On Feb 1, 2010, at 8:20 AM, marlene marchena wrote:
Hi,
Thanks a lot it works. But as Ted remarked it was a simple case of a
more
complex
procedure.
Here is what I trying to do (sorry I was not able to find another
example)
I wrote a fuction that I call bullexe(phi,theta,L) the code of this
function
is at the end of this mail.
I want to run different models with my fuction for different values.
I want
to make a table.
I started to write a code to get the results for L=1,2,3,4, and 5
with phi
and theta fixed.
bull<-
matrix
(data
=NA,nrow=5,ncol=1,dimnames=list(c("m1","m2","m3","m4","m5"),c("be")))
bull
VarLT<-
matrix
(data
=
NA
,nrow=5,ncol=1,dimnames=list(c("m1","m2","m3","m4","m5"),c("VarLT")))
VarLT
for(i in 1:5)
{ m=bullexe(0.75,0.1,i)
bull[i,]<-m[1]
VarLT[i,]<-m[3]
}
est<-matrix(data=c(bull,VarLT),nrow=5,ncol=2)
est
It works fine but I am wondering if I can do it directly to fill
scp<-
matrix
(data
=
NA
,nrow
=
5,ncol=2,dimnames=list(c("m1","m2","m3","m4","m5"),c("bull","VarLT")))
scp
You might want to look at mapply or Vectorize if your function will
not take vector arguments
Might be as simple as:
scp <- mapply(bullexe, list(L= 1:5), MoreArgs=list(phi= 0.75,
theta=0.1) )
Since your are returning: bull=M,VarD=VarD,VarLT=VarLT ...
you could drop the unwanted column
scp[ , -2]
--
David.
How could I solve this problem?
and still how could I address the problem if I want to move the three
parameters at the same time?
I found this post https://stat.ethz.ch/pipermail/r-help/2008-May/163110.html
but there the problem is quite different. Any idea?
Here is the code of my fuction bullexe()
bullexe=function(phi,theta,L)
{ arma=ARMAtoMA(ar=phi, ma=theta, 1000);
VarD=sum(arma^2)+1;
values=ARMAtoMA(ar=phi, ma=theta, L);
total = 0
for (i in 1:L)
{ valsum= sum (values[i:L]);
if( i==1)
{
total = valsum;
}
else
{
total = total + values[i-1] * valsum;
}
}
M=1+2*total/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 <ted.hard...@manchester.ac.uk>
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]<-k[i]
print(k)
}
s
Thanks in advance,
Marlene.
Use s[,i]<-k instead of s[,i]<-k[i] since k=sqrt(i) assigns a
single value to k. Hence k[1] will be the same as k, but for
i>1 k[i] will always be NA.
A better way to do the whole thing (in your example) is simply
s <- sqrt(1:5)
but your example may be a very simple case of a more complex
procedure!
Hoping this helps,
Ted.
--------------------------------------------------------------------
E-Mail: (Ted Harding) <ted.hard...@manchester.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 01-Feb-10 Time: 11:41:46
------------------------------ XFMail ------------------------------
[[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, minimal, self-contained, reproducible code.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
______________________________________________
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, minimal, self-contained, reproducible code.
