similar to: coerce mode list?

I am having a problem understanding the lda package. I have a dataset here: [,1] [,2] [,3] [1,] 2.95 6.63 0 [2,] 2.53 7.79 0 [3,] 3.57 5.65 0 [4,] 3.16 5.47 0 [5,] 2.58 4.46 1 [6,] 2.16 6.22 1 [7,] 3.27 3.52 1 If I do the following; "names(d)<-c("y","x1","x2") d$x1 = d$x1 * 100 d$x2 = d$x2 * 100 g<-lda( y ~ x1 + x2, data=d) v2

Hi, I run this code to get the power of the test for modified bartlett's test..but I'm not really sure that my coding is right.. #normal distribution unequal variance asim<-5000 pv<-rep(NA,asim) for(i in 1:asim) {print(i) set.seed(i) n1<-20 n2<-20 n3<-20 mu<-0 sd1<-sqrt(25) sd2<-sqrt(50) sd3<-sqrt(100) g1<-rnorm(n1,mu,sd1) g2<-rnorm(n2,mu,sd2)

Thanks David, Yes, I am talking about the MASS package.Thank you for pointing out that these scale the same. My question is, how do I get from the V1 data: V1 1 164.4283 2 166.2492 3 170.5232 4 156.5622 5 127.7540 6 136.7704 7 136.3436 to the other set of data: + 1 -2.3769280 + 2 -2.7049437 + 3 -3.4748309 + 4 -0.9599825 + 5 4.2293774 + 6 2.6052193 + 7 2.6820884 On Mon, Sep 28, 2009

... and is there also such a nice tool (like spss.get) for exporting data frames to SPSS? write.table does not keep the data frame labels - neither did the other exporting tools that I found. Thanks! Michael [[alternative HTML version deleted]]

# Peter B. Mandeville kindly offered to send me code for Hotelling's T^2 # Test. Unfortunately there seems to be no route to his machine. # So i'm trying to reach him via the Mailing List. ------------------------------------------------------------------------ Sir, this morning i recieved your message about the availability of the code for Hotelling's Test. I hurried to find out

Hi all; I have two vectors A=c(5,2,2,3,3,2) and B=c(2,3,4,5,6,1,3,2,4,3,1,5,1,4,6,1,4) and I want to make the following matrix using the information I have from the above vectors. 0 1 1 1 1 1 1 0 1 0 0 0 0 1 0 1 0 0 1 0 1 0 1 0 1 0 0 1 0 1 1 0 0 1 0 0 so the first vector says that I have 6 elements therefor I have to make a 6 by 6 matrix and then I have to read 5 elements from the

By accident, I left out the lines defining n1 and n2. Here it is as a function. Peter B. hotelling <- function(d1,d2){ k <- ncol(d1) n1 <- nrow(d1) n2 <- nrow(d2) xbar1 <- apply(d1,2,mean) xbar2 <- apply(d2,2,mean) dbar <- xbar2-xbar1 v <- ((n1-1)*var(d1)+(n2-1)*var(d2))/(n1+n2-2) t2 <- n1*n2*dbar%*%solve(v)%*%dbar/(n1+n2) f <-

Please give me just a reference where I can find something useful. In summary, I need to : - find the median of each row of a matrix - create a new matrix with each value in the first matrix divided by the median of its row - if a value "a" in the second matrix is < 1, I need to substitute it with 1/a I know that for some of you it must be overeasy, but I swear I googled for two

Dear R-helpers, I want to make a series of boxplots on several numeric univariates with two group variables (species and population, population nested in species, and with population as the X-axis). In order to get a proper order of the individual populations in X-axis, I need to assign a wanted order to the factor (population). I used the levels() function to do this assignment, but it seemed

I am needing some expertise with regard to the S-Plus command lmRobMM and its R counterpart rlm(formula,data,method="MM") I have used lmRobMM(formula,data) in S-Plus on the Stackloss data and obtained for my residuals 6.217777 1.150717 6.427946 8.174019 -0.6713005 -1.248641 -0.4236203 0.5763797 -1.057899 0.3593823 11 12 13 14 15 16

