This is my code
BSUPred-(forecast(BSU,h=h)[[2]])
PressurePred-(forecast(Pressure,h=h)[[2]])
Placer-(rep(1,h))
test-as.data.frame(cbind(BSUPred,PressurePred))
test$Placer-rep(1:2,h/12)
test$Placer-i
test-as.data.frame((test[c(Placer,BSUPred,PressurePred)]))
This is not reproducible (see for example [1]), so it is very difficult to know
exactly what the problem is. Also, you need to post on this list using the
plain text format option in your email software, since the HTML format you used
can mess up your code.
I can say that using [ indexing on a
Hi all,
I am trying to use the mle function in R to find the maximum likelihood
estimator. The ll function below is the negative of the log likelihood.
Suppose x0 is the observed values, I want to find the maximum likelihood
for a and b. After running the code below, I get the error message
You do not appear to provide initial values for a and b , i.e. the
start argument for mle.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
Clifford Stoll
On Sun, Nov 2,
Hi list,
I have trying to calculate the covariance/correlation of three elements. I
have vector say,
v - c(700, 800, 1000)
I want to have a 3 by 3 correlation matrix, meaning cor(v1, c2), cor(v1,
c3), cor(v2, v3), etc...
So far I get,
cor(v)
Error in cor(v) : supply both 'x' and 'y' or a
CRAN (and crantastic) updates this week
New packages
* addreg (1.2)
Maintainer: Mark Donoghoe
Author(s): Mark Donoghoe mark.donog...@mq.edu.au
License: GPL (= 2)
http://crantastic.org/packages/addreg
Methods for fitting identity-link GLMs and GAMs to discrete data. The
What is your question? The matrix form is probably what you are looking for,
but you put the same vector in three times so if course it is all ones. I don't
know what you expected to happen when you entered cor(v) since there is nothing
to correlate if you only have one vector.
Please post in
Thanks, Jeff. I had some misunderstanding.
So, I want to calculate the squared exponential of vector v
v = c(700, 800, 1029)
formula is:
k(x_i, x_j)=sigma^2 * exp(-1/(2*l^2) * (x_i - x_j) ^2)
where,
sigma=7, l=100
I used,
v - c(700, 800, 1029)
corr.matrix(cbind(v),scales=0.5)
[,1]
k - sigma^2 * exp( -1/(2*l^2) * outer( v,v,FUN=function(x,y){(x-y)^2}))
but perhaps you should look at the e1071 package instead?
---
Jeff NewmillerThe . . Go Live...
Thanks Bert for the reply. I still get a message when adding the start
argument.
n - 8
x0 - c(2,3)
ll- function(a,b,x=x0,size=n){
+
-sum(log(gamma((n-1)/2+a-1)/(gamma((n-1)/2)*gamma(a))*1/(2*b^a)*(x/2)^((n-1)/2-1)*(1/b+x/2)^(-((n-1)/2+a-1}
fit - mle(ll, start=list(a=3, b=1), nobs =
I think I made an error in my funciton. Now it works.
library(stats4)
n - 8
ll- function(a,b,x){
-sum(log(gamma((n-1)/2+a-1)/(gamma((n-1)/2)*gamma(a))*1/(2*b^a)*(x/2)^((n-1)/2-1)*(1/b+x/2)^(-((n-1)/2+a-1}
fit - mle(ll, start=list(a=3, b=1), fixed=list(x=c(2,3)))
2014-11-02 22:04
On Nov 1, 2014, at 2:58 AM, Upananda Pani wrote:
Dear All,
I am getting the following error when i am using interpNA function from
timeSeries package
# Missing Value Treatment (Linear Interpolation)
spt = interpNA(spt, method = c(linear))
Error in interpNA(spt, method = c(linear)) :
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