Re: [R] Help interpreting density().
You should read the documentation more carefully. The bw is not essentially the sd. To quote the documentation the bw is the smoothing bandwidth to be used. The kernels are scaled such that this is the standard deviation of the smoothing kernel. That is a very different thing. You are confusing the standard deviation of the distribution with the standard deviation of the gaussian smoothing kernels. In the second case, density(rpois(1000, 0)), you are getting the kernel density for a sample of 1000 zeros. So there is just one distinct smoothing kernel and the bw is a default used for this case. If you plot(density(rpois(1000, 0))) you will see what that smoothing kernel looks like. Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: Tuesday, 29 July 2008 2:15 PM To: r-help@r-project.org Subject: [R] Help interpreting density(). I issue the following: d - density(rnorm(1000)) d and get: Call: density.default(x = rnorm(1000)) Data: rnorm(1000) (1000 obs.); Bandwidth 'bw' = 0.2235 x y Min. :-3.5157 Min. :2.416e-05 1st Qu.:-1.6892 1st Qu.:1.129e-02 Median : 0.1373 Median :7.267e-02 Mean : 0.1373 Mean :1.367e-01 3rd Qu.: 1.9639 3rd Qu.:2.693e-01 Max. : 3.7904 Max. :4.014e-01 The documentation indicates that the bw is essentially the sd. Yet I have specified an sd of 1? How am I to interpret the ranges of the values? x ranges almost from -4 to +4 and y ranges from 0 to 0.4. The mean x is .1 which isn't too awfully close to what I would expect (0.0). Then there is: d - density(rpois(1000,0)) d Call: density.default(x = rpois(1000, 0)) Data: rpois(1000, 0) (1000 obs.); Bandwidth 'bw' = 0.2261 x y Min. :-0.6782 Min. :0.01979 1st Qu.:-0.3391 1st Qu.:0.14073 Median : 0. Median :0.57178 Mean : 0. Mean :0.73454 3rd Qu.: 0.3391 3rd Qu.:1.32830 Max. : 0.6782 Max. :1.76436 Here I am getting the mean that I expect from a Poisson distribuition but y ranges from 0 to 1.75. Again I am not sure what these numbers mean. How can I map the output to the standard distirbution description parameters? Thank you. Kevin __ R-help@r-project.org 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. __ R-help@r-project.org 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] Help interpreting density().
Hi Kevin, The documentation indicates that the bw is essentially the sd. d - density(rnorm(1000)) Not so. The documentation states that the following about bw: The kernels are scaled such that this is the standard deviation of the smoothing kernel..., which is a very different thing. The default bandwidth used by density is ?bw.nrd0. Read that documentation carefully and all might be clear. HTH, Mark. rkevinburton wrote: I issue the following: d - density(rnorm(1000)) d and get: Call: density.default(x = rnorm(1000)) Data: rnorm(1000) (1000 obs.); Bandwidth 'bw' = 0.2235 x y Min. :-3.5157 Min. :2.416e-05 1st Qu.:-1.6892 1st Qu.:1.129e-02 Median : 0.1373 Median :7.267e-02 Mean : 0.1373 Mean :1.367e-01 3rd Qu.: 1.9639 3rd Qu.:2.693e-01 Max. : 3.7904 Max. :4.014e-01 The documentation indicates that the bw is essentially the sd. Yet I have specified an sd of 1? How am I to interpret the ranges of the values? x ranges almost from -4 to +4 and y ranges from 0 to 0.4. The mean x is .1 which isn't too awfully close to what I would expect (0.0). Then there is: d - density(rpois(1000,0)) d Call: density.default(x = rpois(1000, 0)) Data: rpois(1000, 0) (1000 obs.); Bandwidth 'bw' = 0.2261 x y Min. :-0.6782 Min. :0.01979 1st Qu.:-0.3391 1st Qu.:0.14073 Median : 0. Median :0.57178 Mean : 0. Mean :0.73454 3rd Qu.: 0.3391 3rd Qu.:1.32830 Max. : 0.6782 Max. :1.76436 Here I am getting the mean that I expect from a Poisson distribuition but y ranges from 0 to 1.75. Again I am not sure what these numbers mean. How can I map the output to the standard distirbution description parameters? Thank you. Kevin __ R-help@r-project.org 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. -- View this message in context: http://www.nabble.com/Help-interpreting-density%28%29.-tp18704955p18706154.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Rf_error crashes entire program.
Rf_error is used so often in R itself and packages that it is almost
certainly not the problem -- rather something else your program has done
has damaged R's internals (e.g. overrrun an array).
'Writing R Extensions' discusses how to debug R code, including foreign
code. For example, I would run under valgrind to look for incorrect
memory accesses.
This was an appropriate topic for R-devel -- please DO study the posting
guide.
On Mon, 28 Jul 2008, Andrew Redd wrote:
I'm having a problem with the error and warning functions. I've tried this
on multiple machine so I'm fairly sure it's not machine dependent and I've
tried it on the latest versions 2.6.0-2.7.1. Whenever my program gets to an
error or warning it crashes the entire program rather than throwing the
error like it should.
Here are the relevant bits.
...
char const * const ExeedsMinVarianceError = PFDA ERR: Near zero
variance encountered. Estimation Unstable. Terminating Estimation.;
if(debug){printf(Da:\n);printmat(DaOld,1,*ka);fflush(stdout);}
daxpy_(ka, mOne, Da, one, DaOld, one);
for(i=0;i*ka;i++)convergenceCriteria+=fabs(DaOld[i]);
if(Da[*ka] MinVariance){
printf(PING);fflush(stdout);
warning(ExeedsMinVarianceError);
break;
}
and the output from running in batch mode that I get is this:
Da:
0.18034.988e-017
PING
and here the program crashes. I've tried this in multiple places and
sometimes the error is thrown sometimes not. Does anyone have an idea of
what is going on. I could not find any discussion of this error yet.
Thank you,
Andrew Redd
[[alternative HTML version deleted]]
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--
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
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[R] How to set the parameters in Trellis Graphics (by Lattice package)
Dear R users I plot the trellis graphics by using the lattice package. Everything is OK. Now, I want set some parameters of the trellis graphics. 1. The tick label site. By default, only two tick labels had been output in x-axis of my plot. I want output four or five tick labels. In the traditional graphics system, it will be very easy. Just not output the axis in plot by use the parameter xaxt=n, and then used axis to add the axis. But it seem useless in the grid graphics system. Then how can I do it? 2. How to change the margin of the figure? 3. Can I change the style of the figure like par in traditional system. For example, I want change all the fontface into bold. Now, I must change the fontface for every item by using command trellis.par.set() like this: trellis.par.set(list(par.xlab.text = gpar(font = 2), par.ylab.text = gpar(font = 2), axis.text = gpar(font = 2), add.text = gpar(font=2) )) is there a method to change this parameter in one time, like par(font=2) thanks G.H.Zuo [[alternative HTML version deleted]] __ R-help@r-project.org 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] product of successive rows
Assuming that the number of rows is even and that your matrix is A,
element-wise product of pairs of rows can be calculated as
A[seq(1,nrow(A),by=2),]*A{seq(2,nrow(A),by=2),]
--- On Mon, 28/7/08, rcoder [EMAIL PROTECTED] wrote:
From: rcoder [EMAIL PROTECTED]
Subject: [R] product of successive rows
To: r-help@r-project.org
Received: Monday, 28 July, 2008, 8:20 AM
Hi everyone,
I want to perform an operation on a matrx that outputs the
product of
successive pairs of rows. For example: calculating the
product between rows
1 2; 3 4; 5 6...etc.
Does anyone know of any readily available functions that
can do this?
Thanks,
rcoder
--
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained,
reproducible code.
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Re: [R] Help interpreting density().
OK. Thank you for pointing out my mistake. I still have my original question. How does the output relate to estimating the parameters of a given density? I read that for a gausian kernal: bw.nrd0 implements a rule-of-thumb for choosing the bandwidth of a Gaussian kernel density estimator. It defaults to 0.9 times the minimum of the standard deviation and the interquartile range divided by 1.34 times the sample size to the negative one-fifth power (= Silverman's ‘rule of thumb’ But how does that relate to say a Poisson distribution or a two-parameter distribution like a normal, beta, or binomial distribution? Thank you. Kevin Mark Difford [EMAIL PROTECTED] wrote: Hi Kevin, The documentation indicates that the bw is essentially the sd. d - density(rnorm(1000)) Not so. The documentation states that the following about bw: The kernels are scaled such that this is the standard deviation of the smoothing kernel..., which is a very different thing. The default bandwidth used by density is ?bw.nrd0. Read that documentation carefully and all might be clear. HTH, Mark. rkevinburton wrote: I issue the following: d - density(rnorm(1000)) d and get: Call: density.default(x = rnorm(1000)) Data: rnorm(1000) (1000 obs.); Bandwidth 'bw' = 0.2235 x y Min. :-3.5157 Min. :2.416e-05 1st Qu.:-1.6892 1st Qu.:1.129e-02 Median : 0.1373 Median :7.267e-02 Mean : 0.1373 Mean :1.367e-01 3rd Qu.: 1.9639 3rd Qu.:2.693e-01 Max. : 3.7904 Max. :4.014e-01 The documentation indicates that the bw is essentially the sd. Yet I have specified an sd of 1? How am I to interpret the ranges of the values? x ranges almost from -4 to +4 and y ranges from 0 to 0.4. The mean x is .1 which isn't too awfully close to what I would expect (0.0). Then there is: d - density(rpois(1000,0)) d Call: density.default(x = rpois(1000, 0)) Data: rpois(1000, 0) (1000 obs.); Bandwidth 'bw' = 0.2261 x y Min. :-0.6782 Min. :0.01979 1st Qu.:-0.3391 1st Qu.:0.14073 Median : 0. Median :0.57178 Mean : 0. Mean :0.73454 3rd Qu.: 0.3391 3rd Qu.:1.32830 Max. : 0.6782 Max. :1.76436 Here I am getting the mean that I expect from a Poisson distribuition but y ranges from 0 to 1.75. Again I am not sure what these numbers mean. How can I map the output to the standard distirbution description parameters? Thank you. Kevin __ R-help@r-project.org 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. -- View this message in context: http://www.nabble.com/Help-interpreting-density%28%29.-tp18704955p18706154.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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. __ R-help@r-project.org 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] Help interpreting density().
Sorry, poor example. I started with normal deviates and jumped without thinking to Poisson. The main crux of the question is how does the output of density relate to the parameters that describe some of the standard distributions (mean and std for normal, lambda for Poisson, n and p for Binomial, alpha and beta for Beta, etc.). Thank you. Kevin [EMAIL PROTECTED] wrote: You should read the documentation more carefully. The bw is not essentially the sd. To quote the documentation the bw is the smoothing bandwidth to be used. The kernels are scaled such that this is the standard deviation of the smoothing kernel. That is a very different thing. You are confusing the standard deviation of the distribution with the standard deviation of the gaussian smoothing kernels. In the second case, density(rpois(1000, 0)), you are getting the kernel density for a sample of 1000 zeros. So there is just one distinct smoothing kernel and the bw is a default used for this case. If you plot(density(rpois(1000, 0))) you will see what that smoothing kernel looks like. Bill Venables CSIRO Laboratories PO Box 120, Cleveland, 4163 AUSTRALIA Office Phone (email preferred): +61 7 3826 7251 Fax (if absolutely necessary): +61 7 3826 7304 Mobile: +61 4 8819 4402 Home Phone: +61 7 3286 7700 mailto:[EMAIL PROTECTED] http://www.cmis.csiro.au/bill.venables/ -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: Tuesday, 29 July 2008 2:15 PM To: r-help@r-project.org Subject: [R] Help interpreting density(). I issue the following: d - density(rnorm(1000)) d and get: Call: density.default(x = rnorm(1000)) Data: rnorm(1000) (1000 obs.); Bandwidth 'bw' = 0.2235 x y Min. :-3.5157 Min. :2.416e-05 1st Qu.:-1.6892 1st Qu.:1.129e-02 Median : 0.1373 Median :7.267e-02 Mean : 0.1373 Mean :1.367e-01 3rd Qu.: 1.9639 3rd Qu.:2.693e-01 Max. : 3.7904 Max. :4.014e-01 The documentation indicates that the bw is essentially the sd. Yet I have specified an sd of 1? How am I to interpret the ranges of the values? x ranges almost from -4 to +4 and y ranges from 0 to 0.4. The mean x is .1 which isn't too awfully close to what I would expect (0.0). Then there is: d - density(rpois(1000,0)) d Call: density.default(x = rpois(1000, 0)) Data: rpois(1000, 0) (1000 obs.); Bandwidth 'bw' = 0.2261 x y Min. :-0.6782 Min. :0.01979 1st Qu.:-0.3391 1st Qu.:0.14073 Median : 0. Median :0.57178 Mean : 0. Mean :0.73454 3rd Qu.: 0.3391 3rd Qu.:1.32830 Max. : 0.6782 Max. :1.76436 Here I am getting the mean that I expect from a Poisson distribuition but y ranges from 0 to 1.75. Again I am not sure what these numbers mean. How can I map the output to the standard distirbution description parameters? Thank you. Kevin __ R-help@r-project.org 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. __ R-help@r-project.org 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] Help interpreting density().
Hi Kevin, I still have my original question. How does the output relate to estimating the parameters of a given density? I read that for a gausian kernal: This isn't the place for such questions: you need to do some _basic_ reading on the subject so that you begin to understand something about the method you are messing about with. Basically (very basically) it's a smoothed out histogram. And you will probably (?) know that a histogram is [still used] to show you how a set of univariate data (random variable) is distributed. Perhaps start with http://en.wikipedia.org/wiki/Kernel_density, then go somewhere else. But since you have access to the web you really should have found something like this yourself. Regards, Mark. rkevinburton wrote: OK. Thank you for pointing out my mistake. I still have my original question. How does the output relate to estimating the parameters of a given density? I read that for a gausian kernal: bw.nrd0 implements a rule-of-thumb for choosing the bandwidth of a Gaussian kernel density estimator. It defaults to 0.9 times the minimum of the standard deviation and the interquartile range divided by 1.34 times the sample size to the negative one-fifth power (= Silverman's ‘rule of thumb’ But how does that relate to say a Poisson distribution or a two-parameter distribution like a normal, beta, or binomial distribution? Thank you. Kevin Mark Difford [EMAIL PROTECTED] wrote: Hi Kevin, The documentation indicates that the bw is essentially the sd. d - density(rnorm(1000)) Not so. The documentation states that the following about bw: The kernels are scaled such that this is the standard deviation of the smoothing kernel..., which is a very different thing. The default bandwidth used by density is ?bw.nrd0. Read that documentation carefully and all might be clear. HTH, Mark. rkevinburton wrote: I issue the following: d - density(rnorm(1000)) d and get: Call: density.default(x = rnorm(1000)) Data: rnorm(1000) (1000 obs.); Bandwidth 'bw' = 0.2235 x y Min. :-3.5157 Min. :2.416e-05 1st Qu.:-1.6892 1st Qu.:1.129e-02 Median : 0.1373 Median :7.267e-02 Mean : 0.1373 Mean :1.367e-01 3rd Qu.: 1.9639 3rd Qu.:2.693e-01 Max. : 3.7904 Max. :4.014e-01 The documentation indicates that the bw is essentially the sd. Yet I have specified an sd of 1? How am I to interpret the ranges of the values? x ranges almost from -4 to +4 and y ranges from 0 to 0.4. The mean x is .1 which isn't too awfully close to what I would expect (0.0). Then there is: d - density(rpois(1000,0)) d Call: density.default(x = rpois(1000, 0)) Data: rpois(1000, 0) (1000 obs.); Bandwidth 'bw' = 0.2261 x y Min. :-0.6782 Min. :0.01979 1st Qu.:-0.3391 1st Qu.:0.14073 Median : 0. Median :0.57178 Mean : 0. Mean :0.73454 3rd Qu.: 0.3391 3rd Qu.:1.32830 Max. : 0.6782 Max. :1.76436 Here I am getting the mean that I expect from a Poisson distribuition but y ranges from 0 to 1.75. Again I am not sure what these numbers mean. How can I map the output to the standard distirbution description parameters? Thank you. Kevin __ R-help@r-project.org 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. -- View this message in context: http://www.nabble.com/Help-interpreting-density%28%29.-tp18704955p18706154.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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. __ R-help@r-project.org 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. -- View this message in context: http://www.nabble.com/Help-interpreting-density%28%29.-tp18704955p18707522.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Help interpreting density().
