similar to: Finding local maxima on a loess surface

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

Dear R users, I have created a Loess surface in R, in which x is relative longitude by miles, y is relative latitude by miles, and z is population density at the neighborhood level. The purpose is to identify some population centers in the region. I'm wondering if there is a way to determine the coordinates (x,y) of each center, so I can know exactly where they are. Let me use the

Hi Folks, I'm probably overlooking some small point, but can't see it. Trying the "Local trend surfaces" example from p.437 of MASS (3rd edn) by V&R, all goes well until the line contour(topo.mar$x, topo.mar$y, topo.lo$fit, levels=seq(700,1000,25), xlab="fi t", ylab="") which produces the response: Error in contour.default(topo.mar$x,

R- Users: Can someone indicate what I am during wrong? This is a script essentially from Venerable & Ripley's text on interpolating a surface with loess function but I can not get it to run. Thanks. John Lewis Professor McGill University Montreal library(MASS) library(modreg) data(topo) par(mfcol=c(2,2), pty="s") topo.loess <- loess(z ~ x * y, topo, degree=2,

I'm reading Ripley-Venables "Modern Applied Statistics with S - Fourth edition" , at the same time trying the examples proposed in the book using R 1.8.1 under linux. Now I'm trying the following code from the book (example code of spatial statistics at page 76) with R : | data(topo) library("spatial") topo.loess<-loess(z ~ x * y, topo, | degree= 2,span=0.25) |

Good morning, I am Marta Colombo,student at Politecnico,Milan. I am studying local regression models and I am using loess function. My problem is that when I have a loess object I don't know how to display the fitted surface; in fact, while in S when you have a loess object you can see it writing plot(object), in R this dosen't work. Also I'd like to know if there is something like the

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, 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.

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

I have a problem either understanding what loess is doing or that loess has a problem itself. As the x-axis variables become more concentrated on a particular point,the estimated loess tends to zero????. the examples below show what i am talking about, why is that? my intution tells me that it should tend to the mean of the variable which is been smoothed. Here's a worked up example x

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 all, I wonder which R algorithm could perform a search of local maxima in an spatial grid, in other words, having the coordinates of a map (x,y,z... up to 6 coordinates) and then the altitude/height at each point (h) (in total 7 numerical variables) I would like to localise the peaks (local maxima) of that topological surface. Which R algorithm can perform this? Thanks a lot, Josep Maria,

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.

I have data in the form of (x,y) pairs and would like to find local minima and maxima (typically the zeros of the 2nd derivative) of the y values. I looked at numericDeriv, but I don't have an "expression" per se. I looked at optim, also, but it looks like it will find only one "global" max or min. I can code up my own piecewise derivatives, but wondered if there is

I''m looking to write a javascript heavy clientside program with a something serverside backend that connects to the free maxima math program. I have extensive knowledge of ruby on rails, so I would prefer to call Maxima with ruby, but I don''t know if this is even possible. Its fairly easy to call Maxima (with a lisp implementation) using ANSI C, it is a little less easy to

Hi. Suppose I have a vector that I partition into disjoint, contiguous subvectors. For example, let v = c(1,4,2,6,7,5), partition it into three subvectors, v1 = v[1:3], v2 = v[4], v3 = v[5:6]. I want to find the maximum element of each subvector. In this example, max(v1) is 4, max(v2) is 6, max(v3) is 7. If I knew that the successive subvector maxima would never decrease, as in the example,

Dear all, suppose I have a bi-variate function f(x,y), I want to find the maxima. I define x and y vector, and get matrix z=f(x,y). how can I get which (x0,y0) makes z become the maxima? I can do two loops to get the x0 and y0, but I think there may exist a function to do this.

> On 26 Aug 2017, at 16:39, niharika singhal <niharikasinghal1990 at gmail.com> wrote: > > Hi, > > Thanks for your mail, and time > > It is not working for some arguments, when mean value is like >6. > > > case > > mc0 <- c(0.08844446,0.1744455,0.1379778,0.1209769,0.1573065,0. > 1134463,0.2074027) > > rv

Hi, Thanks for your mail, and time It is not working for some arguments, when mean value is like >6. case mc0 <- c(0.08844446,0.1744455,0.1379778,0.1209769,0.1573065,0. 1134463,0.2074027) rv <-UnivarMixingDistribution(Norm(486.4255, 53.24133), Norm(664.0713, 3.674773), Norm(669.0484, 4.101381),