similar to: cubic spline smoothers with heterogeneous variances

Hello. I want to estimate the predicted values and standard errors of Y=f(t) and its first derivative at each unique value of t using the smooth.spline function. However, the data (plant growth as a function of time) show substantial heterogeneity of variance since the variance of plant mass increases over time. What is the consequence of such heterogeneity of variance in terms of bias in the

Greetings, I would like to use a cubic spline or smoother to model the fixed effects within nlme. So far the only smoother I have been able to get to run successfully in nlme is smooth(). I tried smooth.spline: fixed=list(lKa~1,lCL~smooth.spline(BSA, df=3)) the error I got was the following. Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid

Hello. I want to fit a smoother spline (or an equivalent local regression method) to a series of data in which the initial value of the 1st derivative (slope) is constrained to a specific value. Is it possible to do this? If so, how? Bill Shipley [[alternative HTML version deleted]]

Dear all, I want to use smooth.spline to construct a cubic smoothing spline and its first derivative to my data. However, the predict.smooth.spline does not seem to provide a SE for both the fitted values and their derivatives. How should I calculate it? Thank you very much, Bingzhang

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 =

Hello,        Dr. Simon Wood told me how to force a cubic spline passing through a point. The code is as following. Anyone  who knows how I can change the code to force the first derivative to be certain value. For example, the first derivative of the constrained cubic spline equals 2 at point (0, 0.6).        I really appreciate your help!        Thanks!                 Best             Victor   

Hello everyone,          Dr. wood told me that I can adapting his example to force cubic spline to pass through certain point.          I still have no idea how to achieve this. Suppose we want to force the cubic spline to pass (1,1), how can I achieve this by adapting the following code? # Penalized example: monotonic penalized regression spline ..... # Generate data from a monotonic truth.

Dear all, Anybody knows about the difference amongst various spline smoothers, specifically in R, 'bs' (by default a cubic spline), 'ns', smooth.spline with roughness penalty along with many other smoothers? I've consulted serveral books like 'S-plus Guide to Statistics' by Mathsoft, Ripley and Venables' Splus book and Eubank's Spline Smoothing, but I was still

Hello. I am trying to fit a non-rectangular hyperbola function to data of photosynthetic rate vs. light intensity. There are 4 parameters that have to be estimated. I find the nls function very difficult to use because it often fails to converge and then gives out cryptic error messages. I have tried playing with the control parameters but this does not always help. Is there another

Suppose I have two var x and y,now I want to fits a natural cubic spline in x to y,at the same time create new var containing the smoothed values of y. How can I get it?

Dear all, I'm a bit stunned by the behaviour of a gam model using cyclic P-spline smoothers. I cannot provide the data, as I have about 61.000 observations from a time series. I use the following model : testgam <- gam(NO~s(x)+s(y,bs="cs")+s(DD,bs="cs")+s(TT),data=Final) The problem lies with the cyclic smoother I use for seasonal trends. The variable Final$y is a

Hi, Everyboday, Does anybody know how to get the first derivative form the cubic natural spline matrix? Thanks !!! Jassy ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Nuoo-Ting (Jassy) Molitor, MS, Junior Statistician Division of Biostatistics Department of Preventive Medicine University of Southern California Tel: (323) 442-2584 1540 Alcazar St., CHP-210F Fax:

Dear R help list, I'm fitting a 'variable coefficient model' in the MGCV package and I want to plot the different smoothers I get for each factor level in one graph. So, I do something like this to fit the gam: Mtest <- gam(outcome ~ s(age, by=as.numeric(gender==0)) + s(age,by=as.numeric(gender==1))+factor(Gender)) Then I can plot the smoother for gender=0: plot(Mtest,select=1)

re: Cubic splines in package "mgcv" I don't have access to Gu (2002) but clearly the function R(x,z) defined on p126 of Simon Wood's book is piecewise quartic, not piecewise cubic. Like Kunio Takezawa (below) I was puzzled by the word "cubic" on p126. As Simon Wood writes, this basis is not actually used by mgcv when specifying bs="cr". Maybe the point is

Hello everone,            Anyone who knows how to force a cubic smoothing spline to pass through a particular point?            I found on website  someone said that we can use "cobs package" to force the spline pass through certain points or impose shape           constraints (increasing, decreasing). However,  this package is using  B-spline and can only do linear and quadratic

Hi, all I am trying to get the basis matrix and penalty matrix for natural cubic splines. In the "splines" package of R,"ns" can generate the B-spline basis matrix for a natural cubic spline. How can I get the basis matrix and penalty matrix for natural cubic spline. Thanks a lot! Lee [[alternative HTML version deleted]]

Hallo, I'm facing a problem and I would really appreciate your support. I have to translate some Matalb code in R that I don't know very well but I would like to. I have to interpolate 5 point with a cubic spline function and then I expect my function returns the Y value as output a specific X value inside the evaluation range. Let's suppose that: 1- *X = [-10, -5, 0, 5, 10]* 2

(corrected version of previous posting) I fit a GAM to turtle growth data following methods in Limpus & Chaloupka 1997 (http://www.int-res.com/articles/meps/149/m149p023.pdf). I want to obtain figures similar to Fig 3 c & f in Limpus & Chaloupka (1997), which according to the figure legend are "expected size-at-age functions" obtained by numerically integrating

I have used LME to fit a mixed effects model on my data. The data has 274 subjects with 1 to 6 observations per subject. Time is not linearly associated with the outcome, so I used ns to fit a natural cubic spline with 3 auto knots. Subject and the natural cubic time of spline are both treated as random effects. This model has run without any problem, but now I would like to plot trajectories for

Dear all, Is the restricted cubic spline function working properly in the glm model? We used glm(y~rcs(x,5), family=binomial) but it seems that for some theoretical reasons the rcs, restricted cubic spline function can not be fitted by a glm function. Is this correct? Regards, Jan ((Originally, we used lrm(y~ rcs(x,5)) but we couldn't find how to derive the AIC value of the fitted model.