Re: [R] R-help Digest, Vol 54, Issue 30
Ron Crump wrote: Hi, I have a dataframe that contains pedigree information; that is individual, sire and dam identities as separate columns. It also has date of birth. These identifiers are not numeric, or not sequential. Obviously, an identifier can appear in one or two columns, depending on whether it was a parent or not. These should be consistent. Not all identifiers appear in the individual column - it is possible for a parent not to have its own record if its parents were not known. Missing parental (sire and/or dam) identifiers can occur. I need to export the data for use in another program that requires the pedigree to be coded as integers, increasing with date of birth (therefore sire and dam always have lower identifiers than their offspring) and with missing values coded as 0. How would I go about doing this? You might look at http://www.qimr.edu.au/davidD/sib-pair.R, specifically the read.pedigree() and wrlink() functions. The former is not very impressive speedwise -- I usually perform these tasks in the my Sib-pair (Fortran) program, which is on the same webpage. It will order the pedigree by generational position, so a DOB is not required to do the sort. Terry Therneau's kinship package does that ordering, but doesn't include output routines for the Linkage format. David Duffy. | David Duffy (MBBS PhD) ,-_|\ | email: [EMAIL PROTECTED] ph: INT+61+7+3362-0217 fax: -0101 / * | Epidemiology Unit, Queensland Institute of Medical Research \_,-._/ | 300 Herston Rd, Brisbane, Queensland 4029, Australia GPG 4D0B994A v __ 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] R-help Digest, Vol 46, Issue 27
On Wednesday 27 December 2006 06:00, [EMAIL PROTECTED] wrote:
jingjiangyan
I agree, you can use 'assign'. To be more explicit, you could use the
following function.
jingjiangyan -
function(formula, data)
{
m - match.call()
%,% - function(x,y)paste(x,y,sep=)
d.nm - as.character(m$data)
y.nm - as.character(formula[[2]])
x.nm - as.character(formula[[3]])
for(i in levels(data[[x.nm]])){
var.name - d.nm %,% . %,% i
var.val - data[[y.nm]][data[[x.nm]]==i]
cmd - var.name %,% - %,% var.val
eval(cmd)
assign(var.name, var.val, globalenv())
}
}
Next, assuming the data.frame listed in the previous posting, 'df'
exists in your workspace, the call
jingjiangyan(bb ~ aa, data=df)
would produce the desired results.
Cheers,
Grant Izmirlian
--
Հրանդ Իզմիրլյան
__
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
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R-help Digest, Vol 39, Issue 13
r-help@stat.math.ethz.ch on Saturday, May 13, 2006 at 6:00 AM -0500 wrote:
lme(biomass~age, random=~woods/age)?
Jörn
Consult Pinheiro and Bates (2000, Mixed-effects models in S and S-Plus,
Springer, ISBN 0-387-98957-0 ref 7 at
http://www.r-project.org/doc/bib/R-books.html ) for how to fit more elaborate
models, but two straightforward ones that might be adequate
are
lme( biomass~age, random=~1|woods )
and
lme( biomass~age, random=~age|woods )
In the lme4 library corresponding syntax is
lmer( biomass~age+(1|woods) )
and
lmer( biomass~age+(age|woods) )
For vignettes on the lme4 library see the mlmRev library and
@ARTICLE{Rnews:Bates:2005,
AUTHOR = {Douglas Bates},
TITLE = {Fitting Linear Mixed Models in {R}},
JOURNAL = {R News},
YEAR = 2005,
VOLUME = 5,
NUMBER = 1,
PAGES = {27--30},
MONTH = {May},
URL = {[ http://CRAN.R-project.org/doc/Rnews/
]http://CRAN.R-project.org/doc/Rnews/}
}
alan
--
Alan B. Cobo-Lewis, Ph.D. (207) 581-3840 tel
Department of Psychology(207) 581-6128 fax
University of Maine
Orono, ME 04469-5742[EMAIL PROTECTED]
http://www.umaine.edu/visualperception
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Re: [R] R-help Digest, Vol 38, Issue 30
Mi nueva dirección de correo es: [EMAIL PROTECTED] New e-mail address: [EMAIL PROTECTED] __ 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
Re: [R] R-help Digest, Vol 38, Issue 19
Mi nueva dirección de correo es: [EMAIL PROTECTED] New e-mail address: [EMAIL PROTECTED] __ 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
Re: [R] R-help Digest, Vol 38, Issue 9
Mi nueva dirección de correo es: [EMAIL PROTECTED] New e-mail address: [EMAIL PROTECTED] __ 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
Re: [R] R-help Digest, Vol 37, Issue 26
Mi nueva dirección de correo es: [EMAIL PROTECTED] New e-mail address: [EMAIL PROTECTED] __ 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
Re: [R] R-help Digest, Vol 37, Issue 15
Mi nueva dirección de correo es: [EMAIL PROTECTED] New e-mail address: [EMAIL PROTECTED] __ 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
Re: [R] R-help Digest, Vol 37, Issue 12
Hi r-users, I would like to know if R have any solution to the Address standardization. The problem is to classify a database of addresses with the real addresses of a streets of Spain. Ideally, I would like to assign Postal code, census data and other geographic information. If this is not possible I would like to know solutions in R about text mining, text classification, distance within text data,... Any help will be appreciate Thanks in advance Ferran Carrascosa __ 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
Re: [R] R-help Digest, Vol 37, Issue 12
From: Ferran Carrascosa Hi r-users, I would like to know if R have any solution to the Address standardization. The problem is to classify a database of addresses with the real addresses of a streets of Spain. Ideally, I would like to assign Postal code, census data and other geographic information. I have no idea about this one... If this is not possible I would like to know solutions in R about text mining, text classification, distance within text data,... RSiteSearch(text mining) produced hits that look relevant. Andy Any help will be appreciate Thanks in advance Ferran Carrascosa __ 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 __ 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
Re: [R] R-help Digest, Vol 37, Issue 1
Mi nueva dirección de correo es: [EMAIL PROTECTED] New e-mail address: [EMAIL PROTECTED] __ 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
Re: [R] R-help Digest, Vol 36, Issue 21
Hello, dear R users. I've already sent a question here, but I'm not sure that it had been read. I need to visualize classification of my numerical data based on 2-3 factors. As I suppose, the best way is a tree. With an orbitrary function at the ends (leaves), or at least with means of my data at the ends. What is the way to do it? As I found, ctree offers binary classification, but it that the only way? Of course, tree is not only way, may be you could offer other ways. Thank you. __ 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
Re: [R] R-help Digest, Vol 36, Issue 21
