Hi,
Most likely you are mixing and matching things that do not play
nicely together.
Since you did not give us details on just what you did ("I have
re-installed", is so vague as to be of no value, please provide accurate
and complete descriptions of what you did when asking for help)
You c
Here is one way of doing it. I put your data in a file and then read
it in my lines and deleted everything upto END and then used
textConnection to read the rest of the data. When I looked at the
first 82 values, then seem to have started at the correct place, but I
did see the value you expected
If what you want to do is to 'remove' from 'x' all the occurrances of
the matching values in 'z' and then create a new sequence, here is a
way of doing it:
> x <- c(3,3,4,4,4,4,5,5,6,7,8,8,9)
> z <- c(1,2,3,4,4,5,5,5)
> #want to remove the matching number in 'z' from 'x'
> # determine which number
On Nov 15, 2007 8:08 PM, Bob Farmer <[EMAIL PROTECTED]> wrote:
> Hi.
> I've got a lattice plot with multiple panels and two groups superimposed
> on each panel. Each panel has an independently scaled y-axis (scales =
> list(relation = "free")).
>
> I've successfully put up 95%CI error bars using p
You might try looking at the tkrplot package, it uses win.metafile and
captures the graphics device into a variable (which is then put into a
tk lable, but you could probably do something else with it).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTEC
And here is a slightly shorter variation that encodes the
sequence number as the last two digits and then
removes it at the end:
xx <- 100 * x + seq(x) - match(x, x)
zz <- 100 * z + seq(z) - match(z, z)
setdiff(xx, zz) %/% 100
On Nov 15, 2007 4:54 PM, Gabor Grothendieck <[EMAIL PROTECTED]> wrote
We can append a row of 0's to handle that case:
with(rle(as.vector(rbind(prova, 0))), table(lengths[values == 1]))
On Nov 15, 2007 11:36 AM, Marc Schwartz <[EMAIL PROTECTED]> wrote:
> Ah...OK. I misunderstood then. I thought that you wanted the number of
> runs of 1's in each column.
>
> This i
From: sigalit mangut-leiba <[EMAIL PROTECTED]>
Date: Nov 15, 2007 3:24 PM
Subject: Re: [R] a repetition of simulation
To: r-help <[EMAIL PROTECTED]>
Hello,
In addition to my question a few days ago,
Now I have a matrix of the coefficients,
how can I see all the P.Values ( Pr(>|z|)) of the covariat
On Nov 15, 2007 2:41 PM, Michael Lawrence <[EMAIL PROTECTED]> wrote:
> This is possible using the cairoDevice package and RGtk2.
>
> Turning an R graphic into a raw vector of bytes:
>
> library(cairoDevice)
> library(RGtk2)
>
> # create a pixmap and tell cairoDevice to draw to it
>
> pixmap <- gdkP
Ah...OK. I misunderstood then. I thought that you wanted the number of
runs of 1's in each column.
This is actually easier, _if_ there is not an overlap of 1's from the
end of one column to the start of the next column:
res <- rle(as.vector(prova))
> res
Run Length Encoding
lengths: int [1:11]
try this:
v <- list(c(1,2), c(3,4), 5, c(-3:3))
sapply(v, "[", i = 2)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32
Kalman filter for general state space models, especially in its naive
version, is known for its numerical instability. This is the reason
why people developed square root filters, based on Cholesky
decomposition of variance matrices. In package dlm the implementation
of Kalman filter is based on
On Nov 15, 2007 7:12 PM, Bert Gunter <[EMAIL PROTECTED]> wrote:
>
> Folks:
>
> delta <- 1:5
>
> I would like to put 5 separate lines of text of the form "10 %+-% delta[i]"
> into a lattice key legend, where ""%+-%" is the plotmath plus/minus symbol
> and delta[i] is the ith value of delta.
>
> The
On 16/11/2007, at 9:44 AM, Greg Snow wrote:
>
>
> [snip]
>>
>> or (I can't resist)
>>
>> unlist(lapply(v,function(x,i){x[i]},i=2)) # For more
>> flexibility.
>
> Well, if we are not resisting the fun ones then try:
>
> sapply( v, `[`, i=2)
Admittedly *much* cooler than my some
I have a data frame "reading" that includes a logical variable "OLT"
along with response variable "Reading" and predictor "True" (BOTH are
numeric variables; it's "True" as in the true value).
