From: Ogbos Okike
> Sent: Mittwoch, 27. Februar 2019 22:53
> To: Meyners, Michael
> Cc: r-help
> Subject: Re: [R] Randomization Test
>
> Dear Kind List,
>
> I am still battling with this. I have, however, made some progress with the
> suggestions of Micheal and othe
Ogbos,
You do not seem to have received a reply over the list yet, which might be due
to the fact that this seems rather a stats than an R question. Neither got your
attachment (Figure) through - see posting guide.
I'm not familiar with epoch analysis, so not sure what exactly you are doing /
Juan,
Your question might be borderline for this list, as it ultimately rather seems
a stats question coming in R disguise.
Anyway, the short answer is that you *expect* to get a different p value from a
permutation test unless you are able to do all possible permutation and
therefore use the
Apologies if I missed any earlier replies - did you check
multcompLetters in package {multcompView}?
It allows you to get connecting letters reports (if that's what you are after,
I didn't check what exactly agricolae is providing here). May have to add some
manual steps to combine this with any
Did you search the internet? At first attempt,
vegdist {vegan}(worked well for me in the past) and
dist.binary {ade4}
seem to offer what you need.
HTH, Michael
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of sreenath
> Sent: Mittwoch, 15. J
Fox [mailto:j...@mcmaster.ca]
> Sent: Dienstag, 7. Juli 2015 03:24
> To: angelo.arc...@virgilio.it
> Cc: Meyners, Michael; r-help@r-project.org
> Subject: Re: [R] R: RE: Bonferroni post hoc test in R for repeated measure
> ANOVA with mixed within and between subjects design
>
&g
Untested, but if anything, your best bet is likely something like
summary(glht(lme_H2H, linfct=mcp(Emotion = "Tukey")),
test=adjusted("bonferroni"))
should work (despite the question why you'd want to use Bonferroni rather than
Tukey
For a reference, see the book on the topic by the package au
ken and working solution found, so all fine :-)
Cheers, Michael
> -Original Message-
> From: Martin Maechler [mailto:maech...@stat.math.ethz.ch]
> Sent: Montag, 8. Juni 2015 16:43
> To: Meyners, Michael
> Cc: r-help@r-project.org
> Subject: Re: [R] mismatch between ma
ple, it does. I must be missing something
fundamental...
Michael
> -Original Message-
> From: Meyners, Michael
> Sent: Montag, 8. Juni 2015 12:02
> To: 'r-help@r-project.org'
> Subject: mismatch between match and unique causing ecdf (well,
> approxfun) to fail
&
All,
I encountered the following issue with ecdf which was originally on a vector of
length 10,000, but I have been able to reduce it to a minimal reproducible
example (just to avoid questions why I'd want to do this for a vector of length
2...):
test2 = structure(list(X817 = 3.39824670255344,
Not sure about JMP 11, but remember that JMP 10 did not run with R version >=
3.0.0
It depends a bit on the changes that come with new R versions; with JMP 10,
several versions of the 2.x series were compatible even though JMP officially
only supported earlier versions. I had hoped that with JM
You don't need a constraint (rbinom won't give x>n), but you need to make sure
you are using the n you want to use: try
x <- cbind(x,rbinom(300,n[i],p[i]))# mind the "[i]" after the n
at the respective line. Furthermore, you need to remove one transformation of x
to make sure you divide by t
Jochen,
a) this is an English-spoken mailing list; other languages are not encouraged
nor will they typically generate a lot of replies...
b) your code is fine, so this is not an R-issue; you are rather stuck with some
of the stats background -- you might want to see a friendly local statisticia
Impossible to say w/o a reproducible example, but to start with let me suggest
looking at the exact= (both functions) and correct= (wilcox.test) arguments.
Experience shows that some change of the default settings allows you to
reproduce results from other software (and the help pages will expla
for taking the time to reply.
Cheers, Michael
> -Original Message-
> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
> Sent: Dienstag, 4. September 2012 16:58
> To: Meyners, Michael
> Cc: r-help
> Subject: Re: [R] unexpected (?) behavior of sort=TRUE in merge function
&g
1]] is, which to me is still "unexpected".
