similar to: Loess with glm ?

Hi, I have a data frame with 6563 observations. I can run a regression with loess using four explanatory variables. If I add a fifth, R crashes. There are no missings in the data, and if I run a regression with any four of the five explanatory variables, it works. Its only when I go from four to five that it crashes. This leads me to believe that it is not an obvious problem with the data,

Hello, I am looking for a measure of linearity in the relationship between two variables. Specifically, I have two variables for which the relationship is reasonably linear over a certain range of values, and then diverges from linearity at either end of the range, as one or other variable "saturates" at a maximum or minimum value. I want to identify the region of linearity, where

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

Dear R community, I am trying to understand how the predict function, specifically, the predict.loess function works. I understand that the loess function calculates regression parameters at each data point in 'data'. lo <- loess ( y~x, data) p <- predict (lo, newdata) I understand that the predict function predicts values for 'newdata' according to the loess regression

For some reason the following code is not plotting as I want it to. I want to plot a "loess" line plotted over a scatter plot. I get a jumble, with lines connecting all the points. I had a similar problem with "lowess". I solved that by dropping "NA" rows from the data columns. Please help. library(stats) attach(gini_pci_wdi_narm) plot(ny_gnp_pcap_pp_kd, si_pov_gini)

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

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 >

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

If a run a LOESS model and then produce a smoothed surface: Is there any way to determine the coordinates of the local maxima on the surface? [[alternative HTML version deleted]]

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

Gurus, I used predict(loess(Y~X)); where Y and X are of the same length. But there are same NA's in both Y and X. Those NA's are in the same locations in Y and X. The following is the error messageI got: Error in "[<-"(*tmp*, , i, value = predict(loess(Y~X))) : number of items to replace is not a multiple of replacement length If I replaced the NA's with a

Dear all, To draw a lowess line on a plot was a piece of cake; to draw a loess line, however, seems not that easy. Is the loess plotting implemented at all in relation to the loess function, or do I have to look in add-on packages? Thanks, Marcin

Hi, I was wondering if anyone knows of a way by which one can estimate values from a contour plot created by using the loess function? I am hoping to use the loess contour plot as a means of interpolation to identify the loess created values at points at pre-defined (x,y) locations. Could anyone point me in the right direction please? Thanks. Laura Quinn Institute of Atmospheric Science School

Hello, I'm currently using a tool that provides a Loess fitting, but I obtained results that are slightly different from those provided by R implementation of the Loess. That's why I would like to know if you could give me a source (bibliography or web) that explains in a clear way each step of the algorithm, with the possible options to choose, etc.. in order for me to understand those

Hi all, I'm fitting a line to my dataset. Later I want to predict missing values that exceed the [min,max] interval of my empirical data, therefore I choose surface="direct" for extrapolation. l1<-loess(y1~x1,span=0.1,data.frame(x=x1,y=y1),control=loess.control(surface="direct")) In my application it is highly important that the fitted line intercepts at the point of

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

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, 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);

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