regards,
Christoph Buser
## R example
postscript(test.eps, width = 14, height = 8,
onefile = FALSE, horizontal=FALSE, paper=special)
plot(1:10)
dev.off()
embedFonts(file = test.eps, outfile = test1.eps)
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,
Christoph
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and provide commented, minimal, self-contained, reproducible code.
Hope this helps
Christoph Buser
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output, the random effects are
## standard deviations and not variance components and you
## should square them to compare them with Montgomery
## 1.307622^2 = 1.71 1.624466^2 = 2.64
## Or you can use
VarCorr(material.lme)
I hope this helps you.
Best regards,
Christoph Buser
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Sebastian Weber writes:
Hello everyone
Christoph Buser
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Regards,
Christoph
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Dear John
?ordered
will help you.
Regards,
Christoph Buser
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Christoph
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Henrik Parn writes:
Dear all
Dear Joseph
Have a look at the questions and answers in the two links
below. There the topic has been discussed.
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/68905.html
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/6943.html
Best regards,
Christoph Buser
.
Regards,
Christoph Buser
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,
Christoph Buser
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(y~habitat, random = ~1|lagoon/habitat, data = dat))
anova(reg2)
Best regards,
Christoph Buser
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habitat.
aov() will not return you the test and warn you about the
singular model.
lme() will estimate a variance component for lagoon, but does
not provide you a test for the fixed factor.
Regards,
Christoph Buser
set.seed(1)
dat - data.frame(y = rnorm(100), lagoon = factor(rep(1:4,each = 25
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COMTE Guillaume writes:
Thks for your answer
do know the gls syntax.
Best regards,
Christoph
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fitted object is passed to the
function. Default is 'TRUE'.
Regards,
Christoph Buser
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in my
understanding of the help page and the R code?
Thank you very much.
Best regards,
Christoph Buser
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results if
the unit is chosen automatically and removed afterwards.
Regards,
Christoph Buser
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Fred J. writes
it.
Regards,
Christoph
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that is under development which needs some time.
Regards,
Christoph
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of your coefficients, especially when you have an
interaction in your model.
Regards,
Christoph Buser
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phone: x
Buser
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Hi Rick
There may be a better way, but the following should work:
attributes(vc.fit)$sc
Regards,
Christoph Buser
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example above.
Regards,
Christoph Buser
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mixed)
Regards,
Christoph Buser
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Hi
try ?aov
That's a good starting point with references to other functions
in R and some literature.
Regards,
Christoph Buser
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Dear Julien
Have a look at the type argument in ?anova.lme
Regards,
Christoph Buser
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coefficients have been
estimated, due to singularities.
It is possible to use other contrasts than contr.treatment,
contr.sum, contr.helmert, contr.poly, but then you have to
specify the correctly in the model.
Regards,
Christoph Buser
://finzi.psych.upenn.edu/R/Rhelp02a/archive/67414.html
Regards,
Christoph Buser
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data.frame first to summarize your repeated values to unique
ones and the proceed with merge, but that depends on your
problem.
Regards,
Christoph
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[EMAIL
Hi
It is not so clear to me what your intention is. Could you
provide a reproducible example to show what you want to do.
Regards,
Christoph Buser
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Dear Claus
You can write
earlySessions[,seconds] instead of earlySessions$seconds and
with that syntax you can also use:
col1 - seconds
earlySessions[,col1]
and you will get what you are looking for.
Best regards,
Christoph Buser
for similar
problems. Lots of answers are already there and only wait to be
found.
So there is a good chance that you will find the information
that you are looking for in these archives.
Best regards,
Christoph Buser
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Hi
Have a look at ?polyroot. This might be helpful.
Regards,
Christoph Buser
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Buser
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if it is a straight line or not
(since the sample is to small) and in this case it is the
correct conclusion that we can not say anything about the
distribution.
I hope this will be helpful.
Regards,
Christoph Buser
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,
Christoph Buser
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)
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Spencer Graves writes:
Others
Hi
If you'd like to fit a fixed effect model without random
effects, you can use lm() or aov() (see ?lm and ?aov). If your
variable is a factor (?factor) then you can specify your model
in lm() without coding all dummy variables.
Regards,
Christoph Buser
Hi
Maybe matplot (see ?matplot) can help you.
Regards,
Christoph Buser
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to the first posts about t tests
and wilcoxon test.
I attached the email below and recommend to read it carefully. It
might be helpful for you, too.
Regards,
Christoph Buser
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,
Christoph Buser
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What about
x-c(0,0,0.28,0.55,1.2,2,1.95,1.85,
1.6,0.86,0.78,0.6,0.21,0.18)
y-c(0,0,0,0.53,1.34,1.79,2.07,1.88,
1.52,0.92,0.71,0.55,0.32,0.19)
sum(((x-y)^2/x)[x!=0])
Regards,
Christoph Buser
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= )), pos = 4)
I do not have an elegant solution for the alignment.
Regards,
Christoph Buser
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with the previous value of the
chain it should work in my context.
Best regards,
Christoph
[EMAIL PROTECTED] writes:
Quoting Christoph Buser [EMAIL PROTECTED]:
Dear R-users
I have a problem choosing initial points for the function arms()
in the package HI
I intend to implement
of your sample, you
know the true means and that they are different. It is only a
question of the sample size and the power of your test, if this
difference is detected.
Is that something you are investigating? Maybe a power
calculation or something similar.
Regards,
Christoph Buser
(z20.05)/1000 ## 0.81
Regards,
Christoph Buser
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]*co[x1] + co[f1b]
## prediction of observation 11
predict(reg, type=c(response))[11]
Regards,
Christoph
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point
y.start. To show the effect I constructed a demonstration
example. It is reproducible without further information.
