similar to: Weighted Mean Confidence Interval

As a very new user of R, and a user of S with very limited experience, I thought I would start humbly with a simple t.test. The response to the command sequence sample <- scan() 4 5 6 7 8 9 t.test(sample,mu=20) rather surprised me, because of the idiosyncratic choice of a confidence interval for the mean of the population from which the sample was drawn. I asked my colleague Bob Henery

I'd like to learn the process of revising R functions & packages and then submitting proposed patches to the R Core team. Would someone be willing to mentor me through one example? For starters, consider an example. I'd like to revise the t.test function to return the stderr value to the user. We only need to change the "rval" in the third-from-the end line of

I have installed Sweave as recommended. http://lifeasclay.wordpress.com/tag/sweave/. Placing a test.Rnw file, for instance, in "/Library/Frameworks/R.framework/Versions/2.15/Resources/library/utils/Sweave" generates test.tex files that generate pdf files with LaTeX. To be honest, placing Rnw files seem to make sense. The "exams" package does require not this step. It uses

R-developers, In version R 2.11.0, weighted.mean was changed such that: > weighted.mean(NA, na.rm=TRUE) [1] 0 rather than NaN as in previous versions of R. I see a note in the NEWS file indicates that weighted.mean was changed "so an infinite value with zero weight does not force an NaN result." In case the side effect of returning 0 rather than NaN in this case was unintentional,

Hi, I am doing double for loops to calculate SDs with some weights and wondering if I can get rid of the outer for loop as well. I made a simple examples which is essentially what I am doing. Thanks for your help! -Young #------------------------------------------------------ # wtd.var is Hmisc package # you can replace the 3 lines inside for loop as # sdx[i,] =

I have found a solution to the new("call") problem that I believe produces the correct behavior for the default call object, and am also reclassifying this as a bug, as I believe the current behavior to be incorrect. Recap, the following error occurs: > new("call") Error in print("<undef>"()) : couldn't find function "<undef>" It looks

R-help, Given a simple linear model, say lm(x ~ y + z), I would like to remove model terms that are factors with only one level. Thus, if 'z' were a factor with only one level, lm(x ~ y + z) becomes lm(x ~ y + 1). Likewise, if both 'y' and 'z' are one-level factors, then the resulting calculation is simply lm(x ~ 1). Unfortunately, I have not been able to come up with an

FYI,=20 I've tried posting the below message twice to the bug tracking system, once by email (below), and the second time 5 days later directly to the bugs.r-project.org website. As far as I can tell, the bug tracking system hasn't picked this up. Also it looks like the latest "incoming" bug is dated 25 May 2008, so perhaps others are having difficulty as well. (cc: r-bugs)

Hello, I wanted to suggest that the below method for split.Date be added to the base library to significantly speed up splits with values of class Date. In the below example I show a speed improvement of 175x for 1000 data points. On a vector of size 1e6, the time difference was 22 minutes for split.default versus 0.3 seconds for the split.Date function below (!). Note that this improvement will

Shouldn't this work? > .a <- 5 > exists(".a") [1] TRUE > getAnywhere(".a") Error in exists(x, envir, mode, inherits) : invalid first argument getAnywhere doesn't seem to like the "." prefix. Is this a bug? Thanks, Robert Robert McGehee Geode Capital Management, LLC 53 State Street, 5th Floor | Boston, MA | 02109 Tel: 617/392-8396

Hello, I was playing around with the newly implemented 'incomparables' argument in 'match' and realized the argument does not behave anything like I expected. Can someone explain what is going on here? Sorry if I'm misreading the documentation. > match(1:3, 1:3, incomparables=1) [1] NA 2 3 # This seems right, the 1 in 'x' is 'incomparable' >

I wish the t.test function in stats would return the standard error. It would be nicer for students if R simply reported the standard error used to calculate the t value. I trolled for this in r-help and got no answers, which I interpreted to mean that this is boring but possibly not wrong. Hopefully. I believe only simple changes are needed. In the source code src/library/stats/t.test.R file:

RE t.test in R I objected a day or two ago to the behaviour of the one-sample t.test in R where it is easy to generate a "confidence interval" for the mean of the population which does not contain the sample mean, in the case where the null hypothesis is rejected. It now appears that the same behaviour is latent in the code for the two-sample version of this test. The relevant lines

Hello, I've decided to take the leap and try my hand at the lattice package, though I am getting stuck at what one might consider a trivial problem, plotting text at a point in a graph. Apologies in advance if (that) I'm missing something extremely basic. Consider in base graphics: > plot(1:10) > text(2, 4, "Text") In the above you will see text centered at the point (2,

R Developers, 1000000000000000000 %% 11 [1] -32 I now understand that integers cannot be larger than .Machine$integer.max, but because the above produces a result than is patently wrong instead of an error, I'm reporting this as a bug. Thank you for the incredible contributions all of you have made in developing the R platform. Best, Robert Robert McGehee Geode Capital Management, LLC 53

R-developers, I've encountered another perl library regex bug that causes a segmentation faults on my Linux/Windows R session. I reduced the script to the snippet below. (Apologies if this was fixed with bug 7479, but this bug seems quite different). string <- paste(rep("=", 10000), collapse = " ") crash <- function(x) { for (i in 1:5) { x <-

I glanced at the .leap.seconds object and noticed that it has not been updated for the most recent leap second that occurred 2005 December 31, 23h 59m 60s. See the IERS bulletin here: http://hpiers.obspm.fr/iers/bul/bulc/bulletinc.dat Moreover, after a more careful glance at the .leap.seconds object, I noticed that there are two incorrect entries. First, there was not a leap second on 1986 June

Hello all, I just upgraded to R 2.7.0 and found that the behavior of 'var' and 'sd' have changed in the presence NAs (this wasn't explicit in the NEWS file, though I see it probably has to do with the change for cor/cov). Anyway, I just want to make sure that it was intentional to produce an error when there was all NAs and na.rm=TRUE, rather than returning an NA (like R

R-devel, I discovered a segfault in my R code that boiled down to my incorrect use of the Recall() function embedded within a with() function. Since segfaults are generally bad things, even when it's the user's fault for writing nonsense code, I thought I'd pass along the offending code. I've tested the crash on R 2.11.1 (on Linux and Mac), but not in devel versions of R. HTH,

Hello, I have a system command that occasionally fails and writes output to standard error, which R will print to the screen when ignore.stderr = FALSE. For example: > system("BadCommand") sh: line 1: BadCommand: command not found I would like to know if the above command fails, and can presumably do this by parsing the stderr message that R prints to the screen. My (hopefully