Hi R users.
I've got a problem in producing the pdf file from Latex with R code. When I
run the code Sweave("example.Rtex") in R it seems working, but when I run
the Latex file it doesn't. The code error shown to me is below:
*Runaway argument?
{echo=FALSE}
data<- read.csv("C:\\Users\\Daniele\\D
Hi everybody.
I'm trying to use R with Sweave but I have a problem perhaps with the
directory path of sweave in R.
The windows path is this:
C:\Program Files (x86)\R\R-2.9.2\share\texmf\Sweave
When I run the latex file with R, the program works well, without any
errors, but when I create the pdf
Thank you very very much... It's perfect to my goal...
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Hi R-users,
I'm trying to built a plot of two series of data, but thees series
result "superimposed". The R-code is like this:
goal <- c(125, 143, 81, 26, 2, 3)
numgoal <- 0:5
lambda <- sum(goal*numgoal)/sum(goal)
plot(numgoal, goal, type="h")
x <- 0:5
y<-dpois(x, lambda)
att<-y*380
The code is like this:
plot(0,0, type="n", xlim=c(1,15), ylim=0:1, xlab="...",
ylab="...", main="Example)
lines(1:15, line1, col=4)
lines(1:15, line2, col=2)
legend("topright", c("alpha = 0.2", "alpha = 0.4"), col=c(4,2), lty=1)
How can I build the vectors line1 and line2 ?
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Dear R-users,
I have to plot a exponential distribution like the plot in the pdf
attached. I've write this code but I don't know how to draw the two lines..
Can anyone help me please?
Thank you very much
Pippo http://r.789695.n4.nabble.com/file/n3394476/exponential_smoothing.pdf
exponentia
Hi to everyone,
if the estimate of the parameter results in 0.196 and his standard
error is 0.426, can I say that this parameter is not significant for the
model?
Thank you very much
Pippo
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Hi R-users,
I've found the error
In these rows:
# Model (a)
testtemp <- lm.bp(doctorco~sex+age+income, prescrib~sex+age+income,
data=bootdata)
betafound <- c(testtemp$beta,testtemp$beta3)
results[i,] <- betafound
betafound must to be equal to:
betafound <- c(testtemp$beta1,testtemp$beta
Thanks for the answer.
The R code is here:
library("bivpois")
data("ex3.health")
# Bivariate Poisson models
ex3.model.a <- lm.bp(doctorco~sex+age+income, prescrib~sex+age+income,
data=ex3.health)
Bootstrap standard errors can be obtained easily using the following script
for model (a):
n <- lengt
Hi,
I have the same problem to find out the standard errors of the parameter
in the same package you have used.
I couldn't find out how to get standard errors and p-values from the
package, so I bootstrapped them.
Can you explain your method to find out the standard errors with the
bootstrap m
Can anyone help me please?
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https://stat.ethz.
when I calculate the bootstrap standard errors the results are summarized in
a matrix with columns equal to the parameters of the model. In my case I
have a matrix like this:
[,1] [,2] [,3] [,4] [,5]
[,6][,7] [,8][,9]
[
I'm sorry it was my mistake. The two vectors are found as:
dpois(num,0.87)
dpois(num,2.08)
and represent the discrete density of 8 intervals (num=0:8).
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Hi,
I have two vector with the marginal distribution like this:
> a
[1] -0.419 -0.364 -0.159 -0.046 -0.010 -0.002 0.000 0.000 0.000
> b
[1] 0.125 0.260 0.270 0.187 0.097 0.041 0.014 0.004 0.001
How can I calculate the joint distribution with R?
Thank you to all
Dan
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Hi everyone,
if I have the bootsrap values of the parameters of my regression model,
how can I calculate the standard error of them to refer to the parameters?
Thank you
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Sent from
Hi everyone,
I'm building a matrix in R with a cycle for like this:
pp_ris2=matrix(NA,6,6)
for(i in 0:6){
for(j in 0:6){
if(i>j){
pp_ris2[i,j]=myfunction}
else if(i==j){
print(c(i,j))
pp_ris2[i,j]=myfunction}
but if in my function
pp_ris2[i,j]=myfunction}
must be the indexes 0-0,0-1,0-2,0-3, ?
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