similar to: formula mungeing

Recursively walk the formula performing the replacement: g <- function(e, ...) { if (length(e) > 1) { if (identical(e[[2]], as.name(names(list(...))))) { e <- eval(e, list(...)) } if (length(e) > 1) for (i in 1:length(e)) e[[i]] <- Recall(e[[i]], ...) } e } g(f, lambdas = 2:3) ## y ~ qss(x, lambda = 2L) + qss(z, 3L) + s On Fri, Oct

I'm having difficulty with an environmental issue: I have an additive model fitting function with a typical call that looks like this: require(quantreg) n <- 100 x <- runif(n,0,10) y <- sin(x) + rnorm(n)/5 d <- data.frame(x,y) lam <- 2 f <- rqss(y ~ qss(x, lambda = lam), data = d) this is fine when invoked as is; x and y are found in d, and lam is found the

I'm in the very initial stage of expanding the formula processing in my quantile regression function rq() to handle additive nonparametric components, say qss(x), or qss(x,z). I need some advice about strategy for formula processing. My initial foray was to use: terms(formula,specials="qss") and then modify the components of the resulting terms.object. But in changing formula

The officially sanctioned way to put the expression "lambda_1 = x" in a title is something like this: title(substitute(lambda[1] == lamb, list(lamb = x))) but suppose I have two lambdas and would like something like "lambda_1 = x_1 , lambda_2 = x_2" to appear. What then? Undoubtedly I'm missing something blindingly obvious with lists, but having tried several

Hi, I need to fit a piecewise linear regression. x = c(6.25,6.25,12.50,12.50,18.75,25.00,25.00,25.00,31.25,31.25,37.50,37.50,50.00,50.00,62.50,62.50,75.00,75.00,75.00,100.00,100.00) y = c(0.328,0.395,0.321,0.239,0.282,0.230,0.273,0.347,0.211,0.210,0.259,0.186,0.301,0.270,0.252,0.247,0.277,0.229,0.225,0.168,0.202) there are two change points. so the fitted curve should look like \ \ /\

Rers: I installed R 2.9.0 from the Debian package manager on our amd64 system that currently has 6GB of RAM -- my first question is whether this installation is a true 64-bit installation (should R have access to > 4GB of RAM?) I suspect so, because I was running an rqss() (package quantreg, installed via install.packages() -- I noticed it required a compilation of the source) and

Hi, I have used package quantreg to estimate a non-linear fit to the lowest part of my data points. It works great, by the way. But I'd like to extract the predicted values. The help for predict.qss1 indicates this: predict.qss1(object, newdata, ...) and states that newdata is a data frame describing the observations at which prediction is to be made. I used the same technique I used

Dear all, My machine is SUN Java Workstation 2100 with 2 AMD Opteron CPUs and 16GB RAM. R is compiled as 64bit by using SUN compilers. I trying to fit quantile smoothing on my data and I got an message as below. > fit1<-rqss(z1~qss(cbind(x,y),lambda=la1),tau=t1) Error in as.matrix.csr(diag(n)) : cannot allocate memory block of size 2496135168 The lengths of vector x and y are

Hello, i have the following data: x=c(0,0.02,0.03,0.04,0.05,0.06,0.07,0.08,0.09,0.1,0.11,0.12,0.13,0.14,0.15,0.16,0.17,0.18,0.19,0.2,0.21,0.22,0.23,0.25,0.26,0.27,0.46,0.47,0.48,0.49) y=c(0.48,0.46,0.41,0.36,0.32,0.35,0.48,0.47,0.55,0.56,0.54,0.67,0.61,0.60,0.54,0.51,0.45,0.42,0.44,0.46,0.41,0.43,0.43,0.48,0.48,0.47,0.39,0.37,0.32,0.29) and tried to get piecewise linear regression. Doing a

Hi - I've been using the option main=bquote(paste(mu==.(mu),", ",lambda==.(lambda),", ",truncation==.(truncation),", ",N[T]==.(n))) to produce a title when using the "plot" command - a title which includes variable names (two Greek) along with their values. The above option, however, does not work within the "boxplot" command. Any

I am experimenting with parallel processing using foreach and seem to be missing something fundamental. Cool stuff. I've gone through the list and seen a couple of closely related issues, but nothing I've tried seems to work. I know that the results from foreach are combined, but what if there is more than one variable within the loop? Below is a snippet (non-functioning) of code that I

Can someone please show me how to smooth time series data that I have in the form of a zoo object? I have a monthly economies series and all I really need is to see a less jagged line when I plot it. If I do something like s <- smooth.spline(d.zoo$Y, spar = 0.2) plot(predict(s,index(d.zoo)), xlab = "Year") # not defined for Date objects and if I do something like

Hi All I would like to know how to access the values of the variable lambda.mu and and see what abs(lambdas[1]) does since lambdas is not a keyword. Snippet of the code: scoreFunction <- function(lambdas) { lambda.mu <- abs(lambdas[1]) sme.em(yi,tmei,Xi,Ni,G,lambda.mu,lambda.v)$AICc } Thanks in advance. Regards Ap -- Aparna Sampath Master of Science (Bioinformatics)

Hi, I am writing some basic smoothers in R for cleaning some spectral data. I wanted to see if I could get close to matlab for speed, so I was trying to compare SparseM with Matrix to see which could do the choleski decomposition the fastest. Here is the function using SparseM difsm <- function(y, lambda, d){ # Smoothing with a finite difference penalty # y: signal to be smoothed #

Thanks John. ?boxcox says: ************************* Arguments object a formula or fitted model object. Currently only lm and aov objects are handled. ************************* I read that as saying that boxcox(lm(z+1 ~ 1),...) should run without error. But it didn't. And perhaps here's why: BoxCoxLambda <- function(z){ b <- MASS:::boxcox.lm(lm(z+1 ~ 1), lambda = seq(-5, 5,

At useR 2006 I mentioned that it would be nice to have a way to specify binomial links that involved free parameters and described some experience with a Gosset link involving a free degrees of freedom parameter, and a Tukey-lambda link with two free parameters. My implementation of this involved some rather kludgey modifications of binomial, make.link and glm that (essentially) added a

Hi Bert, On 2023-07-08 3:42 p.m., Bert Gunter wrote: > Caution: This email may have originated from outside the organization. Please exercise additional caution with any links and attachments. > > > Thanks John. > > ?boxcox says: > > ************************* > Arguments > > object > > a formula or fitted model object. Currently only lm and aov objects

About 2 percent of the time I use fix() to edit a function that is sitting in .RData I get the response: > fix(qss) Error in edit(name, file, editor) : problem with running editor vi when I try to close the editing session. I used to think that these were always cases where there was some syntactical error with the edited file, but this is not the case. I realize that one surefire way to

Is it possible to mix symbols and evaluated objects inside the expression() function ? The following example shows what I am trying to achieve: for (m in 1:3) { plot(1:10); #just a place holder for the real plots title(expression(y = m * lambda)); } I want to actually evaluate the variable m but keep lambda as a symbol in the title. I tried to wrap an eval() around various subparts of

I want to create a list of functions in a for loop, with the index of the loop appearing explicitly in the function code. After quite a bit of thrashing around I figured out how to do it. Here is a toy example: junk <- vector("list",4) for(i in 1:4) { itmp <- i junk[[i]] <- eval(bquote(function(x){42 + .(itmp)*x})) } So I'm *basically* happy, but there's