discriminant rule, ...)?

[R] LDA/PCA and more

Hi!! I was trying to analyze my data by doing a linear discriminants and I encountered problems with collinearity. So I am planning to do a PCA to extract orthogonal components and then analyze them using linear discriminants. I don't know how to relate the orthogonal components back to my

[R] lda (MASS)

hi! this is a question about lda (MASS) in R on a particular dataset. I'm not a specialist about any of this but: First with the well-known iris dataset, I tried using lda to discriminate versicolor from the other to classes and I got approx. 70% of accuracy testing on train set. In iris,

Re: [R] lda (MASS)

Now, I use my real dataset (900 instances, 21 attributes), which 2 classes can be serparated with accuracy no more than 80% (10xval) with KNN, SVM, C4.5 and the like. I thinks these accuracies are based on cross-validation runs. Whereas the 80% accuracy you report using LDA is not based on

[R] LDA: variables seems to be constant

Hi, when I performe LDA on some of my datasets I get the error message: Error in lda.default(x, grouping, ...) : variable(s) 3 appear to be constant within groups I assume the reason is the small values of my varibale 3 ( e.g. 4.530353e-05). Has someone a suggestion how to solve this problem?

[R] LDA discriminant coefficients

Hi! I want to determine the relative contribution (or importance) of the independet variables in a two-class case from the discriminant coefficients given in obj.lda$scaling. Are the discriminant coefficinets standardized or do I have to do that? What about other method like partial F-test or

Re: [R] LDA with previous PCA for dimensionality reduction

Dear Cristoph, David, Torsten and Bjørn-Helge, I think that Bjørn-Helge has made more explicit what I had in mind (which I think is close also to what David mentioned). As well, at the very least, not placing the PCA inside the cross-validation will underestimate the variance in the

Re: [R] LDA with previous PCA for dimensionality reduction

Torsten Hothorn writes: as long as one does not use the information in the response (the class variable, in this case) I don't think that one ends up with an optimistically biased estimate of the error I would be a little careful, though. The left-out sample in the LDA-cross-validation, will

[R] LDA with previous PCA for dimensionality reduction

Dear all, not really a R question but: If I want to check for the classification accuracy of a LDA with previous PCA for dimensionality reduction by means of the LOOCV method: Is it ok to do the PCA on the WHOLE dataset ONCE and then run the LDA with the CV option set to TRUE (runs LOOCV) --

Re: [R] LDA with previous PCA for dimensionality reduction

Dear Cristoph, I guess you want to assess the error rate of a LDA that has been fitted to a set of currently existing training data, and that in the future you will get some new observation(s) for which you want to make a prediction. Then, I'd say that you want to use the second approach. You

Re: [R] LDA with previous PCA for dimensionality reduction

On Wed, 24 Nov 2004, Ramon Diaz-Uriarte wrote: Dear Cristoph, I guess you want to assess the error rate of a LDA that has been fitted to a set of currently existing training data, and that in the future you will get some new observation(s) for which you want to make a prediction. Then, I'd

Re: [R] LDA with previous PCA for dimensionality reduction

Thank you, Torsten; that's what I thought, as long as one does not use the 'class label' as a constraint in the dimension reduction, the procedure is ok. Of course it is computationally more demanding, since for each new (unknown in respect of the class label) observation one has to compute a

Re: [R] Lda versus Rda

Julien Trolet wrote: Hello, I used the lda function from the MASS (VR) package and the rda function from the klaR package. I wanted to compare the result of this two functions by using the same training set. Thus, I used the rda function with lambda=1 an gamma=0, I should emulate the lda

[R] Lda versus Rda

Hello, I used the lda function from the MASS (VR) package and the rda function from the klaR package. I wanted to compare the result of this two functions by using the same training set. Thus, I used the rda function with lambda=1 an gamma=0, I should emulate the lda function and I should

[R] lda

Hi !! I am trying to analyze some of my data using linear discriminant analysis. I worked out the following example code in Venables and Ripley It does not seem to be happy with it. library(MASS) library(stats) data(iris3) ir-rbind(iris3[,,1],iris3[,,2],iris3[,,3])