Hope this clarifies my question.
Any thoughts appreciated.
Michael
> -----Original Message-
> From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
> Sent: Dienstag, 4. September 2012 14:01
> To: Meyners, Michael
> Cc: r-help
&g
All,
I realize from the archive that the sort argument in merge has been subject to
discussion before, though I couldn't find an explanation for this behavior. I
tried to simplify this to (kind of) minimal code from a real example to the
following (and I have no doubts that there are smart peop
Kel,
in addition, and depending on how you define "similarity", you might want to
look into the RV coefficient as a measure of it (it is actually related to a
correlation, so similarity would rather mean "similar information" though not
necessarily small Euclidean distance); coeffRV in FactoMine
No, the authors are correct: the individuals (i.e. the 17 individuals) you have
need to be independent (i.e. no correlation between them, let alone any
individual running through your temporal experiment more than once, as
indicated in the citation), while the *observations* are of course depend
The devil is in the details (and in the arguments in Lukasz code). The defaults
for the two functions are different: wilcox.test uses an exact test (which is
not available in kruskal.test afaik) for your data, and uses the continuity
correction if the normal approximation is requested (neither a
Dan,
It depends on what you want to achieve. I suspect you just want to remove
missing values before summing; if so, consider
sapply(x, sum, na.rm=TRUE)
instead. To make your code running, try
sapply(x, function(x) sum(!is.na(x)))
However, this would just count the number of non-missing value
I’m not aware of any, but if you really want this, it should be possible to
modify the code of any of the functions you propose and delete the part doing
the translation. I’m not sure that this is a good idea, though; either your
matrices are *truly* centered, then it doesn’t make a difference,
elp@r-project.org
Cc: Meyners, Michael
Subject: RE: [R] Pearson chi-square test
Dear Michael,
Thanks very much for your answers!
The purpose of my analysis is to test whether the contingency table x is
different from the contingency table y.
Or, to put it differently, whether there is a significant diffe
t least it works here). Note that
something like
chisq.test(as.vector(x), as.vector(y))
will give a different test, i.e. based on a contingency table of x cross y).
M.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] O
Not sure what you want to test here with two matrices, but reading the manual
helps here as well:
y a vector; ignored if x is a matrix.
x and y are matrices in your example, so it comes as no surprise that you get
different results. On top of that, your manual calculation is not correct
res$nb.recognition = recognize
res$maxML = maxML
res$confusion = confusion
res$minimum = minimum
if (length(preference)>0) res$pref = pref
##res$complete = result
return(res)
}
** end updated code for triangle.test
-Original Message-
From: Francois Husson
Sent: Wed
Vijayan,
I cannot find an error in your code, but I had a look at the code of
triangle.test -- unless I'm missing something, it contains a bug. If you study
the way in which the matrix "pref" is updated, you find that the vector
preference is compared to 1, 2 and 3 instead of "X", "Y" and "Z" a
I suspect you need to give more information/background on the data (though this
is not primarily an R-related question; you might want to try other resources
instead). Unless I'm missing something here, I cannot think of ANY reasonable
test: A permutation (using permtest or anything else) would
John,
Why would you want to fit the model without intercept if you seemingly need it?
Anyway, I assume that the intercept from your first model just moves into the
random effects -- you have intercepts there for worker and day, so any of these
(or both) will absorb it. No surprise that the esti
I assume you installed WinEdt 6. See
https://stat.ethz.ch/pipermail/r-help/2010-May/238540.html
and related messages. RWinEdt does not work (yet) with WinEdt 6, so you'll have
to downgrade back to WinEdt 5.x (or use another editor for the time being).
HTH, Michael
-Original Message-
From
Strange, the following works reproducibly on my machine (Windows 2000
Pro):
options(scipen = 50, digits = 5)
x = c(1e7, 2e7)
?barplot
barplot(x)
while I also get scientific with your code. After I called ?barplot
once, I'm incapable of getting the scientific notation again, though...
