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I have a really long functions, and at the end of the function, I am using a if statement to tag certain keywords based on whether they have certain values contained in them. However, the if statement doesn't seem to work. When I had split up the commands into various functions, it worked fine, but I'm not sure what going on now that it's combined into a single function. myfunc

Hi, We have to servers set up in two different networks. We are able to connect calls via IAX and they work perfectly. We do not see caller ID from clients on either side. Our Grandstream phones say Eri and our XTen phones say Asterisk. We did a debug and I am pasting the output from both servers below. We tried setCallerId in several different ways. We see the value get passed to the

Hello, I have questions regarding penalized Cox regression using survival package (functions coxph() and ridge()). I am using R 2.8.0 on Ubuntu Linux and survival package version 2.35-4. Question 1. Consider the following example from help(ridge): > fit1 <- coxph(Surv(futime, fustat) ~ rx + ridge(age, ecog.ps, theta=1), ovarian) As I understand, this builds a model in which `rx' is

I had no problems running regression models in SPSS and R that yielded the same results for these data. The difference you are observing is from fitting different models. In R, you fitted: res <- lm(DEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=dat) summary(res) The interaction term is the product of ZMEAN_PA and ZDIVERSITY_PA. This is not a standardized variable itself and not the same as

I asked you before, but in case you missed it: Are you looking at the right place in SPSS output? The UNstandardized coefficients should be comparable to R, i.e. the "B" column, not "Beta". -pd > On 5 May 2017, at 01:58 , Nick Brown <nick.brown at free.fr> wrote: > > Hi Simon, > > Yes, if I uses coefficients() I get the same results for lm() and

Hi Nick, I think that the problem here is your use of $coef to extract the coefficients of the ridge regression. The help for lm.ridge states that coef is a "matrix of coefficients, one row for each value of lambda. Note that these are not on the original scale and are for use by the coef method." I ran a small test with simulated data, code is copied below, and indeed the output from

Hi, i have now a retriever of Rock Ridge names from ISO directory records and their eventual Continuation Areas. Further i have a detector for SUSP and Rock Ridge signatures. Both have been tested in libisofs by comparing their results with the Rock Ridge info as perceived by the library. 50 ISO images tested. Some bugs repaired. Now they are in sync. (The macro case

Dear Nick, On 2017-05-05, 9:40 AM, "R-devel on behalf of Nick Brown" <r-devel-bounces at r-project.org on behalf of nick.brown at free.fr> wrote: >>I conjecture that something in the vicinity of >> res <- lm(DEPRESSION ~ scale(ZMEAN_PA) + scale(ZDIVERSITY_PA) + >>scale(ZMEAN_PA * ZDIVERSITY_PA), data=dat) >>summary(res) >> would reproduce the

It seems to me that summary for ridge coxph() prints summary but returns NULL. It is not a big issue because one can calculate statistics directly from a coxph.object. However, for some reason the score test is not calculated for ridge coxph(), i.e score nor rscore components are not included in the coxph object when ridge is specified. Please find the code below. I use 2.9.2 R with 2.35-4 version

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:

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?

Thanks, I was getting to try this, but got side tracked by actual work... Your analysis reproduces the SPSS unscaled estimates. It still remains to figure out how Nick got > coefficients(lm(ZDEPRESSION ~ ZMEAN_PA * ZDIVERSITY_PA, data=s1)) (Intercept) ZMEAN_PA ZDIVERSITY_PA ZMEAN_PA:ZDIVERSITY_PA 0.07342198 -0.39650356

Dear all, I need to calculate likelihood ratio test for ridge regression. In February I have reported a bug where coxph returns unpenalized log-likelihood for final beta estimates for ridge coxph regression. In high-dimensional settings ridge regression models usually fail for lower values of lambda. As the result of it, in such settings the ridge regressions have higher values of lambda (e.g.

Hi Simon, Yes, if I uses coefficients() I get the same results for lm() and lm.ridge(). So that's consistent, at least. Interestingly, the "wrong" number I get from lm.ridge()$coef agrees with the value from SPSS to 5dp, which is an interesting coincidence if these numbers have no particular external meaning in lm.ridge(). Kind regards, Nick ----- Original Message -----

Hi, I'm a beginner in the field, I have to perform the ridge regression with lm.ridge for many datasets, and I wanted to do it in an automatic way. In which way I can automatically choose lambda ? As said, right now I'm using lm.ridge MASS function, which I found quite simple and fast, and I've seen that among the returned values there are HKB estimate of the ridge constant and L-W

Hi John, Thanks for the comment... but that appears to mean that SPSS has a big problem. I have always been told that to include an interaction term in a regression, the only way is to do the multiplication by hand. But then it seems to be impossible to stop SPSS from re-standardizing the variable that corresponds to the interaction term. Am I missing something? Is there a way to perform the

It seems to me that R returns the unpenalized log-likelihood for the ratio likelihood test when ridge regression Cox proportional model is implemented. Is this as expected? In the example below, if I am not mistaken, fit$loglik[2] is unpenalized log-likelihood for the final estimates of coefficients. I would expect to get the penalized log-likelihood. I would like to check if this is as expected.

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)

Cool, looks like that'd do it, almost as if converting an entire record to a character string and comparing strings. -- M. B. Hardy, statistician work: Applied Research Associates, S. E. Div. 8537 Six Forks Rd., # 6000 / Raleigh, NC 27615-2963 (919) 582-3329, fax: 582-3301 home: 1020 W. South St. / Raleigh, NC 27603-2162 (919) 834-1245

The anti_join from the package dplyr might also be handy. install.package("dplyr") library(dplyr) anti_join (x1, x2) You can get help on the different functions by ?function.name(), so ?anti_join() will bring you help - and examples - on the anti_join function. It might be worth testing your approach on a small subset of the data. That makes it easier for you to follow what happens