Greetings, Sorry, the last message was sent by mistake! Here it is again: I encounter a strange problem computing some numerical integrals on [0,oo). Define $$ M_j(x)=exp(-jax) $$ where $a=0.08$. We want to compute the $L^2([0,\infty))$-inner products $$ A_{ij}:=(M_i,M_j)=\int_0^\infty M_i(x)M_j(x)dx $$ Analytically we have $$ A_{ij}=1/(a(i+j)). $$ In the code below we compute the matrix

Full_Name: Kirk Hampel Version: 2.4.1 OS: Windows Submission from: (NULL) (144.53.251.2) Given an AR(p) model, the last p SE's are wrong. The source of the bug is that the C code (ver 2.4.0) assumes *npsi is the length of the psi vector (which is n+p), whilst the predict.ar function in R passes out as.integer(npsi), where npsi <- n-1. Some R code following reproduces the error. Let p=4,

I'm using R on Windows v2.0.1 with the nlme package (v3.1-53) and am finding some unexpected discrepancies in the output of intervals.lme and coef.lme. I've included a toy dataset at the end, but briefly, the data are longitudinal data from couples in marital therapy. Each spouse's relationship satisfaction is measured 4 times; I've fit both linear and quadratic models to the

I have a model object that acts as a list. The position of each object is determined by two factors: number of votes ascending and age of object descending. When a user adds or deletes a vote I need to quickly update the associated model object''s position. Right now this is done with the following method: def update_position position = nil Bug.find(:all, :order =>

Greetings,   I encounter a strange problem computing some numerical integrals on [0,oo). Define $$ M_j(x)=exp(-jax) $$ where $a=0.08$. We want to compute the $L^2([0,\infty))$-inner products $$ A_{ij}:=(M_i,M_j)=\int_0^\infty M_i(x)M_j(x)dx $$ Analytically we have $$ A_{ij}=1/(a(i+j)). $$ In the code below we compute the matrix $A_{i,j}$, $1\leq i,j\leq 5$, numerically and check against the known

Hi, by replacing 'll' with 'wh' in the source code for base::which() one gets ~20% speed up for *named logical vectors*. CURRENT CODE: which <- function(x, arr.ind = FALSE) { if(!is.logical(x)) stop("argument to 'which' is not logical") wh <- seq_along(x)[ll <- x & !is.na(x)] m <- length(wh) dl <- dim(x) if (is.null(dl)

hi, ive been reading up on ferret, acts_as_ferret, and other search plugins for rails. after reading about ferret, i found out about the acts_as_ferrt plugin. my first question about acts_as_ferret: 1. from reading about ferret, do i still need to manually save the IDX and add a IDX column field to my model table for acts_as_ferret to work? they say that acts_as_ferret handles everything,

Hi Zed, I checked in a pure-Ruby URI classifier to Mongrel trunk. Ola''s Java port of the TST had some bug, and I don''t think it''s necessary in the first place. The Ruby classifier is around 25 lines instead of the 400-odd lines for the C extension and the 200-odd for the Java extension. It uses a Regexp which is perhaps shady: @matcher = Regexp.new(routes.map do

Forms without explicit enctype are submitted as application/x-www-form- urlencoded. This is the default behaviour in Rails. However, this enctype does not allow transmission of binary data (files). Would it not make sense to specify the enctype multipart/form-data by default instead? i.e. all the form_for helpers would add this enctype to the form tag, unless overriden by the developer. This

maybe it's in the data? So here it comes. > sv.growth Grouped Data: length ~ meas | box_id meas spec comp water box_id sprouts leaves length long.sprout 1 1 Sv control moist 1 8.800000 37.80 211.2000 60.6 2 1 Sv xfull moist 2 7.000000 8.00 174.8000 62.8 3 1 Sv control moist 3 9.000000