Great, "--vanilla" works!
Can you please tell me what this parameter means?
And, it works with the execuatble(/usr/lib/R/bin/exec/R). I'm wondering
is it possible to do the same with the shell script(/usr/bin/R), anyway
when you type "R" in command line, it means the shell script.
Thanks again!
>
Ian,
pairs() is (normally) a base graphics function, *not* a trellis
graphics function. You can not easily mix the two graphics systems. So
you can not use panel.number(), panel.smooth(), etc with pairs.
There is a trellis graphics equivalent to pairs, called splom (in the
lattice package). I do
On Sat, 01 Nov 2008 22:57:38 +0100
Wacek Kusnierczyk <[EMAIL PROTECTED]> wrote:
> Patrick Burns wrote:
> > Wacek Kusnierczyk wrote:
> >> smells bad design.
> >>
> >
> > Nonsense.
>
> not really, i'm afraid.
>
[...]
> to the point:
>
> is.integer(1) # FALSE
> is.integer(1:1) # TRUE
>
> is no
what is being represented here that is not on a regular spectrogram?
I am not being crass I can't figure out what I am looking at. Maybe a
contour plot- wouldn't be in 3d but would represent the same
information. I am envisioning something like a wavelet time scale
diagram... Just a thought.
O
That's too easy.
The posting guide is your friend. Especially the part about using help
as in
help('normal')
which would reply:
No documentation for 'normal' in specified packages and libraries:
you could try '??normal'
And indeed you could, which leads rather quickl
Is there anyone knowing a function for standard normal cumulative
distribution?
Φ(z=-0.1)=?
also
Φ(z=?)=0.025
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View this message in context:
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Sent from the R help mailing list archive at Nabble.
Its a FAQ:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#How-can-I-turn-a-string-into-a-variable_003f
On Sat, Nov 1, 2008 at 7:54 PM, Lo, Ken <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I know this is an idiotic question, but I am trying to iterate through a
> list of data.frame using ls (I have name
See ?get
However, it sounds like you might look at storing your data.frames in an
object of class "list". Then you can do things like:
alist <- list(df1 = data.frame(a = 1:10, b = 2:11), df2 = data.frame(a =
2:100, b = 3:101))
lapply(alist, summary)
If you wrote your own function to do t
See ?get
And you'll probably also want ?assign
On Sat, Nov 1, 2008 at 7:54 PM, Lo, Ken <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I know this is an idiotic question, but I am trying to iterate through a
> list of data.frame using ls (I have named the data frames in a way that
> lets me subset them w
Hi all,
I know this is an idiotic question, but I am trying to iterate through a
list of data.frame using ls (I have named the data frames in a way that
lets me subset them with ls(pattern="test", for example). But ls
returns a character vector. How to I refer to the data frames by their
names i
This is not the same as the recent thread on a waterfall graph.
I'm thinking about the rolling FFT display used in acoustics and other
spectrum analysis tasks.
Here's an example of a very fancy 3-D waterfall display:
http://www.ultimaserial.com/UltimaWaterfall.html
I was just wondering if th
Patrick Burns wrote:
> Wacek Kusnierczyk wrote:
>> smells bad design.
>>
>
> Nonsense.
not really, i'm afraid.
> One of the key design features of R is that it
> hides implementation details from users. They
> are free to think about the substantive issues with
> their data rather than worryi
On 01-Nov-08 20:59:29, RON70 wrote:
> Still no reply. Is my question not understandable at all?
> RON70 wrote:
>> I am wondering how R calculate p-value for a test. Does R do some
>> "Approximate" integration on the p.d.f of null distribution?
>> How I can see the code for this particular calculati
On Thu, Oct 30, 2008 at 11:14 AM, <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> Does anyone knows the expression for the cumulative distribution of the
> loglogistic distribution used in survreg? (I.e. how are the results related
> to the cumulative distribution?)
I bet that Terry Therneau and Thoma
On Sat, Nov 1, 2008 at 9:59 PM, RON70 <[EMAIL PROTECTED]> wrote:
>
> Still no reply. Is my question not understandable at all?
Patience! 14 minutes in the middle of the night, and a big holiday!
Yes, your question is understandable, although too general to generate
any useful reply. One attemt is:
Still no reply. Is my question not understandable at all?
RON70 wrote:
>
> I am wondering how R calculate p-value for a test. Does R do some
> "Approximate" integration on the p.d.f of null distribution? How I can see
> the code for this particular calculation?
>
> Your help will be highly app
On Fri, Oct 31, 2008 at 2:27 PM, Terry Therneau <[EMAIL PROTECTED]> wrote:
> Use the survreg function.
The survreg function cannot fit left censored data (correct me if I am
wrong!), neither can phreg or aftreg (package eha). On the other hand,
if Borja instead wanted to fit left truncated data (
Hi, a follow up on this as well: I've fixed the below bug in
writeMat(). R.matlab v1.2.4 is now available on CRAN. Please update.