Sorry I tried WikiPedia and only found: Wikipedia does not have an article with this exact name. I will try to find some other sources of information. Kevin Mark Difford [EMAIL PROTECTED] wrote: Hi Kevin, I still have my original question. How does the output relate to estimating the parameters of a given density? I read that for a gausian kernal: This isn't the place for such questions: you need to do some _basic_ reading on the subject so that you begin to understand something about the method you are messing about with. Basically (very basically) it's a smoothed out histogram. And you will probably (?) know that a histogram is [still used] to show you how a set of univariate data (random variable) is distributed. Perhaps start with http://en.wikipedia.org/wiki/Kernel_density, then go somewhere else. But since you have access to the web you really should have found something like this yourself. Regards, Mark. rkevinburton wrote: OK. Thank you for pointing out my mistake. I still have my original question. How does the output relate to estimating the parameters of a given density? I read that for a gausian kernal: bw.nrd0 implements a rule-of-thumb for choosing the bandwidth of a Gaussian kernel density estimator. It defaults to 0.9 times the minimum of the standard deviation and the interquartile range divided by 1.34 times the sample size to the negative one-fifth power (= Silverman's ‘rule of thumb’ But how does that relate to say a Poisson distribution or a two-parameter distribution like a normal, beta, or binomial distribution? Thank you. Kevin Mark Difford [EMAIL PROTECTED] wrote: Hi Kevin, The documentation indicates that the bw is essentially the sd. d - density(rnorm(1000)) Not so. The documentation states that the following about bw: The kernels are scaled such that this is the standard deviation of the smoothing kernel..., which is a very different thing. The default bandwidth used by density is ?bw.nrd0. Read that documentation carefully and all might be clear. HTH, Mark. rkevinburton wrote: I issue the following: d - density(rnorm(1000)) d and get: Call: density.default(x = rnorm(1000)) Data: rnorm(1000) (1000 obs.); Bandwidth 'bw' = 0.2235 x y Min. :-3.5157 Min. :2.416e-05 1st Qu.:-1.6892 1st Qu.:1.129e-02 Median : 0.1373 Median :7.267e-02 Mean : 0.1373 Mean :1.367e-01 3rd Qu.: 1.9639 3rd Qu.:2.693e-01 Max. : 3.7904 Max. :4.014e-01 The documentation indicates that the bw is essentially the sd. Yet I have specified an sd of 1? How am I to interpret the ranges of the values? x ranges almost from -4 to +4 and y ranges from 0 to 0.4. The mean x is .1 which isn't too awfully close to what I would expect (0.0). Then there is: d - density(rpois(1000,0)) d Call: density.default(x = rpois(1000, 0)) Data: rpois(1000, 0) (1000 obs.); Bandwidth 'bw' = 0.2261 x y Min. :-0.6782 Min. :0.01979 1st Qu.:-0.3391 1st Qu.:0.14073 Median : 0. Median :0.57178 Mean : 0. Mean :0.73454 3rd Qu.: 0.3391 3rd Qu.:1.32830 Max. : 0.6782 Max. :1.76436 Here I am getting the mean that I expect from a Poisson distribuition but y ranges from 0 to 1.75. Again I am not sure what these numbers mean. How can I map the output to the standard distirbution description parameters? Thank you. Kevin __ R-help@r-project.org 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. -- View this message in context: http://www.nabble.com/Help-interpreting-density%28%29.-tp18704955p18706154.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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. __ R-help@r-project.org 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. -- View this message in context: http://www.nabble.com/Help-interpreting-density%28%29.-tp18704955p18707522.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
Re: [R] Help interpreting density().
Hi Kevin, Clicking on the link I sent gets me there (?), though things are pretty slow at the moment. Perhaps try this related link, and from it get back to the first one: http://en.wikipedia.org/wiki/Density_estimation You can also get to this via histogram, so search for that in Wiki, and then... HTH, Mark. rkevinburton wrote: Sorry I tried WikiPedia and only found: Wikipedia does not have an article with this exact name. I will try to find some other sources of information. Kevin Mark Difford [EMAIL PROTECTED] wrote: Hi Kevin, I still have my original question. How does the output relate to estimating the parameters of a given density? I read that for a gausian kernal: This isn't the place for such questions: you need to do some _basic_ reading on the subject so that you begin to understand something about the method you are messing about with. Basically (very basically) it's a smoothed out histogram. And you will probably (?) know that a histogram is [still used] to show you how a set of univariate data (random variable) is distributed. Perhaps start with http://en.wikipedia.org/wiki/Kernel_density, then go somewhere else. But since you have access to the web you really should have found something like this yourself. Regards, Mark. rkevinburton wrote: OK. Thank you for pointing out my mistake. I still have my original question. How does the output relate to estimating the parameters of a given density? I read that for a gausian kernal: bw.nrd0 implements a rule-of-thumb for choosing the bandwidth of a Gaussian kernel density estimator. It defaults to 0.9 times the minimum of the standard deviation and the interquartile range divided by 1.34 times the sample size to the negative one-fifth power (= Silverman's ‘rule of thumb’ But how does that relate to say a Poisson distribution or a two-parameter distribution like a normal, beta, or binomial distribution? Thank you. Kevin Mark Difford [EMAIL PROTECTED] wrote: Hi Kevin, The documentation indicates that the bw is essentially the sd. d - density(rnorm(1000)) Not so. The documentation states that the following about bw: The kernels are scaled such that this is the standard deviation of the smoothing kernel..., which is a very different thing. The default bandwidth used by density is ?bw.nrd0. Read that documentation carefully and all might be clear. HTH, Mark. rkevinburton wrote: I issue the following: d - density(rnorm(1000)) d and get: Call: density.default(x = rnorm(1000)) Data: rnorm(1000) (1000 obs.); Bandwidth 'bw' = 0.2235 x y Min. :-3.5157 Min. :2.416e-05 1st Qu.:-1.6892 1st Qu.:1.129e-02 Median : 0.1373 Median :7.267e-02 Mean : 0.1373 Mean :1.367e-01 3rd Qu.: 1.9639 3rd Qu.:2.693e-01 Max. : 3.7904 Max. :4.014e-01 The documentation indicates that the bw is essentially the sd. Yet I have specified an sd of 1? How am I to interpret the ranges of the values? x ranges almost from -4 to +4 and y ranges from 0 to 0.4. The mean x is .1 which isn't too awfully close to what I would expect (0.0). Then there is: d - density(rpois(1000,0)) d Call: density.default(x = rpois(1000, 0)) Data: rpois(1000, 0) (1000 obs.); Bandwidth 'bw' = 0.2261 x y Min. :-0.6782 Min. :0.01979 1st Qu.:-0.3391 1st Qu.:0.14073 Median : 0. Median :0.57178 Mean : 0. Mean :0.73454 3rd Qu.: 0.3391 3rd Qu.:1.32830 Max. : 0.6782 Max. :1.76436 Here I am getting the mean that I expect from a Poisson distribuition but y ranges from 0 to 1.75. Again I am not sure what these numbers mean. How can I map the output to the standard distirbution description parameters? Thank you. Kevin __ R-help@r-project.org 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. -- View this message in context: http://www.nabble.com/Help-interpreting-density%28%29.-tp18704955p18706154.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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. __ R-help@r-project.org 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,
Re: [R] how to add notes to the graph?
On Mon, 2008-07-28 at 06:32 -0700, rlearner309 wrote: Hi, I have a simple graph: x - c(1,2,3) plot(x, pch=16,type=b) I would like to add some notes just beside these 3 dots, and the notes are stored in a vector: a - c(12,54,84) So the result will be: there should be a 12 below the first dot (or next to it, but not replacing the solid dot), a 54 next to the second dot... what if a is not numeric? a - c(a,b,c) Hi rlearner309, Have a look at thigmophobe.labels in the plotrix package. Jim __ R-help@r-project.org 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.
[R] Problem reading a particular file with read.spss()
Hi All: I have a seemingly typical SPSS data file with 219 rows and 486 variables. When I attempt to read this file into R with read.spss() in the foreign package, I consistently get a crash. This is the sequence of events: library(foreign) sessionInfo() R version 2.7.1 Patched (2008-07-24 r46120) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] foreign_0.8-27 read.spss(c:/chuck/mySPSSfile.sav) R for Windows GUI front-end has encountered a problem and needs to close. We are sorry for the inconvenience. It get a similar result trying to load the file under Rterm.exe. The file opens without any trouble in SPSS version 16.0.2 (April 10, 2008). For anyone willing to investigate, I placed a copy of the SPSS file here: http://www.chuckcleland.net/mySPSSfile.sav I would be grateful if someone could help to figure out why this file causes a problem. thanks, Chuck -- Chuck Cleland, Ph.D. NDRI, Inc. (www.ndri.org) 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@r-project.org 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] Is there way to multiple plots on gap.plot?
On Mon, 2008-07-28 at 12:15 -0700, Arthur Roberts wrote:
Hi, all,
Does anyone now of a way to put multiple plots on gap.plot?
Much appreciated,
Hi Art,
You must have read my mind. In solving the problem you had with
gap.plot, I considered including an add argument that would allow the
user to add values to an existing plot. The bad news is that I decided
against it. As the examples show, you can plot several series of data
simultaneously with not very much trouble. However, if you can convince
me that add is a good idea, I might do it. Right now, though, I would
appreciate your testing of the new gap.plot function below.
Jim
gap.plot-function(x,y,gap,gap.axis=y,bgcol=white,breakcol=black,
brw=0.02,xlim,ylim,xticlab,xtics=NA,yticlab,ytics=NA,lty=rep(1,length(x)),
col=rep(par(col),length(x)),pch=rep(1,length(x)),...) {
if(missing(y) !missing(x)) {
y-x
x-1:length(y)
}
if(missing(gap)) stop(gap must be specified)
gapsize-diff(gap)
figxy - par(usr)
xaxl-par(xlog)
yaxl-par(ylog)
xgw-(figxy[2]-figxy[1])*brw
ygw-(figxy[4]-figxy[3])*brw
if(missing(xtics)) xtics-pretty(x)
if(missing(ytics)) ytics-pretty(y)
if(missing(xticlab)) xticlab-xtics
if(missing(yticlab)) yticlab-ytics
if(length(col) length(y)) col-rep(col,length.out=length(y))
if(gap.axis == y) {
littleones-which(y gap[1])
if(length(gapsize) 2) {
middleones-which(y = gap[2] y gap[3])
bigones-which(y = gap[4])
lostones-sum(c(y gap[1] y gap[2], y gap[3] y gap[4]))
if(missing(ylim)) ylim-c(min(y),ygw*2
+max(y)-(gapsize[1]+gapsize[3]))
else ylim[2]-ygw*2+ylim[2]-(gapsize[1]+gapsize[3])
}
else {
middleones-NA
bigones-which(y = gap[2])
lostones-sum(y gap[1] y gap[2])
if(missing(ylim)) ylim-c(min(y),max(y)-gapsize[1])
else ylim[2]-ygw+ylim[2]-gapsize[1]
}
if(lostones) warning(some values of y will not be displayed)
if(missing(xlim)) xlim-range(x)
}
else {
littleones-which(x gap[1])
if(length(gapsize) 2) {
middleones-which(x = gap[2] x gap[3])
bigones-which(x = gap[4])
lostones-sum(c(x gap[1] x gap[2], x gap[3] x gap[4]))
if(missing(xlim)) xlim-c(min(x),xgw*2
+max(x)-(gapsize[1]+gapsize[3]))
else xlim[2]-xgw*2+xlim[2]-(gapsize[1]+gapsize[3])
}
else {
middleones-NA
bigones-which(x = gap[2])
lostones-sum(x gap[1] x gap[2])
if(missing(xlim)) xlim-c(min(x),max(x)-gapsize[1])
else xlim[2]-xgw+xlim[2]-gapsize[1]
}
if(lostones) warning(some values of x will not be displayed)
if(missing(ylim)) ylim-range(y)
}
if(length(lty) length(x)) lty-rep(lty,length.out=length(x))
if(length(col) length(x)) col-rep(col,length.out=length(x))
if(length(pch) length(x)) pch-rep(pch,length.out=length(x))
plot(x[littleones],y[littleones],xlim=xlim,ylim=ylim,axes=FALSE,
lty=lty[littleones],col=col[littleones],pch=pch[littleones],...)
box()
if(gap.axis == y) {
if(!is.na(xtics[1])) axis(1,at=xtics,labels=xticlab)
littletics-which(ytics gap[1])
if(length(gapsize) 2) {
middletics-which(ytics = gap[2] ytics = gap[3])
bigtics-which(ytics = gap[4])
show.at-c(ytics[littletics],ytics[middletics]-gapsize[1],],...)
ytics[bigtics]-(gapsize[1]+gapsize[3]))
show.labels-c(yticlab[littletics],yticlab[middletics],yticlab
[bigtics])),
}
else {
bigtics-which(ytics = gap[2])
show.at-c(ytics[littletics],ytics[bigtics]-gapsize[1])
show.labels-c(ytics[littletics],yticlab[bigtics])
}
axis(2,at=show.at,labels=show.labels)
axis.break(2,gap
[1]-ygw,style=gap,bgcol=bgcol,breakcol=breakcol,brw=brw)
if(length(gapsize) 2) {
axis.break(2,gap[3]-(gapsize[1]+ygw),style=gap,bgcol=bgcol,
breakcol=breakcol,brw=brw)
points(x[middleones],y[middleones]-gapsize[1],
lty=lty[middleones],col=col[middleones],pch=pch[middleones],...)
points(x[bigones],y[bigones]-(gapsize[1]+gapsize[3]),
lty=lty[bigones],col=col[bigones],pch=pch[bigones],...)
}
else points(x[bigones],y[bigones]-gapsize[1],
lty=lty[bigones],col=col[bigones],pch=pch[bigones],...)
}
else {
if(!is.na(ytics[1])) axis(2,at=ytics,labels=yticlab)
littletics-which(xticsgap[1])
if(length(gapsize) 2) {
middletics-which(xtics = gap[2] xtics = gap[3])
bigtics-which(xtics = gap[4])
show.at-c(xtics[littletics],xtics[middletics]-gapsize[1],],...)
xtics[bigtics]-(gapsize[1]+gapsize[3]))
show.labels-c(xticlab[littletics],xticlab[middletics],xticlab[bigtics])
}
else {
bigtics-which(xtics = gap[2])
show.at-c(xtics[littletics],xtics[bigtics]-gapsize[1])
show.labels-c(xticlab[littletics],xticlab[bigtics])
}
axis(1,at=show.at,labels=show.labels)
axis.break(1,gap[1]-xgw,style=gap)
if(length(gapsize) 2) {
axis.break(1,gap[3]-(gapsize[1]+xgw),style=gap)
points(xgw+x[middleones]-gapsize[1],y[middleones],])
lty=lty[middleones],col=col[middleones],pch=pch[middleones],...)
points(x[bigones]-(gapsize[1]+gapsize[3]),y[bigones],
lty=lty[bigones],col=col[bigones],pch=pch[bigones],...)
