I couldn't download this either with install.packages
from two or three mirrors- I haven't tried since thinking that the
binaries were being built for windows and mac ... If this isn't right
I would be interested.
On Tue, Oct 28, 2008 at 6:10 PM, David Huffer <[EMAIL PROTECTED]> wrote:
> I'm try
Hi,
On Tue, Oct 28, 2008 at 2:25 PM, Minho Chae <[EMAIL PROTECTED]> wrote:
> Dear R users,
>
> I am tryting to save an 3d array to a matlab file like the following.
>
> A <- array(1:24, c(2,3,4))
> writeMat(filename, A=A)
>
> But if I load the mat file from Matlab, it is not 3d matrix anymore.
> D
Igor Telezhinsky wrote:
Dear all,
I have recently found out about the R project. Could anyone tell me if it is
possible to make the comparison of two distributions using R packages? The
problem is TRICKY because I have the distributions of measurements and each
measurement in the given distribut
Zitat von David Scott <[EMAIL PROTECTED]>:
> On Tue, 28 Oct 2008, Oliver Bandel wrote:
>
> > Hello,
> >
> > what are they meaning? It could be something that
> > would show a threshhold above which the result is
> > indicating different meanings then just random noise.
> > But there is no descript
Dear R users,
Is it possible to have an overlap histogram plot? For example:
stuff <- data.frame(Mode = c("Land", "Air"), AgeGroup = c("Young", "Adult",
"Old"), Value = sample(1:300))
histogram( ~Value | AgeGroup * Mode, data = stuff, auto.key = TRUE)
Instead of having 2 * 3 panel, I want to hav
On Tue, 28 Oct 2008, Oliver Bandel wrote:
Zitat von David Scott <[EMAIL PROTECTED]>:
On Tue, 28 Oct 2008, Oliver Bandel wrote:
Hello,
what are they meaning? It could be something that
would show a threshhold above which the result is
indicating different meanings then just random noise.
But
Zitat von David Scott <[EMAIL PROTECTED]>:
> On Tue, 28 Oct 2008, Oliver Bandel wrote:
>
> > Zitat von David Scott <[EMAIL PROTECTED]>:
> >
> >> On Tue, 28 Oct 2008, Oliver Bandel wrote:
> >>
> >>> Hello,
> >>>
> >>> what are they meaning? It could be something that
> >>> would show a threshhold a
Try this:
do.call(rbind.data.frame,
lapply(p1, function(i)power.prop.test(p1=i,p2=0.5*i,power=0.8,
sig.level=0.05)))
On Tue, Oct 28, 2008 at 7:16 PM, Farrel Buchinsky <[EMAIL PROTECTED]> wrote:
> I have solved this problem once before but don't recall exactly how.
> Is there a url t
On Tue, Oct 28, 2008 at 4:31 PM, Ferry <[EMAIL PROTECTED]> wrote:
> Dear R users,
>
> Is it possible to have an overlap histogram plot? For example:
>
> stuff <- data.frame(Mode = c("Land", "Air"), AgeGroup = c("Young", "Adult",
> "Old"), Value = sample(1:300))
> histogram( ~Value | AgeGroup * Mode
JohnLi sent the following at 28/10/2008 19:56:
> Thanks a lot ! I learn a lot about R while I solve my problem according to
> those provided informatin.
Me too: thanks all.
This particular question has suddenly become relevant to me as I'm
trying to do a bit of simulating to model power for a r
Hi,
I have a question of aov. e.g.
aov.ex = aov(x~y)
summary(aov.ex)
The aov summary will print to the screen. How can I extract the aov
result, in particular the values of Pr(>F) and F value into a vector
so that I can use them for other use?
Thanks.
--
Waverley @ Palo Alto
__
Waverley
In general, examine the structure of the object and go from there.
