similar to: MARS help?

Hi, I am using function mars() in package mda to find knots in a whole bunch of predictor variables. I hope to be able to replicate all or some of the basis functions that the MARS software from Salford Systems creates. When I ran mars() on a small dataset, I was able to get the knots. However, when I tried running mars() on a larger dataset (145 predictor variables), for a different

Hi, I'm trying to use mars function in R to interpolate nonlinear multivariate functions. However, it seems that mars gives me a fit which uses only very few basis function and it underfits very badly. For example, I have tried the following code to test mars: require("mda") f <- function(x,y) { x^2-y^2 }; #f <- function(x,y) { x+2*y }; # Grid x <-

Hi, I am trying to write a loop to recode my data from -999 to NA in R. What's the most efficient way to do this? Below is what I'm presently doing, which is inefficient. Thanks, Chris dat0 <- read.table("time1.dat") colnames(dat0) <- c("e1dq", "e1arcp", "e1dev", "s1prcp", "s1nrcp", "s1ints",

Hi, is there an R function which can compute the R-squared for a mars model fitted using the mda package? many thanks in advance best regards Marco Girardello

Hi! I have some questions about MARS model's coefficient of determination. I use the MARS method in my master's thesis and I have noticed some problems with the MARS model's R^2. You can see the following example that the MARS model's R^2 is too big when i have used mars() -function for MARS model building, and when I have made MARS-model using a linear regression, it

Hello all, I have read as many merge issues as I possibly could tonight and although I presume this is a small error, I have not found the solution to my problem. I'm trying to merge two data sets: dat0 and TransTable. As you can see below, dat0 has 8000 rows, whereas TransTable has 47296 rows. I would expect when I merge the two data sets, with all.x=F, and all.y=F, that the

Hello all, So I'm doing some data analysis using MARS. I have a matrix of 65 independent variables, from which I'm trying to predict 71 dependent variables. I have 900+ data points (so a 900x136 matrix of data), which I randomly split into training and validation sets, for ~450 data points in each set. Occasionally, this works well, and I get decent predictions. However, quite often

Dear R Helpers, I have noticed that when I use lmer to analyse data, the summary function gives different values for the AIC, BIC and log-likelihood compared with the anova function. Here is a sample program #make some data set.seed(1); datx=data.frame(array(runif(720),c(240,3),dimnames=list(NULL,c('x1','x2','y' )))) id=rep(1:120,2); datx=cbind(id,datx) #give x1 a

Dear R-help, In the rms package, I have fitted an ols model with a variable represented as a restricted cubic spline, with the knot locations specified as a previously defined vector. When I save the model object and open it in another workspace which does not contain the vector of knot locations, I get an error message if I try to predict with that model. This also happens if only one workspace

Dear Sir/Madam, I am a phd student and I am trying to implement a new approach for the knot selection procedure in MARS algorithm. In order to apply and test the efficiency of my method, is it possible to get the code (sourse) of "earth" library? Thank you for your interest and concern. Best Wishes, Elcin Kartal, Research Assistant Department of Statistics Middle East Technical

Dear List, I'm using gam in a multiple imputation framework -- specifying the knot locations, and saving the results of multiple models, each of which is fit with slightly different data (because some of it is predicted when missing). In MI, coefficients from multiple models are averaged, as are variance-covariance matrices. VCV's get an additional correction to account for how

Hi, I have a question about the usage of eval(). Wonder if any experienced user can help me out of it. I use eval() in the following function: semireg.pwl <- function(coef.s=rnorm(1),coef.a=rnorm(1),knots.pos=knots.x,knots.ini.val=knots.val){ knotn <- length(knots.pos) def.par.env <- sys.frame(1) print(def.par.env) print(environment(coef.s)) tg <- eval( (parse(text=

Hello dear R help members. I am trying to understand the anova.rq, and I am finding something which I can not explain (is it a bug?!): The example is for when we have 3 nested models. I run the anova once on the two models, and again on the three models. I expect that the p.value for the comparison of model 1 and model 2 would remain the same, whether or not I add a third model to be compared

Goodmorning, This is a documentation related question about the B-spline function in R. In the help file it is stated that: "df degrees of freedom; one can specify df rather than knots; bs() then chooses df-degree-1 knots at suitable quantiles of x (which will ignore missing values)." So if one were to specify a spline with 6 degrees of freedom (and no intercept) then a basis

Dear R developers, I noticed a bug in the stats4 package, specifically in the confint method applied to ?mle? objects. In particular, when some ?fixed? parameters define the log likelihood, these parameters are stored within the mle object but they are not used by the ?confint" method, which retrieves their value from the global environment (whenever they still exist). Sample code: >

I am trying to find methods for testing and visualizing calibration to Cox models with time-depended coefficients. I have read this nice article <http://journals.sagepub.com/doi/10.1177/0962280213497434>. In this paper, we can fit three models: fit0 <- coxph(Surv(futime, status) ~ x1 + x2 + x3, data = data0) p <- log(predict(fit0, newdata = data1, type = "expected")) lp

Hi all, I have a vector of proportions (post_op_prw) such that >summary(amb$post_op_prw) Min. 1st Qu. Median Mean 3rd Qu. Max. NA's 0.0000 0.0000 0.0000 0.3985 0.9134 0.9962 1.0000 > summary(cut2(amb$post_op_prw,0.0001)) [0.0000,0.0001) [0.0001,0.9962] NA's 1904 1672 1

Hi r-helpers. Please forgive my ignorance, but I would like to plot a smoothing spline (smooth.spline) from package "stats", and show the knots in the plot, and I can't seem to figure out where smooth.spline has located the knots (when I use nknots). Unfortunately, I don't know a lot about splines, but I know that they provide me an easy way to estimate the location of local

Hi , I want to apply the following code fo my data with 400 predictors. I was wondering if there ia an alternative way instead of typing 400 predictors for the following code. I really appreciate your help. fit0<-lm(Y~1, data= mydata) fit.final<- lm(Y~X1+X2+X3+.....+X400, data=mydata) ??? step(fit0, scope=list(lower=fit0, upper=fit.final), data=mydata, direction="forward")

A data set(dat),has 2 variables: x and a, and 100 rows. I wanna add 2 variables,and call the new data set dat1: var1:f = a/median(a) var2:x_new = x*f My solution: dat1<-transform(dat,f = a/median(a),x_new = x*f) But gets error reply which says that "f" is not exits since dat has no variables called "f". So I have to do through 2 steps: