similar to: loess function takes long to estimate

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

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

Full_Name: Jeff Schwarz Version: R1.3.1 OS: Windows 2000 Submission from: (NULL) (129.22.170.115) Error message (using predict and predict.nnet) > predict (smalltest, smallx[-jj,]) Error in matrix(NA, length(keep), nout, dimnames = list(rn, dimnames(object$fitted)[[2]])) : length of dimnames[1] not equal to array extent *** all relevant code and data source is given below *** I

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

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

Please do as we ask (repeatedly) and study the help page before posting. 'family' is a separate argument, not part of loess.control, as the help page correctly documents. If you use cars.lo2 <- loess(dist ~ speed, cars, family = "symmetric", control = loess.control(surface = "direct", iterations = 20)) cars.lo2$pars$iterations it prints *20*, as it is

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 >

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, I was hoping that someone familiar with the implementation details of the loess algorithm might be able to help me resolve some difficulties I am having. I am attempting to reproduce some of the functionality of the loess() function in C++. My primary motivation is that I would like to understand the algorithm in detail. So far I have managed to create a working port in C++ for the

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

I am trying to (semi) calculate the confidence intervals for a loess smoother (function: loess()), but have been thus far unsuccessful. The CI for the loess predicted values, yhat, are apparently yhat +- t*s * sqrt(w^2), where s is the residual sum of squares and w is the weight function Correct me of I'm wrong, but R uses the tricubic function (1-abs(z)^3)^3, where z = (x-xi)/h, where h

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, 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

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

Alright, I apologize for this basic question - I am both an R and loess noob. I am trying to predict the values of column Y in data1 (100000x18 entries) using a loess fit on training (500x18 entries) and columns A B and C. (training are not members of data1) fit <- loess(Y ~ A + B + C, training) predicted <- predict(fit, data1) However, I'm getting such good predictions that I have

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

>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,

Hello, I am interested in performing a 2D loess smooth on microarray data, i.e. log2 ratios on a 2D grid, using different spans in the horizontal and vertical directions (the immediate reason being that replicate spots are laid out in the horizontal direction). Is it possible to do this in R? Functions like loess(stats) seem to apply the same span for all predictors, which carries over to

Hi, I'm not sure what the recommended thing to do here would be: In the lattice library, panel.loess needs to use loess.smooth(), which is in the modreg library. Now should I (1) make lattice depend on modreg, and call require(modreg) in zzz.R OR (2) use autoload("loess", "modreg") I like the second option better, as a particular session might not need panel.loess.