When I suppress the intercept, model.matrix gives me OLTTRUE and
OLTFALSE columns. Why? Can I do anythi
Try:
> w <- ave(x,g)
> n <- ave(x,g, FUN=length)
> wnew <- (w*n - x)/(n-1)
Or
> s <- ave(x,g, FUN=sum)
> n <- ave(x,g, FUN=length)
> wnew <- (s-x)/(n-1)
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> --
On 15/11/2007, at 6:05 PM, Bryan Klingaman wrote:
> Ya, I was wondering if anyone knows how to use R and can do some
> computing with it. I have a problem that involves implementing the
> EM Algorithm for censored normal data. So I was wondering if
> anyone knows how to code a problem inv
For the first question use ave along with mean like Gabor suggested.
For the second question (finding the mean with that value removed) you
can use ave with the sum function ("ave(x, group, FUN=sum)") to find the
sum rather than the average, then use ave again with the length function
to find out
Hi,
if you use the function kmean in the package stats, for example
clust <- kmeans(data, k, iter.max = 10)
where k is the number of desired cluster, kmeans will choose the first
k centers randomly. Because of this random initialization, after
iter.max iteration the solution may converge to diff
?sweep
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of
> [EMAIL PROTECTED]
> Sent: Thursday, November 15, 2007 10:50 AM
> To: r-
Hi Allen,
Its difficult to know what is the problem without knowing what type of
object is 'Disease.FL'. plot() is a generic function and it will act
differently depending on the type of object you are passing to it. As
always, you should provide 'provide commented, minimal, self-contained,
On Thu, 2007-11-15 at 17:53 +0100, A M Lavezzi wrote:
> thank you.
> I did not think about the case of overlapping of
> 1's from the end of one column to the start of the next,
> this would actually be a problem
>
> In the simulations I am running each column
> corresponds to the path followed
Sorry, I wasn't sure what you meant. This way will return more than one
answer, right?
N<-c(1,2,1,3)
R<-c(1.75,3.5,1.75,1.3125)
## get all 126 combinations of five 0's and four 1's for matrix
cbn<-as.matrix(expand.grid( rep( list(0:1), 9)))
cbn<- cbn[rowSums(cbn)==4,]
ans<-list()
ctr<-0
## lo
Carles Fan a écrit :
> Dear all
>
> i have a time series containing trading dates and historical stock prices:
> Date Price
> 10-Jan-2007 100
> 11-Jan-2007 101
> 13-Jan-2007 99
> ..
> ..
> ..
> 10-Nov-2007 200
>
> i want to sample every 21st data of each month:
> 21-Ja
Thomas
> Not sure i explained it good enough. Ill try with an example
>
> say
>
> x=[3,3,4,4,4,4,5,5,6,8]
> z=[3,4,4,5,5]
>
> what i want to get after removing z from x is something like
> x=[3,4,4,6,8]
This will work, but I imagine there are better ways (assuming z is always a
subset of x)
On Fri, 16 Nov 2007, Tom Minka wrote:
> I haven't seen the original code, but the problem with Ray's code is that
> the two projections are not synchronized. Specifically, they are using
> different (default) values for the orientation. To synchronize the
> projections, either specify the orienta
I am writing to verify the syntax that I am using to test a 3-level
> model with a random intercept at the second level (participant in my
> model) versus a model with a random slope and intercept at this
> level. Specifically, I am testing a 3 level model in which time
> (WEEK) is nested in
On Thu, 2007-11-15 at 17:50 +, [EMAIL PROTECTED] wrote:
> Hi,
>
> I've got an array, say with i,jth entry = A_ij, and a vector, say with jth
> entry= v_j. I would like to multiply each column of the array by the
> corresponding vector component, i,e. find the array with i,jth entry
>
> A_ij
On Wed, 14 Nov 2007, Nadine Mugusa wrote:
> Hi everyone,
> Can someone help me with root bisection algorithm?
