Hi
I have a dataset that looks like this (dput'd below):
> head(x, 20)
time status
1 2009-07-02 10:32:37 1
2 2009-07-02 10:32:43 0
3 2009-07-02 10:32:43 1
4 2009-07-02 10:32:44 0
5 2009-07-02 10:32:44 1
6 2009-07-02 10:32:48 0
7 2009-07-02 10:32:48 1
8 2009-07-02 10:32:54 0
9 2009-07-02 10:33:04
Sorry - In my previous mail, I said page 39, when I actually meant page 36.
Cheers
-- Rory
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PLEASE do read the posting guide h
Hi all
On page 39 of this paper [1] by Andrew Lo there is a very interesting
correlation network diagram (sorry I dont have a more convenient link to
the type of diagram I'm talking about). does anyone know of any package in
R that can generate these types of diagrams?
Cheers
-- Rory
[1] h
Hi
I have two irregular time series, which are of different lengths and being
and end at different times. For the common subset of time that they both
span, they should have the same values, but the values may occur at
slightly different time intervals. I am trying to "line up" the identical
Hi guys
Does anyone know if it is possible to index a zoo series by a sequence? For
instance, with the following irregular zoo object, I can calculate the
range of its time-based index:
> r <- range(index(l.zoo))
> r
[1] "2009-06-15 01:44:20.802 GMT" "2009-06-15 16:54:24.124 GMT"
If I just w
Thanks Gabor - I'll check it out.
Actually I just realised I can also do what I am looking for in a
ridiculously simple manner (as the data I have is intra-day):
aggregate(l.zoo, hours(index(l.zoo)), mean)
Cheers
-- Rory
On Jun 16, 2009 2:46pm, Gabor Grothendieck wrote:
> See R News 4/1.
Hi all
I have an irregular zoo series, where the time index looks like the
following:
> head(time(l.zoo))
[1] "2009-06-15 01:44:20.802 GMT" "2009-06-15 01:44:20.812 GMT" "2009-06-15
01:44:20.837 GMT" "2009-06-15 01:44:20.848 GMT" "2009-06-15 06:00:01.320
GMT"
[6] "2009-06-15 06:00:01.330 G
Hi all
Does anyone here know of an efficient algorithm to find the longest
identical subsequence across multiple time series? Say I have two time
series A and B (not necessarily of identical length), and I wish to find the
beginning and ending index of the longest common subsequence across both
se
Thanks Gabor. Using the data.matrix() approach as outlined in the FAQ works
for me.
Cheers
Rory
On Mar 3, 2009 4:00pm, Gabor Grothendieck wrote:
> On Tue, Mar 3, 2009 at 10:37 AM, rory.wins...@gmail.com> wrote:
> > Hi guys
> >
> > I have a reasonably basic question with zoo usage, but I hav
Hi guys
I have a reasonably basic question with zoo usage, but I havent been able
to find a satisfactory workaround yet.
Heres a simple example of what I'm talking about (the following data frame
contains numeric columns that contains NAs):
> head(ebs)
time src tstamp code bid ask
1 2009-03-
thias
>
> Prof Brian Ripley wrote:
>
>> Look at packages distr* : they can do your example and might do what your
>> real applications.
>>
>> On Fri, 26 Dec 2008, Rory Winston wrote:
>>
>> Hi
>>>
>>> Firstly , happy Christmas to R-Help! Sec
Hi
Firstly , happy Christmas to R-Help! Secondly, I wonder if anyone can help
me with the following query: I am trying to reproduce some explicit
probability calculations performed in APPL (a Maple extension for
computational probability). For instance, in APPL, to compute the
probability that the
Great! Thanks for the advice.
Cheers
Rory
--Original Message--
From: Duncan Murdoch
To: Rory Winston
Cc: r-help@r-project.org
Sent: 3 Jul 2008 05:08
Subject: Re: [R] Plotting Prediction Surface with persp()
On 02/07/2008 8:47 PM, Rory Winston wrote:
> Hi all
>
> I have a quest
Hi all
I have a question about correct usage of persp(). I have a simple neural
net-based XOR example, as follows:
library(nnet)
xor.data <- data.frame(cbind(expand.grid(c(0,1),c(0,1)), c(0,1,1,0)))
names(xor.data) <- c("x","y","o")
xor.nn <- nnet(o ~ x + y, data=xor.data, linout=FALSE, size=1)
y$x, type='l',ylim=c(0.5,4))
On Sun, Jun 8, 2008 at 7:07 AM, Rory Winston <[EMAIL PROTECTED]
<mailto:[EMAIL PROTECTED]>> wrote:
Hi
Consider the following graph:
x <- rnorm(1000)
x <- x + exp(-x/2)
layout(matrix(rep(c(1,1,2), 2), 2, 3, byrow=T
Hi
Consider the following graph:
x <- rnorm(1000)
x <- x + exp(-x/2)
layout(matrix(rep(c(1,1,2), 2), 2, 3, byrow=TRUE))
boxplot(x)
rug(jitter(x), side=2)
plot(density(x))
What I would really like to do is to have the density plot rotated by 90
degrees so that I can see it line up with the rug
Hi
Ive looked around but I cant figure out how to do this without a for
loop. I have a vector of neural net weights from coef.nnet(), which
looks like c(3,3,1). I also have a list of weight prefixes, which are
c("x","h","y"). I would like to obtain a vector that looks like
c("x1","x2","x3","h
Hi
Does anyone know how I might pick out diagonal elements of a matrix using a
vector?
If I create a matrix a:
a <- matrix(c(1:16), 4, byrow=TRUE)
and I want to pick out the elements (1,1),(2,2),(3,3), or another arbitrary
diagonal (upper or lower), is there any way I can use a vector to do thi
Thanks Earl
Thats exactly what I was looking for - an extension that uses libgmp and
provides a bignum type that can be combined with standard operators and
numeric variables. Somehow my original search on CRAN missed this one.
Cheers
Rory
Earl F. Glynn wrote:
>
> "Rory Winst
Hi
(If you're wondering, this is a Project Euler question :))
If I wanted to calculate the sum of the digits in the decimal representation
of 2^1000, what would be a good way to go about that? I've tried the
following methods:
# Calculate the sum of digits in the decimal representation of 2^n
Me too. Getting directly spammed like this is really annoying. I dont mind a
general post to the list, but individually spamming each member of the list
is unacceptable. Especially as I have no interest in the stupid product in
question.
Gorden T Jemwa wrote:
>
> Dear R Admins,
>
> I received
Sorry, I should have specified:
windows(height=y, width=x)
works for the general caseand specifying <<...width=x, height=y>>= to
Sweave works perfectly. Thank you!
Cheers
Rory
On 10/29/07, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
>
> On Mon, 29 Oct 2007, Rory Wi
ion for
this, or do I need to manually call pdf() and \includegraphics here?
Thanks
Rory
On 10/29/07, Rory Winston <[EMAIL PROTECTED]> wrote:
>
> Hi
>
> I am having a bit of difficulty with changing the canvas size on a
> trellis/lattice plot. I am plotting two "cubes&
Hi
I am having a bit of difficulty with changing the canvas size on a
trellis/lattice plot. I am plotting two "cubes" of 3-dimensional random
numbers, as follows:
library(gsl)
library(lattice)
q <- qrng_alloc(type="sobol", 3)
npoints <- 200
rs <- qrng_get(q,npoints)
# Plot the normal variates in
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