Hi Professor Ripley,
Thanks for the tip. It worked for me. I'll need to work on packaging
next and maybe a bit more testing of the steps on another build machine.
I just found out that there is already an older version of R binary for
Solaris 10 at
http://apstc.sun.com.sg/popup.php?l1=research
On 10 Nov 2006 23:39:14 +0100, Peter Dalgaard <[EMAIL PROTECTED]> wrote:
> [EMAIL PROTECTED] writes:
>
> > > x = rnorm(100)
> > > b = seq(min(x) - 1, max(x) + 1, length = 11)
> > > b
> > [1] -3.4038769 -2.7451072 -2.0863375 -1.4275678 -0.7687980 -0.1100283
> > [7] 0.5487414 1.2075111 1.8662808
[EMAIL PROTECTED] writes:
> > x = rnorm(100)
> > b = seq(min(x) - 1, max(x) + 1, length = 11)
> > b
> [1] -3.4038769 -2.7451072 -2.0863375 -1.4275678 -0.7687980 -0.1100283
> [7] 0.5487414 1.2075111 1.8662808 2.5250506 3.1838203
> >
> > invisible(hist(x, breaks = b, include.lowest = TRUE, pl
> x = rnorm(100)
> b = seq(min(x) - 1, max(x) + 1, length = 11)
> b
[1] -3.4038769 -2.7451072 -2.0863375 -1.4275678 -0.7687980 -0.1100283
[7] 0.5487414 1.2075111 1.8662808 2.5250506 3.1838203
>
> invisible(hist(x, breaks = b, include.lowest = TRUE, plot = FALSE))
Warning message:
argument 'i
'Freedman' is misspelled (as 'Friedman') in
src/library/graphics/man/hist.Rd. As a result, the help page currently
implies that
breaks = "Fried"
is a valid argument to hist, but results in an error:
> hist(rnorm(100), breaks = "Fried")
Error in match.arg(tolower(breaks), c("sturges", "fd", "free
On Nov 10, 2006, at 2:04 PM, [EMAIL PROTECTED] wrote:
> Full_Name: Hendrik Fuß
> Version: 2.4.0
> OS: Mac OS 10.4.8
> Submission from: (NULL) (201.27.208.230)
>
>
> R 2.4.0 on Mac OS X cannot seem to build shared libraries using R
> CMD SHLIB.
>
> R CMD SHLIB mylib.o yields an error message:
>
Full_Name: Hendrik Fuß
Version: 2.4.0
OS: Mac OS 10.4.8
Submission from: (NULL) (201.27.208.230)
R 2.4.0 on Mac OS X cannot seem to build shared libraries using R CMD SHLIB.
R CMD SHLIB mylib.o yields an error message:
/usr/bin/libtool: unknown option character `m' in: -macosx_version_min
I can
On 11/10/2006 12:52 PM, Romain Francois wrote:
> Duncan Murdoch wrote:
>> On 11/9/2006 5:14 AM, Romain Francois wrote:
>>> Hello,
>>>
>>> What about an `invert` argument in grep, to return elements that are
>>> *not* matching a regular expression :
>>>
>>> R> grep("pink", colors(), invert = TRUE,
Duncan Murdoch wrote:
> On 11/9/2006 5:14 AM, Romain Francois wrote:
>> Hello,
>>
>> What about an `invert` argument in grep, to return elements that are
>> *not* matching a regular expression :
>>
>> R> grep("pink", colors(), invert = TRUE, value = TRUE)
>>
>> would essentially return the same as
On 11/10/2006 6:28 AM, Prof Brian Ripley wrote:
> On Fri, 10 Nov 2006, Duncan Murdoch wrote:
>
>> On 11/9/2006 5:14 AM, Romain Francois wrote:
>>> Hello,
>>>
>>> What about an `invert` argument in grep, to return elements that are
>>> *not* matching a regular expression :
>>>
>>> R> grep("pink", c
On Fri, 10 Nov 2006, Duncan Murdoch wrote:
> On 11/9/2006 5:14 AM, Romain Francois wrote:
>> Hello,
>>
>> What about an `invert` argument in grep, to return elements that are
>> *not* matching a regular expression :
>>
>> R> grep("pink", colors(), invert = TRUE, value = TRUE)
>>
>> would essential
On 11/9/2006 5:14 AM, Romain Francois wrote:
> Hello,
>
> What about an `invert` argument in grep, to return elements that are
> *not* matching a regular expression :
>
> R> grep("pink", colors(), invert = TRUE, value = TRUE)
>
> would essentially return the same as :
>
> R> colors() [ - grep(
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