Evgeniy Kachalin wrote: Hello, dear R users. I've already sent a question here, but I'm not sure that it had been read. I need to visualize classification of my numerical data based on 2-3 factors. As I suppose, the best way is a tree. With an orbitrary function at the ends (leaves), or at least with means of my data at the ends. What is the way to do it? As I found, ctree offers binary classification, but it that the only way? Of course, tree is not only way, may be you could offer other ways. Or the best way of it is to do it with replacement, like a 'heatmap', but with means in the cells instead of colors, if it is possible. Sorry for the second letter. __ 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
Re: [R] R-help Digest, Vol 35, Issue 24
I've been reluctant to step into this topic, but now feel that it may be helpful to make a certain point. On the internet, for the most part, the person behind the email is invisible and intangible. It is therefore possible, when someone puts their foot down, to stamp inadvertently on someone else's already broken toes. A friend of mine, very intelligent, very knowledgeable and creative, very articulate, nevertheless when writing uses spelling which can be a close approximation to random, and some interesting variants of grammar and vocabulary as well. The reason: dyslexia. While most of us hit the wrong keys at times (and when we read back over what we've written tend to see what we intended to write rather than what we did write), and when backed against the wall would admit that we could have got it right if we had paid better attention, there are some people who can't help getting it wrong. But, on the internet, one cannot readily recognise who they are (though in some cases, if one knows the signs, one may guess). Best wishes to all, Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 25-Jan-06 Time: 10:06:35 -- XFMail -- __ 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
Re: [R] R-help Digest, Vol 35, Issue 24
[Gabor Grothendieck] [...] this list is inhabited by some rather rude participants but everyone puts up with them in the hope that they do have some useful remarks. I've been witnessing this list for about one year, and also read *lots* of archived messages. While it is true that a few members do not use white gloves, are rather fond on concise replies, and do express strong opinions at times, they never went overboard insulting people and always kept a reasonable measure, at least so far that I could see (yet who knows, outliers might happen! :-). (*) Our whole society is a bit shy and shivers easily when opinions are expressed nowadays, I often observed than people quickly get insecure, feel attacked, and overreact (by running away or starting a fight). there is even a group of thought that feels it is a justifiable way to keep the list volume under control. This may work because of the starred paragraph above, that is, for wrong reasons. Best is, and this often occurs on the R list, when everything (facts, opinions) is being shared efficiently, without useless arguing. Then, threads quickly fade out. -- François Pinard http://pinard.progiciels-bpi.ca __ 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
Re: [R] R-help Digest, Vol 35, Issue 24
Dear Prof Ripley, First of all, unless you are an english professor, then I do not think you have any business policing language. I'm still very much a student, both in R, and regarding signal analysis. My competence on the subject as compared too your own level of expertise, or my spelling for that matter, may be a contension for you, but it would have been better had you kept that opinion too yourself. There are plenty of other reasons besides laziness or carelessness that people will consistently error in language use, such as learning disorders, head injuries, and/or vertigo. On the contrary, I am aware of the definition of a periodogram, and I know what the unnormalized periodogram in the data I presented looks like. Spec.pgram() is actually normalized too something, because it's discrete integral is not well above the SS amplitude of the signal it computed the periodogram for. In other words, the powers are not in units of around 4,000, which the peak would be if the units were merely the modulus squared of the Fourier coeficients of the data I presented. Alas, the modulus squared of the Fourier coeficients IS the TWO SIDED unnormalized periodogram, ranging from [-fc, fc] | fc=nyquist critical frequency. The definition of the ONE SIDED periodogram IS the modulus squared of the Fourier coeficients ranging over [0, fc], but since the function is even, data points in (0, fc) non-inclusive, need to be multiplied by 2. Thus is according too the definition given by Press, et al (1988, 1992, 2002, c.f. cp 12 13). I'm assuming that R returns an FFT in the same layout as Press, et al describe. Press, et al. are also very clear about the existence of far too many ways of normalizing the periodogram too document, which they stated before delving into particularly how they normalized to the mean squared amplitude of the signal that the periodogram was computed from. In the page before, and perhaps this is where some of the confusion arises from, they document the calculations for MS and SS amplitudes and time integral squared amplitude of the signal in the time domain, not the frequency domain. The page after that, their example only shows how to normalize a periodogram so its sum is equal too the MS amplitude. In short, but starting from SS amplitude: a). sum(a[index=(1:N) or t=(0:N-1)]^2) = SS amplitude calculated in time domain b). 1/N * sum(Mod(fft[-fc:fc])^2) = two sided periodogram that sums too the SS amplitude c). Same as b but over the range [0, fc], and (0, fc) multiplied by 2 is the one sided periodogram, also sums too the SS amplitude For MS amplitude, the procedures are identical, only the time domain is divided by N, and the frequency domain figures are divided by N^2 instead of N. When the periodogram is in power per unit time, as in the above, so that the power is interpretable at N/2+1 independent frequencies, it is a normalized periodogram. spec.pgram() IS normalized, I just do not know what it's normalized too because I can not seem to get spec.pgram to stop tapering (at which point the normalization should be dead on, not just close). By the way, normalized does not automatically mean anything unless to what is stated. I could normalize something arbitrarily to the number of tics on my dogs back side, and still call it normed, or erroneously refer too it as unnormed. If normalized is suposed to mean something specific, then I am confident that more than 90% of undergraduates are not familiar with what the term should mean. Stats and coding and using programs are a human endeavor. This human seems to have made meaning out of terms differently than what those who wrote the documentation seem to have intended. Only, I do not know where the documentation or my understanding may have been missled (R docs, Numerical Recipes, or any other source I looked at since I started). Cheers, KeithC. First, please look up `too' in your dictionary. Second, please study the references on the help page, which give the details. That is what references are for! The references will also answer your question about the reference distribution. The help page does not say it is `normalized' at all: it says it computes the peridogram, and you seem unaware of the definitions of the latter (and beware, there are more than one). On Tue, 24 Jan 2006, Keith Chamberlain wrote: __ 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