> Nadine
>
Well you really should read the posting guide because this is not a very
informative enquiry. However if what you really want to do is find the
root of an equation you shou
On Nov 15, 2007 12:54 PM, Carles Fan <[EMAIL PROTECTED]> wrote:
> Dear all
>
> i have a time series containing trading dates and historical stock prices:
> Date Price
> 10-Jan-2007 100
> 11-Jan-2007 101
> 13-Jan-2007 99
> ..
> ..
> ..
> 10-Nov-2007 200
>
> i want to samp
Hi All,
I am having great trouble doing something pretty simple.
Here is what I did:
> x <- read.table("clipboard")
> dim (x)
[1] 126 10
> typeof(x)
[1] "list"
> w <- array(x)
> typeof(w)
"list"
Q1: How come after constructing an array out of the list, the type of
the array is still "list"?
On 11/15/2007 4:54 PM, Gang Chen wrote:
> I want to identify whether a variable is character(0), but get lost.
> For example, if I have
>
> > dd<-character(0)
>
> the following doesn't seem to serve as a good identifier:
>
> > dd==character(0)
> logical(0)
>
> So how to detect character(0)?
Ya, I was wondering if anyone knows how to use R and can do some computing with
it. I have a problem that involves implementing the EM Algorithm for censored
normal data. So I was wondering if anyone knows how to code a problem
involving the EM Algorithm in R, and then estimate the parameters,
On Nov 15, 2007 4:36 PM, Johan A. Stenberg <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> I'm quite sure that this is a stupid question, but I'll ask anyway.
> I want to perform an ANCOVA with two continuous factors and three
> categorical factors.
>
> Plant population growth rate (GR) = dependent vari
Using your "set", wouldn't it be simpler like this?
t(apply(combn(7,2), 2, function(x) set[x]))
Hth,
Adrian
On Thursday 15 November 2007, [EMAIL PROTECTED] wrote:
> There are a number of packages that do this, but here is a simple
> function for choosing subsets:
>
> subsets <- function(n, r) {
Maybe Joren means that the y axis has values greater than 1? If that
is the case, that is certainly not evidence of any problem; the
density can have values larger than 1 and still integrate to 1. (And,
just as a silly example, try "dnorm(0, mean = 0, sd = 0.1)").
Best,
R.
On Nov 15, 2007 11:0
Your model is fully saturated. It specifies terms that use
up all degrees of freedom. There are no degrees of freedom left
over for a Residual term and therefore there is no denominator for
the tests.
When you drop one term, then those degrees of freedom are left over,
that is they form the Res
Dear Daniel,
May I point you to Thomas Lumley's paper in R News 2001-3 ("Programmer’s
Niche: Macros in R\n Overcoming R’s virtues) and to the defmacro utility
of the gtools package ?
HTH
Emmanuel Charpentier
Daniel Myall a écrit :
> Hi,
>
> I have a lar
On Thursday 15 November 2007, [EMAIL PROTECTED] wrote:
> Actually, (now that I know about combn), a better way is
>
> t(matrix(set[combn(7,2)], nrow = 2))
Indeed, or to avoid transposing:
matrix(set[combn(7,2)], ncol = 2, byrow=T)
Adrian
--
Adrian Dusa
Romanian Social Data Archive
1, Schitu Ma
I think you've read Thomas's request in reverse. and
what he want is:
x[!x %in% z]
Thanks for the %in% approach BTW.
--- Charilaos Skiadas <[EMAIL PROTECTED]> wrote:
>
> On Nov 15, 2007, at 9:15 AM, Thomas Fr��jd
wrote:
>
> > Hi
> >
> > I have three vectors say x, y, z. One of them, x
> c
Just an couple of ideas that might or might not work ...
Josh Tolley a écrit :
> I'm using R embedded in PostgreSQL (via PL/R), and would like to use
> it to create images. It works fine, except that I have to create every
> image in a file (owned by and only readable by the PostgreSQL server),
>
Thanks a lot for all who've provided suggestions!
Gang
On Nov 15, 2007, at 5:09 PM, Duncan Murdoch wrote:
> On 11/15/2007 4:54 PM, Gang Chen wrote:
>> I want to identify whether a variable is character(0), but get
>> lost. For example, if I have
>> > dd<-character(0)
>> the following doesn't
see ?is.character and ?length
something like
#not tested
dd <- character(0)
is.character(dd) & length(dd) == 0
should do it.
