similar to: non linear quantile regression - Median not plotting where it should

Hi, I'm relatively new to R and need to do a quantile regression. Linear quantile regression works, but for my data I need some quadratic function. So I guess, I have to use a nonlinear quantile regression. I tried the example on the help page for nlrq with my data and it worked. But the example there was with a SSlogis model. Trying to write dat.nlrq <- nlrq(BM ~ I(Regen100^2),

I have been quantreg library for a number of projects but have just hit a snag. I am using nlrq to examine an asymptotic relationship between 2 variables at the 99th percentile. It performs as expected, however when I try to extract the coefficients along with se and significance I am running into problems. The problem is that for the nlrq regression Dat.nlrq, summary(Dat.nlrq) reports a different

Hello. I am trying to fit an asymptotic relationship (nonlinear) to some ecological data, and am having problems. I am interested in the upper bound on the data (i.e. if there is an upper limit to 'y' across a range of 'x'). As such, I am using the nonlinear quantile regression package (nlrq) to fit a michaelis mention type model. The errors I get (which are dependant on

Hello, I?m with a problem in using nonlinear quantile regression, the function nlrq. I want to do a quantile regression o nonlinear function in the form a*log(x)-b, the coefficients ?a? and ?b? is my objective. I try to use the command: funx <- function(x,a,b){ res <- a*log(x)-b res } Dat.nlrq <- nlrq(y ~ funx(x, a, b), data=Dat, tau=0.25, trace=TRUE) But a can?t solve de problem,

Please, I have a problem with nonlinear quantile regression. My data shows a large variability and the quantile regression seemed perfect to relate two given variables. I got to run the linear quantile regression analysis and to build the graph in the R (with quantreg package). However, the up part of my data dispersion seems a positive exponential curve, while the down part seems a negative

Dear all, I have a tree data matrix. For some trees I lack info about tree species, but I want to set them to be spruce. For some reason the tree species names on the remaining (non-NA) rows are changed into numbers (that I do not recognise). I guess that ifelse is not the correct function to use, but I have not found any better one in my searches. Thanks in advance! Sincerely, Tord >

Dear List, Apologies if this is a known problem ... I wasn't able to find it on the bug list, but it is a problem that does not seem to occur with a MAC build of R 2.0, so perhaps this problem has already been addressed for the future. I am getting *sporadic* errors when refitting the same model to the same data set, using nlrq() in the nlrq package. The algorithm is not stochastic, so I

Dear r-list, I've got a data base: > HData[1:10,] NumTree Site Species Date Age DBH H IdentTree 1 1 Queige Spruce 2002 184 49 33.5 Queige 1 2 2 Queige Fir 2002 NA 5 4.6 Queige 2 3 3 Queige Fir 2002 25 8 6.6 Queige 3 4 4 Queige Spruce 2002 198 47 32.5 Queige 4 5 5 Queige Fir 2002 200 59 35.3 Queige 5 6 6 Queige

Dear all, I sent an email on Friday asking about nlrq {quantreg}, but I haven't received any answer. I need to estimate the quantile regression estimators of a model as: y = exp(b0+x'b1+u). The model is nonlinear in parameters, although I can linearise it by using log.When I write: fitnl <- nlrq(y ~ exp(x), tau=0.5) I have the following error: Error in match.call(func, call = cll) :

I couldn't really find anything about this in the archives, sorry if it's been beat to death. I've got a Debian Linux box running kernel 2.0.33. The machine has two ethernet cards and does NAT between a LAN and a cable modem link. For some reason, it's got 127.0.0.1 listed for its name: root@spruce:/var/samba>cat wins.dat "SPRUCE#20" 914324043 127.0.0.1 46R

Hello everyone, I am new to R, I have been using SAS for a while. Not surprisingly, I find R much better in graphics, which is publication ready right away. Recently, I have been trying to calculate some basic statistics using R. I have a dataset of multiple rows per subject. For example: Subject Date Factor1 Factor2 Factor3 P1 0.5 1 1 3 P1 1 3 2 5 P1 2 3 5 NA ... P2 0.5 1 6 4 P2 1 2 NA 7 P2

Hello, I have a data set with 9700 records, and 7 parameters. The data were collected for a survey of forest communities. Sample plots (1009) and species (139) are included in this data set. I need to determine how species are accumulated as new plots are considered. Basically, I want to develop a species area curve. I've included the first 20 records from the data set. Point

Hi there, Can anyone please tell me if it is possible to limit parameters in nlrq() to 'upper' and 'lower' bounds as per nls()? If so how?? Many thanks in advance

In "lm" command, we can use "vcov" option to get variance-covariance matrix. Does anyone know how to get variance-covariance matrix in nlrq? Thanks, Kate [[alternative HTML version deleted]]

Dear r-help members I posted this message already yesterday, but don't know whether it reached you since I joined the group only yesterday. I would like to estimate the boxcox transformed model (y^t - 1)/t ~ b0 + b1 * x. Unfortunately, R returns with an error message when I try to perform this with the call nls( I((y^t - 1)/t) ~ I(b0 + b1*x), start = c(t=1,b0=0,b1=0), data = mydataframe)

Hi, I have set up a Fedora Core 3 machine with Samba 3.0.9 to act as a PDC file and print server. Everything works fine but I am having problems adding new machine accounts. When I try to add them manually using "/usr/sbin/useradd -g machines -d /dev/null -c "Spruce" -s /bin/false SPRUCE$", I get the error message: "invalid user name 'SPRUCE$'". I can add

All, This worked: mBW <- function( ... ) ... # matrix-valued function BaconWatts <- function(formula, mmf=mBW, # model matrix function(x, bp, g) data, plot=T, tau=0.5 ) { ... m.nl <- nlrq(y ~ b0 + mBW(x,bp,g) %*% c(b1,b2), tau=tau, start=par0, trace=T )$m ... } For some reason the following reports a failure to find the

Hi. I have data which i would want to assess the degree of agreement between two assays, e.g., to evaluate reproducibility or for inter-rater reliability. I have used the Pearson product-moment correlation coefficient. It looks good ranginging between 0.90 to 0.998. Though this looks good. I am told the Concordance correlation coefficient will give a better picture of how reproducible the assay

Dear R-helpers, This may sound simple to you, but I'm a beginner in this, so please be forgiving. I have a following problem: two analytes were measured in patient's blood on 4 occasions: ProteinA and ProteinB. How to correctly evaluate correlation between ProteinA and ProteinB? I tried: x <- data.frame(Patient.ID=rep(1:10, each=4), Visit=rep(c(1:4),10), ProteinA=rnorm(m=10,

Hi All: Package TSERIES is stated to depend on MVA. However, there is no MVA package to be found under the list of package sources. Best wishes, ANDREW tseries: Package for time series analysis Package for time series analysis with emphasis on non-linear and non-stationary modelling Version: 0.7-6 Depends: ts, mva, quadprog Date: 2001-08-27 Author: Compiled by Adrian