Re: [R] R-help Digest, Vol 35, Issue 24
Dear Mr. Chamberlain: You asked for free consulting, and as near as I can tell, you got pretty good advice. Now you complain that you don't like the packaging. If you can't stand the heat, get out of the kitchen. Professor Brian Ripley has an international reputation based on solid contributions to human knowledge over many years. He is an expert in statistical science, not diplomacy. Professor Ripley has been incredibly generous in donating substantial portions of his time for many years both to help make R what it is today and to answering questions on this listserve. I think he deserves a great deal of respect for not only the time he has devoted to this but to how much he has achieved with that time. What would you like him to do as a result of your email? Retire? Stop contributing to this listserve and to the R project more generally? I sincerely hope he does not consider such. It would be a great loss to humanity if he did. Mr. Chamberlain, if English (or as Prof. Ripley might say, American) is your mother tongue, then your deplorable lack of skill in its use raises serious questions about the standard of academic excellence at the University of Colorado, which I had previously thought was a great university and the finest Colorado had to offer. Of course, if English is a second language for you, then I would not complain. Rather, I would be humbled and honored that you chose to meet the rest of the world in my native tongue. Another question: The web lists you as a senior in psychology. Have you learned anything in your study of psychology? I would think that psychology students should meet a much higher standard for social skills and communications than you have displayed today. Would you like me to forward your correspondence to, say, the editor of the Flatiron News there in Boulder or Prof. W. Edward Craighead, the chair of the Psychology Dept., asking if a degree from the once-great University of Colorado is supposed to imply that the degree holder meets any standard for academic excellence in comportment and the use of language? Sincerely, Spencer Graves [EMAIL PROTECTED] wrote: Dear Prof Ripley, First of all, unless you are an english professor, then I do not think you have any business policing language. I'm still very much a student, both in R, and regarding signal analysis. My competence on the subject as compared too your own level of expertise, or my spelling for that matter, may be a contension for you, but it would have been better had you kept that opinion too yourself. There are plenty of other reasons besides laziness or carelessness that people will consistently error in language use, such as learning disorders, head injuries, and/or vertigo. On the contrary, I am aware of the definition of a periodogram, and I know what the unnormalized periodogram in the data I presented looks like. Spec.pgram() is actually normalized too something, because it's discrete integral is not well above the SS amplitude of the signal it computed the periodogram for. In other words, the powers are not in units of around 4,000, which the peak would be if the units were merely the modulus squared of the Fourier coeficients of the data I presented. Alas, the modulus squared of the Fourier coeficients IS the TWO SIDED unnormalized periodogram, ranging from [-fc, fc] | fc=nyquist critical frequency. The definition of the ONE SIDED periodogram IS the modulus squared of the Fourier coeficients ranging over [0, fc], but since the function is even, data points in (0, fc) non-inclusive, need to be multiplied by 2. Thus is according too the definition given by Press, et al (1988, 1992, 2002, c.f. cp 12 13). I'm assuming that R returns an FFT in the same layout as Press, et al describe. Press, et al. are also very clear about the existence of far too many ways of normalizing the periodogram too document, which they stated before delving into particularly how they normalized to the mean squared amplitude of the signal that the periodogram was computed from. In the page before, and perhaps this is where some of the confusion arises from, they document the calculations for MS and SS amplitudes and time integral squared amplitude of the signal in the time domain, not the frequency domain. The page after that, their example only shows how to normalize a periodogram so its sum is equal too the MS amplitude. In short, but starting from SS amplitude: a). sum(a[index=(1:N) or t=(0:N-1)]^2) = SS amplitude calculated in time domain b). 1/N * sum(Mod(fft[-fc:fc])^2) = two sided periodogram that sums too the SS amplitude c). Same as b but over the range [0, fc], and (0, fc) multiplied by 2 is the one sided periodogram, also sums too the SS amplitude For MS amplitude, the procedures are identical, only the time domain is divided by N, and
Re: [R] R-help Digest, Vol 35, Issue 24
[EMAIL PROTECTED], addressing to Brian Ripley] First of all, unless you are an english professor, then I do not think you have any business policing language. We all do mistakes (English or otherwise). I'm very grateful that people forgive my own errors, and I try to be tolerant to others. (Yet, it happens that people lacking good will ask for stronger reactions.) This is the business of everybody, really, building a better community in every possible aspect, and the means for this go through interaction and collaboration. Let's all be humble enough to ponder the criticism of others, improve ourselves, and so increase the value of our share. -- François Pinard http://pinard.progiciels-bpi.ca __ 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
Re: [R] R-help Digest, Vol 35, Issue 24