Re: [R] lda

On Tue, 2 Nov 2004, T. Murlidharan Nair wrote: Hi !! I am trying to analyze some of my data using linear discriminant analysis. I worked out the following example code in Venables and Ripley It does not seem to be happy with it. What is `it'? If you mean R, which version, and which

Re: [R] lda

T. Murlidharan Nair wrote: Hi !! I am trying to analyze some of my data using linear discriminant analysis. I worked out the following example code in Venables and Ripley It does not seem to be happy with it. library(MASS) library(stats) data(iris3)

[R] lda predict

Dear R-helpers, I have a model created by lda, and I would like to use this model to make predictions for new or old data. The catch is, I want to do this without using the predict function, i.e. only using information directly from the foo.lda object to create my posterior

Re: [R] lda predict

On Tue, 21 Sep 2004, rob foxall (IFR) wrote: Dear R-helpers, I have a model created by lda, and I would like to use this model to make predictions for new or old data. The catch is, I want to do this without using the predict function, i.e. only using information directly from

[R] lda() - again.

Hi. I asked a question about lda() and got some answers. However, one question remains (which is not independent of the earlier ones): What output does lda() produce which I can use to compute the posteriors? I know predict(lda())$posterior will give me precisely the posteriors, but suppose

Re: [R] lda() - again.

I remember doing this some time ago but forgot. Perhaps this might help you MASS:::predict.lda On Tue, 2004-07-13 at 23:56, marzban wrote: Hi. I asked a question about lda() and got some answers. However, one question remains (which is not independent of the earlier ones): What output

[R] lda()

Hello, For a simple problem with 1 predictor (x) and 2 classes (0 and 1), the linear discriminant function should be something like 2(mu_0 - mu_1)/var x+x-independent-terms

Re: [R] lda()

At 08:45 AM 7/12/2004, marzban wrote: Hello, For a simple problem with 1 predictor (x) and 2 classes (0 and 1), the linear discriminant function should be something like 2(mu_0 - mu_1)/var x+x-independent-terms where var is the common variance. Question 1: Why does lda() report only

Re: [R] lda()

Perhaps some reading would be helpful. I suggest you look first at the help file for lda(). Second, I suggest you read Venables and Ripley, MASS, 4th Edition, where lda() is discussed extensively. Third, I suggest you read Ripley's Pattern Recognition and Neural Networks, where the

Re: [R] lda()

I haven't done this in years but I think the `scaling' element in the list returned by lda is the original data matrix multiplied by the rotation matrix from the SVD. Taking a look at getAnywhere(lda.default) will probably answer your question. -roger marzban wrote: Perhaps some reading would

[R] lda

I am trying to write the following code in R. The code works in S+ and i am trying to do the program in R. x=discrim(admit~gpa+gmat,prior=c(uniform),data=data.mm) i wrote the following in R: x=lda(admit~gpa+gmat,data=data.mm) i could not figure out how to write prior=c(uniform) in R. I

Re: [R] lda

i could not figure out how to write prior=c(uniform) in R. I would get an error every time. I think that it has something to do with uniform. Do you know what i use instead of uniform for R? I am trying to do a uniform distribution. try ?runif (random uniform distribution)

Re: [R] lda

There is a help page for lda: please read it for yourself (as the posting guide requests you too). lda in R works the same way in R as it works in S-PLUS: in both it is support software for a book, and the posting guide also asks you to read that book. On Sat, 12 Jun 2004, Martin Willett wrote:

RE: [R] lda

Seems rather straightforward to me. The prior=uniform in discrim() says to use equal prior for each group. You can do the same by explicitly specifying the priors; e.g., x - lda(admit ~ gpa + gmat, data=data.mm, prior=1/nlevels(data.mm$admit)) HTH, Andy From: Martin Willett I am

