to contribute
to the project please do not hesitate to contact me.
Kind regards,
Marlene Marchena
[[alternative HTML version deleted]]
___
R-packages mailing list
r-packa...@r-project.org
https://stat.ethz.ch/mailman/li
t package
for R.
Suggestions, bug reports and other comments are very welcome.
All the best,
Marlene Marchena.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the pos
x, na.rm = TRUE)))
> > k
> [[1]]
> [1] 2 4
>
> [[2]]
> [1] 4
>
> [[3]]
> [1] 4
>
> [[4]]
> [1] 4
>
> > sapply(k, paste, collapse = ', ')
> [1] "2, 4" "4""4""4"
> >
>
>
> On Fr
age for ?min.
> JC
>
> 2011/9/16 marlene marchena :
> > Hi,
> >
> > I need to repor the index of a min value of each row in a matrix, but I
> > don't know how to do that when I have more than one min value.
> >
> > Here is my example
> >
> &
Hi,
I need to repor the index of a min value of each row in a matrix, but I
don't know how to do that when I have more than one min value.
Here is my example
> dat <-
matrix(c(5.4,4.8,5.6,4.8,NA,4.4,4.6,3.4,NA,NA,4,2.4,NA,NA,NA,2),byrow=TRUE,ncol=4)
> dat
[,1] [,2] [,3] [,4]
[1,] 5.4 4.8
Very sorry about that!
2010/7/19 marlene marchena
> Hola amor!
>
> Finalmente llegué a D.C. Ya estoy en el albergue pero el cuarto solo va a
> ser liberado a la una o tal vez dos de la tarde. Pero no importa al menos ya
> estoy aqui, ya tomé desayuno que por cierto es horrib
Hola amor!
Finalmente llegué a D.C. Ya estoy en el albergue pero el cuarto solo va a
ser liberado a la una o tal vez dos de la tarde. Pero no importa al menos ya
estoy aqui, ya tomé desayuno que por cierto es horrible. Voy a poder tomar
baño y usar un locker por un dolar. Entonces voy a tomar baño
Hi R users!
Is it possible to cut the edges from a surface graph without to cut the axes
using persp function?
I am using the following code:
op <- par(bg = "white")
persp(phi1, phi2, z,main="Bullwhip generated with AR(2) demand when L=1",
xlab ="phi1" , ylab ="phi2", zlab ="Bullwhip", t
VarD;
arma1=ARMAtoMA(ar=phi, ma=theta,L);
arma2=arma1^2;
totalLT = 0
for (i in 1:L-1)
{ valsumLT= sum(arma2[1:(L-i)]);
totalLT = totalLT + valsumLT;
}
VarLT=L+totalLT;
sc=c(bull=M,VarD=VarD,VarLT=VarLT)
return(sc)
}
2010/2/1 Ted Harding
> On 01-Feb-10 11:29:40, marlene marchena wrote:
&
Hi R-users
I'm writing a code to run a fuction but I found an error that I can't fix. I
reproduced the error with a simple example.
The correct answer is k but I can't fill my s matrix. What I'm doing wrong?
s<-matrix(data=NA,nrow=1,ncol=5 )
s
for(i in 1:5)
{
k=sqrt(i)
s[,i
Hi Matt,
The ARMA model write the MA part with positive coefficients, therefore the
formula that you write for the variance is wrong.
Here an example where the answer is the same for both methods.
phi=0.75
theta=-0.4
coefvar=(1+theta^2+2*phi*theta)/(1-phi^2)
coefvar
result <- ARMAtoMA(ar=c(0.75
both problems with overwriting subsequent
> x-values and obeying the precedence of R operators.
> For the latter, see ?Syntax and compare
> i<-5
> i-3:i-1
> (i-3):(i-1)
>
> hth.