Its not really you. Its a fact of life that this list is inhabited by some rather rude participants but everyone puts up with them in the hope that they do have some useful remarks. This has been discussed repeatedly on the list and there is even a group of thought that feels it is a justifiable way to keep the list volume under control. On 1/24/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Dear Prof Ripley, First of all, unless you are an english professor, then I do not think you have any business policing language. I'm still very much a student, both in R, and regarding signal analysis. My competence on the subject as compared too your own level of expertise, or my spelling for that matter, may be a contension for you, but it would have been better had you kept that opinion too yourself. There are plenty of other reasons besides laziness or carelessness that people will consistently error in language use, such as learning disorders, head injuries, and/or vertigo. On the contrary, I am aware of the definition of a periodogram, and I know what the unnormalized periodogram in the data I presented looks like. Spec.pgram() is actually normalized too something, because it's discrete integral is not well above the SS amplitude of the signal it computed the periodogram for. In other words, the powers are not in units of around 4,000, which the peak would be if the units were merely the modulus squared of the Fourier coeficients of the data I presented. Alas, the modulus squared of the Fourier coeficients IS the TWO SIDED unnormalized periodogram, ranging from [-fc, fc] | fc=nyquist critical frequency. The definition of the ONE SIDED periodogram IS the modulus squared of the Fourier coeficients ranging over [0, fc], but since the function is even, data points in (0, fc) non-inclusive, need to be multiplied by 2. Thus is according too the definition given by Press, et al (1988, 1992, 2002, c.f. cp 12 13). I'm assuming that R returns an FFT in the same layout as Press, et al describe. Press, et al. are also very clear about the existence of far too many ways of normalizing the periodogram too document, which they stated before delving into particularly how they normalized to the mean squared amplitude of the signal that the periodogram was computed from. In the page before, and perhaps this is where some of the confusion arises from, they document the calculations for MS and SS amplitudes and time integral squared amplitude of the signal in the time domain, not the frequency domain. The page after that, their example only shows how to normalize a periodogram so its sum is equal too the MS amplitude. In short, but starting from SS amplitude: a). sum(a[index=(1:N) or t=(0:N-1)]^2) = SS amplitude calculated in time domain b). 1/N * sum(Mod(fft[-fc:fc])^2) = two sided periodogram that sums too the SS amplitude c). Same as b but over the range [0, fc], and (0, fc) multiplied by 2 is the one sided periodogram, also sums too the SS amplitude For MS amplitude, the procedures are identical, only the time domain is divided by N, and the frequency domain figures are divided by N^2 instead of N. When the periodogram is in power per unit time, as in the above, so that the power is interpretable at N/2+1 independent frequencies, it is a normalized periodogram. spec.pgram() IS normalized, I just do not know what it's normalized too because I can not seem to get spec.pgram to stop tapering (at which point the normalization should be dead on, not just close). By the way, normalized does not automatically mean anything unless to what is stated. I could normalize something arbitrarily to the number of tics on my dogs back side, and still call it normed, or erroneously refer too it as unnormed. If normalized is suposed to mean something specific, then I am confident that more than 90% of undergraduates are not familiar with what the term should mean. Stats and coding and using programs are a human endeavor. This human seems to have made meaning out of terms differently than what those who wrote the documentation seem to have intended. Only, I do not know where the documentation or my understanding may have been missled (R docs, Numerical Recipes, or any other source I looked at since I started). Cheers, KeithC. First, please look up `too' in your dictionary. Second, please study the references on the help page, which give the details. That is what references are for! The references will also answer your question about the reference distribution. The help page does not say it is `normalized' at all: it says it computes the peridogram, and you seem unaware of the definitions of the latter (and beware, there are more than one). On Tue, 24 Jan 2006, Keith Chamberlain wrote: __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the
Re: [R] R-help Digest, Vol 35, Issue 23
summary.aov(aovRes, split=list(interval = list(i1 vs i2 = 1, i2 vs i3 = 2, i3 vs i4 = 3, i4 vs i5 = 4, i5 vs i6 = 5))) try class(aovRes) #- aovlist ! summary.aovlist(aovRes, spit=...) or simply summary(aovRes, spit=...) Hoping this helps, Herwig -- Dr. Herwig Meschke Wissenschaftliche Beratung Hagsbucher Weg 27 D-89150 Laichingen phone +49 7333 210 417 / fax +49 7333 210 418 email [EMAIL PROTECTED] __ 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
Re: [R] R-help Digest, Vol 35, Issue 14
On Sun, 15 Jan 2006, Werner Wernersen wrote: Dear all, Is anybody aware of a tutorial, introduction, overview or alike for cluster analysis with R? I have been searching for something like that but it seems there are only a few rather specialized articles around. As an overview (rather than an introduction or tutorial), the Cluster task view might be helpful to you: http://CRAN.R-project.org/src/contrib/Views/Cluster.html Z I would very much appreciate any hint. Thanks a million, Werner ___ Telefonate ohne weitere Kosten vom PC zum PC: http://messenger.yahoo.de __ 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 __ 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
Re: [R] R-help Digest, Vol 35, Issue 14
Dear all, Is anybody aware of a tutorial, introduction, overview or alike for cluster analysis with R? I have been searching for something like that but it seems there are only a few rather specialized articles around. I would very much appreciate any hint. Thanks a million, Werner ___ Telefonate ohne weitere Kosten vom PC zum PC: http://messenger.yahoo.de __ 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
Re: [R] R-help Digest, Vol 35, Issue 14
On Sun, 15 Jan 2006, Werner Wernersen wrote: Is anybody aware of a tutorial, introduction, overview or alike for cluster analysis with R? I have been searching for something like that but it seems there are only a few rather specialized articles around. Chapter 11 of MASS (the book discussed in the FAQ). -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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
Re: [R] R-help Digest, Vol 35, Issue 7
Uwe Ligges пишет:
Evgeniy Kachalin wrote:
Hello, dear participants!