}
else
[R] keywords
Hi R Gurus! When you build a package, you need to put in keywords in the Rd files. Where would you find the list of keywords, please? TIA, Edna Bell __ R-help@r-project.org 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] list
Is this what you are after: this comes from the example of wilcox.test and shows the structure and how to reference the p-value: z - wilcox.test(x, y, paired = TRUE, alternative = greater) str(z) List of 7 $ statistic : Named num 40 ..- attr(*, names)= chr V $ parameter : NULL $ p.value: num 0.0195 $ null.value : Named num 0 ..- attr(*, names)= chr location shift $ alternative: chr greater $ method : chr Wilcoxon signed rank test $ data.name : chr x and y - attr(*, class)= chr htest z$p.value [1] 0.01953125 On Tue, Jul 29, 2008 at 1:00 AM, Paul Adams [EMAIL PROTECTED] wrote: I have run a wilcox test on a dataframe and I have returned to me data which I am wanting to sort and/or pick the max. I have tried the following code to pick the max: v-wilcox.test.run w-max(v,na.rm=FALSE) I have also tried w-pmax(p-value,na.rm=FALSE) for the second line and this returns the following error:invalid 'type' (list) of argument This error is understandable because I am trying to find the max p-value from a list of 100 with the following output: $g100 Wilcoxon signed rank test data:newX[,i] V=741, p-value=8.051e-08 I want to be able to pull only the maximum p-value from this output and the $g that matches the p-value by using max.Any help would be appreciated. Thanks Paul [[alternative HTML version deleted]] __ R-help@r-project.org 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org 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.
[R] Bug in sd() and var() in handling vectors of NA (R version 2.7.1)?
In the previous versions of R (2.6.1), when a vector of NA was given to the functions 'sd' or 'var' with parameter na.rm = TRUE, it used to return NA. Now (2.7.1) it returns an ERROR : Example in 2.6.1: sd(c(NA, NA, NA, NA), na.rm = TRUE) [1] NA Example in 2.7.1: sd(c(NA, NA, NA, NA), na.rm = TRUE) Error in var(x, na.rm = na.rm) : paires d'éléments incomplètes We are actually wondering if it is a bug to report, or if we have to manage it in our own R scripts. Thanks a lot, best regards, Laetitia This footnote confirms that this email message has been scanned by PineApp Mail-SeCure for the presence of malicious code, vandals computer viruses. [[alternative HTML version deleted]] __ R-help@r-project.org 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] Problem reading a particular file with read.spss()
It works for me, on Windows with that version of foreign. I also tried under Linux with valgrind, and nothing untoward was reported. The warnings are Warning messages: 1: In read.spss(mySPSSfile.sav) : mySPSSfile.sav: File-indicated character representation code (1252) looks like a Windows codepage 2: In read.spss(mySPSSfile.sav) : mySPSSfile.sav: Unrecognized record type 7, subtype 18 encountered in system file The first is innocuous for you (given your locale), and the second indicated that there is info that we don't know about (SPSS 16.x is not common). On Tue, 29 Jul 2008, Chuck Cleland wrote: Hi All: I have a seemingly typical SPSS data file with 219 rows and 486 variables. When I attempt to read this file into R with read.spss() in the foreign package, I consistently get a crash. This is the sequence of events: library(foreign) sessionInfo() R version 2.7.1 Patched (2008-07-24 r46120) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] foreign_0.8-27 read.spss(c:/chuck/mySPSSfile.sav) R for Windows GUI front-end has encountered a problem and needs to close. We are sorry for the inconvenience. It get a similar result trying to load the file under Rterm.exe. The file opens without any trouble in SPSS version 16.0.2 (April 10, 2008). For anyone willing to investigate, I placed a copy of the SPSS file here: http://www.chuckcleland.net/mySPSSfile.sav I would be grateful if someone could help to figure out why this file causes a problem. thanks, Chuck -- Chuck Cleland, Ph.D. NDRI, Inc. (www.ndri.org) 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@r-project.org 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. -- 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@r-project.org 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.
[R] max
Hello everyone, I have a vector of p-values and I am trying to find the max for the vector and add it to a list.I am using a loop to loop through 50 times.I have used the following code but it does not pick out the max and add to the list.Any help would be appreciated. w-c(v[[1]][3],v[[2]][3],v[[3]][3],v[[4]][3],v[[5]][3],v[[6]][3],v[[7]][3],v[[8]][3],v[[9]][3], v[[10]][3],v[[11]][3],v[[12]][3],v[[13]][3],v[[14]][3],v[[15]][3],v[[16]][3],v[[17]][3],v[[18]][3], v[[19]][3],v[[20]][3],v[[21]][3],v[[22]][3],v[[23]][3],v[[24]][3],v[[25]][3],v[[26]][3],v[[27]][3], v[[28]][3],v[[29]][3],v[[30]][3],v[[31]][3],v[[32]][3],v[[33]][3],v[[34]][3],v[[35]][3],v[[36]][3], v[[37]][3],v[[38]][3],v[[39]][3],v[[40]][3],v[[41]][3],v[[42]][3],v[[43]][3],v[[44]][3],v[[45]][3], v[[46]][3],v[[47]][3],v[[48]][3],v[[49]][3],v[[50]][3],v[[51]][3],v[[52]][3],v[[54]][3],v[[55]][3], v[[56]][3],v[[57]][3],v[[58]][3],v[[59]][3],v[[60]][3],v[[61]][3],v[[62]][3],v[[63]][3],v[[64]][3], v[[65]][3],v[[66]][3],v[[67]][3],v[[68]][3],v[[69]][3],v[[70]][3],v[[71]][3],v[[72]][3],v[[73]][3], v[[74]][3],v[[75]][3],v[[76]][3],v[[77]][3],v[[78]][3],v[[79]][3],v[[80]][3],v[[81]][3],v[[82]][3], v[[83]][3],v[[84]][3],v[[85]][3],v[[86]][3],v[[87]][3],v[[88]][3],v[[89]][3],v[[90]][3],v[[91]][3], v[[92]][3],v[[93]][3],v[[94]][3],v[[95]][3],v[[96]][3],v[[97]][3],v[[98]][3],v[[99]][3],v[[100]][3]) w1-max(w,na.rm=FALSE) list-c(list,w1) This runs but it does not pick out the max of the vector nor does it add it to the list.The for loop shuffles the vector then iterates 50 times and there should be 50 values in the list.I created the list outside the for loop with list-NULL.When I look at the list I get the whole vector 50 times(5000 values) so I assume that max is not picking out the maximum value of the vector everytime through the loop. Any help would be appreciated Paul [[alternative HTML version deleted]] __ R-help@r-project.org 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.
[R] FW: Installing BRugs
A funny thing happened when I wanted a student of mine to install Brugs.
Using the InstallPackages in the windows version, firts gives an erro, but
trying again works flawlessly.
R version is 2.7.0 on WinXP.
Any explanation?
Bendix Carstensen
__
Bendix Carstensen
Senior Statistician
Steno Diabetes Center
Niels Steensens Vej 2-4
DK-2820 Gentofte
Denmark
+45 44 43 87 38 (direct)
+45 30 75 87 38 (mobile)
[EMAIL PROTECTED] http://www.biostat.ku.dk/~bxc
-Original Message-
From: ACJS (Anders Christian Jensen)
Sent: 29. juli 2008 11:06
To: BXC (Bendix Carstensen)
Subject:
utils:::menuInstallPkgs()
trying URL
'http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.7/BRugs_0.4-2.zip'
Content type 'application/zip' length 3399130 bytes (3.2 Mb) opened URL
downloaded 3.1 Mb
Error in gzfile(file, r) : cannot open the connection In addition: Warning
messages:
1: In download.file(url, destfile, method, mode = wb, ...) :
downloaded length 3293996 != reported length 3399130
2: In zip.unpack(pkg, tmpDir) : error 1 in extracting from zip file
3: In gzfile(file, r) :
cannot open compressed file 'BRugs/DESCRIPTION', probable reason 'No such
file or directory'
utils:::menuInstallPkgs()
trying URL
'http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.7/BRugs_0.4-2.zip'
Content type 'application/zip' length 3399130 bytes (3.2 Mb) opened URL
downloaded 3.2 Mb
package 'BRugs' successfully unpacked and MD5 sums checked
The downloaded packages are in
c:\temp\RtmpKoBifa\downloaded_packages
updating HTML package descriptions
___
Anders Christian Jensen
Steno Diabetes Center
Niels Steensens Vej 2-4
DK-2820 Gentofte
Denmark
+45 444 22475 (direct)
[EMAIL PROTECTED]
This e-mail (including any attachments) is intended for ...{{dropped:8}}
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] FW: Installing BRugs
BXC (Bendix Carstensen) wrote:
A funny thing happened when I wanted a student of mine to install Brugs.
Using the InstallPackages in the windows version, firts gives an erro, but
trying again works flawlessly.
R version is 2.7.0 on WinXP.
Any explanation?
No, particularly if it worked the second time. Perhaps a broken machine
or a broken internte connection..
Uwe
Bendix Carstensen
__
Bendix Carstensen
Senior Statistician
Steno Diabetes Center
Niels Steensens Vej 2-4
DK-2820 Gentofte
Denmark
+45 44 43 87 38 (direct)
+45 30 75 87 38 (mobile)
[EMAIL PROTECTED] http://www.biostat.ku.dk/~bxc
-Original Message-
From: ACJS (Anders Christian Jensen)
Sent: 29. juli 2008 11:06
To: BXC (Bendix Carstensen)
Subject:
utils:::menuInstallPkgs()
trying URL
'http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.7/BRugs_0.4-2.zip'
Content type 'application/zip' length 3399130 bytes (3.2 Mb) opened URL
downloaded 3.1 Mb
Error in gzfile(file, r) : cannot open the connection In addition: Warning
messages:
1: In download.file(url, destfile, method, mode = wb, ...) :
downloaded length 3293996 != reported length 3399130
2: In zip.unpack(pkg, tmpDir) : error 1 in extracting from zip file
3: In gzfile(file, r) :
cannot open compressed file 'BRugs/DESCRIPTION', probable reason 'No such
file or directory'
utils:::menuInstallPkgs()
trying URL
'http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.7/BRugs_0.4-2.zip'
Content type 'application/zip' length 3399130 bytes (3.2 Mb) opened URL
downloaded 3.2 Mb
package 'BRugs' successfully unpacked and MD5 sums checked
The downloaded packages are in
c:\temp\RtmpKoBifa\downloaded_packages
updating HTML package descriptions
___
Anders Christian Jensen
Steno Diabetes Center
Niels Steensens Vej 2-4
DK-2820 Gentofte
Denmark
+45 444 22475 (direct)
[EMAIL PROTECTED]
This e-mail (including any attachments) is intended for ...{{dropped:8}}
__
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[R] mac install no font found problem in quartz display
Dear list, I was attempted to install 2.7.1 in my mac only recently because of the fantastic quartz display support. I had no problem in building from source. But after installation, I always got font warnings as following when using quartz: Warning messages: 1: In axis(side = side, at = at, labels = labels, ...) ... : no font could be found for family Arial 33: In axis(side = side, at = at, labels = labels, ...) ... : no font could be found for family Arial 34: In title(...) ... : no font could be found for family Arial 35: In title(...) ... : no font could be found for family Arial 36: In title(...) ... : no font could be found for family Arial 37: In title(...) ... : no font could be found for family Arial There is no such problem when using X11 as the display device. I was guessing it might be cairo or fontconfig library? I looked into the configure file and found warnings: ld: warning, duplicate dylib /opt/local/lib/libiconv.2.dylib ld: warning, duplicate dylib /opt/local/lib/libz.1.dylib ld: warning, duplicate dylib /opt/local/lib/libfontconfig.1.dylib ld: warning, duplicate dylib /opt/local/lib/libXrender.1.dylib ld: warning, duplicate dylib /usr/X11/lib/libSM.6.dylib ld: warning, duplicate dylib /usr/X11/lib/libICE.6.dylib ld: warning, duplicate dylib /usr/X11/lib/libX11.6.dylib ld: warning, duplicate dylib /usr/local/lib/libfreetype.6.dylib I used macport to install cairo and pango and gtk before, and I think these libraries are duplicated in /usr/lib somehow. I don't know if R looked into the wrong directory to search for libraries or not. Please advise! Thanks a lot. -- Zhesi He Graham Group, CNAP Department of Biology (Area 7) University of York PO Box 373, York YO10 5YW, UK Phone: +44-(0)1904-328774 Email: [EMAIL PROTECTED] __ R-help@r-project.org 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] product of successive rows
Hi rcoder,
Assuming that the number of rows of your matrix x is even, try also:
x - matrix(1:72,12)
apply(x,2, tapply, rep(1:(nrow(x)/2),each=2),prod)
# or using a function which argument x is your matrix
prod.mat=function(x) {
k=nrow(x)
g=rep(1:(k/2),each=2)
apply(x,2, tapply, g,prod)
}
prod.mat(x)
HTH,
Jorge
On Sun, Jul 27, 2008 at 6:20 PM, rcoder [EMAIL PROTECTED] wrote:
Hi everyone,
I want to perform an operation on a matrx that outputs the product of
successive pairs of rows. For example: calculating the product between rows
1 2; 3 4; 5 6...etc.
Does anyone know of any readily available functions that can do this?
Thanks,
rcoder
--
View this message in context:
http://www.nabble.com/product-of-successive-rows-tp18681259p18681259.html
Sent from the R help mailing list archive at Nabble.com.
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[R] stringdot ?
Dear all, I am using kernlab package in R, and I have amino acid sequences with different lenghts as input for a SVM and I need to go through this sequences using windows (sliding or fixed) of size X. Does anyone has any suggestions about which function I should use? I thought I could use stringdot, but I am not sure whether it will do what I need.., I have defined my stringdot as: mystringdot - stringdot(length = 7, lambda = 0.5, type = sequence, normalized = TRUE) and my svm as: ksvm(mydata[,1],data=mydata,type=C-svc,kernel=mystringdot,C=10) but it doesn't work. Thank you ever so much, Renata - Email sent from www.virginmedia.com/email Virus-checked using McAfee(R) Software and scanned for spam __ R-help@r-project.org 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.
[R] optim fails when using arima
Hi all, I´m using the arima() function to study a time series but it gives me the following error: Error en optim(init[mask], armafn, method = BFGS, hessian = TRUE, control = optim.control, : non-finite finite-difference value [3] I know that I can change the method of the arima() to CSS instead of ML but I'm specially interested in using maximum likelihood. I have read that using the Nelder-Mead method in the optim() function could avoid this error but I think that it is not possible to change the method of optim from arima(). Does anyone have an idea of how could I solve this problem? Thanks and regards, Maider Mateos __ R-help@r-project.org 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.
[R] 'for' loop, two variables
Dear Rusers,
I am still an unexperienced builder of functions and loops, so my question is
very basic: Is it possible to introduce a second variable (j) into my loop.
To examplify:
# This works fine:
fn - function (x) {if (x46 x52) 1 else 0}
res -NULL
for (i in 40:60) res -c(res,fn(i))
res
# But here, there is an error in the for expression:
fn - function (x,y) {if (x46 x52 y12) 1 else 0 }
res -NULL
for (i in 40:60 j in 0:20) res -c(res,fn(i,j))
# How do I have to write the expression i in 40:60 j in 0:20? Or is there
no way to do that, i.e. I have to do the calculation in two steps?
Thanks in advance!
Friderike
[[alternative HTML version deleted]]
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Re: [R] Is there way to multiple plots on gap.plot?
