Hello,
is there any posibility to automatically transform this date notation
"X01.11.2010" into "01.Nov.10" ?
I did not see it before starting my simulation and now I have to deal with
it.
It is no problem for my calculations but for my graphical outputs I would
like
Thank you very much.
It was quite easy to adjust the code to get the background colour I have
always used.
I only tried to get the code using list(qqnorm).
Thanks
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I have tried to build a normal qq plot on my own,
but it does not match with the implemented qqnorm in R.
What am I missing?
As You can see, the points do not match.
qqnorm(c(-2.8,qnorm(seq(0,1,by=(1/200))[2:199],0,1),2.8))
points(qnorm(seq(0,1,by=(1/200))[2:199],0,1),
qnorm(seq(0,1,by=(1/200
Hello,
currently I'm working on a model based on Monte-Carlo-Simulations.
I observed that a generated normal distributed times series using
rnorm(100,mean=0,sd=1)
is far away from being not autocorrelated.
Is there any other gerenator implemented in R, which might solve my problem?
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hHllo,
I'm looking for an algroithm to transform an existing toeplitz matrix
(autocorrelation matrix) to the nearest positive semidefinite toeplitz
matrix.
I merely found an algorithm to transform an correlation matrix via the
function nearcor() based on the algorithm of Higham.
But as I exam
That is exactly what I was looking for.
Thank you
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R-help@r-project.o
I'm having some problems computing a matrix being symmetric on both
diagonals.
Does anyone know a way to get from this matrix
M <- matrix(c(1,0,0,0,2,7,0,0,3,4,0,0,6,0,0,0), ncol=4)
to this one
M_final <- matrix(c(1,2,3,6,2,7,4,3,3,4,7,2,6,3,2,1)
I also tried:
m1 <- length(x)-1
X1<- cbind(x[1:m1],x1[2:length(x)])
X2<- cbind(x[1:m1],x1[2:length(x)])
integral <- function(rho){
m1 <- length(x1)-1
integral <- apply(X2,1,function(y)
apply(X1,1,function(x) pmvno
hello,
I'm trying to improve the speed of my calculation but didn't get to a
satisfying result.
It's about the numerical Integration of a bivariate normal distribution.
The code I'm currently using
x <-
qnorm(seq(.Machine$double.xmin,c(1-2*.Machine$double.eps),by=0.01),
mean=0,sd=1)
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