search for: nsamp

I'm trying to use least trimmed squares using ltsreg with nsamp="exact". When I use the following: rg <- ltsreg(x,y,nsamp="exact") I get: Error in lqs.default(x, y, nsamp = "exact", method = "lts") : NAs in foreign function call (arg 10) In addition: Warning message: NAs introduced by coercion Incidentally,...

...le <- c(floor((n + k + 1)/2), floor((n + k)/2)) quantile <- rep(quantile, length.out = 2) if(is.null(psamp)) psamp <- quantile[1] ## LTS results with robust residuals NameC_lts <- lqs(gpanew~female+female:lastinit+agenew+canadian+mom_ed+yearstudy, quantile = quantile[1], psamp = psamp, nsamp = nsamp) rr <- residuals(NameC_lts)/NameC_lts$scale[2] rr_nok <- abs(rr) > critval[1] ## robust leverage via MCD (or MVE) X <- model.matrix(NameC)[,-1] cv <- cov.rob(X, method = method, quantile = quantile[2], nsamp = dist_nsamp) rd <- sqrt(mahalanobis(X, cv$center, cv$cov)) rd_no...

...24.868, 29.895, 24.200, 23.215, 21.862, 22.274, 23.830, 25.144, 22.430, 21.785, 22.380, 23.927, 33.443, 24.859, 22.686, 21.789, 22.041, 21.033, 21.005, 25.865, 26.290, 22.932, 21.313, 20.769, 21.393) and I would like to fit the following model: mod1 <- lqs(y ~ x1 + x2, method="lms", nsamp="exact") mod1$coefficients (Intercept) x1 x2 35.5293489 0.4422742 -1.2944534 mod1$bestone [1] 12 17 27 Now, instead of using the formula, I want to provide the design matrix and the response, then: X <- cbind(1, x1, x2) mod2 <- lqs.default(X, y, intercept=F,...

...an for each sample. For each sample mean Xbark the 0.95-confidence interval for the mean mew=0 is given by... Ik= ( Xbark plus or minus 1.96/10) Find the number of intervals such that 0 does not belong to Ik. How many of them do you expect to see? Well so far I have come up with... N<-100; Nsamp<-100 A<-matrix(rnorm(N*Nsamp,0,1),ncol=Nsamp) means<-apply(A,2,mean) However I have no idea what I am doing and no idea if that even makes sense. Any help would be greatly appreciated as I have no experience of statistical software whatsoever. Thanks in Advance. Rachel -- View this me...

...ut thereafter the results aren't the same. If I run this code in Splus, the results are the same. Also, if I set the size argument to be the same (or eliminate this line altogether), I get reproducible results.=20=20 =20 I'm using R 2.2.0 =20 Thanks, -Lori =20 test.fun<-function(nsamp, mu,q, seed, nsim=3D2){ set.seed(seed)=20=20=20=20 mnsq<-rep(NA,nsim) for (i in 1:nsim){ y<-rnorm(nsamp,mean=3Dmu) yboot<-sample(y, size=3Dq, replace=3DT) mnsq[i]<-(mean(y)^2) } return(mnsq) } =20 test.fun(50,0,30,12345) test.fun(50,0,10,12345) =20 Lori E. Dodd, Ph.D. Nat...

...f factors. Basically this repates the factors 16 times ### (for each row in the matrix). This results in a data frame with 84*16 ### rows as many columns as there are factors + 2 (probe factor + value ### to be modeled later) probeDf <- function(emat, facts) { df <- NULL n <- 1 nsamp <- ncol(emat) for ( i in 1:nrow(emat) ) { values <- c(t(emat[i,])) df.new <- data.frame(Value = values, facts, probe = rep(n, nsamp)) n <- n + 1 if ( !is.null(df) ) { df <- rbind(df, df.new) } else { df <- df.new...

...runif(n) + 1i * runif(n) dist <- function(x, y) sqrt(sum(Mod(x - y)^2)) perms <- permutations(n) dim(perms) # [1] 40320 8 tmp <- apply(perms, 1, function(ord) dist(x, y[ord])) z <- y[perms[which.min(tmp), ]] # exact solution dist(x, z) # an aproximate random-sampling approach nsamp <- 10000 perms <- t(replicate(nsamp, sample(1:n, size=n, replace=FALSE))) tmp <- apply(perms, 1, function(ord) dist(x, y[ord])) z.app <- y[perms[which.min(tmp), ]] # approximate solution dist(x, z.app) Thanks for any help. Best, Ravi. _____________________________________________...

Hi, I'm completely new to R, I am all at sea with the interface and the confusing help files, so would appreciate some help to do a simple task. Need to present the mean and variance of 100 different samples of poisson distributions (N=1000, with fixed lambda) in a file in two columnns, and then produce histograms. So far I have figured out: > N <- 1000 > x <- rpois(N, 3.1) ,

Hello all I have a feeling this is very simple......but I am not sure how to do it My boss has two variables, one is an average of 4 numbers, the other is an average of 3 of those numbers i.e var1 = (X1 + X2 + X3 + X4)/4 var2 = (X1 + X2 + X3)/3 all of the X variables are supposed to be measuring similar constructs not surprisingly, these are highly correlated (r = .98), the question is how

Hi, I' am doing a stats project using R to work out the size of a t-test and wilcoxon test depending on the distribution and sample size. I just can't get it to work - I want to put my results from the function size() into an array.At the moment I keep getting the error message:Error in res[distribution, test, samplesize] <- results : subscript out of boundsCan anyone tell me where

Hi, I want to calculate the R-squared between two variables. Can you advice me how to identify and remove the outliers before performing R-squared calculation? Thanks, Kan --------------------------------- SBC Yahoo! DSL - Now only $29.95 per month! [[alternate HTML version deleted]]

...ry(doMPI) cl <- startMPIcluster(count=2) #this will be changed later registerDoMPI(cl) #the input parameters! ... #some vecs and matrices for the C program ... #load the C code dyn.load("/export/home/example/Runs/eg.so") Est_Results<-foreach(data_iter = 1:nsamp, .combine="cbind", .inorder=TRUE ) %dopar% { outs=.C("prog1", all of the args ) do more things return( some things) } dyn.unload("/export/home/example/Runs/eg.so") closeCluster(cl) save.image("/export/home/e...

...riginal Rousseeuw et al. program points out that observations 1,2,3,4 and 21 are outliers. This conslusion is arrived at by testing whether the residual is greater than 2.5 * standard error Netx I ran lqs as: m <- lqs(stackloss[,-4], stackloss[,4], method='lms', control=list (psamp=4, nsamp='exact', adjust=TRUE)) (I ran it exhaustively since that was how I ran the original program from Rousseeuw) The coefficients obtained from lqs() are more or less identical to that obtained by the original program. However the scale estimates do not match. I assume that this would be becua...