On Jul 18, 2010, at 4:21 PM, Joe P King wrote:
This is the latest code I have been trying, but it when I try to
turn the
vectors of SD into a vector of variances, it turns into a scalar.
n=c(NULL,13,89,50)
ybar=c(NULL,1.58,1.26,0.80)
sd=c(NULL,2.19,4.18,5.47)
mu=1;sigma=1;p0=0;var1=NULL;p=NULL;pn=NULL
If you are going to passing values to the above using indices then
they should be defined as vectors of the appropriate length. It would
not be absolutely necessary to to so , but if you don't, then you need
to extend them using the c() function, which you do not seem to have
caught onto yet.
for (i in 2:length(n)){
var1[i] = sd[i]^2 #sample variance
p[i] = n[i]/var1[i] #sample precision
p0=p[i-1]
pn[i] = p0[i]+p[i]
mu[i] = ((p0[i])/(pn[i])*mu[i])+((p[i])/(pn[i]))*ybar[i]
sigma[i] = 1/sqrt(pn[i])
mutotals[i,]<-mu[i]
}
Joe King
206-913-2912
j...@joepking.com
"Never throughout history has a man who lived a life of ease left a
name
worth remembering." --Theodore Roosevelt
-----Original Message-----
From: Joe P King [mailto:j...@joepking.com]
Sent: Sunday, July 18, 2010 6:13 AM
To: 'r-help@r-project.org'
Subject: RE: [R] loop troubles
I tried that, this is what I tried, but I only get it to do one
iteration
and then it wont cycle back. I am not sure how to tell it how to
look at the
previous precision to add to the current new sample precision to get
the new
precision. This is my first try at a loop so I am not sure if I am
doing
anything right, sorry about the elementary question.
n=c(5,10,15)
ybar=c(12,13,14)
sd=c(2,3,4)
mutotals<-matrix(nrow=length(n), ncol=1)
mu=1;p0=0
for (i in 1:length(n)){
var = sd[i]^2 #sample variance
p = n[i]/var #sample precision
p0[i]= i
pn = p0[i]+p[i]
mu.out = ((p0[i])/(pn)*mu[i])+((p)/(pn))*ybar[i]
sigma[i] = 1/sqrt(pn[i])
mutotals[i,]<-mu.out[i]
}
Joe King
206-913-2912
j...@joepking.com
"Never throughout history has a man who lived a life of ease left a
name
worth remembering." --Theodore Roosevelt
-----Original Message-----
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Sunday, July 18, 2010 5:36 AM
To: Joe P King
Cc: r-help@r-project.org
Subject: Re: [R] loop troubles
On Jul 17, 2010, at 9:09 PM, Joe P King wrote:
Hi all, I appreciate the help this list has given me before. I have a
question which has been perplexing me. I have been working on doing a
Bayesian calculating inserting studies sequentially after using a
non-informative prior to get a meta-analysis type result. I created a
function using three iterations of this, my code is below. I insert
prior
mean and precision (I add precision manually because I want it to be
zero
but if I include zero as prior variance I get an error cant divide
by zero.
Now my question is instead of having the three iterations below, is
there
anyway to create a loop, the only problem is I want to have elements
from
the previous posterior to be the new prior and now I cant figure out
how to
do the code below in a loop.
Perhaps if you set the problem up with vectors, it would become more
obvious how to do it with indexing or loops:
n <- c(5, 10, 15)
ybar <- c(12,12,14) # and so on...
--
David
The data below is dummy data, I used a starting
mu of 1, and starting precision of 0.
bayes.analysis.treat<-function(mu0,p0){
n1 = 5
n2 = 10
n3 = 15
ybar1 = 12
ybar2 = 13
ybar3 = 14
sd1 = 2
sd2 = 3
sd3 = 4
#posterior 1
var1 = sd1^2 #sample variance
p1 = n1/var1 #sample precision
p1n = p0+p1
mu1 = ((p0)/(p1n)*mu0)+((p1)/(p1n))*ybar1
sigma1 = 1/sqrt(p1n)
#posterior 2
var2 = sd2^2 #sample variance
p2 = n2/var2 #sample precision
p2n = p1n+p2
mu2 = ((p1n)/(p2n)*mu1)+((p2)/(p2n))*ybar2
sigma2 = 1/sqrt(p2n)
#posterior 3
var3 = sd3^2 #sample variance
p3 = n3/var3 #sample precision
p3n = p2n+p3
mu3 = ((p2n)/(p3n)*mu2)+((p3)/(p3n))*ybar3
sigma3 = 1/sqrt(p3n)
posterior<-cbind(rbind(mu1,mu2,mu3),rbind(sigma1,sigma2,sigma3))
rownames(posterior)<-c("Posterior 1", "Posterior 2", "Posterior 3")
colnames(posterior)<-c("Mu", "SD")
return(posterior)}
David Winsemius, MD
West Hartford, CT
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