Henrik
On Wed, Oct 1, 2008 at 12:55 PM, Henrik Bengtsson <[EMAIL PROTECTED]> wrote:
> Hi.
>
> On Sat, Sep 27, 2008 at 5:17 AM, Steele, Dr Douglas <[EMAIL PROTECTED
Hi, a follow up on this: I've fixed the below bug in writeMat().
R.matlab v1.2.4 is now available on CRAN. Please update.
Henrik
On Tue, Oct 28, 2008 at 3:30 PM, Henrik Bengtsson <[EMAIL PROTECTED]> wrote:
> Hi,
>
> On Tue, Oct 28, 2008 at 2:25 PM, Minho Chae <[EMAIL PROTECTED]> wrote:
>> Dear
Hello,
I have made a plot using xYplot, but am having trouble with some final
touching up details. I would appreciate your help...
The below code results in a plot of 4 species, with the species being
the same color. So, the first three points are blue, the second three
are purple, etc...
Fantabulous,
Thanks
On Sat, Nov 1, 2008 at 3:43 PM, jim holtman <[EMAIL PROTECTED]> wrote:
> lapply(names(dataSet), function(.name) plotFunction(dataSet[[.name]],
> main=.name))
>
> On Sat, Nov 1, 2008 at 10:31 AM, Chibisi Chima-Okereke <[EMAIL PROTECTED]>
> wrote:
> > Dear all,
> >
> > I am usi
Rense wrote:
Playing around with my new laptop (Macbook, 2.4Ghz), I encountered
unexplained timings using system.time.
I did the following:
X <- Matrix(rnorm(1e6), 1000)
system.time(for(i in 1:25) X%*%X)
user system elapsed
8.306 0.591 5.031
system.time(for(i in 1:25) solve(X))
user syste
Here is a function that has arguments similar to gsub. The first is the
pattern where the portion to actually be replaced should be in
parentheses and the others are the replacement string and the text:
library(gsubfn)
replace.in.context <- function(pattern, replacement, x, ...) {
gsubfn(
The citation below indicates the authors used the permax package (yet
another Gentleman authored effort) to implement the free step-down
method of W&Y:
Generalized Rank Tests for Replicated Microarray Data:
Mei-Ling Ting Lee, Robert J. Gray, Harry Björkbacka, Mason W. Freeman
http://www.bep
On Nov 1, 2008, at 10:32 AM, hadley wickham wrote:
On Fri, Oct 31, 2008 at 11:25 AM, Philip Twumasi-Ankrah
<[EMAIL PROTECTED]> wrote:
Hi friends,
I need suggestions/directions on how to producing a waterfall plot
for present extend of change in tumour size for a set of
respondents in a stu
Nutter, Benjamin wrote:
** Sorry to repost. I forgot to include a function necessary to make
the example work **
I apologize up front for this being a little long. I hope it's
understandable. Please let me know if I need to clarify anything.
Several months ago I wrote a series of functions
lapply(names(dataSet), function(.name) plotFunction(dataSet[[.name]],
main=.name))
On Sat, Nov 1, 2008 at 10:31 AM, Chibisi Chima-Okereke <[EMAIL PROTECTED]>
wrote:
> Dear all,
>
> I am using lapply to generate plots by applying a plot function to a list of
> dataframes. e.g.
>
> lapply(dataSet,
On Sat, Nov 1, 2008 at 10:32 AM, hadley wickham <[EMAIL PROTECTED]> wrote:
> On Fri, Oct 31, 2008 at 11:25 AM, Philip Twumasi-Ankrah
> <[EMAIL PROTECTED]> wrote:
>> Hi friends,
>> I need suggestions/directions on how to producing a waterfall plot for
>> present extend of change in tumour size for
On Fri, Oct 31, 2008 at 11:25 AM, Philip Twumasi-Ankrah
<[EMAIL PROTECTED]> wrote:
> Hi friends,
> I need suggestions/directions on how to producing a waterfall plot for
> present extend of change in tumour size for a set of respondents in a study.
> Example of use of waterfall plot is in the fo
Dear all,
I am using lapply to generate plots by applying a plot function to a list of
dataframes. e.g.
lapply(dataSet, FUN = plotFunction)
Is there a way to call the list item number inside the plot function so that
I can identify each graph?
Thanks
Chibisi
[[alternative HTML version
Is there a Westfall-Yound algorithm available with which I can test
correlation, 2- and k-sample tests of association?
Thanks, LP
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http:/
On Sat, Nov 1, 2008 at 5:13 AM, baptiste auguie <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I believe you can apply the same procedure as described in Paul Murrell's "R
> graphics" book for arranging lattice plots.
