similar to: finding interpolated values along an empirical parametric curve

A simple question, but I can't find something to do what I want: Given: a vector of numbers, like lambda <- c(0, 0.005, 0.01, 0.02, 0.04, 0.08) Desired: format them in minimal space for use as plot labels, ie, without leading or tailing 0s. For this example: lambdaf <- c("0", .005", ".01", ".02", ".04", ".08") -- Michael

Unlike L1 (lasso) regression or elastic net (mixture of L1 and L2), L2 norm regression (ridge regression) does not select variables. Selection of variables would not work properly, and it's unclear why you would want to omit "apparently" weak variables anyway. Frank maths123 wrote > I have a .txt file containing a dataset with 500 samples. There are 10 > variables. > >

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+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

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

Hello R-list, I am trying to calculate a ridge regression using first the *lm.ridge()* function from the MASS package and then applying the obtained Hoerl Kennard Baldwin (HKB) estimator as a penalty scalar to the *ols()* function provided by Frank Harrell in his Design package. It looks like this: > rrk1<-lm.ridge(lnbcpc ~ lntex + lnbeerp + lnwinep + lntemp + pop, subset(aa,

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:

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

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?

Hi all, I have run a ridge regression on a data set 'final' 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

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)

Readers, Is it possible to create a plot command based upon the indices of missing values in a data set? dataset1<-read.table(text=' 10 2 20 NA 30 5 40 7 50 NA 60 NA 70 2 80 6 90 NA 100 9 ') dataset2<-read.table(text=' 0.2 0.4 0.1 0.9 0.2 0.3 1.1 0.7 0.9 0.6 0.4 ') The 'approx' function is used to obtain the interpolated values for 'NA' in dataset1.

Hello,everybody! I am a beginner of R. And I want to ask a question. If anybody would help me, thank you very much ahead. I want to plot something like a response surface, and I find the "rsm" package. Some commands are like this: #code head library(rsm) CR = coded.data(ChemReact, x1 ~ Time, x2 ~ Temp) CR.rsm = rsm(Yield ~ SO(x1,x2), data = CR) summary(CR.rsm) contour(CR.rsm,x1~x2)

Hi there, I've got this vector: -84 -87 -90 -90 -89 -86 NA NA NA NA NA NA NA NA NA NA NA NA -96 -99 -100 -99 -96 -92 -89 -87 -87 -88 -90 -92 -94 -95 -96 -97 -97 -97 -96 -95 Is there a function in R which replaces the NAs with "interpolated" values between -86 and -96? Thanks, Sven

Dear help list - I have light data with 5-min time-stamps. I would like to insert four 1-min time-stamps between each row and interpolate the light data on each new row. To do this I have come up with the following code: lightdata <- read.table("Test_light_data.csv", header = TRUE, sep = ",") # read data file into object "lightdata" library(chron) mins <-