Hi Art, Ignore the last email, I realized that I had already written the add bit and it will be in plotrix 2.4-5. As far as I can see, both the type and xlim problems are solved. Jim __ R-help@r-project.org 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] 'for' loop, two variables
On 7/29/2008 7:55 AM, Oehler, Friderike (AGPP) wrote:
Dear Rusers,
I am still an unexperienced builder of functions and loops, so my question is
very basic: Is it possible to introduce a second variable (j) into my loop.
To examplify:
# This works fine:
fn - function (x) {if (x46 x52) 1 else 0}
res -NULL
for (i in 40:60) res -c(res,fn(i))
res
# But here, there is an error in the for expression:
fn - function (x,y) {if (x46 x52 y12) 1 else 0 }
res -NULL
for (i in 40:60 j in 0:20) res -c(res,fn(i,j))
# How do I have to write the expression i in 40:60 j in 0:20? Or is there
no way to do that, i.e. I have to do the calculation in two steps?
You need two steps. You probably want either
for (i in 40:60) for (j in 0:20) res -c(res,fn(i,j))
or
i - 40:60
j - 0:20
for (ind in seq_along(i)) res - c(res, fn(i[ind], j[ind]))
(which do quite different loops).
Duncan Murdoch
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Re: [R] 'for' loop, two variables
Dear Frederike,
#Both your functions are vectorized. So you don't need loops. Working
with vectorized functions is much faster than looping.
fn - function (x) {
ifelse(x46 x52, 1, 0)
}
res - fn(40:60)
fn - function (x,y) {
ifelse(x46 x52 y12, 1, 0)
}
datagrid - expand.grid(i = 40:60, j = 0:20)
res - fn(datagrid$i, datagrid$j)
#An other option is to use the functions for the apply-family
fn - function (x) {
ifelse(x46 x52, 1, 0)
}
res - sapply(40:60, fn)
fn - function (x,y) {
ifelse(x46 x52 y12, 1, 0)
}
datagrid - expand.grid(i = 40:60, j = 0:20)
res - apply(datagrid, 1, function(z){
fn(z[i], z[j])
})
#or you can use a nested loop
res -NULL
for (i in 40:60){
for(j in 0:20){
res -c(res,fn(i,j))
}
}
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
[EMAIL PROTECTED]
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Namens Oehler, Friderike (AGPP)
Verzonden: dinsdag 29 juli 2008 13:56
Aan: Oehler, Friderike (AGPP); r-help@r-project.org
Onderwerp: [R] 'for' loop, two variables
Dear Rusers,
I am still an unexperienced builder of functions and loops, so my
question is
very basic: Is it possible to introduce a second variable (j) into my
loop.
To examplify:
# This works fine:
fn - function (x) {if (x46 x52) 1 else 0}
res -NULL
for (i in 40:60) res -c(res,fn(i))
res
# But here, there is an error in the for expression:
fn - function (x,y) {if (x46 x52 y12) 1 else 0 }
res -NULL
for (i in 40:60 j in 0:20) res -c(res,fn(i,j))
# How do I have to write the expression i in 40:60 j in 0:20? Or is
there
no way to do that, i.e. I have to do the calculation in two steps?
Thanks in advance!
Friderike
[[alternative HTML version deleted]]
__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] 'for' loop, two variables
On 29.Jul.2008, at 14:13, ONKELINX, Thierry wrote:
Dear Frederike,
#Both your functions are vectorized. So you don't need loops. Working
with vectorized functions is much faster than looping.
fn - function (x,y) {
ifelse(x46 x52 y12, 1, 0)
}
datagrid - expand.grid(i = 40:60, j = 0:20)
res - apply(datagrid, 1, function(z){
fn(z[i], z[j])
})
or
outer(40:60,0:20,fn2)
which also keeps the matrix structure ...
(use as.vector(...) or as.vector(t(...)) if you need a vector)
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] FW: Installing BRugs
On Tue, 29 Jul 2008, BXC (Bendix Carstensen) wrote:
A funny thing happened when I wanted a student of mine to install Brugs.
Using the InstallPackages in the windows version, firts gives an erro, but
trying again works flawlessly.
There was a problem with the http connection. It happens
R version is 2.7.0 on WinXP.
Any explanation?
Bendix Carstensen
__
Bendix Carstensen
Senior Statistician
Steno Diabetes Center
Niels Steensens Vej 2-4
DK-2820 Gentofte
Denmark
+45 44 43 87 38 (direct)
+45 30 75 87 38 (mobile)
[EMAIL PROTECTED] http://www.biostat.ku.dk/~bxc
-Original Message-
From: ACJS (Anders Christian Jensen)
Sent: 29. juli 2008 11:06
To: BXC (Bendix Carstensen)
Subject:
utils:::menuInstallPkgs()
trying URL
'http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.7/BRugs_0.4-2.zip'
Content type 'application/zip' length 3399130 bytes (3.2 Mb) opened URL
downloaded 3.1 Mb
Error in gzfile(file, r) : cannot open the connection In addition: Warning
messages:
1: In download.file(url, destfile, method, mode = wb, ...) :
downloaded length 3293996 != reported length 3399130
2: In zip.unpack(pkg, tmpDir) : error 1 in extracting from zip file
3: In gzfile(file, r) :
cannot open compressed file 'BRugs/DESCRIPTION', probable reason 'No such file
or directory'
utils:::menuInstallPkgs()
trying URL
'http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.7/BRugs_0.4-2.zip'
Content type 'application/zip' length 3399130 bytes (3.2 Mb) opened URL
downloaded 3.2 Mb
package 'BRugs' successfully unpacked and MD5 sums checked
The downloaded packages are in
c:\temp\RtmpKoBifa\downloaded_packages
updating HTML package descriptions
___
Anders Christian Jensen
Steno Diabetes Center
Niels Steensens Vej 2-4
DK-2820 Gentofte
Denmark
+45 444 22475 (direct)
[EMAIL PROTECTED]
This e-mail (including any attachments) is intended for ...{{dropped:8}}
__
R-help@r-project.org 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.
--
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
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Bug in sd() and var() in handling vectors of NA (R version 2.7.1)?
There was a bug in 2.6.1 which has since been corrected: there is no need to report corrected bugs in obsolete versions. Two different ways to compute the sd of a zero-length vector gave different answers. This is covered by the following NEWS item for 2.7.0 (version from R-patched) o co[rv](use = complete.obs) now always gives an error if there are no complete cases: they used to give NA if method = pearson but an error for the other two methods. (Note that this is pretty arbitrary, but zero-length vectors always give an error so it is at least consistent.) Since sd(na.rm=TRUE) and var(na.rm=TRUE) both call cov(use = complete.obs), this applies also to them. On Tue, 29 Jul 2008, Marisa Laetitia wrote: In the previous versions of R (2.6.1), when a vector of NA was given to the functions 'sd' or 'var' with parameter na.rm = TRUE, it used to return NA. Now (2.7.1) it returns an ERROR : Example in 2.6.1: sd(c(NA, NA, NA, NA), na.rm = TRUE) [1] NA Example in 2.7.1: sd(c(NA, NA, NA, NA), na.rm = TRUE) Error in var(x, na.rm = na.rm) : paires d'?l?ments incompl?tes We are actually wondering if it is a bug to report, or if we have to manage it in our own R scripts. Thanks a lot, best regards, Laetitia This footnote confirms that this email message has been scanned by PineApp Mail-SeCure for the presence of malicious code, vandals computer viruses. [[alternative HTML version deleted]] -- 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@r-project.org 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.
[R] Panel of pie charts
Hi , I am looking to making a panel of pie charts fo some of my dritribution data . I was wondering if there is a way in any R package to write a small script to do so. Thanks, Amin [[alternative HTML version deleted]] __ R-help@r-project.org 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] Negative Binomial Regression
jcarmichael jcarmichael314 at gmail.com writes: Hello. I am attempting to duplicate a negative binomial regression in R. SAS uses generalized estimating equations for model fitting in the GENMOD procedure. proc genmod data=mydata (where=(gender='F')); by agegroup; class id gender type; model count = var1 var2 var3 /dist=NB link=log offset=lregtm; repeated subject=id /type=exch; run; Since my dataset has several observations for each subject, I need the REPEATED statement in order to indicate dependence among observations with the same subject ID and independence amongst those with distinct subject IDs. The TYPE statement goes on to specify the structure of the correlation matrix to be used (exchangeable in this case). I would try glmmPQL in the MASS package. I don't think you can *quite* get negative binomial regression this way, but you can definitely get a quasipoisson model. I think exchangeable correlation corresponds to correlation=corCompSymm() in your glmmPQL command. Ben Bolker __ R-help@r-project.org 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] How to set the parameters in Trellis Graphics (by Lattice package)
On Tue, Jul 29, 2008 at 5:37 PM, G.H. Zuo [EMAIL PROTECTED] wrote: Dear R users I plot the trellis graphics by using the lattice package. Everything is OK. Now, I want set some parameters of the trellis graphics. 1. The tick label site. By default, only two tick labels had been output in x-axis of my plot. I want output four or five tick labels. In the traditional graphics system, it will be very easy. Just not output the axis in plot by use the parameter xaxt=n, and then used axis to add the axis. But it seem useless in the grid graphics system. Then how can I do it? You would generally use scales=list(ticks=5) or scales=list(x=list(ticks=5)). See the entry for scales in ?xyplot However, the default (suggested) number of ticks is 4 or 5, so there may be something odd about your case. How about a reproducible example? 2. How to change the margin of the figure? Why do you want to do that? Grid graphics automatically allocates enough space for the objects to be displayed. If you really want to, there are ways to do it, but I suggest you do not. You are more likely looking for the aspect argument. 3. Can I change the style of the figure like par in traditional system. For example, I want change all the fontface into bold. Now, I must change the fontface for every item by using command trellis.par.set() like this: trellis.par.set(list(par.xlab.text = gpar(font = 2), par.ylab.text = gpar(font = 2), axis.text = gpar(font = 2), add.text = gpar(font=2) )) is there a method to change this parameter in one time, like par(font=2) You can try trellis.par.set(fontsize=list(text=14)) which is similar. Otherwise you could write a function to set the font parameters; this could be modified from latticeExtra::custom.theme thanks G.H.Zuo [[alternative HTML version deleted]] __ R-help@r-project.org 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. -- Felix Andrews / 安福立 PhD candidate Integrated Catchment Assessment and Management Centre The Fenner School of Environment and Society The Australian National University (Building 48A), ACT 0200 Beijing Bag, Locked Bag 40, Kingston ACT 2604 http://www.neurofractal.org/felix/ 3358 543D AAC6 22C2 D336 80D9 360B 72DD 3E4C F5D8 __ R-help@r-project.org 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.
[R] combining zoo series with an overlapping index?
day-structure(c(7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.71, 7.71, 7.7, 7.7, 7.7, 7.7, 7.69, 7.68, 7.68, 7.67, 7.67, 7.67, 7.66, 7.65, 7.65, 7.65, 7.64, 7.64, 7.63, 7.63, 7.63, 7.62, 7.62, 7.62, 7.62, 7.63, 7.63, 7.63, 7.63, 7.63, 7.64, 7.64, 7.65, 7.65, 7.65, 7.66, 7.66, 7.67, 7.67, 7.67, 7.68, 7.68, 7.69, 7.69, 7.69, 7.69, 7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.7, 7.71, 7.71, 7.7, 7.7, 7.7, 7.7, 7.7, 7.7, 7.7, 7.69, 7.68, 7.68, 7.68, 7.67, 7.67, 7.66, 7.66, 7.66, 7.66, 7.65, 7.65, 7.65, 7.65, 7.66, 7.66, 7.66, 7.67, 7.67, 7.68, 7.68, 7.68, 7.68, 7.69, 7.69, 7.69), index = structure(c(13307.000694, 13307.01, 13307.021528, 13307.031944, 13307.042361, 13307.052778, 13307.063194, 13307.073611, 13307.084028, 13307.09, 13307.104861, 13307.115278, 13307.125694, 13307.136111, 13307.146528, 13307.156944, 13307.167361, 13307.18, 13307.188194, 13307.198611, 13307.209028, 13307.219444, 13307.229861, 13307.240278, 13307.250694, 13307.26, 13307.271528, 13307.281944, 13307.292361, 13307.302778, 13307.313194, 13307.323611, 13307.334028, 13307.34, 13307.354861, 13307.365278, 13307.375694, 13307.386111, 13307.396528, 13307.406944, 13307.417361, 13307.427778, 13307.438194, 13307.448611, 13307.459028, 13307.469444, 13307.479861, 13307.490278, 13307.500694, 13307.51, 13307.521528, 13307.531944, 13307.542361, 13307.552778, 13307.563194, 13307.573611, 13307.584028, 13307.59, 13307.604861, 13307.615278, 13307.625694, 13307.636111, 13307.646528, 13307.656944, 13307.667361, 13307.68, 13307.688194, 13307.698611, 13307.709028, 13307.719444, 13307.729861, 13307.740278, 13307.750694, 13307.76, 13307.771528, 13307.781944, 13307.792361, 13307.802778, 13307.813194, 13307.823611, 13307.834028, 13307.84, 13307.854861, 13307.865278, 13307.875694, 13307.886111, 13307.896528, 13307.906944, 13307.917361, 13307.927778, 13307.938194, 13307.948611, 13307.959028, 13307.969444, 13307.979861, 13307.990278), format = structure(c(m/d/y, h:m:s), .Names = c(dates, times)), origin = structure(c(1, 1, 1970), .Names = c(month, day, year)), class = c(chron, dates, times)), class = zoo) #Mullholland ER daytime # extract section of interest w - window(day, start=chron(6/8/6, 4:16:0), end=chron(6/8/6, 20:31:0)) # zap all interior points with NA's and then fill back in using linear interpolation lin - na.approx(replace(w, time(w) start(w) time(w) end(w), NA)) # plot the 3 series on one screen #plot(merge(day, w, lin), col = 1:3, screen = 1) #can I do this with a command (generically?)? I am going to use this inside of a function, and below is the result I would like. d- c(day[1:17,], lin, day[84:96,]) plot(d) -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis [[alternative HTML version deleted]] __ R-help@r-project.org 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] keywords
On Jul 29, 2008, at 5:24 AM, Edna Bell wrote: Hi R Gurus! When you build a package, you need to put in keywords in the Rd files. Where would you find the list of keywords, please? Simplest way is to google for r keywords. First hit is: http://www.stat.ucl.ac.be/ISdidactique/Rhelp/doc/keywords.html Also, if you do help.start() and follow the link Search Engine Keywords, you will find the list. TIA, Edna Bell Haris Skiadas Department of Mathematics and Computer Science Hanover College __ R-help@r-project.org 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] combining zoo series with an overlapping index?