Could you tip me, is there any simple and nice way to build
scatter-plot for three different types of data (, and o and * - signs,
for example) with legend.
Now i can guess only that way:
plot(x~y,data=subset(mydata,factor1=='1'), pch='.',col='blue')
points(x~y,data=subset(mydata,factor1=='2'), pch='*',col='green')
points( etc
What is the simple and nice way?
Thank you very much for your kindness and help.
Example:
with(iris,
plot(Sepal.Length, Sepal.Width, pch = as.integer(Species)))
with(iris,
legend(7, 4.4, legend = unique(as.character(Species)),
pch = unique(as.integer(Species
Uwe, sorry for my stupid question. You mean that when pch=factor , plot
can recycle the factor and use it for subscripts or marks.
Then pch=as.integer(Species) results in c(1,2,3) for 3 factor levels.
And I need symbols 15,16,17 and colors red, blue, green.
So then I do:
iris$Species-spec.symb
iris$Species-spec.col
levels(spec.symb)-c(15,16,17)
levels(spec.col)-c('red','green','blue')
That's the only way?
More of that!!! 'Plot' does not like factors in 'pch'. So it must be so:
plot(x~y,data, pch=as.integer(as.character(spec.symb))).
That's totally crazy...
--
Evgeniy
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Re: [R] R-help Digest, Vol 35, Issue 7
Hi!
Just use your factors for indexing c(15,16,17) and
c(red,green,blue). So, with the iris data:
with(iris, plot(Sepal.Length, Sepal.Width,
pch=c(15,16,17)[as.integer(Species)],
col=c(red,green,blue)[as.integer(Species)] ))
Best regards,
Kyosti Kurikka
Evgeniy Kachalin wrote:
Hello, dear participants!
Could you tip me, is there any simple and nice way to build
scatter-plot for three different types of data (, and o and * - signs,
for example) with legend.
Now i can guess only that way:
plot(x~y,data=subset(mydata,factor1=='1'), pch='.',col='blue')
points(x~y,data=subset(mydata,factor1=='2'), pch='*',col='green')
points( etc
What is the simple and nice way?
Thank you very much for your kindness and help.
Example:
with(iris,
plot(Sepal.Length, Sepal.Width, pch = as.integer(Species)))
with(iris,
legend(7, 4.4, legend = unique(as.character(Species)),
pch = unique(as.integer(Species
Uwe, sorry for my stupid question. You mean that when pch=factor , plot
can recycle the factor and use it for subscripts or marks.
Then pch=as.integer(Species) results in c(1,2,3) for 3 factor levels.
And I need symbols 15,16,17 and colors red, blue, green.
So then I do:
iris$Species-spec.symb
iris$Species-spec.col
levels(spec.symb)-c(15,16,17)
levels(spec.col)-c('red','green','blue')
That's the only way?
More of that!!! 'Plot' does not like factors in 'pch'. So it must be so:
plot(x~y,data, pch=as.integer(as.character(spec.symb))).
That's totally crazy...
--
Evgeniy
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Re: [R] R-help Digest, Vol 35, Issue 7
Evgeniy Kachalin wrote:
Uwe Ligges пишет:
Evgeniy Kachalin wrote:
Hello, dear participants!
Could you tip me, is there any simple and nice way to build
scatter-plot for three different types of data (, and o and * -
signs, for example) with legend.
Now i can guess only that way:
plot(x~y,data=subset(mydata,factor1=='1'), pch='.',col='blue')
points(x~y,data=subset(mydata,factor1=='2'), pch='*',col='green')
points( etc
What is the simple and nice way?
Thank you very much for your kindness and help.
Example:
with(iris,
plot(Sepal.Length, Sepal.Width, pch = as.integer(Species)))
with(iris,
legend(7, 4.4, legend = unique(as.character(Species)),
pch = unique(as.integer(Species
Uwe, sorry for my stupid question. You mean that when pch=factor , plot
can recycle the factor and use it for subscripts or marks.
Yes, it can recycle, but in the example above it does not recycle but
takes the whole Species vector.
Then pch=as.integer(Species) results in c(1,2,3) for 3 factor levels.
And I need symbols 15,16,17 and colors red, blue, green.
What about adding 14 as in as.integer(Species)+14, or 1 for the colors,
respectively?
So then I do:
iris$Species-spec.symb
iris$Species-spec.col
levels(spec.symb)-c(15,16,17)
levels(spec.col)-c('red','green','blue')
That's the only way?
This is one qay of many.
More of that!!! 'Plot' does not like factors in 'pch'. So it must be so:
plot(x~y,data, pch=as.integer(as.character(spec.symb))).
That's totally crazy...
You can set up your own pch variable of course, if you don't like it
this fast and easy way.
