Displaying 20 results from an estimated 50000 matches similar to: "testInstalledBasic / testInstalledPackages"
2008 Feb 13
0
RFC for package PopCon: a popularity contest for R and packages
Hello all,
I've developed a prototype package called PopCon (short for popularity
contest), a package for tracking the popularity of R and its packages.
I'd like this work to be similar in spirit to the Debian package
popularity-contest: http://popcon.debian.org/.
Once Popcon is loaded, it captures two kinds of information from the
user and stores it into a cache: the names of the
2008 Feb 14
0
RFC for package PopCon: a popularity contest for R and packages
(I posted this to the R-devel list yesterday, but I thought others on
this list would be interested, so sorry for those who get it twice.)
Hello all,
I've developed a prototype package called PopCon (short for popularity
contest), a package for tracking the popularity of R and its packages.
I'd like this work to be similar in spirit to the Debian package
popularity-contest:
2024 Oct 17
2
Consider getNamespaceVersion() returning a numeric_version
I mean the `numeric_version` object not a numeric (double/int).
Basically to protect me from myself I'd prefer not to have to remember
to wrap `getNamespaceVersion()` with `as.package_version()`.
I suspect a grep of CRAN may highlight others who are erroneously
comparing character objects rather than a comparison between a
`numeric_version` object and a character.
Tim
On 17/10/2024
2024 Oct 17
1
Consider getNamespaceVersion() returning a numeric_version
On 17/10/2024 13:42, Tim Taylor wrote:
> I mean the `numeric_version` object not a numeric (double/int).
> Basically to protect me from myself I'd prefer not to have to remember
> to wrap `getNamespaceVersion()` with `as.package_version()`.
>
> I suspect a grep of CRAN may highlight others who are erroneously
> comparing character objects rather than a comparison between
2013 Oct 03
1
version comparison puzzle
Can anyone explain what I'm missing here?
max(pp1 <- package_version(c("0.99999911.3","1.0.4","1.0.5")))
## [1] ?1.0.4?
max(pp2 <- package_version(c("1.0.3","1.0.4","1.0.5")))
## [1] ?1.0.5?
I've looked at ?package_version , to no avail.
Since max() goes to .Primitive("max")
I'm having trouble figuring out
2010 Jun 17
0
Modifyiing R working matrix within "gee" source code
Dear all,
I am working on modifying the R working matrix to commodate some other
correlations that not included in the package. I am having problem to locate
where the R matrix are defined for regular matrices, i.e. independence,
exchangeable, AR and unstructure. it might have something within
.C("Cgee",but don't understand it well enough to know. Can you anyone
help?
/*gee source
2024 Oct 17
1
Consider getNamespaceVersion() returning a numeric_version
On 17 October 2024 at 12:38, Tim Taylor wrote:
| Would R-Core be receptive to having getNamespaceVersion() return a
| numeric_version object instead of a named character?
Is this good enough? What's your actual issue a 'numeric' would address?
> as.package_version(getNamespaceVersion("base")) < "4.5.0"
[1] TRUE
>
>
2009 Jun 28
1
ERROR: system is computationally singular: reciprocal condition number = 4.90109e-18
Hi All,
This is my R-version information:---
> version
_
platform i486-pc-linux-gnu
arch i486
os linux-gnu
system i486, linux-gnu
status
major 2
minor 7.1
year 2008
month 06
day 23
svn rev 45970
language R
version.string R version 2.7.1 (2008-06-23)
While calculating partial
2009 Jun 17
1
Inverting a square matrix using solve() with LAPACK=TRUE (PR#13762)
Full_Name: Ravi Varadhan
Version: 2.8.1
OS: Windows
Submission from: (NULL) (162.129.251.19)
Inverting a matrix with solve(), but using LAPACK=TRUE, gives erroneous
results:
Here is an example:
hilbert <- function(n) { i <- 1:n; 1 / outer(i - 1, i, "+") }
h5 <- hilbert(5)
hinv1 <- solve(qr(h5))
hinv2 <- solve(qr(h5, LAPACK=TRUE))
all.equal(hinv1, hinv2) #
2023 Apr 08
0
Time to add is.formula() to 'stats'?
I know that it has been discussed in the past, but I wanted to ask
to revisit the idea of exporting
is.formula <- function(x) inherits(x, "formula")
from 'stats', parallel to is.data.frame() in 'base', given how
widely formulae are used these days in conjunction with data frames,
even outside of model fitting functions (e.g., for split-apply).
