= 10, p = .7, list = FALSE)
Thanks a lot for your help.
Regards,
Trafim
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R
Dear all,
seems that easy question but cannot find the function for that.
How to get the elements of the minor diagonal of the matrix?
Thanks a lot.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch
m(y ~ x, DF)
> predict(m, data.frame(x = seq(-13, 13, 0.5)))
>
>
> I hope it helps.
>
> Best,
> Dimitris
>
>
>
> On 8/16/2010 2:37 PM, Trafim Vanishek wrote:
>
>> Dear all,
>>
>> I have an error in the simple prediction function for lm().
>> M
)
Thanks a lot,
Trafim
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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, m
Dear all,
I produced a part of code which does what needed but by estimations I would
need to wait for that for 300 hours!!!
Maybe someone can give a me a glance advice or something.
Thanks a lot!
I need to define maximum adjusted R-squares from regression model: returns
as dependent variable( he
Dear all,
Are there any functions in R to reverse the order of the string.
smth like reverse("abc") to get "cba"?
Thanks a lot.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-h
Thanks everybody for referring me to FAQ 7.31 but I don't see how to solve
it.
I am giving a concrete number and I need to get 58 not 57. Seems, there is
no way?
On Fri, Jul 9, 2010 at 4:05 PM, David Winsemius wrote:
>
> On Jul 9, 2010, at 9:46 AM, Trafim Vanishek wrote:
&g
Dear all,
might seem and easy question but I cannot figure it out.
floor(100*(.58))
[1] 57
where is the trick here? And how can I end up with the right answer?
Thanks a lot everybody for your help.
Trafim
[[alternative HTML version deleted
Dear all,
My question how is it possible to define a matrix A with 10 rows 1 column,
so that its elements are vectors of undefined length.
I need to have a possibility later to add elements like A[1,1] <-
c(A[1,1],3,4,5)
Thanks a lot for the help!
[[alternative HTML version deleted]]
_
Hi everybody,
would like to ask if there is any possibility to get SOME BEST result for
nls fitting if the function itself doesn't produce it.
I already increased number of iterations, reduced minfactor and still it
doesn't give me the solution.
Thanks a lot.
[[alternative HTML version d
Dear all,
I have a question regarding performing test if the data fits chi-squared
distribution.
For example, using ks.test()
I found in the examples how to fit it to gamma or weibull
x<-rnorm(100)
ks.test(x, "pweibull", shape=2,scale=1)
for the gamma, pgamma can be used
But I cannot find the v
Hi everybody,
Does anyone know what problem may be with this test.
I am applying 5 different normality tests and use p-values for them, but for
some reason S-W gives me NA, while sample size is 100.
Any ideas?
Thanks a lot!
[[alternative HTML version deleted]]
_
This is exactly what I mean.
I need to find integrate(dnorm(mean=8,sd=1)*log(dnorm(mean=8,sd=1)), -Inf,
Inf)
Which doesn't work like that, because it says:
Error in dnorm(mean = 8, sd = 1) : element 1 is empty;
the part of the args list of '.Internal' being evaluated was:
(x, mean, sd, log)
Dear all,
How is it possible in R to calculate the following integral:
Integral(-Inf, Inf)[log(dnorm(mean = 3, sd = 1))]
how can I define that the density dnorm is taken on (-Inf, Inf)
Thanks a lot!
[[alternative HTML version deleted]]
__
R-h
Dear all,
I would like to ask how to extract the p-value for the whole model from
summary(lm).
This didn't help a lot summary.lm
summary(lm(speed~dist, cars))
Thanks a lot!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing
Thanks a lot everybody!
I see how it works now.
But now the other question: I would like to see the Cook's distance plot
which is more understandable for me not the Residuals vs Leverage.
The one it shows in the examples
http://www.personality-project.org/r/r.lm.html
But for some reason it doesn't
Dear all,
does anybody ever encountered the problem with diagnostic plots?
x<-rnorm(100)
y<-rnorm(100)
plot(lm(x~y))
It gives the following message "Waiting to confirm page change..." and
nothing happens.
Thanks a lot!
[[alternative HTML version deleted]]
_
Dear all,
I have a simple question for which I cannot find the answer.