Gang Chen wrote:
> I want to identify whether a variable is character(0), but get lost.
> For example, if I have
>
> > dd<-character(0)
>
> the following doesn't s
I am not sure what you mean when you say it does not integrate to 1.
Here are a couple of cases, and it seems fine to me:
> x <- density(1:30)
> str(x)
List of 7
$ x: num [1:512] -11.0 -10.9 -10.8 -10.7 -10.6 ...
$ y: num [1:512] 6.66e-05 7.22e-05 7.84e-05 8.49e-05 9.20e-05 ...
Try this. xx and zz are the same as x and z except
they have a sequence number appended. We then do
a setdiff and remove the sequence numbers.
> xx <- paste(x, seq(x) - match(x, x))
> zz <- paste(z, seq(z) - match(z, z))
> dd <- setdiff(xx, zz)
> as.numeric(sub(" .*", "", dd))
[1] 3 4 4 6 7 8 8
is.character(dd) && length(dd) == 0
should do it i think.
Gabor
On Thu, Nov 15, 2007 at 04:54:45PM -0500, Gang Chen wrote:
> I want to identify whether a variable is character(0), but get lost.
> For example, if I have
>
> > dd<-character(0)
>
> the following doesn't seem to serve as a good
I want to identify whether a variable is character(0), but get lost.
For example, if I have
> dd<-character(0)
the following doesn't seem to serve as a good identifier:
> dd==character(0)
logical(0)
So how to detect character(0)?
Thanks,
Gang
__
This works on Windows:
dev.control(displaylist="enable") # enable display list
plot(1:10)
myplot <- recordPlot() # load displaylist into variable
You can now store myplot and later fetch it and play it
back via
myplot
# or if not from the console
print(myplot)
On Nov 15, 2007 4:07 PM, Josh Tol
This is possible using the cairoDevice package and RGtk2.
Turning an R graphic into a raw vector of bytes:
library(cairoDevice)
library(RGtk2)
# create a pixmap and tell cairoDevice to draw to it
pixmap <- gdkPixmapNew(w=500, h=500, depth=24)
asCairoDevice(pixmap)
# make a dummy plot
plot(1:1
Dear all
I would like to show my audience that some variables are homogenous inside
groups but different outside. I can use by with summary for all variables
by(iris[,1:4], iris$Species, summary)
what can be quite messy in case of more than few variables and about 8
groups
or densityplot fo
A little more information might be useful. If your matrix is numeric,
then a single copy will require about 250MB of memory. What type of
system are you on and how much memory do you have? When you say you
are having problems, what are they? Is it a problem reading the data
in? Are you getting
On Thu, 15 Nov 2007, Tao Shi wrote:
>
> Hi List,
>
> I'm running R2.5.1 on WinXP. Downloaded RMySQL_0.6-0.zip from
> http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.6/ and the
> installation seemed fine. However, when I tried to load the package, the
> error occured:
>
>
>> utils:::m
Thanks Brian. I installed libf2c for the x86_64 server, and then R and
Bioconductor both installed without error, so apparently it was missing the
library files. Back in business.
Richard Casey, PhD
Rocky Mountain Regional Center of Excellence
fo
I'm using R embedded in PostgreSQL (via PL/R), and would like to use
it to create images. It works fine, except that I have to create every
image in a file (owned by and only readable by the PostgreSQL server),
and then use PostgreSQL to read from that file and return it to the
client. It would be
Dear Brian,
sorry, library(lattice) is loaded, when I start R, so I forgot to add this.
I get "Ingo's title" if I plot directly to the screen. However, I do not get
it if I
use png() or I lose it if I save from the plot (screen).
Ingo
On 15 Nov 2007 at 10:30, Prof Brian Ripley wrote:
> Wo
Hi,
I have a large dataframe than I'm writing functions to explore, and to
reduce cut and paste I'm trying to write a function that does a subset
and then a plot.
Firstly, I can write a wrapper around a plot:
plotwithfits <- function(formula, data, xylabels=c('','')) {
xyplot(formula, data
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Geoffrey Zhu
> Sent: Thursday, November 15, 2007 10:45 AM
> To: r-help@r-project.org
> Subject: [R] HELP: How to subtract a vector out of each row
> of a matrix or array
>
> Hi All,
>
> I am having g
Thank you for all your comments,
Sigalit.