Uwe Ligges
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Re: [R] R-help Digest, Vol 35, Issue 7
Hello, dear participants! Could you tip me, is there any simple and nice way to build scatter-plot for three different types of data (, and o and * - signs, for example) with legend. Now i can guess only that way: plot(x~y,data=subset(mydata,factor1=='1'), pch='.',col='blue') points(x~y,data=subset(mydata,factor1=='2'), pch='*',col='green') points( etc What is the simple and nice way? Thank you very much for your kindness and help. -- Evgeniy Kachalin __ 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
Re: [R] R-help Digest, Vol 35, Issue 7
Evgeniy Kachalin wrote: Hello, dear participants! Could you tip me, is there any simple and nice way to build scatter-plot for three different types of data (, and o and * - signs, for example) with legend. Now i can guess only that way: plot(x~y,data=subset(mydata,factor1=='1'), pch='.',col='blue') points(x~y,data=subset(mydata,factor1=='2'), pch='*',col='green') points( etc What is the simple and nice way? Thank you very much for your kindness and help. Example: with(iris, plot(Sepal.Length, Sepal.Width, pch = as.integer(Species))) with(iris, legend(7, 4.4, legend = unique(as.character(Species)), pch = unique(as.integer(Species Uwe Ligges __ 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
Re: [R] R-help Digest, Vol 34, Issue 14
Guten Tag, Ich bin vom 12. bis 23. Dezember 2005 im Militär-WK. Ich werde die Mails somit nur verzögert beantworten können. Für dringende Fälle: Während diesen zwei Wochen bin ich via Natel (am besten per SMS) erreichbar unter der Nummer 079 438 27 68. Mit freundlichem Gruss Dominik Schaub __ 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
Re: [R] R-help Digest, Vol 33, Issue 27
From: Duncan Murdoch [EMAIL PROTECTED] I'd recommend using the RWinEdt package instead for a different way to integrate winedit with R. winedit and winedt are two different editors, last I checked. best, -tony [EMAIL PROTECTED] Muttenz, Switzerland. Commit early,commit often, and commit in a repository from which we can easily roll-back your mistakes (AJR, 4Jan05). __ 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
Re: [R] R-help Digest, Vol 32, Issue 26
r-help@stat.math.ethz.ch on Wednesday, October 26, 2005 at 6:00 AM -0500 wrote: Ronaldo, Try Harold's suggestion. The df still won't agree, because lmer (at least in its current version) just puts an upper bound on the df. But that should be OK, because all those t tests are approximations anyways, and you can get better confidence intervals (credible intervals, whatever) by using the mcmcsamp() function that works with lmer() alan Doran, Harold [EMAIL PROTECTED] responded: There is an issue with implicit nesting in lmer. In your lme() model you nest block/irrigation/density/fertilizer. In lmer you need to do something like (I dind't include all of your variables, but I think the makes the point) lmer(yield~irrigation*density*fertilizer+(1|fertilizer:density)+(1|density), data) Which notes that fertilizer is nested in density. Try this and then compare the results. Ronaldo Reis-Jr. [EMAIL PROTECTED], wrote: I make the correct model with aov, lme do compare with lmer. But I cant make a correct model in lmer. Look that the aov and lme results are similars, but very different from lmer. In aov and lme is used the correct DF for each variable, in lmer it use a same DF for all? Denom=54. -- Alan B. Cobo-Lewis, Ph.D. (207) 581-3840 tel Department of Psychology(207) 581-6128 fax University of Maine Orono, ME 04469-5742[EMAIL PROTECTED] http://www.umaine.edu/visualperception __ 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
Re: [R] R-help Digest, Vol 32, Issue 26
In addition to the response below, Doug Bates has talked about this on
this list previously. I did
RSiteSearch('bates degrees of freedom lmer')
The first one that came up has Doug's response to this question as well
Harold
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Alan Cobo-Lewis
Sent: Wednesday, October 26, 2005 8:53 AM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] R-help Digest, Vol 32, Issue 26
r-help@stat.math.ethz.ch on Wednesday, October 26, 2005 at 6:00 AM -0500
wrote:
Ronaldo,
Try Harold's suggestion. The df still won't agree, because lmer (at
least in its current version) just puts an upper bound on the df. But
that should be OK, because all those t tests are approximations anyways,
and you can get better confidence intervals (credible intervals,
whatever) by using the mcmcsamp() function that works with lmer() alan
Doran, Harold [EMAIL PROTECTED] responded:
There is an issue with implicit nesting in lmer. In your lme() model
you nest block/irrigation/density/fertilizer. In lmer you need to do
something like (I dind't include all of your variables, but I think
the makes the point)
lmer(yield~irrigation*density*fertilizer+(1|fertilizer:density)+(1|den
sity), data)
Which notes that fertilizer is nested in density.
Try this and then compare the results.
Ronaldo Reis-Jr. [EMAIL PROTECTED], wrote:
I make the correct model with aov, lme do compare with lmer.
But I cant make a correct model in lmer. Look that the aov and lme
results are similars, but very different from lmer. In aov and lme is
used the correct DF for each variable, in lmer it use a same DF for
all? Denom=54.
--
Alan B. Cobo-Lewis, Ph.D. (207) 581-3840 tel
Department of Psychology(207) 581-6128 fax
University of Maine
Orono, ME 04469-5742[EMAIL PROTECTED]
http://www.umaine.edu/visualperception
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Re: [R] R-help Digest, Vol 31, Issue 30
With lme4, use of mcmcsamp can be insightful. (Douglas Bates drew my attention to this function in a private exchange of emails.) The distributions of random effects are simulated on a log scale, where the distributions are much closer to symmetry than on the scale of the random effects themselves. As far as I can see, this is a straightforward use of MCMC to estimate model parameters; it is not clear to me the results from the lmer() fit are used. John Maindonald. On 30 Sep 2005, at 8:00 PM, [EMAIL PROTECTED] wrote: From: Roel de Jong [EMAIL PROTECTED] Date: 29 September 2005 11:19:38 PM To: r-help@stat.math.ethz.ch Subject: [R] standard error of variances and covariances of the randomeffects with LME Hello, how do I obtain standard errors of variances and covariances of the random effects with LME comparable to those of for example MlWin? I know you shouldn't use them because the distribution of the estimator isn't symmetric blablabla, but I need a measure of the variance of those estimates for pooling my multiple imputation results. Regards, Roel. John Maindonald email: [EMAIL PROTECTED] phone : +61 2 (6125)3473fax : +61 2(6125)5549 Centre for Bioinformation Science, Room 1194, John Dedman Mathematical Sciences Building (Building 27) Australian National University, Canberra ACT 0200. __ 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
Re: [R] R-help Digest, Vol 31, Issue 9
?summary.lm and check the Value section.
