similar to: Closest fit data to a particular formula

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 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

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

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

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

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

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

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 everyone. I'm trying to break the y axis on a plot. For instance, I have 2 series (points and a loess). Since the loess is a "continuous" set of points, it passes in the break section. However, with gap.plot I cant plot the loess because of this (I got the message "some values of y will not be displayed"). Here's my code: library(plotrix); #generate some data x

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 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 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

Hello List, I have a longitudinal samples on multiple individuals, for initial analysis, i am doing some plots to see how the variable changes with the age. In some of the individuals one or two data points are missing, IL1Ra Age.at.Sample.Collection Subject.ID 6.575466 2.004106 00709-0 NA 2.162902

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

>From what I understand, loess in R uses the standard tri-cube function. SAS/INSIGHT offers loess with Gaussian weights. Is there a function in R that does the same? Also, can anyone offer any references comparing properties between tri-cube and Gaussian weights in LOESS? Thanks. - Andr? -- View this message in context:

Hi everyone. I'm trying to do a loess with missing value on independant variable. doc = c(2.27904, 2.59536, 7.44696, NA, 6.24264, 4.58400, 5.79192, 5.39502, 7.41216, 4.09440, 4.22868, 4.24620, 5.43804, 1.95528); distance = c(26.5,56.5, 90.3, 123.0, 147.5, 176.0, 215.7, 229.3, 252.0, 325.3, 362.0, 419.3, 454.6, 470.0); myloess = loess(doc ~ distance, na.action = na.omit); plot(distance,

Hi, I have a scatter plot of the variables GNI and Lifeexp (Gross National Income and Life Expectancy, both metric). So I plotted them and I want to add a regression line and a lowess line. I use lowess and not loess because I have missing values. My code: plot(GNI,Lifeexp) abline(lm(Lifeexp~GNI), col="red") y.loess<-loess(Lifeexp~GNI,na. action = na.exclude)