similar to: Installing smooth.spline command

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

Hey, R-listers I was recommended to try using smooth.spline function for estimating 2-Dimensinal curve given a data set. So will you please tell me where to get this R function? Or which package provides this function? Thanks for your point. Fred

Hi, I have three original curves as follows, n<-seq(20,200,by=10) t<-c(0.1138, 0.1639, 0.2051, 0.2473, 0.2890, 0.3304, 0.3827, 0.4075, 0.4618, 0.4944, 0.5209, 0.5562, 0.5935, 0.6197, 0.6523, 0.6771, 0.6984, 0.7209, 0.7453) es<-c(0.3682, 0.4268, 0.5585, 0.6095, 0.7023, 0.7534, 0.8225, 0.8471, 0.8964, 0.9098, 0.9371, 0.9514, 0.9685, 0.9747, 0.9812, 0.9859, 0.9905, 0.9923, 0.9940)

I use smooth.spline(x, y) in package modreg and I would like to get value of penalized log likelihood and preferable also its two parts. To make clear what I am asking for (and make sure that I am asking for the right thing) I clarify my problem trying to use the same notation as in help(smooth.spline): I want to find the natural cubic spline f(x) such that L(f) = \sum_{k=1}{n} w[k](y[k] -

I've had a play with this and, due to my own short-comings, remain none the wiser. In particular, I'm not sure what value of 'spar' is consistent with the magic lambda=1/1600 for quarterly data. I initially interpreted spar as lambda and tried setting spar=1/1600. This results in almost no smoothing while spar=1600 causes an error. The smooth.spline function seems to want

Recent sensible changes to the dynload mechanism have made an old problem resurface: how should we deal with packages which contain Fortran code and may need to be linked against additional libraries such as -lf2c? The current consensus is that extra Fortran libraries maybe needed are handled via the make variable FLIBS, and that `-lf2c' or `-lg2c' are added by default if g77 is used.

Hi there, I've successfully installed und updated some contributed R packages (R version 1.2.2 on Debian Linux, logged in as a normal user), but I'm wondering about this: [...] Could not open /usr/lib/R/doc/html/packages.html at /usr/lib/R/share/perl/R/Rdlists.pm line 238. /usr/lib/R/bin/INSTALL: /usr/lib/R/doc/html/search/index.txt: Permission denied /usr/lib/R/bin/INSTALL: LibIndex:

I would appreciate getting a clarification of /usr/lib/R/library vs /usr/local/lib/R/site-library. I am running R 1.8 on Debian Linux. On one occasion doing update.packages() resulted in versions of one or more libraries being placed in one of these directories without removing the old version, and the old version took precedence over the new. I'm sorry I did not save the steps I used. I

Hello I am running RH7.2 with samba-common-2.2.1a-4 and samba-client-2.2.1a-4 and for printing I use a windows printer (SMB share). Previously this printserver was running under WinNT, and for the installation of this printer I used the installation wizard under RH7.2. Furthermore no modifications were made to the SMBclient (smb.conf) nor the WinNT printserver. Recently, they upgraded the

Readers, Using version 251 I tried the following command: lm(y~a+b,data=datafile) Resulting in, inter alia: ... coefficients (intercept) a 1.2 3.4 Packages installed: acepack ace() and avas() for selecting regression transformations adlift An adaptive lifting scheme algorithm akima Interpolation of irregularly spaced

Hi all, I have created a new edit mode file for the editor JEdit (http://www.jedit.org) based on the file from Zed Shaw. The difference between his and my file is that mine can be used as a spell checker. It does not highlight all functions but provides syntax highlighting for the base package (keyword types 1 and 2) and 26 additional packages (boot, class, cluster, ctest, eda, foreign, grid,

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 have the feeling that this is a dumb question but oh well ... When I have an object x with several attributes, y <- x will make a copy of x with all the attributes included. What if I just want y to get the value assigned but nothing else (no class, names, dim or other attributes)? Johann -- Phone/FAX/Adr/PGPkey ... at http://www.ai.univie.ac.at/~johann

Hello, I've noticed that SPLUS seems to have a function for evaluating derivative matrices of splines. I've found the R function that evaluates matrices from 'smooth.spline'; maybe someone has written something to do the same with smooth.basis? regards, s

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

I like what smooth.spline does but I am unclear on the output. I can see from the documentation that there are fit.coef but I am unclear what those coeficients are applied to.With spline I understand the "noraml" coefficients applied to a cubic polynomial. But these coefficients I am not sure how to interpret. If I had a description of the algorithm maybe I could figure it out but as it

Is there any way to identify or infer the inflection points in a smooth spline object? I am doing a comparison of various methods of time-series analysis (polynomial regression, spline smoothing, recursive partitioning) and I am specifically interested in obtaining the julian dates associated with the inflection points inferred by the various models. Tyler e.g.

Hello. I'm getting an unexpected result when running smooth.spline(). Here is a simple example that replicates the error I'm getting: > aa <- c(1, 2, 3, 8, 8, 8, 8, 8, 8, 8, 8, 8, 12, 13, 14) > bb <- 1:length(aa) > plot(aa, bb) > smooth.spline(aa, bb) Error in smooth.spline(aa, bb) : need at least four unique 'x' values As you can see from the example, my

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 =

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