On 11/15/07, Johannes Hüsing <[EMAIL PROTECTED]> wrote:
>
> Excuse me, but I think your code deserves some comments. Unfortunately,
> the history of postings is in reverse order, so I'll address your
> first question first:
>
> > > >>> The simulation look
[snip]
>
> or (I can't resist)
>
> unlist(lapply(v,function(x,i){x[i]},i=2)) # For more
> flexibility.
Well, if we are not resisting the fun ones then try:
sapply( v, `[`, i=2)
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
summary(log_v)
Julian
sigalit mangut-leiba wrote:
> Hello,
> In addition to my question a few days ago,
> Now I have a matrix of the coefficients,
> how can I see all the P.Values (Pr(>|z|)) of the covariates from the 1000
> iterations?
> I tried names(log_v) and couldn'n find it.
> Thank you,
>
On 11/15/2007 12:55 PM, Loren Engrav wrote:
> With the Mac and R.app, there is a window to the right of the console
> wherein all commands are display and can be re-chosen by double clicking.
>
> Does a similar feature exist with Windows and Rgui? Or have I missed it
> somewhere?
history() will
Actually, (now that I know about combn), a better way is
t(matrix(set[combn(7,2)], nrow = 2))
Bill V.
-Original Message-
From: Adrian Dusa [mailto:[EMAIL PROTECTED]
Sent: Thursday, 15 November 2007 11:55 PM
To: Venables, Bill (CMIS, Cleveland)
Cc: [EMAIL PROTECTED]; r-help@r-project.or
MARK LEEDS-3 wrote:
>
> I was just curious if anyone knew of an alternative model to logistic
> regression where the probabilities seems pretty linear to the predictor
> rather than having that S shape that probit and logit assume.
>
>
Well, the logistic curve is very close to linear over
Excuse me, but I think your code deserves some comments. Unfortunately,
the history of postings is in reverse order, so I'll address your
first question first:
> > >>> The simulation looks like this:
> > >>>
> > >>> z <- 0
> > >>> x <- 0
> > >>> y <- 0
> > >>> aps <- 0
> > >>> tiss <- 0
> > >>> fo
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of
> [EMAIL PROTECTED]
> Sent: Thursday, November 15, 2007 12:04 PM
> To: r-help@r-project.org
> Cc: [EMAIL PROTECTED]
> Subject: [R] not R question : alternative to logistic regression
>
> I was just cu
Hi List,
I'm running R2.5.1 on WinXP. Downloaded RMySQL_0.6-0.zip from
http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.6/ and the
installation seemed fine. However, when I tried to load the package, the error
occured:
> utils:::menuInstallLocal()
package 'RMySQL' successfully unpa
jim holtman wrote:
> Lets take a look at your solution:
>
>> mat1 <- matrix(0, nrow=10, ncol=3)
>> dimnames(mat1) <- list(paste('row', 1:10, sep=''), LETTERS[1:3])
>> mat2 <- matrix(1:3, ncol=1, dimnames=list(c('row3', 'row7', 'row5'), "B"))
>> mat2
> B
> row3 1
> row7 2
> row5 3
>> mat1[rown
Hi.
I've got a lattice plot with multiple panels and two groups superimposed
on each panel. Each panel has an independently scaled y-axis (scales =
list(relation = "free")).
I've successfully put up 95%CI error bars using panel.arrows (and some
help from the mailing list). My question is whet
I was just curious if anyone knew of an alternative model to logistic
regression where the probabilities seems pretty linear to the predictor rather
than having that S shape that probit and logit assume.
Maybe there is there some kind of other GLM that could accomplish that. Any
textbook refere
Bert Gunter said the following on 11/15/2007 1:12 PM:
> Folks:
>
> delta <- 1:5
>
> I would like to put 5 separate lines of text of the form "10 %+-% delta[i]"
> into a lattice key legend, where ""%+-%" is the plotmath plus/minus symbol
> and delta[i] is the ith value of delta.
