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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. > > help("==")...

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 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...

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:

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("[&q...

Greetings, I have been using lm to perform weighted linear regressions and resid to extract the residuals. I happened upon some class notes on the internet that described how one can specify type="pearson" in resid to extract the weighted residuals. Where is this option documented? And if you know that, what is the best way to find such documentation if you don't know that the

Is there a built-in function in R that will generate simultaneous confidence and prediction bands for linear regression? Tom -- View this message in context: http://www.nabble.com/Simultaneous-Confidence-Prediction-Bands-tp17941537p17941537.html Sent from the R help mailing list archive at Nabble.com.

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,

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

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] &

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"

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) ?

After installing rggobi, I get the following error when I try to load it: > local({pkg <- select.list(sort(.packages(all.available = TRUE))) + if(nchar(pkg)) library(pkg, character.only=TRUE)}) Loading required package: RGtk2 Error in dyn.load(x, as.logical(local), as.logical(now)) : unable to load shared library

The function wtd.var(x,w) in Hmisc calculates the weighted variance of x where w are the weights. It appears to me that wtd.var(x,w) = var(x) if all of the weights are equal, but this does not appear to be the case. Can someone point out to me where I am going wrong here? Thanks. Tom La Bone [[alternative HTML version deleted]]

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:

...rs, I wonder if there is any function which would render the value of the neighbors of the given element [i,j] of a matrix. Thanks, Rostam [[alternative HTML version deleted]] ------------------------------ Message: 9 Date: Sun, 3 Aug 2008 08:04:04 -0700 (PDT) From: Tom La Bone <booboo at gforcecable.com> Subject: [R] Will This Computer Run 64 bit Ubuntu/R? To: r-help at r-project.org Message-ID: <18799376.post at talk.nabble.com> Content-Type: text/plain; charset=us-ascii After doing some reading about 64-bit systems and software I am still somewhat uncertain about some things....

Greetings, In the following one-way ANOVA I am attempting to calculate the means of each treatment along with their 95% Tukey confidence intervals for the data shown below using a routine from the HH package. library(HH) options(digits=10) # load data treat voltage 1 130 1 74 1 155 1 180 2 150 2 159 2 188 2 126 3 138 3 168 3 110 3 160 4 34

Minitab can perform a "Parametric Distribution Analysis - Arbitrary Censoring" with one of eight distributions (e.g., weibull), giving the maximum likelihood estimates of the parameters in the distribution for a given dataset. Does R have a package that provides equivalent functionality? Thanks for any advice you can offer. Tom La Bone [[alternative HTML version deleted]]

When I run this calculation library(ISwR) library(simple.boot) data(thuesen) fit <- lm(thuesen$short.velocity~thuesen$blood.glucose) summary(fit) fit.sb <- lm.boot(fit,R=1000,rows=F) summary(fit.sb) I get the following error from the lm.boot routine: newdata' had 100 rows but variable(s) found have 24 rows I don't know what this error means. Also, this example is

Greetings, I would like to use the "boot" function to generate a bootstrap confidence interval for the slope in a SLR that has a zero intercept. My attempt to do this is shown below. Is this the correct implementation of the boot function to solve this problem? In particular, should I be doing anything with the residuals in the "bs" function (e.g., using weighted residuals)?