Yup, and see also http://had.co.nz/ggplot2/book/grid.pdf
Hadley
--
http://had.co.nz/
On Nov 1, 2008, at 3:30 AM, (Ted Harding) wrote:
On 01-Nov-08 02:51:37, David Winsemius wrote:
Do you want the count of remaining elements which are strictly
greater than the first element?
length(which(a[1] < a[2:10]))
[1] 4
or perhaps a bit more deviously:
sum( a[1]
[1] 4
No need to
Hi,
I have to draw samples from an asymmetric-Laplace-Normal distribution:
f(u|y, x, beta, phi, sigma, tau) \propto exp( - sum( ( abs(lo) +
(2*tau-1)*lo )/(2*sigma) ) - 0.5/phi*u^2), where lo = (y - x*beta) and
y=(y_1, ..., y_n), x=(x_1, ..., x_n)
-- sorry for this huge formula --
A WinBUGS Gibbs s
Playing around with my new laptop (Macbook, 2.4Ghz), I encountered
unexplained timings using system.time.
I did the following:
> X <- Matrix(rnorm(1e6), 1000)
> system.time(for(i in 1:25) X%*%X)
user system elapsed
8.306 0.591 5.031
> system.time(for(i in 1:25) solve(X))
user system elapsed
8.93
Here is one possibility:
# random data
set.seed(1)
growth <- sort(runif(215, -100, 100), decreasing = TRUE)
col <- factor(1 + (growth < -25) + (growth < 25),
lab = c("black", "grey50", "grey80"))
barplot(growth, col = levels(col)[col], ylab = "% Growth",
main = "Random Data", xla
Hi,
I have to draw samples from an asymmetric-Laplace-Normal distribution:
f(u|y, x, beta, phi, sigma, tau) \propto exp( - sum( ( abs(lo) +
(2*tau-1)*lo )/(2*sigma) ) - 0.5/phi*u^2), where lo = (y - x*beta) and
y=(y_1, ..., y_n), x=(x_1, ..., x_n)
-- sorry for this huge formula --
A WinBUGS Gibbs s
Thomas Petzoldt wrote:
Philip Twumasi-Ankrah schrieb:
Hi friends,
I need suggestions/directions on how to producing a waterfall plot
for present extend of change in tumour size for a set of respondents
in a study. Example of use of waterfall plot is in the following
slides presented at ASCO
Patrick Burns wrote:
Peter Dalgaard wrote:
megh wrote:
Is there anything like "goto" loop, which exists in most computer
programs?
e.g. I am looking for this kind of stuff :
if(i < 6) goto "step-02"
Any idea?
Regards,
It doesn't exist, but it can always be replaced by if() {} else
Philip Twumasi-Ankrah wrote:
Hi friends,
I need suggestions/directions on how to producing a waterfall plot for present extend of change in tumour size for a set of respondents in a study. Example of use of waterfall plot is in the following slides presented at ASCO 2007 by Axel Grothey.
Hi Ph
Patrick Burns wrote:
> Peter Dalgaard wrote:
>> megh wrote:
>>
>>> Is there anything like "goto" loop, which exists in most computer
>>> programs?
>>> e.g. I am looking for this kind of stuff :
>>>
>>> if(i < 6) goto "step-02"
>>>
>>> Any idea?
>>>
>>> Regards,
>>>
>>
>> It doesn't exist, bu
Hi,
I believe you can apply the same procedure as described in Paul
Murrell's "R graphics" book for arranging lattice plots.
library(grid)
library(ggplot2)
?grid.layout
df <- data.frame(x=rnorm(100), y=rnorm(100))
df2 <- data.frame(x <- rnorm(100), y=runif(x))
p <- qplot(x,y, data=df)
p2 <
Peter Dalgaard wrote:
megh wrote:
Is there anything like "goto" loop, which exists in most computer programs?
e.g. I am looking for this kind of stuff :
if(i < 6) goto "step-02"
Any idea?
Regards,
It doesn't exist, but it can always be replaced by if() {} else {}
constructs. (You do
Dear UseRs,
For various reasons I need to plot multiple ggplots on one device
(preferably pdf). Is there a way to achieve that?
par(mfrow), split.screen() and layout() seem not to do the job.
Thanks,
Vitalie.
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R-help@r-project.org mailing list
ht
David,
Yes, it is what I want. It is a great help. Thank you,
Jin
David Winsemius wrote:
>
> Do you want the count of remaining elements which are strictly
> greater than the first element?
>
> > length(which(a[1] < a[2:10]))
> [1] 4
>
> or perhaps a bit more deviously:
>
> > sum( a[1
On 01-Nov-08 02:51:37, David Winsemius wrote:
> Do you want the count of remaining elements which are strictly
> greater than the first element?
>
> > length(which(a[1] < a[2:10]))
> [1] 4
>
> or perhaps a bit more deviously:
>
> > sum( a[1] [1] 4
No need to be devious! Simply
sum(a[1] <
I am wondering how R calculate p-value for a test. Does R do some
"Approximate" integration on the p.d.f of null distribution? How I can see
the code for this particular calculation?
Your help will be highly appreciated.
Regards,
--
View this message in context:
http://www.nabble.com/How-R-cal
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