similar to: improved pairs.formula?

Hello, I have posted this mail a few days ago but I did it wrong, I hope is right now: I have the following doubts related with lm.ridge, from MASS package. To show the problem using the Longley example, I have the following doubts: First: I think coefficients from lm(Employed~.,data=longley) should be equal coefficients from lm.ridge(Employed~.,data=longley, lambda=0) why it does not happen?

Dear r-devel list members, On a couple of occasions I've encountered the issue illustrated by the following examples: --------- snip ----------- > mod.1 <- lm(Employed ~ GNP.deflator + GNP + Unemployed + + Armed.Forces + Population + Year, data=longley) > mod.2 <- update(mod.1, . ~ . - Year + Year) > all.equal(mod.1, mod.2) [1] TRUE > > f <-

I want to apply Nyblom-Hansen test with the strucchange package, but I don't know how is the correct way and what is the difference between the following two approaches (leeding to different results): data("longley") # 1. Approach: sctest(Employed ~ Year + GNP.deflator + GNP + Armed.Forces, data = longley, type = "Nyblom-Hansen") #results in: # Score-based CUSUM

Dear all I'm familiarising myself with Ridge Regressions in R and the following is bugging me: How does one get p-values for the coefficients obtained from MASS::lm.ridge() output (for a given lambda)? Consider the example below (adapted from PRA [1]): > require(MASS) > data(longley) > gr <- lm.ridge(Employed ~ .,longley,lambda = seq(0,0.1,0.001)) > plot(gr) > select(gr)

Hello dear R help group, I have the following setup to analyse: We have about 150 subjects, and for each subject we performed a pair of tests (under different conditions) 18 times. The 18 different conditions of the test are complementary, in such a way so that if we where to average over the tests (for each subject), we would get no correlation between the tests (between subjects). What we wish

Dear Terry, Even the behaviour of lm() and glm() isn't entirely consistent. In both cases, singularity results in NA coefficients by default, and these are reported in the model summary and coefficient vector, but not in the coefficient covariance matrix: ---------------- > mod.lm <- lm(Employed ~ GNP + Population + I(GNP + Population), + data=longley) >

>>>>> Fox, John <jfox at mcmaster.ca> >>>>> on Thu, 14 Sep 2017 13:46:44 +0000 writes: > Dear Martin, I made three points which likely got lost > because of the way I presented them: > (1) Singularity is an unusual situation and should be made > more prominent. It typically reflects a problem with the > data or the

>>>>> Martin Maechler <maechler at stat.math.ethz.ch> >>>>> on Thu, 14 Sep 2017 10:13:02 +0200 writes: >>>>> Fox, John <jfox at mcmaster.ca> >>>>> on Wed, 13 Sep 2017 22:45:07 +0000 writes: >> Dear Terry, >> Even the behaviour of lm() and glm() isn't entirely consistent. In both cases,

[This question is hopefully straight-forward, but difficult to provide reproducible code.] I'm doing a multivariate bootstrap, using boot::boot(), where the output of the basic computation is a k x p matrix of coefficients, representing a tuning constant x variable, as shown in the $t0 component from my run, giving a 3 x 6 matrix > lboot$t0 GNP Unemployed Armed.Forces

For an application of ridge regression, I need to get the covariance matrices of the estimated regression coefficients in addition to the coefficients for all values of the ridge contstant, lambda. I've studied the code in MASS:::lm.ridge, but don't see how to do this because the code is vectorized using one svd calculation. The relevant lines from lm.ridge, using X, Y are:

>>>>> Fox, John <jfox at mcmaster.ca> >>>>> on Wed, 13 Sep 2017 22:45:07 +0000 writes: > Dear Terry, > Even the behaviour of lm() and glm() isn't entirely consistent. In both cases, singularity results in NA coefficients by default, and these are reported in the model summary and coefficient vector, but not in the coefficient covariance

Dear Martin, I made three points which likely got lost because of the way I presented them: (1) Singularity is an unusual situation and should be made more prominent. It typically reflects a problem with the data or the specification of the model. That's not to say that it *never* makes sense to allow singular fits (as in the situations you mentions). I'd favour setting

I'm looking for a function for the 'multiple correlation' but can not figure out what it is called in R by using the html search function. Maybe it is called in another way in english? I only know the german term? Can anyone help me? Thomas Pesl -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Dear Terry, It's not surprising that different modeling functions behave differently in this respect because there's no articulated standard. Please see my response to Martin for my take on the singular.ok argument. For a highly sophisticated user like you, singular.ok=TRUE isn't problematic -- you're not going to fail to notice an NA in the coefficient vector -- but I've

Dear Berwin, Simply stacking the problems and treating the resulting observations as independent will give you the correct coefficients, but incorrect coefficient variances and artificially zero covariances. The approach that I suggested originally -- testing a linear hypothesis using the coefficient estimates and covariances from the multivariate linear model -- seems simple enough. For

--MP_/kvy20nVajVG/n.8m=_ZjLAX Content-Type: text/plain; charset=US-ASCII Content-Transfer-Encoding: 7bit Content-Disposition: inline On Tue, 7 Oct 2008 19:31:03 +0800 Berwin A Turlach <berwin at maths.uwa.edu.au> wrote: > The attached patch (against the current SVN version of R) implements > the latter strategy. With this patch applied, "make check > FORCE=FORCE" passes

G'day all, I guess it is still early enough in the year to wish everybody a happy and successful new year. I thought I should report that the installation of the CRAN package rstan regularly fails on my machine (a 64 bit linux box running Xubuntu 15.10). The reason being that I have the 32-bit and the 64-bit architecture of R installed, and my /tmp file is on a partition with about 1Gb

Thanks for the replies to my prior question. My problem is that R always says object not found when I enter a variable name into a command. I converted a Stata file into an Rdata file by first loading the foreign package by entering require(foreign) Then I asked R to read the Stata file by entering pol572a1<- read.dta("C:\\alex\\Graduate Coursework\\Pol 572\\pol572a1.dta") So

I can plot a time series with: gnp <- ts(cumsum(1+round(rnorm(100), 2)), start=c(1954,7), frequency=12) plot(gnp) But I want to plot more time series in the same plot. How can I do it? =-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-= r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or

Thanks to Andy and Thomas, Reading help(hclust) more carefully would have done it but sometimes you do not see the wood for the trees... So hc$merge does exactly what I want. I have never been aware of the command str to get the structure of an R-object. It seems pretty useful to me. Thanks, Arne > -----Original Message----- > From: Liaw, Andy [mailto:andy_liaw at merck.com] >