I need to make an easy plot, but for the x axis I need to be able to specify
by myself the division of x axis from x[,1] either every single observation,
or every 5th, or 10th or 20th
x <- matrix(data=NA, nrow=100, ncol=2)
x[
Dear all,
I am using GARCH (1,1) model to simulate volatility.
But seems that I am missing something about how it works in R.
The following code produces negative results, though vola cannot be.
What is wrong here?
library("fSeries")
library("fGarch")
spec = garchSpec(model = list(omega = 0.01,
Dear all,
I cannot find to explicitly get the R-squared or adjusted R-squared from
summary(lm())
Thanks a lot!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do rea
Dear all,
I have a question is there a possibility to plot points with different
symbols like stars, crosses?
I looked at different types for plot command and didn't find anything like
that.
Thanks a lot.
[[alternative HTML version deleted]]
_
Dear all,
Does anybody know the probable reason why <= gives false when it should give
true?
These two variables are of the same type, and everything works in the cycle
but then it stops when they are equal.
this is the output result
> Rk[47] <= RB[21]
[1] FALSE
> Rk[47]
[1] 0.002842007
> RB[21
Dear all,
I currently experience the problem with nicely plotting price data against
the dates.
Data <- read.csv("C:/IBM.csv", header = TRUE, sep = ",")
plot(Data[,1], Data[,2])
I cannot find the way how can I choose the # of breaks for the x axis -
dates in this case?
Thanks a lot
_
Dear all,
I would like to ask if there is a simple was in R to fit the chi-squared
distribution to the empirical data?
Thanks a lot!
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinf
Dear all,
I would like to ask you if there is a possibility in R to give the names to
graphs which are not const.
For example,
How to name each plot, or to add notes like a=x[i], b=y[i] in this cycle
x<-c(1,2,3,4,5,6)
y<-c(7,8,9,10,11,12)
par(mfrow=c(2,3))
for (i in 1:6){
plot(x,y)
}
Thank y
Dear all,
I have the following question.
Is it possible in R to call for the p-value directly in a model like this
a <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
b <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
summary(lm(a~(b+0)))
or even in this
a <- c(4.81,4.17,4.41,3.59,
Hi everybody,
I would like to ask if it is possible using pdf function or some other to
print plots in cycle such that every new plot is on new page.
like this
pdf(file="D:/Plot.pdf",width = 9.25,height=9.25,
family="Helvetica",pointsize=8,bg="white")
for (i in 1:10){
x <- seq(1,40,1)
y <- 2*x
I missed number of bootstrap replicates R
boot(Reg, bootcoeff, R=10)
but still it doesn't work
Error in statistic(data, original, ...) : unused argument(s) (original)
On Wed, Dec 9, 2009 at 2:46 PM, Trafim Vanishek wrote:
> Dear all,
>
> I have some error trying to bootstrap
Dear all,
I have some error trying to bootstrap from a matrix. The error message is
"Error in sample(n, n * R, replace = TRUE) : element 2 is empty;
the part of the args list of '*' being evaluated was: (n, R)"
vv <- c(0.5,3.2,5.4,1.1,1.4,1.2,2.3,2.0)
Reg <- matrix(data=vv, nrow = 4, ncol = 2)
I would like to ask if there is some simple way for building a hist from the
following data.
mm <-c(2,3,0,4,5,0,2,9,0)
mmm <- matrix(mm,ncol=3, nrow=3)
x <- c(0,1.45,2.9)
in mmm there are frequencies (each column is separate histogram and each row
corresponds to value of x).
So that for the firs
t;
> DM
>
> On Wed, Dec 2, 2009 at 2:14 AM, Trafim wrote:
>
>> Dear all,
>>
>> Please, look at the following code:
>>
>> attach(geyser)
>> f1 <- kde2d(duration, waiting, n = 5)
>>
>> a <- 0
>> for (i in 1:5){
>>
Hi everybody,
well, I definitely don't understand anything.
Why the hist function with freq=FALSE gives such a strange result???
R <- c(-1.10, 0.79, -1.17, -0.53, -0.26, -0.22, 0.29, -0.26, -0.26, 0.39)
hist(R, freq=FALSE, breaks=10)
Thanks everybody
[[alternative HTML version delete
Dear all,
Please, look at the following code:
attach(geyser)
f1 <- kde2d(duration, waiting, n = 5)
a <- 0
for (i in 1:5){
for (j in 1:5){
a <- a + f1$z[i,j]
}
}
As far as I understood from Help kde2d returns matrix elements of which are
values of joint probability mass function Pr(X=x,Y=y) t
Thanks a lot!
