search for: smoothly

Hello, can anyome tell me how to access the full script of the panel.smooth function so that I can change the thickness of the smoothing line or its colour ? All I could access is : > panel.smooth function (x, y, col = par("col"), bg = NA, pch = par("pch"), cex = 1, col.smooth = "red", span = 2/3, iter = 3, ...) { points(x, y, pch = pch, col = col, bg =

Hello, I come from a non statistics background, but R is available to me, and I needed to test an implementation of smoothing spline that I have written in c++, so I would like to match the results with R (for my unit tests) I am following http://www.nabble.com/file/p25569553/SPLINES.PDF SPLINES.PDF where we have a list of points (xi, yi), the yi points are random such that: y_i = f(x_i) +

If I have, say, five scatter points and want to connect them together into a smooth curve. I did plot(x,y,type="l"), but the graph is five segments connecting with each other, but not a smooth curve. I wonder if there is a line type that is a curve. Thanks! -- View this message in context: http://r.789695.n4.nabble.com/connecting-points-into-a-smooth-curve-tp3021796p3021796.html Sent

Hello, I have been using scatter.smooth and been trying to format my plots and been unable to adjust the line type or color of the smooth curve. When I specify col in scatter.smooth, I can adjust the color of the scatter plot points, but the line always stays black and solid. lty has no effect. When I specify par prior to calling scatter.smooth, col will change the color of the points,

Dear list, if I do smooth.spline(tmpSec, tmpT, all.knots=T) with the attached data, I get this error-message: Error in smooth.spline(tmpSec, tmpT, all.knots = T) : smoothing parameter value too small If I do smooth.spline(tmpSec[-single arbitrary number], tmpT[-single arbitrary number], all.knots=T) it works! I just don't see it. It works for hundrets other datasets, but not for

Hi, I have modified the USJudgeRatings data (available in R) to illustrate my question. # Use the first 4 variables of USJudgeRatings and add a group variable with two levels USJudgeRatings <- USJudgeRatings[,1:4] USJudgeRatings$group <- factor(c(rep(1, 22), rep(0, 21))) # I can draw a pairs graph where members of each group are drawn in different colors: pairs(USJudgeRatings[,1:4], col

I have 3 sets of Cartesian data, one is 'original' data and the other 2 are "smoothed"data. The smoothed data is the result of applying a smoothing algorithm to the original.One set of smoothed data is the 'old' algorithm and the other set is the 'new' algorithm. Does R have the capability of telling me which data is "smoother"? Example data (subsets

Hi, I am using GAMs (package mgcv) to smooth event rates in a penalized regression setting and I was wondering if/how one can select the order of the derivative penalty. For my particular problem the order of the penalty (parameter "m" inside the "s" terms of the formula argument) appears to have a larger effect on the AIC/deviance of the estimated model than the

Dear All, I've been?searching for?appropriate codes to compute the rate of change and the curvature?of ?nonparametric regression model whish was denoted by a smooth function?but?unfortunately?don't manage to?do?it. I presume that such characteristics from a smooth curve can be determined by the first and second derivative operators. The following are the example of fitting a

I have a set of data (in a matrix). I spliced a column out and parsed it as.ts (time series). I then plotted the time series but I found that it was very noisy. I wanted to smooth it out. However, I am having some problems smoothing and plotting the smoothed version. > A <- as.ts(read.table(choose.files())) > x <- as.ts(A[,10]) > plot(x) > > plot(smooth(x))

Hi, I am working with gam models in the mgcv library. My response variable (Y) is binary (0/1), and my dataset contains repeated measures over 110 individuals (same number of 0/1 within a given individual: e.g. 345-zero and 345-one for individual A, 226-zero and 226-one for individual B, etc.). The variable Factor is separating the individuals in three groups according to mass (group 0,1,2),

I have a simple question that irritatingly I haven't been able to figure out on my own. It seems that some functions from the "Pspline" package are successfully installed while others are not. The code with which I'm working is more complicated, but the following highlights my problem. If I run the following code > tt <- seq (0,1,length=20) > xt <- tt^3 > fit

Hi there, How may I smooth spline two vectors with the smoothing parameter selected by generalized maximum likelihood (GML) .? Thanks a lot. -- View this message in context: http://r.789695.n4.nabble.com/Smoothing-spline-with-smoothing-parameters-selected-by-generalized-maximum-likelihood-tp4210679p4210679.html Sent from the R help mailing list archive at Nabble.com.

I rewrote scatter.smooth to handle missing values, but I have a question about a move I had to make. Here's the code: Mscatter.smooth<-function (x, y, span = 2/3, degree = 1, family = c("symmetric", "gaussian"), xlab = deparse(substitute(x)), ylab = deparse(substitute(y)), ylim = range(y, prediction$y), evaluation = 50, ...) { if (inherits(x,

Dear all, I am new to R, and I am a biotechnologist, I want to fit a smoothing spline with smoothing parameter selected by generalized maximum likelihood. I was wondering what function implement this, and, if possible how I can find the fitted results for a certain point (or predict from the fitted spline if this is the correct language) -- View this message in context:

Dear List, Having had my appetite sufficiently whetted by Prof. Ripley's email about the new graphics capabilities in Unixes, I wanted to try them out. I updated to svn r44608, configured with the following options: R is now configured for x86_64-unknown-linux-gnu Source directory: .. Installation directory: /usr/local C compiler: gcc -O3 -g -std=gnu99

I have noticed a slightly puzzling behaviour exhibited by smooth.spline(). If I do sss <- smooth.spline(x,y) for a certain pair of data vectors x and y, and then do length(sss$x) I get the result ``18''. However if I do length(unique(x)) I get ``27''. Trying to force smooth.spline() to use more knots I tried sss <- smooth.spline(x,y,all.knots=TRUE) but again

Hi, I want to use the result from smooth.spline outside R. I take my data ,which is 180 point stored in x and y s <- smooth(x,y) I can know use to e.g. find the interpolated value at e.g. x=500 predict (s,500) My problem is, that i don't know how to implement the predict function. I have looked at literature, but i cannot connect the output of the smooth.spline() to an actual spline

I'm using R 1.7.0 on linux. With this version of R the package modreg is automatically loaded at start of session. However attempting to use predict.smooth.spline() produces Error: couldn't find function predict.smooth.spline. The function smooth.spline() is OK. What am I missing? ====================================== I.White ICAPB, University of Edinburgh Ashworth Laboratories, West

Can anyone tell me the trick for obtaining the smoother matrix from smooth.spline when there are non-unique values for x. I have the following code but, of course, it only works when all values of x are unique. ## get the smoother matrix (x having unique values smooth.matrix = function(x, df){ n = length(x); A = matrix(0, n, n); for(i in 1:n){ y = rep(0, n); y[i]=1; yi =