Try: window(d, time(lin)) - coredata(lin) On Tue, Jul 29, 2008 at 9:01 AM, stephen sefick [EMAIL PROTECTED] wrote: day-structure(c(7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.71, 7.71, 7.7, 7.7, 7.7, 7.7, 7.69, 7.68, 7.68, 7.67, 7.67, 7.67, 7.66, 7.65, 7.65, 7.65, 7.64, 7.64, 7.63, 7.63, 7.63, 7.62, 7.62, 7.62, 7.62, 7.63, 7.63, 7.63, 7.63, 7.63, 7.64, 7.64, 7.65, 7.65, 7.65, 7.66, 7.66, 7.67, 7.67, 7.67, 7.68, 7.68, 7.69, 7.69, 7.69, 7.69, 7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.7, 7.71, 7.71, 7.7, 7.7, 7.7, 7.7, 7.7, 7.7, 7.7, 7.69, 7.68, 7.68, 7.68, 7.67, 7.67, 7.66, 7.66, 7.66, 7.66, 7.65, 7.65, 7.65, 7.65, 7.66, 7.66, 7.66, 7.67, 7.67, 7.68, 7.68, 7.68, 7.68, 7.69, 7.69, 7.69), index = structure(c(13307.000694, 13307.01, 13307.021528, 13307.031944, 13307.042361, 13307.052778, 13307.063194, 13307.073611, 13307.084028, 13307.09, 13307.104861, 13307.115278, 13307.125694, 13307.136111, 13307.146528, 13307.156944, 13307.167361, 13307.18, 13307.188194, 13307.198611, 13307.209028, 13307.219444, 13307.229861, 13307.240278, 13307.250694, 13307.26, 13307.271528, 13307.281944, 13307.292361, 13307.302778, 13307.313194, 13307.323611, 13307.334028, 13307.34, 13307.354861, 13307.365278, 13307.375694, 13307.386111, 13307.396528, 13307.406944, 13307.417361, 13307.427778, 13307.438194, 13307.448611, 13307.459028, 13307.469444, 13307.479861, 13307.490278, 13307.500694, 13307.51, 13307.521528, 13307.531944, 13307.542361, 13307.552778, 13307.563194, 13307.573611, 13307.584028, 13307.59, 13307.604861, 13307.615278, 13307.625694, 13307.636111, 13307.646528, 13307.656944, 13307.667361, 13307.68, 13307.688194, 13307.698611, 13307.709028, 13307.719444, 13307.729861, 13307.740278, 13307.750694, 13307.76, 13307.771528, 13307.781944, 13307.792361, 13307.802778, 13307.813194, 13307.823611, 13307.834028, 13307.84, 13307.854861, 13307.865278, 13307.875694, 13307.886111, 13307.896528, 13307.906944, 13307.917361, 13307.927778, 13307.938194, 13307.948611, 13307.959028, 13307.969444, 13307.979861, 13307.990278), format = structure(c(m/d/y, h:m:s), .Names = c(dates, times)), origin = structure(c(1, 1, 1970), .Names = c(month, day, year)), class = c(chron, dates, times)), class = zoo) #Mullholland ER daytime # extract section of interest w - window(day, start=chron(6/8/6, 4:16:0), end=chron(6/8/6, 20:31:0)) # zap all interior points with NA's and then fill back in using linear interpolation lin - na.approx(replace(w, time(w) start(w) time(w) end(w), NA)) # plot the 3 series on one screen #plot(merge(day, w, lin), col = 1:3, screen = 1) #can I do this with a command (generically?)? I am going to use this inside of a function, and below is the result I would like. d- c(day[1:17,], lin, day[84:96,]) plot(d) -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis [[alternative HTML version deleted]] __ R-help@r-project.org 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. __ R-help@r-project.org 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] Panel of pie charts
Have a look at mfcol in ?par. --- On Tue, 7/29/08, Amin Momin [EMAIL PROTECTED] wrote: From: Amin Momin [EMAIL PROTECTED] Subject: [R] Panel of pie charts To: r-help@r-project.org Received: Tuesday, July 29, 2008, 7:34 AM Hi , I am looking to making a panel of pie charts fo some of my dritribution data . I was wondering if there is a way in any R package to write a small script to do so. Thanks, Amin [[alternative HTML version deleted]] __ R-help@r-project.org 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. __ [[elided Yahoo spam]] __ R-help@r-project.org 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] combining zoo series with an overlapping index?
when I plot this it gives me the coredata(lin) (Ithink). I would like d- c(day[1:17,], lin, day[84:96,]) to be the result On Tue, Jul 29, 2008 at 9:18 AM, Gabor Grothendieck [EMAIL PROTECTED] wrote: Try: window(d, time(lin)) - coredata(lin) On Tue, Jul 29, 2008 at 9:01 AM, stephen sefick [EMAIL PROTECTED] wrote: day-structure(c(7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.71, 7.71, 7.7, 7.7, 7.7, 7.7, 7.69, 7.68, 7.68, 7.67, 7.67, 7.67, 7.66, 7.65, 7.65, 7.65, 7.64, 7.64, 7.63, 7.63, 7.63, 7.62, 7.62, 7.62, 7.62, 7.63, 7.63, 7.63, 7.63, 7.63, 7.64, 7.64, 7.65, 7.65, 7.65, 7.66, 7.66, 7.67, 7.67, 7.67, 7.68, 7.68, 7.69, 7.69, 7.69, 7.69, 7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.7, 7.71, 7.71, 7.7, 7.7, 7.7, 7.7, 7.7, 7.7, 7.7, 7.69, 7.68, 7.68, 7.68, 7.67, 7.67, 7.66, 7.66, 7.66, 7.66, 7.65, 7.65, 7.65, 7.65, 7.66, 7.66, 7.66, 7.67, 7.67, 7.68, 7.68, 7.68, 7.68, 7.69, 7.69, 7.69), index = structure(c(13307.000694, 13307.01, 13307.021528, 13307.031944, 13307.042361, 13307.052778, 13307.063194, 13307.073611, 13307.084028, 13307.09, 13307.104861, 13307.115278, 13307.125694, 13307.136111, 13307.146528, 13307.156944, 13307.167361, 13307.18, 13307.188194, 13307.198611, 13307.209028, 13307.219444, 13307.229861, 13307.240278, 13307.250694, 13307.26, 13307.271528, 13307.281944, 13307.292361, 13307.302778, 13307.313194, 13307.323611, 13307.334028, 13307.34, 13307.354861, 13307.365278, 13307.375694, 13307.386111, 13307.396528, 13307.406944, 13307.417361, 13307.427778, 13307.438194, 13307.448611, 13307.459028, 13307.469444, 13307.479861, 13307.490278, 13307.500694, 13307.51, 13307.521528, 13307.531944, 13307.542361, 13307.552778, 13307.563194, 13307.573611, 13307.584028, 13307.59, 13307.604861, 13307.615278, 13307.625694, 13307.636111, 13307.646528, 13307.656944, 13307.667361, 13307.68, 13307.688194, 13307.698611, 13307.709028, 13307.719444, 13307.729861, 13307.740278, 13307.750694, 13307.76, 13307.771528, 13307.781944, 13307.792361, 13307.802778, 13307.813194, 13307.823611, 13307.834028, 13307.84, 13307.854861, 13307.865278, 13307.875694, 13307.886111, 13307.896528, 13307.906944, 13307.917361, 13307.927778, 13307.938194, 13307.948611, 13307.959028, 13307.969444, 13307.979861, 13307.990278), format = structure(c(m/d/y, h:m:s), .Names = c(dates, times)), origin = structure(c(1, 1, 1970), .Names = c(month, day, year)), class = c(chron, dates, times)), class = zoo) #Mullholland ER daytime # extract section of interest w - window(day, start=chron(6/8/6, 4:16:0), end=chron(6/8/6, 20:31:0)) # zap all interior points with NA's and then fill back in using linear interpolation lin - na.approx(replace(w, time(w) start(w) time(w) end(w), NA)) # plot the 3 series on one screen #plot(merge(day, w, lin), col = 1:3, screen = 1) #can I do this with a command (generically?)? I am going to use this inside of a function, and below is the result I would like. d- c(day[1:17,], lin, day[84:96,]) plot(d) -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis [[alternative HTML version deleted]] __ R-help@r-project.org 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. -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis [[alternative HTML version deleted]] __ R-help@r-project.org 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] writing the plots
?png ?write.table --- On Mon, 7/28/08, Rajasekaramya [EMAIL PROTECTED] wrote: From: Rajasekaramya [EMAIL PROTECTED] Subject: [R] writing the plots To: r-help@r-project.org Received: Monday, July 28, 2008, 10:54 AM hi there, I want to write the plots in the pdfs and the details about the graph in a seperate notepad. plot(as.numeric(lapply(resultgenes,length)), main= Geneset.gene#.bias.test,xlab=Top.Ranked.Genesets, ylab=gene.number.per.geneset) lines(loess.smooth(c(1:1000),as.numeric(lapply(resultgenes,length)), span = 2/3, degree = 1, family = gaussian, evaluation = 50),col=2, lwd=3) i want this graph in the pdf and some notes regarding the graph in a notepad. Kindly help me Ramya -- View this message in context: http://www.nabble.com/writing-the-plots-tp18689539p18689539.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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. __ [[elided Yahoo spam]] __ R-help@r-project.org 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.
[R] Howto Draw Bimodal Gamma Curve with User Supplied Parameters
Hi, Suppose I have the following vector (data points): x [1] 36.0 57.3 73.3 92.0 300.4 80.9 19.8 31.4 85.8 44.9 24.6 48.0 [13] 28.0 38.3 85.2 103.6 154.4 128.5 38.3 72.4 122.7 123.1 41.8 21.7 [25] 143.6 120.2 46.6 29.2 44.8 25.0 57.3 96.4 29.4 62.9 66.4 30.0 [37] 24.1 14.8 56.6 102.4 117.5 90.4 37.2 79.6 27.8 17.1 26.6 16.3 [49] 41.4 48.9 24.1 23.3 9.9 11.5 15.0 23.6 29.3 27.0 19.2 18.7 [61] 4.1 13.0 3.3 5.5 38.2 8.5 39.6 39.2 16.1 35.3 23.3 31.5 [73] 38.8 51.5 28.4 18.8 24.1 25.4 28.8 32.8 31.0 28.8 33.3 55.5 [85] 39.2 21.0 43.7 16.3 50.6 34.6 66.3 50.5 59.4 46.7 51.9 125.6 [97] 69.8 43.7 86.8 50.6 132.4 56.0 6.1 4.9 7.1 7.1 12.8 12.1 [109] 164.2 69.3 15.6 11.4 34.3 9.2 17.6 21.7 19.2 30.7 61.1 35.8 [121] 185.8 118.4 13.0 9.6 19.1 45.2 94.5 248.0 56.3 24.4 13.8 12.8 [133] 35.0 31.6 22.5 50.1 18.7 22.1 28.3 39.5 48.2 33.1 43.5 35.1 [145] 37.4 30.3 15.8 13.9 15.3 16.1 12.7 11.4 13.0 13.8 31.5 25.3 [157] 65.2 39.5 And the following parameter set (2 component) for gamma function. comp.1comp.2 alpha (shape) 2.855444 2.152056 beta (scale) 10.418785 39.296224 These params are predefined/precalculated by user. My question is how can I create a bimodal gamma curve - based on the two parameter set - on top of the histogram of data points above? - Gundala Viswanath Jakarta - Indonesia __ R-help@r-project.org 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.
[R] About clustering techniques
Hello R users It's some time I am playing with a dataset to do some cluster analysis. The data set consists of 14 columns being geographical coordinates and monthly temperatures in annual files latitutde - longitude - temperature 1 -. - temperature 12 I have some missing values in some cases, maybe there are 8 monthly valid values at some points with four non valid. I don't want to supress the whole row with 8 good/4 bad values as I wanna try annual and monthy analysis. I first tried kmeans but found a problem with missing values. When trying without omitting missing values kmeans gives an error and when excluding invalid data too many values are excluded in some years of the data series. Now I have been reading about pam, pamk and clara, I think they can handle missing values. But can't find out the way to perform the analysis with these functions. As I'm not an statistics nor an R expert the fpc or cluster package documentation is not enough for me. If you know about a website or a tutorial explaining the way to use that functions, with examples to check if possible, please post them. Any other help or suggestion is greatly appreciated. Thanks in advance Paco -- _ El ponent la mou, el llevant la plou Usuari Linux registrat: 363952 --- Fotos: http://picasaweb.google.es/pacomet [[alternative HTML version deleted]] __ R-help@r-project.org 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] Coarsening the Resolution of a Dataset
Unfortunately, when I get to the 'myCuts' line, I receive the following error: Error: evaluation nested too deeply: infinite recursion / options(expressions=)? ...and I also receive warnings about memory allocation being reached (even though I've already used memory.limit() to maximise the memory) - this is a fairly sizeable dataset afterall, 2160 rows by 4320 columns. Therefore I was wondering if there are any alternative ways of coarsening a dataset? Or are there any packages/commands built for this sort of thing? Any advice would be much appreciated! Thanks again, Steve _ Find the best and worst places on the planet __ R-help@r-project.org 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] About clustering techniques
Hi Paco, I got the same problem with you before. Thus, I just impute the missing values For example: newdata-as.matrix(impute(olddata, fun=random)) then I believe that you could analyze your data. Hopefully it helps. Chunhao Quoting pacomet [EMAIL PROTECTED]: Hello R users It's some time I am playing with a dataset to do some cluster analysis. The data set consists of 14 columns being geographical coordinates and monthly temperatures in annual files latitutde - longitude - temperature 1 -. - temperature 12 I have some missing values in some cases, maybe there are 8 monthly valid values at some points with four non valid. I don't want to supress the whole row with 8 good/4 bad values as I wanna try annual and monthy analysis. I first tried kmeans but found a problem with missing values. When trying without omitting missing values kmeans gives an error and when excluding invalid data too many values are excluded in some years of the data series. Now I have been reading about pam, pamk and clara, I think they can handle missing values. But can't find out the way to perform the analysis with these functions. As I'm not an statistics nor an R expert the fpc or cluster package documentation is not enough for me. If you know about a website or a tutorial explaining the way to use that functions, with examples to check if possible, please post them. Any other help or suggestion is greatly appreciated. Thanks in advance Paco -- _ El ponent la mou, el llevant la plou Usuari Linux registrat: 363952 --- Fotos: http://picasaweb.google.es/pacomet [[alternative HTML version deleted]] __ R-help@r-project.org 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. __ R-help@r-project.org 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] About clustering techniques
Dear Paco, in order to use the methods in the cluster package (including pam), look up the help page of daisy, which is able to compute dissimilarity matrices handling missing values appropriately (in most situations). A good reference is the Kaufman and Rousseeuw book cited on that help page. Christian On Tue, 29 Jul 2008, pacomet wrote: Hello R users It's some time I am playing with a dataset to do some cluster analysis. The data set consists of 14 columns being geographical coordinates and monthly temperatures in annual files latitutde - longitude - temperature 1 -. - temperature 12 I have some missing values in some cases, maybe there are 8 monthly valid values at some points with four non valid. I don't want to supress the whole row with 8 good/4 bad values as I wanna try annual and monthy analysis. I first tried kmeans but found a problem with missing values. When trying without omitting missing values kmeans gives an error and when excluding invalid data too many values are excluded in some years of the data series. Now I have been reading about pam, pamk and clara, I think they can handle missing values. But can't find out the way to perform the analysis with these functions. As I'm not an statistics nor an R expert the fpc or cluster package documentation is not enough for me. If you know about a website or a tutorial explaining the way to use that functions, with examples to check if possible, please post them. Any other help or suggestion is greatly appreciated. Thanks in advance Paco -- _ El ponent la mou, el llevant la plou Usuari Linux registrat: 363952 --- Fotos: http://picasaweb.google.es/pacomet [[alternative HTML version deleted]] __ R-help@r-project.org 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. *** --- *** Christian Hennig University College London, Department of Statistical Science Gower St., London WC1E 6BT, phone +44 207 679 1698 [EMAIL PROTECTED], www.homepages.ucl.ac.uk/~ucakche __ R-help@r-project.org 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] combining zoo series with an overlapping index?