Please note that my target density is not logconcave.
Thanks for all comments or ideas.
Christoph Buser
## R Code:
library(HI)
## parameter for the distribution
para - 0.1
it on FALSE if you do not want to
apply Yates continuity correction.
Regards,
Christoph Buser
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very much for the solution.
Regards,
Christoph Buser
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There is also the package maptools if you want or need to read
ESRI shapefiles.
Regards,
Christoph Buser
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it generally.
Hope this helps
Christoph Buser
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will get the values that are behind the original
alphabetical ordering, meaning sp12 is 1, sp2 is 2, etc.
You can change this, too if necessary, using as.character() and
as.numeric() as well.
I hope this is helpful fro your problem.
Regards,
Christoph Buser
. Therefore it will be better to use your own,
hopefully better programmed function in optimize.
Regards,
Christoph Buser
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regards,
Christoph Buser
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minor1.1
year 2005
month06
day 20
language R
Best regards,
Christoph Buser
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Down to 17 if we carry on your idea using the matching property
on the left hand side, too. :-)
p$h=pmax(p$h,p$s)
Gabor Grothendieck writes:
Agree that this definitely should be pursued. :) In fact,
we can shave off several keystrokes by
- replacing p[[1]] with p[1] on the left hand
If you have to deal with missing values, you might be interested
in na.omit().
?na.omit
Regards,
Christoph
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)])^2))
Regards,
Christoph Buser
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text within the plot
See ?legend?text
Regards,
Christoph
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2001-12-07 10:32:00 CET
[3] 2001-11-16 13:58:00 CET 2001-11-28 14:00:00 CET
[5] 2001-11-02 15:22:00 CET 2001-11-26 11:15:00 CET
Regards,
Christoph
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check those to
find out which one you prefer.
Best regards,
Christoph Buser
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Hi
the following works for me:
j - box
z - plot
jpeg(file=paste(YOURPATH,j,z,.jpg, sep = ))
boxplot(rnorm(100))
dev.off()
Regards,
Christoph Buser
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)
major2
minor1.0
year 2005
month03
day 22
language R
Regards,
Christoph Buser
and got 1 or
2.
I recommend to use graphical methods (e.g. normal plots) to
validate the normal distribution of your data instead of testing
it.
See also ?qqnorm or ?qqplot.
Regards,
Christoph Buser
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the rank sum of the smaller group is
sum(1:n1) = sum(1:10) = 10*(10+1)/2 = n1*(n1+1)/2
If you compare this to the test statistics, you'll observe that
in this case the test statistic is 0.
Regards,
Christoph Buser
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bit more about this issue
for my better understanding.
I have one last point to propose:
You could include (as interim solution) a warning (that there might
be an in efficiency loss) in se.contrast() if one uses
non-orthogonal contrasts and a multi-stratum model.
Best regards,
Christoph Buser
)
in this nice balanced design, I am suspicious, especially if you
real design is more complicated.
Even if you get no error message for your data, I'd calculate
the analysis using for example lme for a second time to control
the results.
I wish you all the best for your analysis.
Christoph Buser
-nb.temp+1
}
}
nb[j]=nb.temp/n
}
return(nb)
}
Attention: I didn't look in detail what you did so you sould
check my changes if it is really what you're looking for !!
Best regards,
Christoph
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.
Ripley was right and you have another design. Then you just can ignore this
and your life is much more easier :-)
Best regards,
Christoph Buser
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Hi Jim
I'm not sure if I understand your problem correctly. Is this a solution?
(li - list(a=1,b=200.44))
as.numeric(paste(unlist(li), collapse = ))
Best regards,
Christoph Buser
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Hi Eric
If you want produce a vector with names, you can use
v - rnorm(20)
names(v) - paste(Lab,1:20, sep=)
Regards,
Christoph
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different. As result of the balanced design the
explanatory variables are orthogonal and you can decompose the
sum of squares on the different variables.
But if you have covariables you must be careful. I wouldn't try
to assign r square values for each Variable.
Regards,
Christoph Buser
,
Christoph Buser
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kolluru ramesh writes:
Can we do Cholesky Decompositon in R for any matrix
[1] 115 135
d - scan(cn.x, n=1, sep = )
Read 1 items
d
[1] 175
Regards,
Christoph Buser
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.agr - aggregate(dat2$y, by = list(lot = dat2$lot), FUN = mean)
names(dat2.agr)[2] - y2
dat.mer - merge(dat1, dat2.agr)
str(dat.mer)
Be careful about merging dataframes. There should always be a
control that the right cases are merged together.
Regards,
Christoph Buser
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- estimated parameters
I think this test is more powerful than the KS test,
particularly if you must estimate the parameters from data.
Regards,
Christoph
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__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Best,
Christoph
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Hi Robin
There is a function factorize() in the package conf.design
library(conf.design)
factorize(60)
[1] 2 2 3 5
Hope this helps
Christoph
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ETH (Federal Inst. Technology) 8092 Zurich SWITZERLAND
phone: x-41-1-632-5414
must use the lme
syntax in another way for the lme4 package.
Regards,
Christoph Buser
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
references for more details, too.
Hope this will help you
Christoph
--
Christoph Buser [EMAIL PROTECTED]
Seminar fuer Statistik, LEO C11
ETH (Federal Inst. Technology) 8092 Zurich SWITZERLAND
phone: x-41-1-632-5414 fax: 632-1228
http://stat.ethz.ch/~buser
3264.9 256.8
##- Litter 3 63.6 3328.5 252.0 0.3508 0.78870
##- Mother 3 775.1 4040.0 263.8 4.2732 0.00886 **
##- ---
##- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1
##- Litter stays not signifikant.
Best regards
Christoph Buser
--
Christoph
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