similar to: Wilks' Lambda for lda?!

In the meantime I found out that it should be possible to compute wilks' lambda from the singular values that are returned by "lda". But since I'm only a biologist I got totally stuck in formulas I don't understand ... Can anybody help me to get equivalent outputs of wilks' lambda in R as in SPSS? Yours, Christoph. -- Christoph Lange

I wonder if it was possible (and desireable) to implement means to protect data objects (and functions) from overwriting them. So for instance: > x <- 5 > protect(x) > x <- 8 Error: x is read only Or would that be against the philosophy of R to be compatible to S?! Just a RFC ;-) Yours, Christoph. ps. Your probably know the situations that let you think of such a feature

Hello, using the lda-method from MASS-package I was wondering whether there is a built-in method for figuring out Wilks' Lambda? Searching the Web I found in the r-help archive a thread form june 2002, but it didn't help me. I understand I can use manova and its summary-method to get Wilks' Lambda on the screen, but I don't see the connection to lda from MASS. Or does

The first: if I have a vector as (1,1,3,2,1,1), which is the command that gives all the positions of the min value? From the vector of the example a would like to obtain a new vector as (1,2,5,6) that give me all the positions of the minimum value 1. The second: if I have a matrix "A" and I want to obtain a new matrix deleting a column or a row of A, what have I to do? Thank you.

Hi everybody, I was wondering if you can help me about a module. In fact, I'm looking for a package or module about Wilks' lambda criterion in R environment. I didn't find it in R website ( http://cran.cict.fr/web/packages/index.html#available-packages-W or http://search.cpan.org/faq.html). If this module exists, could you show me the command line. Thank you very much for your help,

Dear helpeRs, the following data set comes from Johnson/Wichern: Applied Multivariate Statistical Analysis, 6th ed, pp. 304-306. /X <- structure(c(9, 6, 9, 3, 2, 7), .Dim = as.integer(c(3, 2))) Y <- structure(c(0, 2, 4, 0), .Dim = as.integer(c(2, 2))) Z <- structure(c(3, 1, 2, 8, 9, 7), .Dim = as.integer(c(3, 2)))/ I would like to compute Wilk's Lambda in R, which I know is 0.0385.

Hi all, I can't figure out how to compute Wilks Lambda in a one way repeated measure design. My matrix looks like: > t2.m Blank ECR ENC UEA UED 1 -0.15 0.14 0.16 0.09 0.14 2 0.30 0.08 0.14 0.14 0.14 [...] where each row is a case and the columns are levels of one factor (named trial): > t2.fit <- manova(t2.m ~ 1) > summary(t2.fit, intercept=T,

Dear R-users, I'm wondering how to obtain Wilks-lambda values when discriminant analyses have only one discriminant function (i.e. 2 categories to discriminate between). The use of manova(predictions~groups, test="Wilks") asks for multiple response and the use of anova(lm(predictions~groups), test="Wilks") simply does not consider the last term. (The alternative

I have used the greedy.wilks to stepwise discriminant analysis, but it doesn't work with my dataset. I don't understand which is the problem Could you help me please? Thanks in advance Marta > str(data_indiciN2) 'data.frame': 200 obs. of 36 variables: $ gruppo: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ... $ I001N2: num 19.32 8.22 28.35 7.24

I don't understand which is the problem Could you help me please? Thanks in advance Marta > str(data_indiciN2) 'data.frame': 200 obs. of 36 variables: $ gruppo: Factor w/ 2 levels "0","1": 2 2 2 2 2 2 2 2 2 2 ... $ I001N2: num 19.32 8.22 28.35 7.24 14.7 ... $ I002N2: num 2.92 2.54 0.11 1.6 7.12 ... $ I003N2: num -22.362 -0.222 -19.291

Hi all, I am newbie in using R software and also doing statistical test. I want to know if my data in in normal distribution. I have 2 groups of data and I did calculate Shapiro Wilks using R software. Here is the results: Group 1: W = 0.9206, p-value = 0.01683 Group 2: W = 0.9626, p-value = 0.4694 I am not quite sure what default confidence level (CF) is used in calculating Shapiro Wilks.

hello list, when I use method lda of the MASS package I experience a warning: variables are collinear in: lda.default(data[train, ], classes[train]) Is there an easy way to recover from this issue within the MASS package? Or how can I tell how severe this issue is at all? I understand that I shouldn't use lda at all with collinear data and should use "quadratische" (squared?)

Hello, I have a question about multivariate Shapiro-Wilks test. I tried to analyze if the data I have are multivariate normal, or how far they are from being multivariate normal. However, any time I did >mshapiro.test(mydata) I get the message: Error in solve.default(R %*% t(R), tol = 1e-18) : system is computationally singular: reciprocal condition number = 5.38814e-021 I tried

Dear All, I would be very appreciative of your help with the following 1). I am running multivariate multiple regression through the manova() function (kindly suggested by Professor Venables) and getting two different answers for test=c("Wilks","Roy","Pillai") and tests=c("Wilks","Roy",'"Pillai") as shown below. In the

Fellow R Users: I'm not extremely familiar with lda or R programming, but a recent editorial review of a manuscript submission has prompted a crash cousre. I am on this forum hoping I could solicit some much needed advice for deriving a classification equation. I have used three basic measurements in lda to predict two groups: male and female. I have a working model, low Wilk's lambda,

Hi Since a stepwise procedure for variable selection (as e.g. in SPSS) for a LDA is not implemented in R and anyway I cannot be sure, that all the required assumptions for e.g. a procedure using a statistic based on wilks' lambda, hold (such as normality and variance homogeneity) I would like to ask you, what you would recommend me: shall I e.g. define a criterion such as the error-rate

Hallo! How please to remove a win2k machine account. I tried smbpasswd -x machine$ without succes. The thing is that I changed the machine domaine to workgroup. And now I'm trying to put again the machine in the domaine but that can't be done. I'm getting an error message that the logon informations are in conflict with existing logon informations. Thank a lot for helping

Reported-by: Jim Davis <jim.epost at gmail.com> Signed-off-by: Ilia Mirkin <imirkin at alum.mit.edu> --- drivers/gpu/drm/nouveau/nouveau_hwmon.c | 1 - 1 file changed, 1 deletion(-) diff --git a/drivers/gpu/drm/nouveau/nouveau_hwmon.c b/drivers/gpu/drm/nouveau/nouveau_hwmon.c index 38a4db5..4aff04f 100644 --- a/drivers/gpu/drm/nouveau/nouveau_hwmon.c +++

Hi, is it possible to draw a box around the following barplot; using "box=TRUE" won't work. ---------- pc<-c(1,2,5,29,27) barplot(pc,ylim=c(0,30),yaxs="r",xaxs="r",ylab="Anzahl" ,names.arg=c("Mac","286er","386er","486er","Pentium I") ,axisnames=T,col="gray") ---------- Thanks in

hi all, I've got to import CSV-datasets (with variable-names in the first line) into data.frames. each is about 12MB (or more!) with 1823 columns and about 500 rows. the first 22 columns are in "character"-mode, the rest is "numeric". I run R 1.4.1 on a Windows 2000 system. First I tried read.table() which works fine for a low number of cases (say, 40). with all cases