similar to: lm.ridge

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 would like to suggest changing the pairs.formula command such that a command like pairs(GNP ~ . - Year - GNP.deflator, longley) would behave in a similar fashion as lm(GNP ~ . - Year - GNP.deflator, longley) i.e., make a pairwise scatterplot of GNP and all other variables in the (longley) dataframe except for Year and GNP.deflator. The above command, with the

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)

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) >

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

>>>>> 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,

>>>>> 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

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

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

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

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] >

Dear R-users, I tried to receive the merge details of a clustering by using the summary function of hclust. For illustration I use the Longley data as done by Prof Ripley (Wed 11 Apr 2001) d <- dist(longley.y) d <- d/max(d) hc <- hclust(d, "ave") But instead of getting a matrix for $merge I get: >summary(hc) Length Class Mode merge 30 -none- numeric

Can you please tell me the minimum system spec for R on a windows machine? Thanks _______________________________________ Matthew Longley IT Administrator Zoological Society of London Regent's Park London NW1 4RY Charity No: 208728 Tel: 020 74496412 Fax: 020 74496294 email: Matthew.Longley at zsl.org WWW: www.zsl.org _______________________________________

[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

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it means that it is advisable to

Hi all, I just cannot think of how to do it: I want to take the first variable (column) of a data frame and regress it against all other variables. bla <- function (dat) { reg <- lm(whateverthefirstofthevariablenamesis ~., data=dat) return(reg) } What kind of function do I have to take instead of the whateverthefirstofthevariablenamesis, eval(), substitute(), get(), ... to

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