>
> marlene marchena schrieb:
>
>> Dear R-user
>>
>> Suppose I have
Hi Petr,
Thanks for your suggestion. It woks, but now I have other problem the
positions of the values changed. I need the NA values in the three first
positions.
y=c(2,1,5,8,11,3,1,7,50,21,33,7,60)
x=as.zoo(y)
> x
1 2 3 4 5 6 7 8 9 10 11 12 13
2 1 5 8 11 3 1 7 50 21 33 7 60
va
Dear R-user
Suppose I have the following data
y=c(2,1,5,8,11,3,1,7,50,21,33,7,60)
x=data.frame(y)
for(i in 4:nrow(x)) x[i,] =var(x[i-3:i-1,])
I'm trying to get a new variable with the variance of the 3 previous values
(just an example) and with NA in the three first positions. I know that my
Hi,
A long time I have some problems to run a SVM - regression. Here an example
with the Ozone data that represents very well my own data.
data(Ozone, package = "mlbench")
#I cut the three first variables and splite the data in two parts
Ozone<- na.omit(Ozone[, -(1:3)])
index <- 1:nrow(O
5
-0.978089239 1.331999124
8 -0.683566350 -0.588683840 1.708596139 0.141129535 -0.656754703
-0.051154801 -0.970752111
9 -0.028332192 -0.687146676 -0.590083898 1.644118288 0.113678155
-0.653662187 -0.037204315
10 0.659663673 -0.030727774 -0.688598758 -0.594375599 1.562091927
0.11878
Dear R users,
I'm running a SVR in package e1071 but I did not able to calculate the
parameters w and b of the regression. I don't know how to do that and if it
is possible to do it with this package.
Someone have some idea. Any help would be much appreciated.
Marlene
[[alternative HTML
Hi there,
I take advantage of this chat to ask other question related to logistic
regression. This is my first time as well.
I have data that I want to model but Im not sure if glm() is the correct
function to use. My problem is as follow, I used Oxford Instability Score of
the shoulder (OIS,
>
> Hi all,
>
> I need some help about how to calculate w in a SVR in package e1071.
>
> I have a regression y_i=f(x_i)+e
>
> where f(*x*)=(w,phi(x))+b
>
> then go on with the SVR calculation I know that w*=Sum_i=1^n [(á_i -
> á*_i)K(x,x_i) ] where á_i and á*_i are the lagrangian multipliers of the
Hi R user,
I need some help about how to calculate w in a SVR in package e1071.
I have a regression y_i=f(x_i)+e
where f(*x*)=(w,phi(x))+b
then go on with the SVR calculation I know that w*=Sum_i=1^n [(á_i -
á*_i)K(x,x_i) ] where á_i and á*_i are the lagrangian multipliers of the
dual form.
o.
So christofer proved (carry on the same procedure infinite times) that the
forecast of the variance converge to the long run variance when a+b<1
E(Sigma[t+2]^2) = w/(1-a-b) + [(a+b)^t]* Sigma[t+1]^2
Therefore when you predict n.ahead = 20 it must to converge to the long run
variance. It can be
Thanks a lot Andrea, it works!
Marlene.
2009/5/29 Andrea Weidacher
> Hi Marlene,
>
> try this:
>
> x[which(x==Inf)] <- 0
>
> Andrea.
>
> ------
> *Von:* marlene marchena
> *An:* R-help@r-project.org
> *Gesendet:* Freitag, den 29.
Hi R users,
Someone knows how to replace Infinite value by zero. I have a vector with
some Inf value and I want to substitute these values by zero to get the mean
of the components of the vector.
Any idea?
Many thanks,
Marlene.
[[alternative HTML version deleted]]
Hi,
> Is the variable st character or a factor? What does str(train$st) show?
> str(train$st)
Factor w/ 208 levels "0,000","0,0058643",..: 132 134 41 29 42 151 195 195
196 207 ...
Thank you very much Max with your help I found my error, now it works.
Marlene.
Hi R users,
I'm trying to run a SVM - regression using e1071 package but the function svm()
all the time apply a classification method rather than a regression.
svm.m1 <- svm(st ~ ., data = train, cost = 1000, gamma = 1e-03)
Parameters:
SVM-Type: C-classification
SVM-Kernel:
25 matches
Mail list logo