[R] lda() called with data=subset() command

Hi I have a data.frame with a grouping variable having the levels C, mild AD, mod AD, O and S since I want to compute a lda only for the two groups 'C' and 'mod AD' I call lda with data=subset(mydata.pca,GROUP == 'mod AD' | GROUP == 'C') my.lda - lda(GROUP ~ Comp.1 + Comp.2 + Comp.3 +

Re: [R] lda() called with data=subset() command

I presume is lda from the uncredited package MASS and you ignored the advice to ask the maintainer? The short answer is `don't ignore the warning', and set up a proper data frame with just the groups you actually want. As a quick fix, look in lda.default and alter the line that looks like

Re: [R] lda source code

Frank Gibbons wrote: Wei Geng, I asked the same question about six weeks ago, so let me try to answer it. The source for the entire package 'MASS' is in a single file, I believe (at least this is true on my Linux setup). The thread gets boring, but let me correct this belief: NO! There is

[R] lda source code

I am new to R. Trying to find out how lda() {in MASS R1.8.0 Windows} was implemented in R. Does anyone know where to find out lda source code ? Thanks. Wei __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help

Re: [R] lda source code

Wei Geng wrote: I am new to R. Trying to find out how lda() {in MASS R1.8.0 Windows} was implemented in R. Does anyone know where to find out lda source code ? Thanks. Here: http://cran.r-project.org Hint: MASS is a *package*. You want to view its *source*. Same with most other R packages.

Re: [R] lda source code

Consider the following: library(MASS) lda function (x, ...) UseMethod(lda) environment: namespace:MASS methods(lda) [1] lda.data.frame lda.defaultlda.formulalda.matrix Now type lda.data.frame or lda.default, etc., at a command prompt to see the corresponding R code. Is this

RE: [R] lda source code

Hi Jason, Spencer, Thanks for the prompt response. The strange thing about MASS is that it's not in Package Sources as most of other R packages are. It seems to come with the binary R installation. I checked out the Rxx/library/MASS on my laptop, there are source code (script) for Venables

Re: [R] lda source code

Wei Geng, I asked the same question about six weeks ago, so let me try to answer it. The source for the entire package 'MASS' is in a single file, I believe (at least this is true on my Linux setup). The exact location of that file you'll have to determine by searching the directory/folder

Re: [R] lda source code

With R 1.7.1 on Windows 2000, I got fine R source code for each of the 4 options. What version of R and what exactly did you do? I can't reproduce your error. hope this helps. spencer graves Wei Geng wrote: Hi Jason, Spencer, Thanks for the prompt response. The strange thing about MASS

Re: [R] lda source code

Wei Geng wrote: Does anyone know where to find out lda source code ? Try typing lda.default at the prompt. That should get you started. Also see: methods(lda) as lda.default isn't the only bit of code used in lda() Alternatively, grab the source from CRAN and read it at your leisure. HTH Gav

Re: [R] lda source code

Wei Geng wrote: Hi Jason, Spencer, Thanks for the prompt response. The strange thing about MASS is that it's not in Package Sources as most of other R packages are. It seems to come with the binary R installation. I checked out the Rxx/library/MASS on my laptop, there are source code (script)

Re: [R] lda source code

Wei Geng wrote: Hi Jason, Spencer, Thanks for the prompt response. The strange thing about MASS is that it's not in Package Sources as most of other R packages are. It seems to come with the binary R installation. I checked out the Rxx/library/MASS on my laptop, there are source code (script) for

Re: [R] lda source code

You are using a *beta* version of R 1.8.0 and in that version lda.default is not visible to the user (hidden in a namespace). You can access it though by using the ::: (triple colon) operator, as in library(MASS) MASS:::lda.default Actually, the first library() call is not necessary. -roger

[R] LDA in R: how to extract full equation, especially constant term

Hi, Having dipped my toe into R a few times over the last year or two, in the last few weeks I've been using it more and more; I'm now a thorough convert. I've just joined the list, because although it's great, I do have this problem... I'm using linear discriminant analysis for binary

[R] LDA: normalization of eigenvectors (see SPSS)