One could argue
2009 Jun 18
0
Inverting a square matrix using solve() with LAPACK=TRUE (PR#13765)
rvaradhan at jhmi.edu wrote:
> Full_Name: Ravi Varadhan
> Version: 2.8.1
> OS: Windows
> Submission from: (NULL) (162.129.251.19)
>
>
> Inverting a matrix with solve(), but using LAPACK=TRUE, gives erroneous
> results:
Thanks, but there seems to be a much easier fix.
Inside coef.qr, we have
coef[qr$pivot, ] <-
.Call("qr_coef_real", qr, y, PACKAGE =
2009 Jul 14
1
LAPACK package
Hi All,
Can someone tell me if solve function shown below for my version of R is
proper or not?
I am using R 2.7.2 .Wherever i have used this function ,i got results which
were different from the expected results as computed using SPSS.
Description of this function says:--
Solve a System of EquationsDescription
This generic function solves the equation a %*% x = b for x, where b can be
either
2013 Feb 26
2
Efficient way to perform linear regressions
Hi All,
I have millions of regression lines to fit. So I am looking for the
most efficient approach in R.
Details:
I have a large desing matrix X. The dimension is n by p.
Each time when fitting the model, select rows from this matrix X and
form a new design matrix, called X_current.
There is another binary matrix M, with dim m by n, and each row is a
1*n vector. It helps to determin X_current.
2000 Mar 14
1
qr.solve (fwd)
Two friend reported me a problem, which I can't solve:
(I run R-1.0.0, Debian Linux)
They hava a function "corr.matrix" (see end of mail), and when they
create a 173x173 matrix with this function
V <- corr.matrix(0.3, n=173)
V1 <- qr.solve(V)
reports:
Error in qr(a, tol = tol) : NA/NaN/Inf in foreign function call (arg 1)
For n < 173, qr.solve returns the correct
2006 Aug 31
1
NaN when using dffits, stemming from lm.influence call
Hi all
I'm getting a NaN returned on using dffits, as explained
below. To me, there seems no obvious (or non-obvious reason
for that matter) reason why a NaN appears.
Before I start digging further, can anyone see why dffits
might be failing? Is there a problem with the data?
Consider:
# Load data
dep <-
2011 Apr 12
0
all.equal(data.frame(package_version()), ...) infinite recursion
With R-2.12.2 on Linux:
> z <- data.frame(Version=package_version(c("0.1")),
row.names=c("pkgA"))
> all.equal(z, z) # expect TRUE
Error: evaluation nested too deeply: infinite recursion /
options(expressions=)?
> traceback()
... lots of lines in a 3-cycle ...
6: all.equal.list(target, current, ...)
5: all.equal.default(target[[i]], current[[i]],
2009 Jun 18
1
Inverting a square... (PR#13762)
Refiling this. The actual fix was slightly more complicated. Will soon
be committed to R-Patched (aka 2.9.1 beta).
-p
rvaradhan at jhmi.edu wrote:
> Full_Name: Ravi Varadhan
> Version: 2.8.1
> OS: Windows
> Submission from: (NULL) (162.129.251.19)
>=20
>=20
> Inverting a matrix with solve(), but using LAPACK=3DTRUE, gives erroneo=
us
> results:
Thanks, but there seems
2006 Mar 13
0
wishlist: function mlh.mlm to test multivariate linear hypotheses of the form: LBT'=0 (PR#8680)
Full_Name: Yves Rosseel
Version: 2.2.1
OS:
Submission from: (NULL) (157.193.116.152)
The code below sketches a possible implementation of a function 'mlh.mlm' which
I think would be a good complement to the 'anova.mlm' function in the stats
package. It tests a single linear hypothesis of the form H_0: LBT'= 0 where B is
the matrix of regression coefficients; L is a matrix
2012 Apr 26
2
How does .Fortran "dqrls" work?
Hi, all.
I want to write some functions like glm() so i studied it.
In glm.fit(), it calls a fortran subroutine named "dqrfit" to compute least
squares solutions
to the system
x * b = y
To learn how "dqrfit" works, I just follow how glm() calls "dqrfit" by my
own example, my codes are given below:
> qr <-
>
2011 Jun 29
0
Error in testInstalledBasic
Hi,
I am running R 2.13.0 on a Windows 7 machine.
I ran the script:
testInstalledBasic('devel')
and received the following warning message:
running tests of consistency of as/is.*
creating ?isas-tests.R?
running code in ?isas-tests.R?
comparing ?isas-tests.Rout? to ?isas-tests.Rout.save? ...running tests of random deviate generation -- fails occasionally
running code in