The code I gave replaces that portion of d that overlaps with lin with lin. Is that not what you wanted? (please try to use minimal examples as shown here) library(zoo) d - zoo(1:10) + 100 lin - - head(d, 4) window(d, time(lin)) - coredata(lin) d 123456789 10 -101 -102 -103 -104 105 106 107 108 109 110 On Tue, Jul 29, 2008 at 9:48 AM, stephen sefick [EMAIL PROTECTED] wrote: when I plot this it gives me the coredata(lin) (Ithink). I would like d- c(day[1:17,], lin, day[84:96,]) to be the result On Tue, Jul 29, 2008 at 9:18 AM, Gabor Grothendieck [EMAIL PROTECTED] wrote: Try: window(d, time(lin)) - coredata(lin) On Tue, Jul 29, 2008 at 9:01 AM, stephen sefick [EMAIL PROTECTED] wrote: day-structure(c(7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.71, 7.71, 7.7, 7.7, 7.7, 7.7, 7.69, 7.68, 7.68, 7.67, 7.67, 7.67, 7.66, 7.65, 7.65, 7.65, 7.64, 7.64, 7.63, 7.63, 7.63, 7.62, 7.62, 7.62, 7.62, 7.63, 7.63, 7.63, 7.63, 7.63, 7.64, 7.64, 7.65, 7.65, 7.65, 7.66, 7.66, 7.67, 7.67, 7.67, 7.68, 7.68, 7.69, 7.69, 7.69, 7.69, 7.7, 7.7, 7.7, 7.7, 7.7, 7.71, 7.7, 7.7, 7.71, 7.71, 7.7, 7.7, 7.7, 7.7, 7.7, 7.7, 7.7, 7.69, 7.68, 7.68, 7.68, 7.67, 7.67, 7.66, 7.66, 7.66, 7.66, 7.65, 7.65, 7.65, 7.65, 7.66, 7.66, 7.66, 7.67, 7.67, 7.68, 7.68, 7.68, 7.68, 7.69, 7.69, 7.69), index = structure(c(13307.000694, 13307.01, 13307.021528, 13307.031944, 13307.042361, 13307.052778, 13307.063194, 13307.073611, 13307.084028, 13307.09, 13307.104861, 13307.115278, 13307.125694, 13307.136111, 13307.146528, 13307.156944, 13307.167361, 13307.18, 13307.188194, 13307.198611, 13307.209028, 13307.219444, 13307.229861, 13307.240278, 13307.250694, 13307.26, 13307.271528, 13307.281944, 13307.292361, 13307.302778, 13307.313194, 13307.323611, 13307.334028, 13307.34, 13307.354861, 13307.365278, 13307.375694, 13307.386111, 13307.396528, 13307.406944, 13307.417361, 13307.427778, 13307.438194, 13307.448611, 13307.459028, 13307.469444, 13307.479861, 13307.490278, 13307.500694, 13307.51, 13307.521528, 13307.531944, 13307.542361, 13307.552778, 13307.563194, 13307.573611, 13307.584028, 13307.59, 13307.604861, 13307.615278, 13307.625694, 13307.636111, 13307.646528, 13307.656944, 13307.667361, 13307.68, 13307.688194, 13307.698611, 13307.709028, 13307.719444, 13307.729861, 13307.740278, 13307.750694, 13307.76, 13307.771528, 13307.781944, 13307.792361, 13307.802778, 13307.813194, 13307.823611, 13307.834028, 13307.84, 13307.854861, 13307.865278, 13307.875694, 13307.886111, 13307.896528, 13307.906944, 13307.917361, 13307.927778, 13307.938194, 13307.948611, 13307.959028, 13307.969444, 13307.979861, 13307.990278), format = structure(c(m/d/y, h:m:s), .Names = c(dates, times)), origin = structure(c(1, 1, 1970), .Names = c(month, day, year)), class = c(chron, dates, times)), class = zoo) #Mullholland ER daytime # extract section of interest w - window(day, start=chron(6/8/6, 4:16:0), end=chron(6/8/6, 20:31:0)) # zap all interior points with NA's and then fill back in using linear interpolation lin - na.approx(replace(w, time(w) start(w) time(w) end(w), NA)) # plot the 3 series on one screen #plot(merge(day, w, lin), col = 1:3, screen = 1) #can I do this with a command (generically?)? I am going to use this inside of a function, and below is the result I would like. d- c(day[1:17,], lin, day[84:96,]) plot(d) -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis [[alternative HTML version deleted]] __ R-help@r-project.org 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. -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide
Re: [R] About clustering techniques
A quick comment on this: imputation is an option to make things technically work, but it is not necessarily good. Imputation always introduces some noise, ie, it fakes information that is not really there. Whether it is good depends strongly on the data, the situation and the imputation method (random often not being a very sensible choice). Christian On Tue, 29 Jul 2008, [EMAIL PROTECTED] wrote: Hi Paco, I got the same problem with you before. Thus, I just impute the missing values For example: newdata-as.matrix(impute(olddata, fun=random)) then I believe that you could analyze your data. Hopefully it helps. Chunhao Quoting pacomet [EMAIL PROTECTED]: Hello R users It's some time I am playing with a dataset to do some cluster analysis. The data set consists of 14 columns being geographical coordinates and monthly temperatures in annual files latitutde - longitude - temperature 1 -. - temperature 12 I have some missing values in some cases, maybe there are 8 monthly valid values at some points with four non valid. I don't want to supress the whole row with 8 good/4 bad values as I wanna try annual and monthy analysis. I first tried kmeans but found a problem with missing values. When trying without omitting missing values kmeans gives an error and when excluding invalid data too many values are excluded in some years of the data series. Now I have been reading about pam, pamk and clara, I think they can handle missing values. But can't find out the way to perform the analysis with these functions. As I'm not an statistics nor an R expert the fpc or cluster package documentation is not enough for me. If you know about a website or a tutorial explaining the way to use that functions, with examples to check if possible, please post them. Any other help or suggestion is greatly appreciated. Thanks in advance Paco -- _ El ponent la mou, el llevant la plou Usuari Linux registrat: 363952 --- Fotos: http://picasaweb.google.es/pacomet [[alternative HTML version deleted]] __ R-help@r-project.org 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. __ R-help@r-project.org 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. *** --- *** Christian Hennig University College London, Department of Statistical Science Gower St., London WC1E 6BT, phone +44 207 679 1698 [EMAIL PROTECTED], www.homepages.ucl.ac.uk/~ucakche __ R-help@r-project.org 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.
[R] Most often pairs of chars across grouping variable
Hi list, is there a package or function to compute the frequencies of pairs of chars in a variable across a grouping variable? Eg: d - data.frame(ID=gl(2,3), F=c(A,B,C,A,C,D)) d ID F 1 1 A 2 1 B 3 1 C 4 2 A 5 2 C 6 2 D Now I want to summarize the frequencies of all pairs A-B, A-C, A-D, B-C, B-D, C-D across ID: A B C D A - 1 2 1 B - - 1 0 C - - - 1 here, the combination A-C is most frequent. The real problem behind that is that 'F' codes diagnoses and I search for the most often pairs of diagnoses. Thanks, Sven __ R-help@r-project.org 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] Negative Binomial Regression
On Tue, 29 Jul 2008, Ben Bolker wrote: jcarmichael jcarmichael314 at gmail.com writes: Hello. I am attempting to duplicate a negative binomial regression in R. SAS uses generalized estimating equations for model fitting in the GENMOD procedure. proc genmod data=mydata (where=(gender='F')); by agegroup; class id gender type; model count = var1 var2 var3 /dist=NB link=log offset=lregtm; repeated subject=id /type=exch; run; Since my dataset has several observations for each subject, I need the REPEATED statement in order to indicate dependence among observations with the same subject ID and independence amongst those with distinct subject IDs. The TYPE statement goes on to specify the structure of the correlation matrix to be used (exchangeable in this case). I would try glmmPQL in the MASS package. I don't think you can *quite* get negative binomial regression this way, but you can definitely get a quasipoisson model. I think exchangeable correlation corresponds to correlation=corCompSymm() in your glmmPQL command. The problem here is that GLMM and GEE are not fitting the same model -- in one the coefficients are subject-specific and in the other population-average (see MASS4 or Diggle, Liang, Zeger +/- Heagarty). There are several R packages for GEE, including gee, yags, geepack. The documentation of geeglm (geepack) claims it can be used with families as in glm(), so you could try it with MASS's negative.binomial family. -- 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@r-project.org 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] Most often pairs of chars across grouping variable
on 07/29/2008 09:51 AM [EMAIL PROTECTED] wrote: Hi list, is there a package or function to compute the frequencies of pairs of chars in a variable across a grouping variable? Eg: d - data.frame(ID=gl(2,3), F=c(A,B,C,A,C,D)) d ID F 1 1 A 2 1 B 3 1 C 4 2 A 5 2 C 6 2 D Now I want to summarize the frequencies of all pairs A-B, A-C, A-D, B-C, B-D, C-D across ID: A B C D A - 1 2 1 B - - 1 0 C - - - 1 here, the combination A-C is most frequent. The real problem behind that is that 'F' codes diagnoses and I search for the most often pairs of diagnoses. Thanks, Sven I suspect that there might be something over in Bioconductor, but here is one approach: table(data.frame(t(do.call(cbind, tapply(d$F, d$ID, function(x) combn(as.character(x), 2)) X2 X1 B C D A 1 2 1 B 0 1 0 C 0 0 1 See ?combn to create the initial pairs from the data. This is done on a per ID basis using tapply. The result is transposed into a data frame and then table() is used to create the cross tabulation of the results. HTH, Marc Schwartz __ R-help@r-project.org 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] Case statements in R
Dear all, may I suggest to include this quotation of Patrick Burns in the fortunes package? :-) Best, Roland Patrick Burns wrote: A good reason to use '' rather than '' is if evaluating whatever is on the right will create an error if what is on the left is FALSE. '' and '||' stop if they already know the answer from seeing the left side. So they are likely faster by a few nanoseconds, but it's the ability to stop before you go off a cliff that is probably the key feature. __ R-help@r-project.org 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.
[R] subscript out of bounds error.
(mean.std.s2n.loss.gain[[1]])
Probe.Set.ID rho_prime r ho_prime_sdpom Expr1
matchinggenes Mean std_dev
29 SNP_A-190846347 2.47 0.75 0 PRKCZ -
0.34560. 1344 30 SNP_A-213137044 2.61
0.58 0 PRKCZ -0.32700.1517
31 SNP_A-220300945 2.67 0.58 0 PRKCZ
-0.3242.1555
32 SNP_A-209854643 2.81 0.58 1 PRKCZ
-0.3212 0.155
33 SNP_A-226456542 2.90 0.58 1 PRKCZ
0.3400.1782
34 SNP_A-178872841 2.84 0.58 1 PRKCZ
0.339 .1778
1398 SNP_A-1834906 25 1.90 0.42 0 FRAP1
-0.2710 0.03668
1399 SNP_A-2146504 25 1.90 0.42 0 FRAP1
-0.2700.03627
1400 SNP_A-230730925 1.90 0.42 0 FRAP1
-0.061 0.0820
1401 SNP_A-226165827 2.07 0.42 0 FRAP1
-0.276 0.04101
1402 SNP_A-427448129 2.21 0.42 0 FRAP1
-0.277 0.0412
1403 SNP_A-205307029 2.21 0.42 0 FRAP1
- 0.285 0.0501
1404 SNP_A-427638229 2.21 0.42 0 FRAP1
-0.30060.0523
mean.std.s2n.loss.gain is a l;ist object of length 1000.for sample i have
provided with 1 element in the list.
I want to calculate the mean and std dev for the unique genes.I wrote a code
for that.but it is throwing me an error subscript out of bounds for
.unique.genes[[i]]
genes.info.unique - lapply(mean.std.s2n.loss.gain,function(.unique){
.unique.genes-unique(.unique[,6])
.unique.genes-as.vector(as.matrix(.unique.genes))
snpnumber-list()
mean.genes-list()
sd.genes-list()
class(.unique.genes)
for(i in 1:length(.unique.genes))
{
snpnumber[[i]]- length(.unique[.unique[,6] %in% .unique.genes[[i]],1])
mean.genes[[i]]-mean(.unique[.unique[,6] %in% .unique.genes[[i]],7])
sd.genes[[i]]-sd(.unique[.unique[,6] %in% .unique.genes[[i]],7])
}
cbind(.unique.genes,unlist(snpnumber),unlist(mean.genes),unlist(sd.genes))
})
kindly help me
Ramya
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[R] lattice question
Dear friends - I have a plot of simulated data to compare to the observed and want the simulated to appear accumulated so that a darker grey corresponds to more lines of simulated data, but on top of that I want the measured values (Na) to be very visible. The code below meets only few of the desires. Hope someone will give me help to master the panel formation. Best wishes Troels xyplot(Na+yrep[,1:100]~time|as.factor(ID),yrepbug,type=c(g,o), span=0.75,ylab=[Na]) -- Troels Ring - - Department of nephrology - - Aalborg Hospital 9100 Aalborg, Denmark - - +45 99326629 - - [EMAIL PROTECTED] __ R-help@r-project.org 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.
[R] subscript out of bounds help
(mean.std.s2n.loss.gain[[1]])
Probe.Set.ID rho_prime rho_prime_sd pom Expr1 matchinggenes
Meanstd_dev
29 SNP_A-190846347 2.47 0.75 0 PRKCZ
-0.345616170.13443676
30 SNP_A-213137044 2.61 0.58 0 PRKCZ
-0.327046920.15171799
31 SNP_A-220300945 2.67 0.58 0 PRKCZ
-0.324239380.15554016
32 SNP_A-209854643 2.81 0.58 1 PRKCZ
-0.321298220.15939414
33 SNP_A-226456542 2.90 0.58 1 PRKCZ
0.340551500.17828152
34 SNP_A-178872841 2.84 0.58 1 PRKCZ
0.339537440.17782981
35 SNP_A-184250934 2.59 0.58 1 PRKCZ
-0.297423980.14795647
36 SNP_A-425038933 2.60 0.50 1 PRKCZ
-0.294941910.14550642
1398 SNP_A-183490625 1.900.42 0 FRAP1
-0.271049420.03668130
1399 SNP_A-214650425 1.900.42 0 FRAP1
-0.270215920.03623132
1400 SNP_A-230730925 1.900.42 0 FRAP1
-0.061870950.08202400
1401 SNP_A-226165827 2.070.42 0 FRAP1
-0.276791000.04101169
1402 SNP_A-427448129 2.210.42 0 FRAP1
-0.277134330.0412
1403 SNP_A-205307029 2.210.42 0 FRAP1
-0.285844830.05010136
1404 SNP_A-427638229 2.21 0.420 FRAP1
-0.300633420.05233933
mean.std.s2n.loss.gain is a list object of length 1000.
I want to calculate the mean and std dev for the unique genes using the
column 7 values.I wrote a code but it showing me subscript out of bounds
error for .unique.genes[[i]].Could any body help me fixing the error.
genes.info.unique - lapply(mean.std.s2n.loss.gain,function(.unique){
.unique.genes-unique(.unique[,6])
.unique.genes-as.vector(as.matrix(.unique.genes))
snpnumber-list()
mean.genes-list()
sd.genes-list()
class(.unique.genes)
for(i in 1:length(.unique.genes))
{
snpnumber[[i]]- length(.unique[.unique[,6] %in% .unique.genes[[i]],1])
mean.genes[[i]]-mean(.unique[.unique[,6] %in% .unique.genes[[i]],7])
sd.genes[[i]]-sd(.unique[.unique[,6] %in% .unique.genes[[i]],7])
}
cbind(.unique.genes,unlist(snpnumber),unlist(mean.genes),unlist(sd.genes))
})
kindly help me
Ramya
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[R] Removing script file
Colleagues, (Running R 2.7.0) I have a script that I want to delete as it completes execution. The penultimate line of the script (before the quit command) is: file.remove(Scriptname) The script is executed as: R --no-save Scriptname In OS X and Linux this is successful and returns: file.remove(x) [1] TRUE and the file is deleted In Windows XP, it returns: [1] FALSE and the file is not deleted. I realize that this is an OS issue rather than an issue with execution of the command. Is there some means to delete the script from within R? Dennis Dennis Fisher MD P (The P Less Than Company) Phone: 1-866-PLessThan (1-866-753-7784) Fax: 1-415-564-2220 www.PLessThan.com [[alternative HTML version deleted]] __ R-help@r-project.org 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.
