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