Wuming
On 9/10/05, Ping Yao [EMAIL PROTECTED] wrote:
Hi:
I use lm (linear model) to analyze 47 variables , 8 responses
So I use loop to finish it .
I want the program to show the results that P-value is less than 0.05.
How can I cite the P-valus from lm result ?
Ping
The code:
#using LM to model general fati
for (j in 48:52) {
for (i in 3:46){
gen.fat-y_x[,j]
gen.fat-as.numeric(gen.fat)
snp_marker-y_x[,i]
x-colnames(y_x)
#snp_marker-as.matrix(snp_marker)
#mode(snp_marker)
cat(phenotype is = ,x[j] , \n)
cat(snp marker is = ,x[i] , \n)
zz-summary(lm.D9 - lm(gen.fat~snp_marker))
print(zz)
return
}
}
[[alternative HTML version deleted]]
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Re: [R] R-help Digest, Vol 31, Issue 9
Hi Ping,
You can use zz$coefficients[,4] to get the p values for each estimated
coefficients in your context.
Wuming
On 9/11/05, Ping Yao [EMAIL PROTECTED] wrote:
Wuming:
Thanks for your help.
I use the fuction:
call(fstatistic,zz)
call(p-value,zz)
I can get each variable P-values,but I can't get P-value of the model.
How can I do ?
one of the results is following :
Call:
lm(formula = gen.fat ~ snp_marker)
Residuals:
Min 1Q Median 3Q Max
-10.5455 -3.0481 0.4545 3.9519 6.9519
Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept)13.0481 0.4518 28.881 2e-16 ***
snp_markerallele2 0.5107 0.9102 0.561 0.5753
snp_markerBoth 1.4974 0.6927 2.162 0.0318 *
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 4.607 on 212 degrees of freedom
Multiple R-Squared: 0.02166,Adjusted R-squared: 0.01244
F-statistic: 2.347 on 2 and 212 DF, p-value: 0.0981
I use the code :
zz-summary(lm.D9 - lm(gen.fat~snp_marker))
coe-coef(lm.D9)# the bare coefficients
if (coe[2]=.05||coe[3]=.05||coe[4]=.05||coe[5]=.05) {
cat(phenotype is = ,x[j] , \n)
cat(snp marker is = ,x[i] , \n)
sign-call(fstatistic,zz)
call(p-value,zz)
#print(coe)
print(zz)
}
On 9/10/05, Wuming Gong [EMAIL PROTECTED] wrote:
?summary.lm and check the Value section.
Wuming
On 9/10/05, Ping Yao [EMAIL PROTECTED] wrote:
Hi:
I use lm (linear model) to analyze 47 variables , 8 responses
So I use loop to finish it .
I want the program to show the results that P-value is less than 0.05.
How can I cite the P-valus from lm result ?
Ping
The code:
#using LM to model general fati
for (j in 48:52) {
for (i in 3:46){
gen.fat-y_x[,j]
gen.fat-as.numeric(gen.fat)
snp_marker-y_x[,i]
x-colnames(y_x)
#snp_marker-as.matrix(snp_marker)
#mode(snp_marker)
cat(phenotype is = ,x[j] , \n)
cat(snp marker is = ,x[i] , \n)
zz-summary( lm.D9 - lm(gen.fat~snp_marker))
print(zz)
return
}
}
[[alternative HTML version deleted]]
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Re: [R] R-help Digest, Vol 31, Issue 9
Hi:
I use lm (linear model) to analyze 47 variables , 8 responses
So I use loop to finish it .
I want the program to show the results that P-value is less than 0.05.
How can I cite the P-valus from lm result ?
Ping
The code:
#using LM to model general fati
for (j in 48:52) {
for (i in 3:46){
gen.fat-y_x[,j]
gen.fat-as.numeric(gen.fat)
snp_marker-y_x[,i]
x-colnames(y_x)
#snp_marker-as.matrix(snp_marker)
#mode(snp_marker)
cat(phenotype is = ,x[j] , \n)
cat(snp marker is = ,x[i] , \n)
zz-summary(lm.D9 - lm(gen.fat~snp_marker))
print(zz)
return
}
}
[[alternative HTML version deleted]]
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Re: [R] R-help Digest, Vol 30, Issue 26
Dear R helpers, For me ( i.e. R 2.1.1 on Mac OS X), using trellis.device (postscript, onefile = F, etc ... with the lattice library within a R function works fine to obtain the desired graph as an EPS file , provided that : 1) the command dev.off() is not included in this function 2) and it is issued at the command level after the function has been exited I would like to know if there is a way to close the EPS file within the function itself, freeing the user to issue the closing command (I already tried trellis.device (), and trellis.device (null) without any success). Regards, J.-M. Jean-Marc Ottorini LERFoB, UMR INRA-ENGREF 1092 email [EMAIL PROTECTED] INRA - Centre de Nancy voice +33-0383-394046F54280 - Champenoux fax+33-0383-394034 France __ 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
Re: [R] R-help Digest, Vol 30, Issue 6
On Fri, 5 Aug 2005 Julia Reid wrote: Subject: [R] GAP pointer I am trying to do a simple segregation analysis using the GAP package. I have the documentation for pointer but I desperately need an example so that I can see how to format the datfile and the jobfile. For each individual, I have FamilyId, SubjectId, FatherId, MotherId, and AffectedStatus (0/1). I would like to obtain the likelihood ratio statistic for transmission. I would greatly appreciate any help on this subject. Best to all, Julia Reid I wouldn't use Pointer myself (there are lots of more recent packages*), but look at the examples in http://cedar.genetics.soton.ac.uk/pub/PROGRAMS/pointer/pointer.tar.Z and the manual, which is in the book: Morton N.E., Rao D.C Lalouel J-M (1983). Methods in Genetic Epidemiology. Karger PO Box, CH-4009 Basel (Switzerland). ISBN 3-8055-3668-2 which you will find in many academic libraries. David Duffy. * Don't you use Pap or JPap at Myriad? __ 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
Re: [R] R-help Digest, Vol 28, Issue 28