Thus:
str(summary(aov.ex))
will reveal that summary(aov.ex) is a list of 1. Thus you probably want
to examine that single element in more detail:
str(summary(aov.ex)[[1]]) # the [[1]] to get the first element of the
l
Dear Waverley,
Try
# Data
x=rnorm(10)
y=rep(1:2,5)
aov.ex = aov(x~y)
tail(unlist(summary(aov.ex)),2)
# or
anova(lm(x~y))$"Pr(>F)"
HTH,
Jorge
On Tue, Oct 28, 2008 at 8:48 PM, Waverley <[EMAIL PROTECTED]>wrote:
> Hi,
>
> I have a question of aov. e.g.
>
> aov.ex = aov(x~y)
> summary(aov.ex
Any insight into the behavior of "by" in the following case would be
appreciated. There is a note in the help details for "by" about
documenting behavior since v2.7 but I don't entirely understand what it
is saying. I'm using R2.7.2 Windows. I'm interested if the following
behavior was a cha
On 29/10/2008, at 2:04 PM, Jeff Laake wrote:
Any insight into the behavior of "by" in the following case would be
appreciated. There is a note in the help details for "by" about
documenting behavior since v2.7 but I don't entirely understand
what it
is saying. I'm using R2.7.2 Windows. I'
On Tue, 28 Oct 2008 18:04:57 -0700,
Jeff Laake <[EMAIL PROTECTED]> wrote:
> Any insight into the behavior of "by" in the following case would be
> appreciated. There is a note in the help details for "by" about
> documenting behavior since v2.7 but I don't entirely understand what
> it is saying.
#this is my stab at - I am sure that I am missing something. If this
doesn't make sense then please ask for more details. #This may show
my low level of programing knowledge
hester. <- c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4)
value <- rnorm(16)
x <- data.frame(value, hester.)
z <-
(if(x[,"hester."]==
On 29/10/2008, at 2:41 PM, stephen sefick wrote:
#this is my stab at - I am sure that I am missing something. If this
doesn't make sense then please ask for more details. #This may show
my low level of programing knowledge
hester. <- c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4)
value <- rnorm(16)
x <-
Thanks to those that replied. It was reproducible on my system but it
was nested within other code which would have been obvious had I left
the browse> portions in it. The reason for the behavior was obvious
when I used str(samples). Even though Effort looked like a vector in
samples, it was
Hello. I am relatively new to R and trying to figure out how go about
graphing something like this: http://www.math.yorku.ca/SCS/Gallery/images/NYweather.jpg
Specifically, I am having a hard time graphing the mean, min, and max
temperatures as they are depicted on that graph. When I plot th
Hello, R subscribers.
I would like to define a function like this one:
Fun <- function(x)sin(x)/x
with value 1 at x=0. How can I do that?
Is there a way to plot it "symbolically", without using x as a vector?
e.g.
x <- seq(from=-10, to=10, length=100)
plot(x, Fun(x))
Thank you in advance.
___
Dear George,
Perhaps redefining Fn as
Fun2 <- function(x) ifelse(x==0,1,sin(x)/x)
could help. Now, for the symbolic part (I'm not sure if understood
correctly), try ?curve as follows:
curve(Fun2(x),-10,10)
HTH,
Jorge
On Wed, Oct 29, 2008 at 12:21 AM, KARAVASILIS GEORGE <[EMAIL PROTECTED]>wr
G'day George,
On Wed, 29 Oct 2008 06:21:37 +0200
KARAVASILIS GEORGE <[EMAIL PROTECTED]> wrote:
> Hello, R subscribers.
> I would like to define a function like this one:
> Fun <- function(x)sin(x)/x
> with value 1 at x=0. How can I do that?
Fun <- function(x) ifelse(x==0, 1, sin(x)/x)
> Is
This certainly involves arranging multiple plots and plotting multiple
variables within the same plot (in the upper graph). The lower plots looks
rather standard and just have to be arranged right. Further, you have to put
the labels (if they are desired). There is a book on R-graphics which may
he
Thank you Deepayan.
Still on lattice, can the same overlap function be applied to categorical
variable plot such as barchart? I mean, can I create an overlap barchart
plot?
On Tue, Oct 28, 2008 at 5:41 PM, Deepayan Sarkar
<[EMAIL PROTECTED]>wrote:
> On Tue, Oct 28, 2008 at 4:31 PM, Ferry <[EMAIL
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