similar to: How does predict.loess work?

Hi, I am segfaulting when using predict.loess() (checked with r62092). I've traced the source with the help of valgrind (output pasted below) and it appears that this is due to int overflow when allocating an int work array in loess_workspace(): liv = 50 + ((int)pow((double)2, (double)D) + 4) * nvmax + 2 * N; where liv is an (global) int. For D=1 (one x variable), this overflows at

Dear altogether, I tried local regression with the following data. These data are a part of a bigger dataset for which loess is no problem. However, the plot shows extreme values and by looking into the fits, it reveals very extreme values (up to 20000 !) although the original data are > summary(cbind(x,y)) x y Min. :1.800 Min. :2.000 1st Qu.:2.550

Hi! In a current project, I am fitting loess models to subsets of data in order to use the loess predicitons for normalization (similar to what is done in many microarray analyses). While working on this I ran into a problem when I tried to predict from the loess models and the data contained NAs or NaNs. I tracked down the problem to the fact that predict.loess will not return a value at all

Hi all, I am doing a projection pursuit regression using the ppr() function from modreg. I would also like to use predict.ppr(). However, I cannot find any information about it in the help files. There is a link to predict.ppr in the index for modreg, but that link is to the help for ppr(). Has predict.ppr() not been implemented? If not, does anyone have a suggestion as to how to implement

Hi, I wonder why my attempt to extend an existing loess fit to a new data set is producing error. I was trying the following: dat = read.csv(choose.files()) x = dat[,2]; y = dat[,1] x.sort = sort(x) y.loess = loess(y~x, span=0.75) # For testing the above fit with a new dataset: test = read.csv(choose.files()) # test data new_x = test [,1]; new_y = test[,2] new_x.sort = sort(new_x) predicted

Hi, For some reason I have been unable to use the predict function when I desire the standard error to be calculated too. For example, when I try the following: l<- loess(d~x+y, span=span, se=TRUE) p<- predict(l, se=TRUE) I get the following error message: Error in vector("double", length) : vector size cannot be NA In addition: Warning message: In N * M1 : NAs produced by

I'm updating the loess routines to allow for, among other things, arbitrary local polynomial degree and number of predictors. For now, I've given the updated package its own namespace. The trouble is, panel.loess still calls the original code in package:stats instead of the new loess package, regardless of whether package:loess or package:lattice comes first in the search list. If I

Thanks, John. However, loess.smooth() is producing a very different curve compared to the one that results from applying predict() on a loess(). I am guessing they are using different defaults. Correct? On Thu, 23 Mar 2023 at 20:20, John Fox <jfox at mcmaster.ca> wrote: > Dear Anupam Tyagi, > > You didn't include your data, so it's not possible to see exactly what >

Hello, I need help with the details of loess prediction algorithm. I would like to get it implemented as a part of a measurement system programmed in LabView. My job is provide a detailed description of the algorithm. This is a simple one-dimensional problem - smoothing an (x, y) data set. I found quite a detailed description of the fitting procedure in the "white book". It is also

I have a question about running an optimization function on an existing LOESS function defined in R. I have a very large dataset (1 million observations) and have run a LOESS regression. Now, I want to run a Newton-Raphson optimization to determine the point at which the slope change is the greatest. I am relatively new to R and have tried several permutations of the maxNR and nlm functions with

Dear Anupam Tyagi, You didn't include your data, so it's not possible to see exactly what happened, but I think that you misunderstand the object that loess() returns. It returns a "loess" object with several components, including the original data in x and y. So if pass the object to lines(), you'll simply connect the points, and if x isn't sorted, the points

Hello Masters, I run the loess() function to obtain local weighted regressions, given lowess() can't handle NAs, but I don't improve significantly my situation......, actually loess() performance leave me much puzzled.... I attach my easy experiment below #------SCRIPT---------------------------------------------- #I explore the functionalities of lowess() & loess() #because I have

Dear all, I'm trying to construct confidence intervals for a LOWESS estimation (by not using bootstrapping). I have checked previous posts and other material online and I understand that the main procedure is: my.count<- seq(...) fit<- loess (y ~ x, data=z) pred<- pred(fit, my.count, se=TRUE) and then the plotting. However, it's not working; as confidence

Not sure what I'm doing wrong. Can't seem to get loess values. It looks like loess is returning the same values as the input. j <-loess(x1$total~as.numeric(index(x1) plot(x1$total,type='l', ylab='M coms/y global',xlab='') lines(loess(total~as.numeric(index(x1)),x1)) The plot statement works fine No errors with the "lines" statement But I don't

Is there an R implementation of a scheme for automatic smoothing parameter selection with loess, e.g., by minimizing one of the AIC/GCV statistics discussed by Hurvich, Simonoff & Tsai (1998)? Below is a function that calculates the relevant values of AICC, AICC1 and GCV--- I think, because I to guess from the names of the components returned in a loess object. I guess I could use

Hi Folks, I'm looking to do Confidence bands around LOESS smoothing curve. If found the older post about using the Standard error to approximate it https://stat.ethz.ch/pipermail/r-help/2008-August/170011.html Also found this one http://www.r-bloggers.com/sab-r-metrics-basics-of-loess-regression/ But they both seem to be approximations of confidence intervals and I was wonder if there was

Hi, I am trying to create a loess smooth from hydrologic data. My goal is to create a smooth line that describes discharge at a certain point in time. I have done this using the 'lowess' function and had no problem, but I'm having some difficulty with loess. I am inputting the date ('date') and discharge ('q') values using the 'scan' function, then inputting

Dear R-community, maybe someone can help me with this: I've been using the loess() smoother for quite a while now, and for the matter of documentation I'd like to resolve the acronym LOESS. Unfortunately there's no explanation in the help file, and I didn't get anything convincing from google either. I know that the predecessor LOWESS stands for "Locally Weighted

Hi, I posted a message earlier entitled "How to fit a line through the "Mountain crest" ..." I figured loess is probably the best way, but it seems that the problem is the robustness of the fit. Below I paste an example to illustrate the problem: tmp=rnorm(2000) X.background = 5+tmp; Y.background = 5+ (10*tmp+rnorm(2000)) X.specific = 3.5+3*runif(1000);

Dear R experts, I have a problem that is a related to the question raised in this earlier post https://stat.ethz.ch/pipermail/r-help/2007-January/124064.html My situation is different in that I have only 2 predictors (coordinates x,y) for local regression but a number of global ("parametric") offsets that I need to consider. Essentially, I have a spatial distortion overlaid over a