David: Brilliant! Thanks very much.
As Berend pointed out, I was not precise in the query, sorry. Please note in
the example that we have a run of three state 2 after 0, and later another run
of two state 2 after 0.
0 2 2 2 0 0 2 2
Whenever the state moves from 0 to 2, I want to compute t
Folks,
I'm trying to get stats from a matrix for each transition from one state to
another.
I have a matrix x as below.
structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0,
0, 2, 2, 0.21, -0.57, -0.59, 0.16, -1.62, 0.18, -0.81, -0.19,
-0.76, 0.74, -1.51, 2.79, 0.41, 1.63, -0.86, -0.
Interesting variety of solutions! Thanks very much.
Murali
-Original Message-
From: Henrique Dallazuanna [mailto:www...@gmail.com]
Sent: 31 March 2011 18:26
To: Menon Murali
Cc: r-help@r-project.org
Subject: Re: [R] choosing best 'match' for given factor
Try this:
bestMatch <- function
Folks,
I have a 'matching' matrix between variables A, X, L, O:
> a <- structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58,
0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list(
c("A", "X", "L", "O"), c("A", "X", "L", "O")))
> a
A X L O
A 1.00 0.4
Folks,
I have the following dataframe:
> x <- structure(list(name = c("EU B", "EU B", "EU B", "EU B", "EU B",
"EU B", "AU A", "AU A", "AU A", "AU A", "AU A", "AU A"), date = c("2010-10-11",
"2010-10-12", "2010-10-13", "2010-10-14", "2010-10-15", "2010-10-18",
"2010-10-11", "2010-10-12", "2010-
Hiya,
Thanks for this. It's the height parameters that baffle me. Why are they 1, 1,
1/3, 1/3, 1/3?
Which subscreens do these heights correspond to?
I did it like this:
nf <- layout(cbind(c(1,1,1,3,3,3,5,5,5),c(2,2,2,4,4,4,6,7,8)))
mainly because I'm not clear about how the heights work.
Tha
Folks,
I'm battling the layout() functionality in graphics, and getting a bit mixed
up. I'd like to create subscreens like so:
_ _
| | |
|1|2|
|_| |
| | |
|3|4|
|_|_|
| |6
Hi folks,
I want to sort a matrix row-by-row and create a new matrix that contains the
corresponding colnames of the original matrix.
E.g.
> set.seed(123)
> a <- matrix(rnorm(20), ncol=4); colnames(a) <- c("A","B","C","D")
> a
A B C D
[1,] -0.56047565
Bert,
I expect you are correct, burrito notwithstanding (wasn't Taco Bell, was it?
:-)
The full model adds differences and lags, and incorporates non-zero covariances
in the innovations. I only simplified to get an idea of how to implement in R.
For anyone interested, I'm looking at the Balv
Hi Duncan,
Thanks for your response.
Indeed, independent normal errors were what I had in mind. As for variances, if
I assume they are the same, would a 'pooled model' apply in this case? Is that
equivalent to your suggestion of concatenating x(1,t) and x(2,t)?
Cheers,
Murali
-Original M
Hi folks,
Not sure what this sort of estimation is called. I have a 2-column time-series
x(i,t) [with (i=1,2; t=1,...T)], and I want to do the following 'simultaneous'
regressions:
x(1,t) = (d - 1)(x(1, t-1) - mu(1))
x(2,t) = (d - 1)(x(2, t-1) - mu(2))
And I want to determine the coefficients
folks,
does anyone know if the SEMIFAR model has been implemented in R? i see that
there's a S-FinMetrics function SEMIFAR() that does the job, but I have no
access to that software. essentially, this semiparametric fractional
autoregressive model introduces a deterministic trend to the FARIMA
David, Stephen,
You're right - it's the time zone conventions that threw me as well. I tried
those round() operations earlier, but inevitably ended up being either an hour
behind. Even when I specified my time zone, it didn't make any difference. So
there's something else that I'm missing. I'll
Hi folks,
I've got a POSIXct datum as follows:
> Sys.time()
[1] "2010-07-23 11:29:59 BST"
I want to convert this to the nearest half-hour, i.e., to "2010-07-23 11:30:00
BST"
(If the time were "11:59:ss", I want to convert to "12:00:00").
How to achieve this?
Thanks,
Murali
Folks,
I've got a matrix x as follows:
> x <- matrix(c(1,2,3,5,3,4,3,2,1), ncol = 3, byrow = TRUE)
> x
[,1] [,2] [,3]
[1,]123
[2,]534
[3,]321
In each row of x, I want to replace the minimum value by -1, the maximum
value by +1 and all other values by 0.
Folks,
I have some weekly dataseries that I convert to monthly xts (with
yearmon indices), and obtain the two following extracts:
> str(sig)
An 'xts' object from Apr 1998 to Sep 1998 containing:
Data: num [1:6, 1] 0.0083 0.2799 -0.2524 -0.0119 0.18 ...
- attr(*, "dimnames")=List of 2
..$ :
hi folks,
if i have an xts object as follows:
library(xts)
> dd <- as.POSIXct(strptime(c("2009-06-01 08:00:00", "2009-06-01
08:30:00","2009-06-01 09:00:00","2009-06-02 08:00:00", "2009-06-03
08:00:00", "2009-06-03 08:30:00"),"%Y-%m-%d %H:%M:%S"))
> a <- xts(1:6,dd)
> a
[[alternativ
Jim, Gabor, William,
Thanks very much. Works a treat.
Cheers,
Murali
-Original Message-
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: 18 June 2009 18:27
To: MENON Murali; r-help@r-project.org
Subject: RE: [R] Replace zeroes in vector with nearest non-zero value
approx() almost d
Folks,
If I have a vector such as the following:
x <- c(0, -1, -1, -1, 0, 0, 1, -1, 1, 0)
and I want to replace the zeroes by the nearest non-zero number to the
left, is there a more elegant way to do this than the following loop?
y <- x
for (i in 2 : length(x))
{
if (y[i] == 0) {
Hmm, silly of me. I have the solution, after some mucking about with
coercion and so on:
> xx <- zoo(mm[as.character(as.yearmon(dd))], order.by = dd)
> xx
2008-11-03 2008-11-05 2009-01-04 2009-02-02 2009-02-17 2009-03-13
2009-03-14 2009-03-18 2009-03-26
15 15 12 1
Folks,
If I have a series mm of, say, monthly observations, and a series dd of
daily dates, what's a good way of expanding mm such that corresponding
to each day in dd within the corresponding month in mm, the values of mm
are repeated?
So e.g., if I have mm:
mm <- c(15, 10, 12, 13, 11)
names
Benjamin, Dimitris,
Thanks very much. Neat work!
Murali
-Original Message-
From: Nutter, Benjamin [mailto:nutt...@ccf.org]
Sent: 27 March 2009 13:52
To: MENON Murali; r-help@r-project.org
Subject: RE: [R] adding matrices with common column names
Shucks, Dimitris beat me to it. And his c
folks,
if i have three matrices, a, b, cc with some colnames in common, and i
want to create a matrix which consists of the common columns added up,
and the other columns tacked on, what's a good way to do it? i've got
the following roundabout code for two matrices, but if the number of
matrices
Folks,
I put up a brief note describing my naive attempts to compute Goldbach
partitions, starting with a brute-force approach and refining
progressively.
http://jostamon.blogspot.com/2009/02/goldbachs-comet.html
I'd welcome your suggestions on improvements, alternatives, other
optimisations
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