But I admi
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Uwe Dippel
> Sent: Freitag, 29. Januar 2010 11:57
> To: r-help@r-project.org
> Subject: [R] Explanation w.r.t. rbind, please!
>
> This is what I tried:
>
> > num.vec <- c(12.3
(i) you EITHER correct the p value (by multiplying by 8 in your case) OR
you use the Bonferroni-threshold of 0.05/8, not both. If you correct the
p values, your threshold remains 0.05. If you use 0.05/8, you use the
original p values.
(ii) Yes, if the p value is 0.15, then the corrected one for 8 t
?p.adjust
Apply that to a vector containing all p values you get from
wilcox.exact. Alternatively, multiply each p value by the number of
comparisons you perform, see any textbook for that. You might want to
consider a less conservative correction, though.
HTH, Michael
> -Original Message-
Well, seems it's an assignment, so you should REALLY get them on your
own and not enquire the list. Foolish me...
M.
> -Original Message-
> From: einohr2...@web.de [mailto:einohr2...@web.de]
> Sent: Mittwoch, 20. Januar 2010 19:32
> To: Meyners,Michael,LAUSANNE,AppliedMa
Sorry, wrong button. Below a hopefully more helpful solution...
Etienne,
I don't see the point in avoiding some 'special' packages. If you are
willing to change your mind in this regard, try one of the following
solutions that work for me:
library(combinat)
apply(mat, 2, function(x) unique(permn(
Etienne,
I don't see the point in avoiding some 'special' packages. If you are
willing to change your mind in this regard, try something like
library()
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Etienne Stockhausen
> S
y
library(asdf)
with "asdf" replaced by the correct package name, and mind the spelling!
Michael
> -Original Message-
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: Donnerstag, 7. Januar 2010 14:22
> To: Meyners,Michael,LAUSANNE,AppliedMathemat
Are you sure you called
library(Quantreg)
before calling any function?
M.
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Gough Lauren
> Sent: Donnerstag, 7. Januar 2010 13:44
> To: r-help@r-project.org
> Subject: [R] Quantr
Moreno,
I leave the discussion on the mixed models to others (you might consider
the SIG group on mixed models as well for this), but try a few hints to
make your code more accessible:
* The "." in updating a formula is substituted with the respective old
formula (depending on the side), but is
Or, more general (if you need to include more than just one variable from
TestData), something like
by(TestData, LEAID, function(x) median(x$RATIO))
Agreed, this is less appealing for the given example than Ista's code, but
might help to better understand by and to generalize its use to other
Daniel,
quick guess: There has been a major change in the package some time ago,
with quite a few functions vanishing and others added. So I guess your
"recycled" code was based on an older version of multcomp. My current
version (1.1-0) does not have csimtest anymore either. I guess you want
to lo
Julia, see
http://www.r-project.org/ -> Documentation -> Manuals (-> An introduction to R)
(or use: http://cran.r-project.org/manuals.html)
for a starting point. In addition, you might want to check the annotated list
of books to see which one might best fit your needs:
http://www.r-project.o
FAQ 7.31:
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-the
se-numbers-are-equal_003f
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Peter Tillmann
> Sent: Mittwoch, 4. November 2009 13:50
> To: r-help
You don't tell us which function you use, but fixing the zlim argument
in image or related functions should do the trick.
M.
