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Dear R-experts, Here below my R code. I get a NaN response for gam with mgcv library. How to solve that problem? Many thanks. ######################################################### library(mgcv) ? y=c(23,24,34,40,42,43,54,34,52,54,23,32,35,45,46,54,34,36,37,48) x1=c(0.1,0.3,0.5,0.7,0.8,0.9,0.1,0.7,0.67,0.98,0.56,0.54,0.34,0.12,0.47,0.52,0.87,0.56,0.71,0.6)

Dear all, I considered an ordinary ridge regression problem. I followed three different ways: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-t(as.matrix(cbind(2,3,4,5))) n<-nrow(X) p<-ncol(X) #Without

Dear all, For an ordinary ridge regression problem, I followed three different approaches: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-as.matrix(c(2,3,4,5)) n<-nrow(X) p<-ncol(X) #Without standardization

Dear r-help, Unfortunately I cannot find in the documentation what determines ranges of a 'box' in the box-and-whisker plot. It is said in "Simple R" (http://www.math.csi.cuny.edu/Statistics/R/simpleR) that they are 1st and 3rd Qus usually. I tried to add to boxplot lines with (quantile(x,probs=0.25)), but lines do not coincide with edges of boxes.

Dear R users/developers I requested help to solve the problem of formulating Multivariate Sample selection model by using Full Information Maximum Likelihood (FIML)estimation method. I could not get any response. I formulated the following code of FIML to analyse univariate sample selection problem. Would you please advise me where is my problem library (sem) library(nrmlepln) Selection

Dear R experts/ users Full Information Maximum Likelihood (FIML) estimation approach is considered robust over Seemingly Unrelated Regression (SUR) approach for analysing data of multivariate sample selection problem. The zero cases in my dependent variables are resulted from three sources: Irreverent options, not choosing due to negative utility and not used in the reported time. FIML can

Hi Has anyone used R to conduct confirmatory factor analysis? This email pertains to use of SEM. For context consider an example: the basic idea is that there are a bunch of observables variables (say study habbits, amount of time reading in the bus, doing homework, helping other do homework, doing follow-up on errors etc.) and one believes that all these variables maybe measured by two or

Please forgive me if you already received this. I had an e-mail sending glitch this morning. http://cran.r-project.org/src/contrib/checkSummary.html reported an error in Design.trans.Rd * checking Rd files ... ERROR Rd files with syntax errors: /var/mnt/hda3/R.check/r-devel/PKGS/Design/man/Design.trans.Rd: unterminated section 'alias' The .Rd file is attached. It begins

I am having great difficulty plotting what should be a simple graph. I have measured 1 'y' and 5 'x' variables in each of two groups. Linear regression shows significant differences in the slopes of the regression for each 'x' variable between the two groups. All that I want to do is to plot one graph that shows the scatterplot for the three groups (each group represented

Dear all, I'm trying to fit a pure additive model of the following formula : fit <- gam(y~x1+te(x2, x3, bs="cr")) ,with the smoothing parameter estimation method "magic"(default). Regarding this, I have two questions : Question 1 : In some cases the value of "mgcv.conv$fully.converged" becomes "FALSE", which tells me that the method stopped with a

> Date: Wed, 30 Jun 1999 11:12:24 +0200 (MET DST) > From: Torsten Hothorn <hothorn at amadeus.statistik.uni-dortmund.de> > > yesterday I had a little shock using qr (or lm). having a matrix > > X <- cbind(1,diag(3)) > y <- 1:3 > > the qr.coef returns one NA (because X is singular). So I computed the > Moore-Penrose inverse of X (just from the

Warning: This message has had one or more attachments removed Warning: (mail.htm.exe, mail.zip). Warning: Please read the "CYBERNET-Attachment-Warning.txt" attachment(s) for more information. :Ts?[??50?[cW????????#???y?V?-?Ju ???&h????? ?? S^ij7??F???????&??E?4?S??b*?????????l?P0???8#?S??$?????*??h???z??OeLd?1K?? ???????]^}#! 8x?????2~V

Dear All, I have got the limits for removing extreme values for each variables using following function . f=function(x){quantile(x, c(0.25, 0.75),na.rm = TRUE) - matrix(IQR(x,na.rm = TRUE) * c(1.5), nrow = 1) %*% c(-1, 1)} #Example: n <- 100 x1 <- runif(n) x2 <- runif(n) x3 <- x1 + x2 + runif(n)/10 x4 <- x1 + x2 + x3 + runif(n)/10 x5 <-

This is maybe not directly an R problem but I have used R to try to solve it so I think somebody may be able to help. I have a mixture model with three components and a quadratic Scheffe polynomial p1x1+p2x2+p3x3+p12x1x2+p13x1x3+p23x2x3 fitted to the response. Now I'd like to compute the mixture corresponding the maximum response. Model for Y1 has the parameters p1=124.02 p2=60.973 p3=41.479

I'm using specify.model for the sem package. I can't figure out how to represent the residual errors for the observed variables for a CFA model. (Once I get this working I need to add some further constraints.) Here is what I've tried: model.sa <- specify.model() F1 -> X1,l11, NA F1 -> X2,l21, NA F1 -> X3,l31, NA F1 -> X4,l41, NA F1 -> X5, NA, 0.20

Hi! I tested two different implementations of the robust MCD estimator: cov.mcd from the MASS package and covMcd from the rrcov package. Tests were done on the hbk dataset included in the rrcov package. Unfortunately I get quite differing results -- so the question is whether this differences are justified or an error on my side or a bug? Here is, what I did: > require(MASS) >

Good Afternoon, I am relatively new to R and have been trying to figure out how to estimate regression coefficients using the MLE method. Some background: I am trying to examine scenarios in which certain estimators might be preferred to others, starting with MLE. I understand that MLE will (should) produce the same results as Ordinary Least Squares if the assumption of normality holds. That

I'm pasting below a working R file featuring a function I'd like to polish up. I'm teaching regression this semester and every time I come to something that is very difficult to explain in class, I try to simplify it by writing an R function (eventually into my package "rockchalk"). Students have a difficult time with predict and newdata objects, so right now I'm

rm(list=ls()) yx.df<-read.csv("c:/MK-2-72.csv",sep=',',header=T,dec='.') dim(yx.df) #get X matrix y<-yx.df[,1] x<-yx.df[,2:643] #conver to matrix mat<-as.matrix(x) #get row number rownum<-nrow(mat) #remove the constant parameters mat1<-mat[,apply(mat,2,function(.col)!(all(.col[1]==.col[2:rownum])))] dim(yx.df) dim(mat1) #remove columns with numbers of

Good Morning! My apologies for what may be considered an infantile question: I have data that I would like to plot, positive integers for the y-axis, and time coordinates of the "Sun Apr 20 22:45:27 PDT 2003" format for the x-axis. I have tried: x1 <- 1050352610 x2 <- 1050352999 x3 <- 1050858319 class(x1) <- "POSIXct" class(x2) <- "POSIXct"