[R] tensor product of equi-spaced B-splines in the unit square
Dear all, I need to compute tensor product of B-spline defined over equi-spaced break-points. I wrote my own program (it works in a 2-dimensional setting) library(splines) # set the break-points Knots = seq(-1,1,length=10) # number of splines M = (length(Knots)-4)^2 # short cut to splineDesign function bspline = function(x) splineDesign(Knots,x,outer.ok = T) # bivariate tensor product of bspline btens = function(x) t(bspline(x[1]))%*%bspline(x[2]) # numebr of points to plot ng = 51 # create vectors for plotting xgr = seq(-1,1,length=ng) xgr2= expand.grid(xgr,xgr) # generate random coef. of linear combination bet = rnorm(M) # create matrix for contour-type plot Bx = apply(xgr2,1,btens) Bmat= matrix(t(Bx)%*%bet,ng) # plot the result contour(xgr,xgr,Bmat) persp(xgr,xgr,Bmat,theta=15) any of you have a better idea (ie more efficient)? Thanks in advance, Patrizio Frederic +- | Patrizio Frederic | Research associate in Statistics, | Department of Economics, | University of Modena and Reggio Emilia, | Via Berengario 51, | 41100 Modena, Italy | | tel: +39 059 205 6727 | fax: +39 059 205 6947 | mail: [EMAIL PROTECTED] +- __ R-help@r-project.org 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] Negative Binomial Regression
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Prof Brian Ripley wrote: | On Tue, 29 Jul 2008, Ben Bolker wrote: | | jcarmichael jcarmichael314 at gmail.com writes: | | | | Hello. | | I am attempting to duplicate a negative binomial regression in R. | SAS uses | generalized estimating equations for model fitting in the GENMOD | procedure. | | proc genmod data=mydata (where=(gender='F')); | by agegroup; | class id gender type; | model count = var1 var2 var3 /dist=NB link=log offset=lregtm; | repeated subject=id /type=exch; | run; | | Since my dataset has several observations for each subject, I need the | REPEATED statement in order to indicate dependence among observations | with | the same subject ID and independence amongst those with distinct subject | IDs. The TYPE statement goes on to specify the structure of the | correlation | matrix to be used (exchangeable in this case). | | I would try glmmPQL in the MASS package. I don't think you | can *quite* get negative binomial regression this way, but | you can definitely get a quasipoisson model. I think exchangeable | correlation corresponds to correlation=corCompSymm() in your | glmmPQL command. | | The problem here is that GLMM and GEE are not fitting the same model -- | in one the coefficients are subject-specific and in the other | population-average (see MASS4 or Diggle, Liang, Zeger +/- Heagarty). | | There are several R packages for GEE, including gee, yags, geepack. The | documentation of geeglm (geepack) claims it can be used with families as | in glm(), so you could try it with MASS's negative.binomial family. | ~ Point taken (although I guess I was pointing the original poster to a way to do a reasonable analysis, not necessarily to duplicate the SAS analysis as requested). Will the negative.binomial family really work for this, since it seems to require a fixed theta (overdispersion) parameter? ~ If I very naively do the following: library(geepack) data(dietox) mf2 - formula(Weight~Cu*Time+I(Time^2)+I(Time^3)) gee2 - geeglm(mf2, data=dietox, id=Pig, ~ family=poisson(identity),corstr=ar1) library(MASS) gee2 - geeglm(mf2,data=dietox,id=Pig,family=negative.binomial(theta=100),corstr=ar1) ~ gives an error variance invalid -- ~ so the whole thing would seem to take a bit of troubleshooting ~ (geeglm also gives warnings about non-integer Poisson values -- I don't know why a Poisson link is being used in this example for a non-integer Weight value ... ?) ~ Ben Bolker -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.6 (GNU/Linux) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org iD8DBQFIj0Euc5UpGjwzenMRAhs/AJ9EOu8RpJibN9LMLbI0p1tCTls5xQCfRLnf 6hkinpQu0PMQgh4f/Sl5BaM= =y7CX -END PGP SIGNATURE- __ R-help@r-project.org 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.
[R] Bootstraping GAMs for confidence intervales calculation
Dear R-Users,
I am resending this message just to reminder my question regarding the
calculation of a bootstrap confidence intervals for a GAM plot.
I am trying to apply a bootstrap to a GAM in order to calculate the 95%
confidence intervals for a smooth curve obtained by the “plot.gam”
function of the mgcv package. Nonetheless, I am getting some
difficulties in transposing the results for the graphs.
I used the following commands in R, “mgcv” and “boot” packages:
* attach(bbvc_11Jul08)*
* model.boot-function(data,indices){*
*+ sub.data-data[indices,]*
*+
model-gam(asin(sqrt(Cov0_30))~s(age,k=4,bs=cr),family=gaussian,data=sub.data)*
*+ coef(model)}*
* gam.boot-boot(bbvc_11Jul08,model.boot,R=1000)*
* *
* gam.boot*
ORDINARY NONPARAMETRIC BOOTSTRAP
Call:
boot(data = bbvc_11Jul08, statistic = model.boot, R = 1000)
Bootstrap Statistics :
original bias std. error
t1* 0.40370112 0.0002762674 0.02017378
t2* 0.04715501 -0.0144132280 0.07648348
t3* 0.14590936 -0.0135820501 0.05161244
t4* 0.13520098 -0.0096405733 0.04470793
I obtained 4 indexes in the “bootstrap” output. I know that for a lm
function, with an independent variable, t1 is the index of the intercept
and t2 is the index of the variable, but for GAMs I don't find what each
indexes really mean!
Moreover, to calculate the confidence intervals of 95% I proceed as follows:
* *
*boot.ci(gam.boot,conf=0.95) *
BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
Based on 1000 bootstrap replicates
CALL :
boot.ci(boot.out = gam.boot, conf = 0.95)
Intervals :
Level Normal Basic Studentized
95% ( 0.3639, 0.4430 ) ( 0.3635, 0.4434 ) ( 0.3583, 0.4507 )
Level Percentile BCa
95% ( 0.3640, 0.4439 ) ( 0.3623, 0.4434 )
Calculations and Intervals on Original Scale
Warning message:
In sqrt(tv[, 2]) : NaNs produced
My first doubt is when I make boot.ci, which of the four should I
choose? By default, R calculates index=1:min(2,length(boot.out$t0) but
I don't know if this is the correct form to do it.
Finally, I don’t know how to introduce the 95% confidence intervals in
the plot.gam.
I hope you can help me. Thanks in advance.
Luís Reino
--
Luis Reino, PhD Student
Centro de Estudos Florestais
Departamento de Engenharia Florestal
Instituto Superior de Agronomia
Universidade Técnica de Lisboa
1349-017 Lisboa
Portugal
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[R] more environment questions
Hi R Gurus:
Here is some code that I was experimenting with, please:
f1 - function(x) {
+ e1 - new.env(parent=.GlobalEnv)
+ environment(e1)
+ print(environment())
+ return(mean(x))
+ }
f1(1:15)
environment: 0x02525444
[1] 8
My question: why isn't the environment within the function set to e1, please?
Thanks,
Edna Bell
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Re: [R] Panel of pie charts
I am looking to making a panel of pie charts fo some of my
dritribution data . I was wondering if there is a way in any R
package to write a small script to do so.
pie() will do you a one-off pie chart, but there is no equivalent using
grid/ lattice graphics. You could write a panel.pie function to draw
them, but be warned: pie charts are almost never the best option for
displaying data. (See
http://www.perceptualedge.com/articles/visual_business_intelligence/save_the_pies_for_dessert.pdf
for example.) Have you considered using barcharts instead?
Regards,
Richie.
Mathematical Sciences Unit
HSL
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This message contains privileged and confidential inform...{{dropped:20}}
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Re: [R] more environment questions
e1 - ...
creates a new environment e1
environment(e1)
does nothing
print(environment(e1))
print environment e1
By the way, if you are doing a lot of manipulations of environments
you might want to look at the proto package which reframes the
whole thing in terms of object oriented programming.
On Tue, Jul 29, 2008 at 12:12 PM, Edna Bell [EMAIL PROTECTED] wrote:
Hi R Gurus:
Here is some code that I was experimenting with, please:
f1 - function(x) {
+ e1 - new.env(parent=.GlobalEnv)
+ environment(e1)
+ print(environment())
+ return(mean(x))
+ }
f1(1:15)
environment: 0x02525444
[1] 8
My question: why isn't the environment within the function set to e1, please?
Thanks,
Edna Bell
__
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[R] Question regarding statisticians
Hi Everyone, I'm apologize for asking this in the R-general list, but I'm unsure of where else to ask this burning question of mine: Where do statisticians talk on the internet about professional/ career developement/ issues? I've found many a list (much like this one) that specailize in talking about statistics, but not about *being* a statistician. Any pointers to a message board or mailing list would be very helpful. The closest I've found is a mailing list of statisticians in the UK, but it seems to be spammed with job offers from the UK and the odd question about an area of statistics. The closest thing I've found, which is for actuaries are actuary.com or http://www.actuarialoutpost.com/actuarial_discussion_forum Thanks in Advance, -Max __ R-help@r-project.org 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.
[R] table questions
Hi again! Suppose I have the following: xy - round(rexp(20),1) xy [1] 0.1 3.4 1.6 0.4 1.0 1.4 0.2 0.3 1.6 0.2 0.0 0.1 0.1 1.0 2.0 0.9 2.5 0.1 1.5 0.4 table(xy) xy 0 0.1 0.2 0.3 0.4 0.9 1 1.4 1.5 1.6 2 2.5 3.4 1 4 2 1 2 1 2 1 1 2 1 1 1 Is there a way to set things up to have 0 - 0.4 0.5 - 0.9 etc. please? I know there is the cut functions, but breaks are required. If you don't have breaks, what should you do, please? Would using the breaks from the hist function work appropriately, please? thanks Edna Bell __ R-help@r-project.org 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] more environment questions
Is there a way to set the environment within a function,, please?
On Tue, Jul 29, 2008 at 11:25 AM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
e1 - ...
creates a new environment e1
environment(e1)
does nothing
print(environment(e1))
print environment e1
By the way, if you are doing a lot of manipulations of environments
you might want to look at the proto package which reframes the
whole thing in terms of object oriented programming.
On Tue, Jul 29, 2008 at 12:12 PM, Edna Bell [EMAIL PROTECTED] wrote:
Hi R Gurus:
Here is some code that I was experimenting with, please:
f1 - function(x) {
+ e1 - new.env(parent=.GlobalEnv)
+ environment(e1)
+ print(environment())
+ return(mean(x))
+ }
f1(1:15)
environment: 0x02525444
[1] 8
My question: why isn't the environment within the function set to e1,
please?
Thanks,
Edna Bell
__
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] more environment questions
No but look at proto since I suspect the creation
of proto objects is basically what you are trying
to do through the back door. Home page:
http://r-proto.googlecode.com
On Tue, Jul 29, 2008 at 12:29 PM, Edna Bell [EMAIL PROTECTED] wrote:
Is there a way to set the environment within a function,, please?
On Tue, Jul 29, 2008 at 11:25 AM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
e1 - ...
creates a new environment e1
environment(e1)
does nothing
print(environment(e1))
print environment e1
By the way, if you are doing a lot of manipulations of environments
you might want to look at the proto package which reframes the
whole thing in terms of object oriented programming.
On Tue, Jul 29, 2008 at 12:12 PM, Edna Bell [EMAIL PROTECTED] wrote:
Hi R Gurus:
Here is some code that I was experimenting with, please:
f1 - function(x) {
+ e1 - new.env(parent=.GlobalEnv)
+ environment(e1)
+ print(environment())
+ return(mean(x))
+ }
f1(1:15)
environment: 0x02525444
[1] 8
My question: why isn't the environment within the function set to e1,
please?
Thanks,
Edna Bell
__
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
[R] Graphics function question
Hello
I have created a graph using the following commands:
startBReP3O1T - diffs$BReP3O1T - diffs$diff_BReP3O1T
endBReP3O1T - diffs$BReP3O1T
x - seq(47,89, length = 10)
ymin - min(min(startBReP3O1T), min(endBReP3O1T))
ymax - max(max(startBReP3O1T), max(endBReP3O1T))
y - seq(ymin, ymax, length = 10)
plot(x,y, type = 'n', xlab = 'Age', ylab = 'BReP3O1T', main = 'Age, decline and
BReP3O1T')
segments(x0 = startage, x1 = endage, y0 = startBReP3O1T, y1 = endBReP3O1T,
col = decline)
legend('topleft', legend = c('Stable', 'Decline'), lty = 1, col = c(1,2))
I would like to make this into a function. The only thing that changes is
BReP3O1T.
The problem is that sometimes this occurs as text (e.g. in ylab and main)
sometimes after a $ (e.g. in the
first two assigment statements), and sometimes it refers to the values (e.g. in
ymin and ymax)
Any help appreciated.
Peter
Peter L. Flom, PhD
Brainscope, Inc.
212 263 7863 (MTW)
212 845 4485 (Th)
917 488 7176 (F)
[[alternative HTML version deleted]]
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Re: [R] combining zoo series with an overlapping index?
plot(replace(day, is.list(index(lin)), coredata(lin))) [[alternative HTML version deleted]] __ R-help@r-project.org 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.
[R] try question
Hi still yet again! I have the following code: try(log(rnorm(25)),silent=TRUE) [1] -0.26396185 NaN NaN -0.13078069 -2.44997193 -2.15603971 NaN 0.94917495 0.07244544 NaN [11] -1.06341127 -0.42293099 -0.53769569 0.95134763 0.93403340 NaN -0.10502078 NaN 0.30283262 NaN [21] -0.11696872 -3.84122332 NaN NaN -0.12808690 Warning message: In log(rnorm(25)) : NaNs produced I thought that putting the silent = TRUE would suppress the warnings, please. What should I do instead, please? Thanks, Edna Bell __ R-help@r-project.org 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.
[R] accessing list elements
Hello everyone, I have a list which I am trying to calculate a max value.I have the list as w-c(v[[1]][1],...v[[100]][1]). The problem I am getting is that the function max is saying the list is an invalid type (list) of argument.When I show the element v[[1]][1] it shows as $statistic,V,736.I am only wanting to use the number 736 from v[[1]][1] but am not sure how to access that number only?I believe if I just use the number then I should be able to calculate the max. Any help would be appreciated Paul [[alternative HTML version deleted]] __ R-help@r-project.org 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] try question
try(log(rnorm(25)),silent=TRUE)
[1] -0.26396185 NaN NaN -0.13078069 -2.44997193
-2.15603971 NaN 0.94917495 0.07244544 NaN
[11] -1.06341127 -0.42293099 -0.53769569 0.95134763 0.93403340
NaN -0.10502078 NaN 0.30283262 NaN
[21] -0.11696872 -3.84122332 NaN NaN -0.12808690
Warning message:
In log(rnorm(25)) : NaNs produced
I thought that putting the silent = TRUE would suppress the
warnings, please. What should I do instead, please?
