search for: gforcec

Displaying 20 results from an estimated 43 matches for "gforcec".

Did you mean: gforce
2017 Oct 25
4
Problem Subsetting Rows that Have NA's
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("==")...
2017 Oct 24
5
Problem Subsetting Rows that Have NA's
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
2017 Oct 25
0
Problem Subsetting Rows that Have NA's
> 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...
2009 May 21
4
Re placing a "+" in a string
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:
2017 Oct 25
0
Problem Subsetting Rows that Have NA's
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...
2007 Oct 17
1
Documentation for Pearson Residuals
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
2008 Jun 17
1
Simultaneous Confidence/Prediction Bands
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.
2009 Feb 01
2
Extracting Coefficients and Such from mle2 Output
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,
2008 Sep 24
1
How do I Convert "<1" to the number 1?
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
2008 Oct 14
6
Doing a Task Without Using a For Loop
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] &
2009 Jan 23
2
Dates in Common
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"
2010 Mar 28
3
Ellipse that Contains 95% of the Observed Data
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) ?
2007 Mar 18
2
Problem Loading rggobi package
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
2007 May 31
3
Problem with Weighted Variance in Hmisc
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]]
2008 Mar 12
7
Specifying relative position of text in a plot
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:
2008 Aug 05
0
P values in non linear regression and singular gradients using nls
...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....
2007 Apr 16
1
Difficulties Using glht.mmc to Calculate Tukey Intervals for Means
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
2007 Jul 25
1
Minitab Parametric Distribution Analysis in R
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]]
2007 Oct 29
1
lm.boot function gives error
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
2007 Nov 06
0
Bootstrap CI of Slope in a Weighted Simple Linear Regression
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)?