>
> The construct
On 16/11/2007, at 7:05 AM, carol white wrote:
> Hi,
> How is it possible to extract athe elements of a list of vectors in
> a fixed position? suppose that I have a list of 2-element vectors,
> how can I extract the 2nd element of all vectors in the list? Can
> it be done with indexing and n
Hello,
I have a data set of about 300.000 measurements made by an STM which should
apporximately fix a normal (Gaussian) distribution.
I have imported the data in R and used plot(density()) to get a nice plot of
the distribution which in fact looks like a real Gaussian.
However, the integral over
I think you have an object x which doensn't allow to give names.
If you use names(z) it will work.
To see what kind of object x is: class(x)
Regards
Bart
Schiller Judith 1541 EB wrote:
>
> hi,
>
> after installing R-2.6.0 the function "names" doesn't work anymore on my
> windows xp machine.
Hi everyone,
Can someone help me with root bisection algorithm?
Nadine
-
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEA
Hi,
How is it possible to extract athe elements of a list of vectors in a fixed
position? suppose that I have a list of 2-element vectors, how can I extract
the 2nd element of all vectors in the list? Can it be done with indexing and
not by element name?
Thanks
carol
So in this example, I wa
Hello, I'm new using R.
I'm trying to develop a K-means Clustering with R for some data I have,
however each time I use that instruction with the same data my cluster
means, clustering vector and within cluster sum of square change and I don't
understand why because I use the same parameters and
Folks:
delta <- 1:5
I would like to put 5 separate lines of text of the form "10 %+-% delta[i]"
into a lattice key legend, where ""%+-%" is the plotmath plus/minus symbol
and delta[i] is the ith value of delta.
The construct:
lapply(delta,function(d)bquote(10%+-%.(d)))
appears to produce a li
Jim,
The one issue with those is that they remove duplicate elements in each
vector before applying their logic. Thus, would not likely work here:
x <- c(3,3,4,4,4,4,5,5,6,8)
z <- c(3,4,4,5,5)
In effect, you end up with:
> unique(x)
[1] 3 4 5 6 8
> unique(z)
[1] 3 4 5
Thus:
> setdiff(x, z)
On Thu, 15 Nov 2007, Loren Engrav wrote:
> With the Mac and R.app, there is a window to the right of the console
> wherein all commands are display and can be re-chosen by double clicking.
>
> Does a similar feature exist with Windows and Rgui? Or have I missed it
> somewhere?
You have. history
> > You can live with it, but
> > be aware that it is there. My suggestion is to start the optimization
> > from several different initial values and compare maximized values of
> > the likelihood. Simulated annealing may be used to better explore the
> > parameter space.
>
> Yes. Are you aware
You can also check out the 'set' operations: setdiff, intersect, union.
On Nov 15, 2007 12:08 PM, John Kane <[EMAIL PROTECTED]> wrote:
> I think you've read Thomas's request in reverse. and
> what he want is:
> x[!x %in% z]
>
> Thanks for the %in% approach BTW.
>
> --- Charilaos Skiadas <[EMAIL PR
If i understand your question, you can do:
x <- matrix(1:10, 2)
y <- sample(10,5)
apply(x, 1, function(.x)mapply(y, .x, FUN="*"))
On 15/11/2007, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I've got an array, say with i,jth entry = A_ij, and a vector, say with jth
> entry= v_j. I would
With the Mac and R.app, there is a window to the right of the console
wherein all commands are display and can be re-chosen by double clicking.
Does a similar feature exist with Windows and Rgui? Or have I missed it
somewhere?
Thank you.
[[alternative HTML version deleted]]
___
Thanks Gabor. Nice solution.
Marc
On Thu, 2007-11-15 at 12:35 -0500, Gabor Grothendieck wrote:
> We can append a row of 0's to handle that case:
>
> with(rle(as.vector(rbind(prova, 0))), table(lengths[values == 1]))
>
>
>
> On Nov 15, 2007 11:36 AM, Marc Schwartz <[EMAIL PROTECTED]> wrote:
>
?sweep
b
On Nov 15, 2007, at 12:50 PM, [EMAIL PROTECTED] wrote:
> Hi,
>
> I've got an array, say with i,jth entry = A_ij, and a vector, say with
> jth
> entry= v_j. I would like to multiply each column of the array by the
> corresponding vector component, i,e. find the array with i,jth entry
>
On Nov 15, 2007 12:50 PM, <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I've got an array, say with i,jth entry = A_ij, and a vector, say with jth
> entry= v_j. I would like to multiply each column of the array by the
> corresponding vector component, i,e. find the array with i,jth entry
>
> A_ij * v_j
>
>
Dear all
i have a time series containing trading dates and historical stock prices:
Date Price
10-Jan-2007 100
11-Jan-2007 101
13-Jan-2007 99
..
