similar to: NAs and weighted.mean

I came across the below mis-feature/bug using match with POSIXlt objects (from strptime) in R 2.11.1 (though this appears to be an old issue). > x <- as.POSIXlt(Sys.Date()) > table <- as.POSIXlt(Sys.Date()+0:5) > length(x) [1] 1 > x %in% table # I expect TRUE [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE > match(x, table) # I expect 1 [1] NA NA NA NA NA NA NA NA

FYI, 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 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

?"=", ?"==", ?"!=", ?">=", and ?"<=" sends me to the documentation for ?help on Windows, while returning the correct documentation on Linux. Robert > version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor

Hello, I'm looking to calculate a 95% confidence interval about my estimate for a sample's weighted mean, where the calculated confidence interval would equal the t-test confidence interval of the sample in the case when all of the weights are equal. My initial thought was to simply implement a modified version of the t-test function but substituting the weighted variance and mean for the

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-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,

Is it correct (by its lack of mention in the R-Language Definition Manual) that it is impossible to create a user-defined unary operator? Ex: (This doesn't work, but it's an example of what I'm looking for) > "%PLUSONE%" <- function(x) x + 1 > %PLUSONE% 2 [1] 3 And if the above is impossible, am I limited to only the + - ~ ! unary operators for overloading? On

R-help, I'm the primary developer for an increasingly large R package with over three thousand lines of code. Unfortunately, do the complexity of the code, I sometimes am required to change several interoperating parts of the package before testing for bugs and performance. And sometimes unnoticed syntax errors slip in that cause Rcmd check / INSTALL to fail with such messages as: Error in

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-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

I'm seeing some inconsistent behavior when re-assigning values in a data frame. The first assignment turns all of the 0s in my data frame to 2s, the second fails to do so. > df1 <- data.frame(a = c(NA, 0, 3, 4)) > df2 <- data.frame(a = c(NA, 0, 0, 4)) > df1[df1 == 0] <- 2 ## Works > df2[df2 == 0] <- 2 Error: NAs are not allowed in subscripted assignments Checking an

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

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,

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' >

Greetings, I'm having some difficulties with evaluating very long expressions (Windows/Linux 2.0.1), as seen below, and would greatly appreciate any help, thoughts or work arounds. Let's say that I wanted to see what I would get if I added 1 to itself 498 times. One way of doing this would be to evaluate the expression 1+1+1+... > eval(parse(text = paste(rep(1, 498), collapse =

I'd like to report a bug (buffer overflow?) in the function sub(..., perl = TRUE) I wanted to implement the familiar perl function for removing white spaces before and after a character string: sub trimwhitespace($) { my $string = shift; $string =~ s/^\s+//; $string =~ s/\s+$//; return $string; } So in R this would (presumably) become: trimwhitespace <- function(x) { 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, I'm trying to convert times in the EST/EDT (New York) format to times in the GMT/BST (London) and UTC+9 format (Tokyo). That is, if I know what time it is in New York, what is local time in London and Tokyo? Ex: Here's the conversion from New York EST/EDT time to London's GMT/BST time zone for three days in 2007. Note that the US and London change to daylight savings on

I was recently browsing through CRAN's Finance task view to remind myself of the publicly available packages relevant to my work. As the reference manuals are all online, I am able to flip through the available functions to get an idea of the package's scope before downloading. That said, many authors have taken the time to additionally provide a useful vignette which provides a better