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Marion Dumas
> Sent: Dienstag, 3. November 2009 14:38
Or you open a new graphics window / device as part of the loop, see, e.g.,
?windows. Alternatively, you may write the content of the graphics device to a
file for each iteration, see, e.g., ?savePlot (but you'd want to make sure that
you have a different filename in each iteration, otherwise, yo
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Christoph Heuck
> Sent: Donnerstag, 15. Oktober 2009 17:51
> To: r-help@r-project.org
> Subject: [R] calculating p-values by row for data frames
>
> Hello R-users,
> I am look
ael
> -Original Message-
> From: Robert Kalicki
> Sent: Mittwoch, 14. Oktober 2009 14:11
> To: Meyners,Michael,LAUSANNE,AppliedMathematics
> Subject: RE: [R] post-hoc test with kruskal.test()
>
> Hi Michael,
> Thank you very much for your clear and prompt answ
Robert,
you can do the corresponding paired comparisons using wilcox.test. As far as I
know, there is no such general correction as Tukey's HSD for the
Kruskal-Wallis-Test. However, if you have indeed only 3 groups (resulting in 3
paired comparisons), the intersection-union principle and the th
It's always worthwhile to look at the articles by Pitman (and maybe the
textbook by Fisher, if you have access to it); Welch is a nice paper,
too, but might be pretty technical to "learn" about the area. I don't
know any of the textbooks except Edgington (which is in its 4th edition
now with co-aut
?paste
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Dr. Arie Ben David
> Sent: Mittwoch, 30. September 2009 10:22
> To: r-help@r-project.org
> Subject: [R] How do I do simple string concatenation in R?
>
> Dear R gurus
>
bution is always equal to 0.
I would strongly suggest to contact a local statistician to clarify these
issues.
HTH, Michael
From: Lina Rusyte [mailto:liner...@yahoo.co.uk]
Sent: Dienstag, 29. September 2009 18:03
To: Meyners,Michael,LAUSAN
Lina, check whether something like
data.frame(density(rnorm(10))[1:2])
contains the information you want. Otherwise, try to be (much) more specific in
what you want so that we do not need to guess (and of course provide minimal,
self-contained, reproducible code). That has a higher chance to t
Let's assume you have just three observations, and x-- = 1:3 for your
observations.
Predictor 1:y = x^2
Predictor 2:y = 1 if x=1
y = 4 if x=2
y = 9 if x=3
y = 0 elsewhere
These predictors are obviously not the same,
Alex,
It's mainly speculation, as I cannot check the Excel add-in nor Vassar, but
I'll give it a try.
For the Friedman-test: Results of R coincide with those reported by Hollander &
Wolfe, which I'd take as a point in favor of R. In any case, my guess is that
ties are handled differently (aver
See the respective help files. The continuity correction only affects
the normal approximation in wilcox.test. With this small samples sizes,
the default evaluation is exact, so it doesn't change anything. In
contrast, kruskal.test is incapable to compute exact values but always
uses the chi-square
Taking the example from ?biplot.princomp with reproducible code (hint,
hint):
biplot(princomp(USArrests))
biplot(princomp(USArrests), col=c(2,3), cex=c(1/2, 2))
clearly changes the color and font size on my system. For changing the
"point symbols" (which are text by default), try setting xlabs a
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Alex Roy
> Sent: Freitag, 14. August 2009 12:05
> To: r-help@r-project.org
> Subject: [R] Permutation test and R2 problem
>
> Hi,
>
>
> I have optimized the shrinkage parame
Nancy (?),
see ?chisq.test (in particular the examples and the comment on "expected
frequency" there).
A rule of thumb (see any basic text book) for the chisquared
approximation being okay is that the expected value in each cell is at
least 5. The warning tells you that this does not hold true for
Hi "unknown",
As a quick patch, try something like
mydata$Set <- 1
mydata$Set[c(sample(mydata$ID[mydata$Type=="A"],
ifelse(runif(1)<1/2,2,3)), sample(mydata$ID[mydata$Type=="B"], 3) )[1:5]
] <- 2
HTH, Michael
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@
SD in y is more than 15 times (!!!) larger than in x and z,
respectively, and hence SD of the mean y is also larger. 100,000 values
are not enough to stabilize this. You could have obtained 0.09 or even
larger as well. Try y with different seeds, y varies much more than x
and z do. Or check the var
Robert, Tom, Peter and all,
If I remember correctly (don't have my copy at hand right now),
Edgington and Onghena differentiate between randomization tests and
permutation tests along the following lines:
Randomization test: Apply only to randomized experiments, for which we
consider the theoret
Rolf,
as you explicitly asked for a comment on your proposal: It is generally
equivalent to McNemar's test and maybe even more appropriate because of
the asymptotics involved in the chi-squared distribution, which might
not be too good with n=12. In some more detail:
McNemar's test basically cons
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