similar to: Unable to run smoother in qplot() or ggplot() - complains about knots

Hi, I am using ggplot2 to with the following code: gmathk2 <- qplot(time,math,colour=Kids,data=kids.ach.lm.k5,geom="smooth",method="lm",formula=y~ns(x,1)) + opts(title="Smoother Plot: Math K-5") + xlab("Time") + ylab("Math") + scale_colour_brewer(pal="Set1"); gmathk2 This plots all the smoother for all the x values. What I'd like

Dear List, I'm using GAMs in a multiple imputation project, and I want to be able to combine the parameter estimates and covariance matrices from each completed dataset's fitted model in the end. In order to do this, I need the knots to be uniform for each model with partially-imputed data. I want to specify these knots based on the quantiles of the unique values of the non-missing

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, all I want to choose same knots in GAM for 10 different studies so that they has the same basis function. Even though I choose same knots and same dimensions of basis smoothing, the basis representations are still not same. My command is as follows: data.gam<-gam(y~s(age,bs='cr',k=10)+male,family=binomial,knots=list(age=seq(45,64,length=10))) What is my mistake for choice of

Hi I was wondering if anyone knew how to work out the number of knots that should be applied to each variable when using gams in the mgcv library? Any help or references would be much appreciated. Thanks Kathryn Baldwin

Hi, I have fit a series of ols() models, by group, in this manner: l <- ols(y ~ rcs(x, 4)) ... where the series of 'x' values in each group is the same, however knots are not always identical between groups. The result is a table of 'coefs' derived from the ols objects, by group: group Intercept top top' top'' 1 6.864 0.01 2.241 -2.65

Hi r-helpers. Please forgive my ignorance, but I would like to plot a smoothing spline (smooth.spline) from package "stats", and show the knots in the plot, and I can't seem to figure out where smooth.spline has located the knots (when I use nknots). Unfortunately, I don't know a lot about splines, but I know that they provide me an easy way to estimate the location of local

Hi All, I am trying to figure out how to get the position of the knots in a pspline used in a cox model. my.model = coxph(Surv(agein, ageout, status) ~ pspline(x), mydata) # x being continuous How do I find out where the knot of the spline are? I would like to know to figure out how many cases are there between each knot. Best, Federico -- Federico C. F. Calboli Department of Epidemiology

Hi all, Cameron and Trivedi in their 1998 Regression Analysis of Count Data refer to NB1 and NB2 NB1 is the negative binomial model with variance = mu + (alpha * mu^1) yielding (1+alpha)*mu NB2 sets the power to 2; hence, variance = mu + (alpha*mu^2) I think that NB2 can be requested via negbin2<-glm(hhm~sex+age,family=quasi(var="mu^2",link="log")) Is

Hi, Let's say I have the following data: > a=matrix(c(1,2,4,4,2,1,1,2,4),nrow=3,byrow=T) > a [,1] [,2] [,3] [1,] 1 2 4 [2,] 4 2 1 [3,] 1 2 4 What syntax should I use to get R to tell me the column that corresponds to the maximum value for each row? For my example, I would like to get a vector that says 3, 1, 3 because the maximum value for row 1 is

Hi, I am having trouble with syntax for a for loop. Here is what I am trying to do. class=c(rep(1,3),rep(2,3),rep(3,3)) out1=rnorm(length(class)) out2=rnorm(length(class)) out3=rnorm(length(class)) data=data.frame(class,out1,out2,out3) dat.split=split(data,data$class) for(i in 1:3){ sub[i]=dat.split[i] } However, the for loop doesn't work. I want to assign each split to a different

Hhm. OK, but it doesn't work. The ups goes offline and starts back immediately. Even if ups.delay.start is set 0. I will try another test circuit after using the upsrw command. -----Ursprüngliche Nachricht----- Von: Nut-upsuser Im Auftrag von Roger Price Gesendet: Sonntag, 27. Oktober 2019 11:11 An: Nut-upsuser at alioth-lists.debian.net Betreff: Re: [Nut-upsuser] Problems with NUT on

Hi, I am trying to write a loop to recode my data from -999 to NA in R. What's the most efficient way to do this? Below is what I'm presently doing, which is inefficient. Thanks, Chris dat0 <- read.table("time1.dat") colnames(dat0) <- c("e1dq", "e1arcp", "e1dev", "s1prcp", "s1nrcp", "s1ints",

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

* Please cc me if you reply as I am a digest subscriber * Hi, I am wondering how I can run a multilevel survival model in R? Below is some of my data. > head(bi0.test) childid famid lifedxm sex age delta 1 22.02 22 CONTROL MALES 21.36893 0 2 13.02 13 MAJOR MALES 21.18001 0 3 64.02 64 CONTROL MALES 20.09377 0 4 5.02 5 CONTROL FEMALES

Hi, I am trying to fit the following model: sr.reg.s4.nore <- survreg(Surv(age_sym4,sym4), as.factor(lifedxm), data=bip.surv) Where age_sym4 is the age that a subject develops clinical thought problems; sym4 is whether they develop clinical thoughts problems (0 or 1); and lifedxm is mother's diagnosis: BIPOLAR, MAJOR DEPRESSION, or CONTROL. I am interested in whether or not

How can I do automatic knot selection while fitting piecewise linear splines to two variables x and y? Which package to use to do it simply? I also want to visualize the splines (and the scatter plot) with a graph. Anupam [[alternative HTML version deleted]]

Hi, I have my data the following way: y A B C 0 1 1 2 0 1 2 1 1 1 1 2 0 1 1 2 1 1 1 2 1 1 2 1 0 1 2 2 . . . And so on. How can I make my data look like the following: y A B C 2 1 1 2 1 1 2 1 0 1 2 2 . . . In other words how can I change my ungrouped cases into grouped cases? Thanks! Chris [[alternative

>>>>> Anupam Tyagi >>>>> on Tue, 9 Jul 2024 16:16:43 +0530 writes: > How can I do automatic knot selection while fitting piecewise linear > splines to two variables x and y? Which package to use to do it simply? I > also want to visualize the splines (and the scatter plot) with a graph. > Anupam NB: linear splines, i.e. piecewise

Hi, I would like to fit an (interpolating) spline to data where the derivatives at the endpoints of the interval are nonzero, thus the natural spline endpoint-specification does not make sense. Books (de Boor, etc) suggest that in this case I use not-a-knot splines. I know what not-a-knot splines are (so if I were solving for the coefficients directly I knew how to do this), but I don't