Hi dear R-users I try to reproduce the steps included in a LDA. Concerning the eigenvectors there is a difference to SPSS. In my textbook (Bortz) it says, that the matrix with the eigenvectors V usually are not normalized to the length of 1, but in the way that the following holds (SPSS does

Re: [R] LDA: normalization of eigenvectors (see SPSS)

The following satisfies some of your constraints but I don't know if it satisfies all of them. Let V = eigenvectors normalized so t(V) %*% V = I. Also, let D.5 = some square root matrix, so t(D.5) %*% D.5 = Derror, and Dm.5 = solve(D.5) = invers of D.5. The Choleski decomposition

Re: [R] LDA: normalization of eigenvectors (see SPSS)

Hi, Christoph: 1. I didn't see in your original email that you wanted V to be orthogonal, only that it's columns have length 1. You have a solution satisfying the latter constraint, but not the former. 2. I don't have time now to sort out the details, and I don't have them on the top

[R] lda

Hello I have been using r to classify fish into groups using the lda formula and predict(object,)$class. I am having problems finding an output that simultaneously shows me classification function coefficients for predictors and groups (such as is given in spss under classify:linear discriminant

[R] lda: how to get the eigenvalues

Dear R-users How can I get the eigenvalues out of an lda analysis? thanks a lot christoph -- Christoph Lehmann [EMAIL PROTECTED] __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help

Re: [R] lda: how to get the eigenvalues

On Tue, 3 Jun 2003, Christoph Lehmann wrote: How can I get the eigenvalues out of an lda analysis? It uses singular values not eigenvalues: see ?lda for a description of the output, and the print method for one way to use them. -- Brian D. Ripley, [EMAIL PROTECTED] Professor

Re: [R] lda: how to get the eigenvalues

On Tue, 3 Jun 2003, Christoph Lehmann wrote: How can I get the eigenvalues out of an lda analysis? It uses singular values not eigenvalues: see ?lda for a description of the output, and the print method for one way to use them. the function discrimin ofthe ade4 package performs discriminat

Re: [R] lda: how to get the eigenvalues

let's compare lda and discrimin (ade4) using the iris data: with lda I get: lda1 - lda(iris[,1:4],iris[,5]) lda1$svd [1] 48.642644 4.579983 with discrimin: discrimin1 - discrimin(dudi.pca(iris[,1:4],scan=F),iris[,5],scan=F) discrimin1 eigen values: 0.9699 0.222 so where and how is the

[R] lda of MASS library

Hi, it seems that the lda function in MASS library doesn't give out the constant for the linear discriminant function under the situation that we don't use standardized variable, anyone knows how to obtain the constant in order to construct the linear discriminant function? I understand that

Re: [R] LDA

On Tue, 1 Apr 2003, array chip wrote: I used the lda function in the MASS library of S-Plus (R) to do a linear discriminant analysis, and got the linear coefficients, say b1 and b2 for the 2 predictors x1 and x2. I have trouble to calculate the discrimiant scores for each observation, I

Re: [R] lda on curves

Hi, I recently work about linear dimension reduction for classification. There is a research report on ftp://ftp.stat.math.ethz.ch/Research-Reports/108.html In this report I discuss nine methods for linear dimension reduction, five of which are new. Four of the methods do not perform internal

[R] lda on curves

I'm working on a rather interesting consulting problem with a client. A number of physical variables are measured on a number of cricket bowlers in the performance of a delivery. An example variable might be a directional component of angular momentum for a particular joint measured at a large

Re: [R] lda

R per se does not have lda() function. Package MASS does, and MASS (the book) describes it in detail. If you use a package supporting a book (there are several) do expect to read the book for the fine details On Mon, 10 Feb 2003, Luis Silva wrote: There are some versions of lda. I would

Re: [R] LDA newbie question

At 01:10 PM 2/1/2003, Roland Goecke wrote: Hi, Is there a simple way to get the discriminant score or do I have to manually multiply the coefficients with the data? predict.lda will generate an object with a scores component, among other things. Try ?predict.lda