I was afraid smth is wrong with my R syntaxis.
On Tue, Dec 1, 2009 at 3:58 PM, Barry Rowlingson <
b.rowling...@lancaster.ac.uk> wrote:
> On Tue, Dec 1, 2009 at 2:01 PM, Trafim wrote:
> > Hi everybody,
> >
> > I have the following problem, the following
Hi everybody,
I have the following problem, the following code seems to run only once for
i and j and for k from one to M.
Doesn't R for increase the argument by itself?
for (i in 1:N){
for (j in 1:(Tk-1)){
if((XGrid[i] < Xk[j+1])&(Xk[j] <= XGrid[i])){
for (k in 1:M){
if ((R
suggestions.
On Mon, Nov 30, 2009 at 5:09 PM, Trafim wrote:
> Also what if the grid matrix is not squared? How can I then find this
> kernel density matrix/
>
> Thanks a lot.
>
>
> On Mon, Nov 30, 2009 at 4:56 PM, Trafim wrote:
>
>> You are right.
>> I brought thi
Also what if the grid matrix is not squared? How can I then find this kernel
density matrix/
Thanks a lot.
On Mon, Nov 30, 2009 at 4:56 PM, Trafim wrote:
> You are right.
> I brought this example just to see how to use its with two time series. In
> reality I have price process an
, Nov 30, 2009 at 4:48 PM, David Winsemius wrote:
>
> On Nov 30, 2009, at 9:30 AM, Trafim wrote:
>
> Unfortunately, it doesn't work.
>> Can you, please, help me with it?
>>
>> In order to support the notion of a 2 dimensional distribution, you need
> a function
Dear all,
I cannot find a function which would allow drawing hist and density on the
same graph.
x <- seq(1,40,1)
y <- 2*x+1+5*rnorm(length(x))
hist(y,freq = FALSE)
plot(density(y))
thanks a lot for the help
[[alternative HTML version deleted]]
Unfortunately, it doesn't work.
Can you, please, help me with it?
Thanks a lot.
On Mon, Nov 30, 2009 at 2:53 PM, Trafim wrote:
> Seems that I found it - kde2d
>
>
> On Mon, Nov 30, 2009 at 2:36 PM, Trafim wrote:
>
>> Hi everybody,
>>
>> I am looking for
Seems that I found it - kde2d
On Mon, Nov 30, 2009 at 2:36 PM, Trafim wrote:
> Hi everybody,
>
> I am looking for the possibility in R to estimate joint density, just for
> example:
>
> x <- seq(1,40,1)
> y <- 2*x+1+5*rnorm(length(x))
> y1 <- x^3+.5*rnorm
Hi everybody,
I am looking for the possibility in R to estimate joint density, just for
example:
x <- seq(1,40,1)
y <- 2*x+1+5*rnorm(length(x))
y1 <- x^3+.5*rnorm(length(x))
Is there a way to approximate the density function s.t. I will later be able
to calculate f(Y=y, Y1=y1)
Thanks a lot
I plot only
hist(y,freq = FALSE)
lines(density(y))
and the graph of density goes higher then hist, but i didn't find any lim
parameter for y axes for lines command.
Thanks a lot
On Mon, Nov 30, 2009 at 1:51 PM, Hrishi Mittal wrote:
>
> Trafim,
>
> If you are plotting more t
Thanks a lot Ted!
I use R help but also cannot find how to expand the graph, i.e. sometimes
the top of it is cut for some reason. Maybe there are some limits imposed on
y axes, but cannot find it.
On Mon, Nov 30, 2009 at 12:27 PM, Ted Harding
wrote:
> On 30-Nov-09 11:09:12, Trafim wr
Dear all,
I cannot find a function which would allow drawing hist and density on the
same graph.
x <- seq(1,40,1)
y <- 2*x+1+5*rnorm(length(x))
hist(y,freq = FALSE)
plot(density(y))
thanks a lot for the help
[[alternative HTML version deleted]]
45 matches
Mail list logo