..
..
10-Nov-2007 200
i want to sample every 21st data of each month:
21-Jan-2007 101
21-Feb-2007 111
21-Mar-2007 131
..
Hi,
I've got an array, say with i,jth entry = A_ij, and a vector, say with jth
entry= v_j. I would like to multiply each column of the array by the
corresponding vector component, i,e. find the array with i,jth entry
A_ij * v_j
This seems so basic but I can't figure out how to do it without a lo
Giovanni Petris wrote:
> Kalman filter for general state space models, especially in its naive
> version, is known for its numerical instability. This is the reason
> why people developed square root filters, based on Cholesky
> decomposition of variance matrices. In package dlm the implementati
Hi -
I'm reading in a tab delimited file that is causing issues with
read.delim. Specifically, for a specific set of lines, the last entry
of the line is misread and considered to be the first entry of a new row
(which is then padded with 'NA's' ). Specifically:
tmp <- read.delim( "troub
[EMAIL PROTECTED] wrote:
> Hi,
>
> Following convention below:
> y(t) = Ax(t)+Bu(t)+eps(t) # observation eq
> x(t) = Cx(t-1)+Du(t)+eta(t) # state eq
>
> I modified the following routine (which I copied from:
> http://www.stat.pitt.edu/stoffer/tsa2/Rcode/Kall.R) to accommodate u(t), an
> exoge
thank you.
I did not think about the case of overlapping of
1's from the end of one column to the start of the next,
this would actually be a problem
In the simulations I am running each column
corresponds to the path followed by an agent
across states of a stochastic process,
so I would like t
Here is another approach:
> tmp <- expand.grid( 1:nrow(B), 1:nrow(A) )
> out <- cbind( A[tmp[,2],], B[tmp[,1],] )
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL P
I am tryindo to do a very simple thing but cannont find how to do it
anywhere. I need to formap part of my title as subscript ans superscript.
How can I do it?
Thanks a lot in advance
José
--
MSc José Alberto F. Monteiro
Botanisches Institut
Universität Basel
[[alternative HTML version d
Here are some sites that I found with maps of europe (or more, but
eourope is included). I don't know how up to date these are, so check
them out for yourself:
http://www.vdstech.com/map_data.htm
http://openmap.bbn.com/data/shape/timezone/
http://arcdata.esri.com/data_downloader/DataDownloader?p
Thomas Frööjd wrote:
>Not sure i explained it good enough. Ill try with an example
>
>say
>
>x=[3,3,4,4,4,4,5,5,6,8]
>z=[3,4,4,5,5]
>
>what i want to get after removing z from x is something like
>x=[3,4,4,6,8]
>
>
>On Nov 15, 2007 3:29 PM, Charilaos Skiadas <[EMAIL PROTECTED]> wrote:
>
>
>>On N
Dear list,
I have a question about using plot().
I tried the code:
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Dear Marc
thank you so much!
One thing: writing xx=[1,2,1,1] is not a typo: I
read it as the count of runs of different length starting from 1.
In "prova" I have 1 run of length one, 2 runs of
length two, 1 run of length three and 1 run of length four.
Can I abuse of your time and ask how to
Dear all,
I am locking for a quadratic solver in R beeing able to solve the following
sort of problem:
minimize || y - Xmatrix u ||^2
subject to:
Amatrix u <= Bbound and u >= lowbound
While the functions
pcls{mgcv} and solve.QP{quadprog} can handle the first part of the con
Not sure i explained it good enough. Ill try with an example
say
x=[3,3,4,4,4,4,5,5,6,8]
z=[3,4,4,5,5]
what i want to get after removing z from x is something like
x=[3,4,4,6,8]
On Nov 15, 2007 3:29 PM, Charilaos Skiadas <[EMAIL PROTECTED]> wrote:
>
>
> On Nov 15, 2007, at 9:15 AM, Thomas Fröö
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