similar to: Simultaneous Confidence/Prediction Bands

I know this is easy, but I am stumped: > gsub("0","K","8.00+00") [1] "8.KK+KK" > gsub("+","K","8.00+00") Error in gsub("+", "K", "8.00+00") : invalid regular expression '+' In addition: Warning message: In gsub("+", "K", "8.00+00") : regcomp error:

To my understanding, a confidence interval typically covers a single valued parameter. In contrast, a confidence band covers an entire line with a band. In regression, it is quite common to construct confidence and prediction bands. I have found that many people are connecting individual confidence/prediction interval values produced with predict(object,sd.fit=T,type="conf/pred") and

This has every appearance of being a bug. If it is not a bug, can someone tell me what I am asking for when I ask for "x[x[,2]==0,]". Thanks. > #here is the toy dataset > x <- rbind(c(1,1),c(2,2),c(3,3),c(4,0),c(5,0),c(6,NA), + c(7,NA),c(8,NA),c(9,NA),c(10,NA) + ) > x [,1] [,2] [1,] 1 1 [2,] 2 2 [3,] 3 3 [4,] 4 0 [5,] 5 0

On 10/25/2017 4:38 AM, Ista Zahn wrote: > On Tue, Oct 24, 2017 at 3:05 PM, BooBoo <booboo at gforcecable.com> wrote: >> This has every appearance of being a bug. If it is not a bug, can someone >> tell me what I am asking for when I ask for "x[x[,2]==0,]". Thanks. > You are asking for elements of x where the second column is equal to zero. > >

Assume that I have the dataframe "data1", which is listed at the end of this message. I want count the number of lines that each person has for each year. For example, the person with ID=213 has 15 entries (NinYear) for 1953. The following bit of code calculates NinYear: for (i in 1:length(data1$ID)) { data1$NinYear[i] <- length(data1[data1$Year==data1$Year[i] &

> On Oct 25, 2017, at 6:57 AM, BooBoo <booboo at gforcecable.com> wrote: > > On 10/25/2017 4:38 AM, Ista Zahn wrote: >> On Tue, Oct 24, 2017 at 3:05 PM, BooBoo <booboo at gforcecable.com> wrote: >>> This has every appearance of being a bug. If it is not a bug, can someone >>> tell me what I am asking for when I ask for "x[x[,2]==0,]". Thanks.

The mle2 function (bbmle library) gives an example something like the following in its help page. How do I access the coefficients, standard errors, etc in the summary of "a"? > x <- 0:10 > y <- c(26, 17, 13, 12, 20, 5, 9, 8, 5, 4, 8) > LL <- function(ymax=15, xhalf=6) + -sum(stats::dpois(y, lambda=ymax/(1+x/xhalf), log=TRUE)) > a <- mle2(LL,

I'm using the package 'locfit' for nonparametric regression. This package contains the function 'scb' to compute simultaneous confidence bands. The variance of the data is unknown. Up to now I compute a fit with 'locfit'. Afterwards an estimate of the residual variance is computed by the function 'rv'. The weights in the 'scb'-function are set 1/sigma^2

I have two collections of dates and I want to figure out what dates they have in common. This is not giving me what I want (I don't know what it is giving me). What is the best way to do this? Tom > data1 [1] "1948-02-24 EST" "1949-04-12 EST" "1950-05-29 EDT" "1951-05-21 EDT" [5] "1951-12-20 EST" "1953-01-22 EST"

Is there an elegant way in R to change a number reported as a less-than number in text format, "<1" for example, to the numeric equivalent 1? I have been trying to use as.numeric, but have not come up with anything clever yet. Tom -- View this message in context: http://www.nabble.com/How-do-I-Convert-%22%3C1%22-to-the-number-1--tp19651018p19651018.html Sent from the R help

Dear list, I'm a bit perplexed why the 95% confidence bands for the predicted probabilities for units where x=0 and x=1 overlap in the following instance. I've simulated binary data to which I've then fitted a simple logistic regression model, with one covariate, and the coefficient on x is statistically significant at the 0.05 level. I've then used two different methods to

I can take the results of a simulation with one random variable and generate an empirical interval that contains 95% of the observations, e.g., x <- rnorm(10000) quantile(x,probs=c(0.025,0.975)) Is there an R function that can take the results from two random variables and generate an empirical ellipse that contains 95% of the observations, e.g., x <- rnorm(10000) y <- rnorm(10000) ?

On Tue, Oct 24, 2017 at 3:05 PM, BooBoo <booboo at gforcecable.com> wrote: > This has every appearance of being a bug. If it is not a bug, can someone > tell me what I am asking for when I ask for "x[x[,2]==0,]". Thanks. You are asking for elements of x where the second column is equal to zero. help("==") and help("[") explain what happens when

I have two questions related to plotting predicted values for a linear mixed model using xyplot: 1: With a groups= argument, I can't seem to get the key to appear inside the xyplot. (I have the Lattice book, but don't find an example that actually does this.) 2: With lme(), how can I generate confidence bands or prediction intervals around the fitted values? Once I get them, I'd

Hello, I am trying to plot confidence bands on the mean and prediction bands for the following nonlinear regression, using maximum likelihood via optim. A toy example with data and code of what I am trying to accomplish is: VOL<-c(0.01591475, 1.19147935 ,6.34102460, 53.68809287, 91.90143074, 116.21397007, 146.41843056, 215.64535337, 256.53149673, 315.73609232) Age <-c(1.622222, 2.833333

I'm using nonparametric regression (packeges ksmooth and ks). My question: is there any way to compute confidence bands (or variability bands) with R. Confidence bands for functions are intervals [CLO(x);CUP(x)] such that with probability 1-alpha the true curve is covered by the band [CLO(x);CUP(x)]. Thanks very much for any help you can offer. Michael G??lger

What is the simplest way to specify the location of text in a scatter plot (created using the plot function) in relative terms rather than specific x-y coordinates? For example, rather than putting text at (300,49) on a plot, how do I put it 1/10 of the way over from the y axis and 1/2 of the way up from the x axis? Thanks. Tom -- View this message in context:

Hello Fellow R Users,I have spent the last week trying to find a work around to this problem and I can't seem to solve it. I simply want to plot my GEE model result with 95% confidence bands. I am using the geepack package to run a basic GEE model involving nestling weights, to a Gaussian distribution, with "exchangeable" error structure. I am examining how nestling weight varies

Dear all, I am trying to plot KM curves with confidence bands with function survplot under package rms. However, the following codes do not seem to work. The KM curves are produced, but the confidence bands are not there. Any insights? Thanks in advance. library(rms) ########data generation############ n <- 1000 set.seed(731) age <- 50 + 12*rnorm(n) label(age) <- "Age"

I am trying to find a confidence band for a fitted non-linear curve. I see that the predict.nls function has an interval argument, but a previous post indicates that this argument has not been implemented. Is this still true? I have tried various ways to extract the interval information from the model object without success. My code is: Model.predict <- predict(My.nls.model,