On Tuesday 28 June 2005 15:30, [EMAIL PROTECTED] wrote: Re : 37. Re: A. Mani : colours in Silhouette (Mulholland, Tom) Message: 37 Date: Tue, 28 Jun 2005 09:08:24 +0800 From: Mulholland, Tom [EMAIL PROTECTED] Subject: Re: [R] A. Mani : colours in Silhouette To: [EMAIL PROTECTED], r-help@stat.math.ethz.ch Message-ID: [EMAIL PROTECTED] Content-Type: text/plain; charset=iso-8859-1 It's not so much a problem, as not working the way you expected. cluster:::plot.partition is used to do the plotting. If you look at the code for this you can see the difficulty in putting every possible permutation into the code. If for example you want the silhouette plot to be red using col = red is not intuitive as the cluster plot (which comes up first) has more than one colour. If you have a look at methods(plot) (assuming that you have loaded the cluster package) you will see that there is a specific piece of code in the form of plot.silhouette. It has an asterisk next to it so you need to use cluster:::plot.silhouette to see the code. It has what you need. args(cluster:::plot.silhouette) function (x, nmax.lab = 40, max.strlen = 5, main = NULL, sub = NULL, xlab = expression(Silhouette width * s[i]), col = gray, do.col.sort = length(col) 1, border = 0, cex.names = par(cex.axis), do.n.k = TRUE, do.clus.stat = TRUE, ...) data(ruspini) pr4 - pam(ruspini, 4) si - silhouette(pr4) plot(si,col = red) I tried that before with many more options and got a blank image. It must have been due to the options. The issue is that whenever code is written there is always a choice as to what functionality is put in place. Just because something can be done, does not mean it will or in some cases should be done. In this case the help for plot.partition notes that For more flexibility, use 'plot(silhouette(x), ...)', see 'plot.silhouette'. Tom Thanks for that I found out something I will find useful in the future. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Behalf Of A. Mani Sent: Tuesday, 28 June 2005 4:30 AM To: r-help@stat.math.ethz.ch Subject: [R] A. Mani : colours in Silhouette Hello, In cluster analysis with cluster, how does one colour the silhouette plots ? For example in using pam. There seems to be some problem there. Everything else can be coloured. Thanks, A. Mani Member, Cal. Math. Soc __ 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
Re: [R] R-help Digest, Vol 28, Issue 11
Dear all, I'm new using R and in (geo)statistics. I have a problem with solving my homework questions. We are working with variograms and trying to write down basic equations for different models (spherical, exponential, Gaussian). I tried to use the 'gstat' and 'geoR' packages to solve the questions but as I said before I'm new in R and always encountered with some syntax errors (I can send some specific examples later). If one of you used this packages and could help me, I will be very glad. Best Wishes, Emre Duran __ 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
Re: [R] R-help Digest, Vol 28, Issue 11
dwfu wrote: Dear all, I'm new using R and in (geo)statistics. I have a problem with solving my homework questions. We are working with variograms and trying to write down basic equations for different models (spherical, exponential, Gaussian). I tried to use the 'gstat' and 'geoR' packages to solve the questions but as I said before I'm new in R and always encountered with some syntax errors (I can send some specific examples later). If one of you used this packages and could help me, I will be very glad. Please read the posting guide which tells you: - Use an informative subject line. - Basic statistics and classroom homework: R-help is not intended for these. - Provide small reproducible examples. Uwe Ligges Best Wishes, Emre Duran __ 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 __ 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
[R] R-help Digest
Hi folks, I have to create my own time series, Is it possible to generate ARIMA time series, where i can define the range of the values in the y axis. (e.g: Values only between 0 and 1) Best regards Sebastian __ 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
Re: [R] R-help Digest
On Fri, 6 May 2005, Sebastian Schoenherr wrote: Hi folks, I have to create my own time series, Is it possible to generate ARIMA time series, where i can define the range of the values in the y axis. (e.g: Values only between 0 and 1) No. Take a look at the definition of an ARIMA process. Suppose e.g. you have an AR(1) process. Then if innovations are positive and the coefficient is positive the value can be arbitrarily large. You can construct all sorts of similar counter-examples. This isn't a real problem is it? -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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
Re: [R] R-help Digest, Vol 4, Issue 27 ( -Reply)
Leo Wang-Kit Cheung [EMAIL PROTECTED] writes: Hi, I am out of town and will get back to you on the 13th of July. Leo [EMAIL PROTECTED] 06/27/03 00:32 Send R-help mailing list submissions to [EMAIL PROTECTED] To subscribe or unsubscribe via the World Wide Web, visit https://www.stat.math.ethz.ch/mailman/listinfo/r-help or, via email, send a message with subject or body 'help' to [EMAIL PROTECTED] You can reach the person managing the list at [EMAIL PROTECTED] When replying, please edit your Subject line so it is more specific than Re: Contents of R-help digest... Today's Topics: 1. create help files ([EMAIL PROTECTED]) ...and a full week of digested messages gets quoted back to the list. Let's hope that not every single digest-subscriber does likewise when he/she goes on holiday! -- O__ Peter Dalgaard Blegdamsvej 3 c/ /'_ --- Dept. of Biostatistics 2200 Cph. N (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
