Displaying 20 results from an estimated 7000 matches similar to: "new("call") problem"
2005 Jan 12
2
?"=" (Windows) (PR#7504)
?"=", ?"==", ?"!=", ?">=", 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
2010 Jul 06
1
with(x, Recall()) Crash
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,
2005 Jan 07
1
Creating unary operators
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
2005 Jan 10
1
new("call") problem (PR#7490)
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
2004 Dec 09
1
Modulus Bug (PR#7409)
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
2006 May 09
0
Typo in getAllMethods() (PR#8848)
The function getAllMethods in the methods package uses the non-existent
function packageName where I believe the function getPackageName was
intended.
For example:
> getAllMethods("formula")
Error in sprintf(gettext(fmt, domain = domain), ...) :
could not find function "packageName"
The patch to RMethodUtils.R is here:
238c238
<
2005 Jan 05
2
Rcmd check Error help
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
2005 Jan 12
0
Function Body / Formals Bug (PR#7495)
I'd like to report two bugs in the R function definitions, the first
slightly problematic, the second likely unimportant. Both of these are
reproducible on both Windows and Linux, R 2.0.1
First Bug:
> a <- new("function")
> formals(a) <- alist(a=, b=, c=, d=)
> a
function (a, b, c) ## the d argument disappeared!
> formals(a) <- alist(a=, b=, c=, d=)
> a
2010 Jun 23
1
NAs and weighted.mean
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,
2010 Jun 29
2
POSIXlt matching bug
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
2005 Jan 08
1
sub( , perl = TRUE) overflow (PR#7479)
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 <-
2006 Nov 29
1
Removing terms from formula
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
2005 Jan 24
6
Very Long Expressions
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 =
2008 Jul 08
1
split.Date
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
2008 Sep 12
1
match and incomparables
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'
>
2006 Sep 29
1
Plotting text with lattice
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,
2008 May 16
1
var/sd and NAs in R2.7.0
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
2006 Jan 12
1
.leap.seconds
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
2008 Apr 28
2
time zone conversion
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
2006 Sep 20
1
seq.Date not accepting NULL length.out (PR#9239)
There seems to be a bug in seq.Date such that it will not allow the user
to pass in length.out =3D NULL, despite the fact that this is the =
default
argument.
For example:
> dt1 <- as.Date("2004-12-31")
> dt2 <- as.Date("2005-12-31")
> seq.Date(dt1, dt2, length.out =3D NULL, by =3D "month")
Error in seq.Date(dt1, dt2, length.out =3D NULL, by =3D