Hi R Users,
I'm going to estimate via. ML the parameters in Poisson Lognormal
model. The model is:
x | lambda ~ Poisson(lambda)
lambda ~ Lognormal(a,b)
Unfortunately, I haven't found a useful package allowing for such
estimation. I tried to use "poilog" package, but there is no equations
and it's
2010/3/26 Charles C. Berry :
> On Fri, 26 Mar 2010, Robert Ruser wrote:
> So this is the generalized linear model with a poisson family, log link, and
> a Gaussian random effect in the linear predictor.
>
> Take a look at lme4, MASS (glmmPQL), and try searching CRAN packages for
&
Chuck thank you very much for help. I will follow your advice and see
what happen.
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Hello,
I want to use the free distribution of R (R REvolution 3.2) and Tinn-R
editor as well. Unfortunately they don't cooperate. In Tinn-R
commands: send selection, send line etc. don't work. Do you have any
idea how to resolve this problem?
Best,
Robert
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> www.r-statistics.com (English)
> --
>
>
>
>
> On Mon, Apr 19, 2010 at 6:50 PM, Robert Ruser
> wrote:
>>
>> Hello,
>> I want to use the free distribution of R (R REvolution 3.2) and Tinn-R
>> editor as well. Unfortunately they don't cooperate.
I use lattice package and 'barchart' to build a chart. I have a
problem with setting different x-axes. Some x categories are missing
but they are display and I don't want. I use scales = list(y =
"free",x="free") but it works only for y-axes. Simple example:
package(lattice)
barchart(yield ~ varie
Hello R Users,
I'm wondering if there exists any elegant and simple way to do the
following: I have a data.frame X fills in numbers. I have a vector y with
numbers as well. Every value in X that is equal to any values in y should
be replaced by e.g. 1. Any idea?
Robert
___
Thank you vary much Ben and Erik.
It's exactly what I want. Below is my a little modified example.
set.seed(12345)
X = sample(c(40:60),40,replace=TRUE)
x = matrix(X,nc=5)
y = c(40,43,55,60)
x[x %in% y] <- -1
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Dear R Users,
I'm looking for a package that allows to test hypothesis about a
homogeneity of odds ratio in k 2x2 tables. I know that Breslow-Day is
suitable but does anybody could me point out a package? I found diffR,
but as far as I see this package is for IRT theory.
Best,
Robert
Hello R Users,
I have vectors
x <- c("a2","b7","c8")
y1 <- c(1,2,3,2)
y2 <- c(4,2,7,5,4,3,8)
y3 <- c(1:10)
and I want to assign values form vector y1 to a new variable which
name comes from the 1st value of the vector x etc. How to do it using
only vector x. As a result I should have:
a2 <- y1
b7
uot;c8")
> dat
> # example
> dat$a2
> dat[["a2"]]
>
>
> ## or using your variables
> dat2 <- list(y1, y2, y3)
> names(dat2) <- x
>
> Cheers,
>
> Josh
>
>
> On Fri, Nov 19, 2010 at 2:57 AM, Robert Ruser wrote:
>> Hello R
Joshua, Dennis ans Henrique I'm immensely indebted to you. Thanks to
you I resolved my problem - a code works perfectly. I put the code,
maybe someone finds it's helpful.
f1 <- readLines(file("file1.txt"))
f2 <- readLines(file("file2.txt"))
for ( i in seq_along(f1)){
assign(f1[i],as.numeric(unli
Hello,
Let's assume that one has a list:
my.list <- list(a=c(1,2,3,4),b=4,c=c(1:7))
I want to save my.list as a matrix in .txt file. Because of different
length we can put NA or 0.
Effect (row.names a, b and c are not necessary)
a 1 2 3 4 0 0 0
b 4 0 0 0 0 0 0
c 1 2 3 4 5 6 7
How to do it?
Than
Hi Mohamed,
It works. Thank you very much.
Best,
Robert
2010/11/20 :
> Hi Robert,
>
> Try to do this
>
> Len <- max(sapply(my.list, length))
>
> fun1 <- lapply(my.list, function(x){
> c(x, rep(0, Len))[1:Len]
> })
> do.call(rbind, fun1)
>
> M
> Regards
Hello,
I'm wondering how to set a value of mar ( par( mar=c(...)) ) in
order to allow labels to be visible in barplot. Is there any relation
between the number of characters in a label and the second value of
mar? Look at my example.
x <- seq(20, 100, by=15)
ety <- rep( "Effect on treatment gr
2010/12/21 Gerrit Eichner :
> par( mar = c( 3, 13, 2, 1), cex = 0.8)
>
> barplot( x, names.arg = NULL, horiz = TRUE, axes = FALSE)
>
> axis( side = 1, at = c( seq( 0, 80, by = 20), 95))
>
> axis( side = 2, at = 1:length(ety), line = -1, las = 1, tick = FALSE,
> labels = ety)
Thank you very mu
2010/12/21 Dieter Menne :
> Standard graphics has fallen a bit out of favor because of these quirks. Try
> lattice:
>
> library(lattice)
> x <- seq(20, 100, by=15)
> ety <- paste("Effect on treatment group",1:length(x))
> barchart(ety~x)
>
> Note that the ety labels must be different to make this w
Dear R Users,
Using lm() function with categorical variable R use contrasts. Let
assume that I have one X independent variable with 3-levels. Because R
estimate only 2 parameters ( e.g. a1, a2) the coef function returns
only 2 estimators. Is there any function or trick to get another a3
values. I
em.
Robert
2011/6/12 Prof Brian Ripley :
> ?dummy.coef
>
> (NB: 'R' does as you tell it, and if you ask for the default contrasts you
> get coefficients a2 and a3, not a1 and a2. So perhaps you did something
> else and failed to tell us? And see the comment in ?dumm
HTH,
> Jorge
>
> On Sun, Jun 12, 2011 at 4:40 PM, Robert Ruser <> wrote:
>>
>> Prof. Ripley, thank you very much for the answer but wanted to get
>> something else. There is an example and an explanation:
>>
>> options(contrasts=c("contr
Hi Weidong,
thank you very much. It really works fine.
Robert
2011/6/12 Weidong Gu :
> this may work.
> X<-data.frame(sapply(X,function(x) as.factor(x)))
> reg3=lm(Y~.,data=X)
> dummy.coef(reg3)
>
> Weidong Gu
>
> On Sun, Jun 12, 2011 at 4:55 PM, Robert Ruser wrote:
Dear R Users,
I'm wondering how is it possible to get |factor loadings| <1 in factor
analysis model (factanal) performing maximum-likelihood estimation. Is
it caused by some constraints? If so, what kind of constraints are
placed and when (during the estimation or after)?
Robert
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