similar to: loess problems

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 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

Dear R users, I am having trouble devising an efficient way to run a loess() function on all columns of a data.frame (with the x factor remaining the same for all columns) and I was hoping that someone here could help me fix my code so that I won't have to resort to using a for loop. (You'll find the upcoming lines of code in a single block near the end of the message.) Here's a

Hello, I have been struggling to do a plot in ggplot(2) that's of lattice equivalent. The following code shows the lattice plot.

Hi all, I was unable to obtained a smoothed line using the loess function. I used the following code reported in the examples of R documentation: cars.lo <- loess(dist ~ speed, cars) Then I tried to plot both the data and the smoothed line plot(cars) lines(cars.lo) but what I obtained is simply a broken line joining all the data points. I tried with different spans, but the results did not

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]]

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 >

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

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

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

Hi, suppose I have a vector of values in 'A'. Suppose I use: scatter.smooth(A) to add in a smooth curve. Is there a way to obtain the maximum value of the smoothed curve? Anyways, any help will be greatly appreciated! -- View this message in context: http://www.nabble.com/Loess-question-tp24738574p24738574.html Sent from the R help mailing list archive at Nabble.com.

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! 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, 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);

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