The 'silent' argument to try supresses errors, not warnings. You can
convert warnings to errors with options(warn=2), or ignore warnings with
options(warn=-1). See ?options for more details.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and confidential inform...{{dropped:20}}
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Re: [R] Removing script file
On Tue, 29 Jul 2008, Dennis Fisher wrote: Colleagues, (Running R 2.7.0) I have a script that I want to delete as it completes execution. The penultimate line of the script (before the quit command) is: file.remove(Scriptname) The script is executed as: R --no-save Scriptname In OS X and Linux this is successful and returns: file.remove(x) [1] TRUE and the file is deleted In Windows XP, it returns: [1] FALSE and the file is not deleted. I realize that this is an OS issue rather than an issue with execution of the command. Is there some means to delete the script from within R? Not in an OS that does not allow open files to be deleted. (Not the whole truth for WIndows, but true of the mechanism used to implement redirection in the compiler used for R.) You could modify Rscript to do so (it does not use re-direction). Dennis Dennis Fisher MD P (The P Less Than Company) Phone: 1-866-PLessThan (1-866-753-7784) Fax: 1-415-564-2220 www.PLessThan.com [[alternative HTML version deleted]] __ R-help@r-project.org 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. -- 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@r-project.org 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] Panel of pie charts
On Tue, 2008-07-29 at 17:23 +0100, [EMAIL PROTECTED] wrote: I am looking to making a panel of pie charts fo some of my dritribution data . I was wondering if there is a way in any R package to write a small script to do so. pie() will do you a one-off pie chart, but there is no equivalent using grid/ lattice graphics. You could write a panel.pie function to draw them, but be warned: pie charts are almost never the best option for displaying data. (See http://www.perceptualedge.com/articles/visual_business_intelligence/save_the_pies_for_dessert.pdf for example.) Have you considered using barcharts instead? All concerns about pie charts aside, Deepayan has already done this for you, as part of his excellent book, Lattice: Multivariate data Visualization with R (2008) Springer, part of the UseR series. Check out the book's website for the code to reproduce all the figures: http://lmdvr.r-forge.r-project.org/figures/figures.html in particular you should look at chapter 14, figure 14.5 and the associated code, including panel functions, to reproduce this figure. HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org 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] Graphics function question
You can do it very easily using subsetting and a bit of paste().
I assumed you didn't need startdata and enddata after the plot
has been made, but if you do you can change the last line of
the function to return them.
Sarah
myfunction - function(dataname)
{
# where dataname is a string, eg myfunction(BReP3O1T)
startdata - diffs[, dataname] - diffs[, paste(diff, dataname, sep=_)]
enddata - diffs[, dataname[
x - seq(47,89, length = 10)
ymin - min(min(startdata), min(enddata))
ymax - max(max(startdata), max(enddata))
y - seq(ymin, ymax, length = 10)
plot(x,y, type = 'n', xlab = 'Age', ylab = dataname, main =
paste(Age, decline and, dataname)
segments(x0 = startage, x1 = endage, y0 = startdata, y1 = enddata,
col = decline)
legend('topleft', legend = c('Stable', 'Decline'), lty = 1, col = c(1,2))
invisible()
}
On Tue, Jul 29, 2008 at 12:38 PM, Peter Flom [EMAIL PROTECTED] wrote:
Hello
I have created a graph using the following commands:
startBReP3O1T - diffs$BReP3O1T - diffs$diff_BReP3O1T
endBReP3O1T - diffs$BReP3O1T
x - seq(47,89, length = 10)
ymin - min(min(startBReP3O1T), min(endBReP3O1T))
ymax - max(max(startBReP3O1T), max(endBReP3O1T))
y - seq(ymin, ymax, length = 10)
plot(x,y, type = 'n', xlab = 'Age', ylab = 'BReP3O1T', main = 'Age, decline
and BReP3O1T')
segments(x0 = startage, x1 = endage, y0 = startBReP3O1T, y1 = endBReP3O1T,
col = decline)
legend('topleft', legend = c('Stable', 'Decline'), lty = 1, col = c(1,2))
I would like to make this into a function. The only thing that changes is
BReP3O1T.
The problem is that sometimes this occurs as text (e.g. in ylab and main)
sometimes after a $ (e.g. in the
first two assigment statements), and sometimes it refers to the values (e.g.
in ymin and ymax)
Any help appreciated.
Peter
--
Sarah Goslee
http://www.functionaldiversity.org
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] accessing list elements
What's v? And w? And what exactly do you want to do? A small reproducible example would be very helpful. Without knowing what your data look like, it's hard to make helpful suggestions. Sarah On Tue, Jul 29, 2008 at 12:56 PM, Paul Adams [EMAIL PROTECTED] wrote: Hello everyone, I have a list which I am trying to calculate a max value.I have the list as w-c(v[[1]][1],...v[[100]][1]). The problem I am getting is that the function max is saying the list is an invalid type (list) of argument.When I show the element v[[1]][1] it shows as $statistic,V,736.I am only wanting to use the number 736 from v[[1]][1] but am not sure how to access that number only?I believe if I just use the number then I should be able to calculate the max. Any help would be appreciated Paul -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Graphics function question
on 07/29/2008 11:38 AM Peter Flom wrote:
Hello
I have created a graph using the following commands:
startBReP3O1T - diffs$BReP3O1T - diffs$diff_BReP3O1T
endBReP3O1T - diffs$BReP3O1T
x - seq(47,89, length = 10)
ymin - min(min(startBReP3O1T), min(endBReP3O1T))
ymax - max(max(startBReP3O1T), max(endBReP3O1T))
y - seq(ymin, ymax, length = 10)
plot(x,y, type = 'n', xlab = 'Age', ylab = 'BReP3O1T', main = 'Age, decline and
BReP3O1T')
segments(x0 = startage, x1 = endage, y0 = startBReP3O1T, y1 = endBReP3O1T,
col = decline)
legend('topleft', legend = c('Stable', 'Decline'), lty = 1, col = c(1,2))
I would like to make this into a function. The only thing that changes is
BReP3O1T.
The problem is that sometimes this occurs as text (e.g. in ylab and main)
sometimes after a $ (e.g. in the
first two assigment statements), and sometimes it refers to the values (e.g. in
ymin and ymax)
Any help appreciated.
Peter
Hey Peter, LTNS! Job change it looks like from the e-mail address?
Here is one approach. It is not clear from the above, where 'startage',
'endage' and 'decline' come from, so I am passing them as arguments:
In your example above, the function call would be:
MyPlot(diffs$BReP3O1T, diffs$diff_BReP3O1T, startage, endage, decline)
MyPlot - function(x, y, startage, endage, decline)
{
start - x - y
end - x
s1 - seq(47, 89, length = 10)
# Don't need to use min/max on each vector separately
ymin - min(start, end, na.rm = TRUE)
ymax - max(start, end, na.rm = TRUE)
s2 - seq(ymin, ymax, length = 10)
# Turn 'x' into a label, stripping anything before $ if present
label - gsub(^.+\\$, , deparse(substitute(x)))
plot(s1, s2, type = n, xlab = Age, ylab = label,
main = paste(Age, decline and, label)
segments(startage, endage, start, end, col = decline)
legend(topleft, legend = c(Stable, Decline), lty = 1,
col = c(1, 2))
}
HTH,
Marc Schwartz
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Re: [R] Negative Binomial Regression
On Tue, 29 Jul 2008, Ben Bolker wrote:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Prof Brian Ripley wrote:
| On Tue, 29 Jul 2008, Ben Bolker wrote:
|
| jcarmichael jcarmichael314 at gmail.com writes:
|
| Hello.
|
| I am attempting to duplicate a negative binomial regression in R.
| SAS uses
| generalized estimating equations for model fitting in the GENMOD
| procedure.
|
| proc genmod data=mydata (where=(gender='F'));
| by agegroup;
| class id gender type;
| model count = var1 var2 var3 /dist=NB link=log offset=lregtm;
| repeated subject=id /type=exch;
| run;
|
| Since my dataset has several observations for each subject, I need the
| REPEATED statement in order to indicate dependence among observations
| with
| the same subject ID and independence amongst those with distinct subject
| IDs. The TYPE statement goes on to specify the structure of the
| correlation
| matrix to be used (exchangeable in this case).
|
| I would try glmmPQL in the MASS package. I don't think you
| can *quite* get negative binomial regression this way, but
| you can definitely get a quasipoisson model. I think exchangeable
| correlation corresponds to correlation=corCompSymm() in your
| glmmPQL command.
|
| The problem here is that GLMM and GEE are not fitting the same model --
| in one the coefficients are subject-specific and in the other
| population-average (see MASS4 or Diggle, Liang, Zeger +/- Heagarty).
|
| There are several R packages for GEE, including gee, yags, geepack. The
| documentation of geeglm (geepack) claims it can be used with families as
| in glm(), so you could try it with MASS's negative.binomial family.
|
~ Point taken (although I guess I was pointing the original poster
to a way to do a reasonable analysis, not necessarily to duplicate
the SAS analysis as requested). Will the negative.binomial family
really work for this, since it seems to require a fixed theta
(overdispersion) parameter?
I was answering 'I am trying to duplicate': I don't know how SAS estimates
the parameter by GEE. The best guess I have is that theta is estimated
from the initial glm fit and fixed in the GEE phase, but that is only
interpolation from very vague descriptions.
~ If I very naively do the following:
library(geepack)
data(dietox)
mf2 - formula(Weight~Cu*Time+I(Time^2)+I(Time^3))
gee2 - geeglm(mf2, data=dietox, id=Pig,
~ family=poisson(identity),corstr=ar1)
library(MASS)
gee2 -
geeglm(mf2,data=dietox,id=Pig,family=negative.binomial(theta=100),corstr=ar1)
~ gives an error variance invalid --
~ so the whole thing would seem to take a bit of troubleshooting
I wasn't placing much faith on geeglm actually being as general as it says
it is ('claims' ... 'could try').
~ (geeglm also gives warnings about non-integer Poisson values -- I
don't know why a Poisson link is being used in this example for
a non-integer Weight value ... ?)
'poisson' _family_, I presume?
--
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
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Re: [R] try question
On Tue, 29 Jul 2008, Edna Bell wrote: Hi still yet again! I have the following code: try(log(rnorm(25)),silent=TRUE) [1] -0.26396185 NaN NaN -0.13078069 -2.44997193 -2.15603971 NaN 0.94917495 0.07244544 NaN [11] -1.06341127 -0.42293099 -0.53769569 0.95134763 0.93403340 NaN -0.10502078 NaN 0.30283262 NaN [21] -0.11696872 -3.84122332 NaN NaN -0.12808690 Warning message: In log(rnorm(25)) : NaNs produced I thought that putting the silent = TRUE would suppress the warnings, please. What should I do instead, please? No, it supresses the error message (and you have no error here). Use suppressWarnings(). -- 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@r-project.org 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] accessing list elements
Hello - Paul Adams wrote: Hello everyone, I have a list which I am trying to calculate a max value.I have the list as w-c(v[[1]][1],...v[[100]][1]). The problem I am getting is that the function max is saying the list is an invalid type (list) of argument.When I show the element v[[1]][1] it shows as $statistic,V,736.I am only wanting to use the number 736 from v[[1]][1] but am not sure how to access that number only?I believe if I just use the number then I should be able to calculate the max. Any help would be appreciated Paul Please see footer of this message, which states ...provide commented, minimal, self-contained, reproducible code. You do not do this. Is your problem like this? a - list(20, 30) max(a) ##error max(unlist(a)) ##no error [[alternative HTML version deleted]] __ R-help@r-project.org 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. __ R-help@r-project.org 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] Negative Binomial Regression
Prof Brian Ripley wrote: 'poisson' _family_, I presume? oops, yes. __ R-help@r-project.org 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] Major Cloud Computing Testbed Announced at University of Illinois
*R we Wishing on a cloud-* I would like to Test R as an cloud analytical tool. remote submit csv files to the web for data mining , get data crunching with scalable processing power and disk storage , display Silverlight graphs of summarization, downloadable in pdf's. data mining on demand, on function. rather than pay costly annual analytical softwares license fees. analytics softwares charge by cpu count on terminal servers. r is free,open source and has the largest pool of statisticians and advanced packages available (like Linux of data mining) www.r-project.org Advanced stats packages limited in say weather forecasting etc because parallel processing /super-computing PC 's are fragmented. would do it myself, but i am the little guy. regards, Ajay Ohri www.decisionstats.com http://www.linkedin.com/in/ajohri (Networking openly.) (cc-ed to R help list) On Tue, Jul 29, 2008 at 10:46 PM, Suman Chaudhuri [EMAIL PROTECTED] wrote: With our partnership with Amazon, we at Patni (www.patni.com) are thinking of doing a prototype test suite for building apps in the cloud that would address something similar to what Reuven was talking about. I would be interested in hearing your thoughts on what features this test suite in the cloud can provide to anyone who would be interested in using this service). If anyone here has thoughts that they would like to share, please do so. We were thinking more so from an application/website perspective (acceptance testing, unit testing, distributed functional testing, testing Ruby-on-Rails, etc), but if there are other needs, like testing out the actual platform, maybe we can have some discussions around that. Suman Date: Tue, 29 Jul 2008 10:06:38 -0700 Subject: Re: Major Cloud Computing Testbed Announced at University of Illinois From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Something like this (i haven't used it, only found it via google search, i'm sure there are others): http://it.tmcnet.com/topics/it/articles/33908-alertsite-debuts-new-web-site-load-testing-solution.htm Seems a not very difficult task to build some EC2 service that will add instances to hammer your site until some specified criteria are met like successful requests/second drops below a threshold or 500 errors crop up. Anyone doing that yet? - randall On Jul 29, 9:49 am, Reuven Cohen [EMAIL PROTECTED] wrote: We've been working with Intel for the last 18 months on several cloud projects and our reoccurring problem has been how to do you test a cloud platform designed to manage upwards of 100,000 servers without access to a real world test bed. This is a good first step, but with less then 1000 servers it still doesn't adequately address the problem. I wonder how others developing large scale cloud platforms are handling scalability testing? Reuven On Tue, Jul 29, 2008 at 12:06 PM, Ameed Taylor [EMAIL PROTECTED] wrote: The University of Illinois announced a Cloud Computing Research Initiative with Yahoo, HP and Intel today. View the press release here.http://tinyurl.com/6z5cpo The fact that this testbed will be Global in nature makes it notable in my opinion. Sites will include The University of Illinois, the National Science Foundation, the Infocomm Development Authority of Singapore (IDA), and the Karlsruhe Institute of Technology in Germany. The press release mentions the testbed will include a 1,024-core HP system with 200 TB of disk space and will run Apache Hadoop and the Pig parallel programming language. Hopefully someone from this group will submit a proposal to take part in the Testbed as this highly visible project could help dispel any lingering doubts about the viability of Cloud Computing especially as it relates to extremely data intensive requirements. Ameed Taylor President Applation LLC www.applation.com Blog -www.ondemandbeat.com -- -- Reuven Cohen Founder Chief Technologist, Enomaly Inc.www.enomaly.comhttp://inc.www.enomaly.com/:: 416 848 6036 x 1 skype: ruv.net // aol: ruv6 blog www.elasticvapor.com - Get Linked inhttp://linkedin.com/pub/0/b72/7b4 --~--~-~--~~~---~--~~ You received this message because you are subscribed to the Google Groups Cloud Computing group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] To post job listing, send email to [EMAIL PROTECTED] (position title, employer and location in subject, description in message body) or visit http://www.cloudjobs.net For more options, visit this group at http://groups.google.ca/group/cloud-computing?hl=en?hl=en Posting guidelines: http://groups.google.ca/group/cloud-computing/web/frequently-asked-qu... To invite your colleague
Re: [R] Graphics function question
One clarification: I did the dirty and completely unprofessional thing, and assumed that this function would only be run in the workspace with all your data. I use functions like this frequently for something I need to do multiple times for a particular project, but will never do anywhere else, but it isn't proper programming, and will fail if it can't find diffs and the other items referenced. If you need this to be portable, you will need to pass them all as arguments to the function. Sarah -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org 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.
[R] List names help
hi , Is there anyways to delet the list names once created. i tried rm(names(mylist)) i didnt work kindly help me Ramya -- View this message in context: http://www.nabble.com/List-names-help-tp18717742p18717742.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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.
[R] biplot_group_colours_and_point_symbols
hi, i am seeking for a solution to create a biplot that shows: a) data points instead of labels and that b) shows two different groups (field: site with two factor levels: forest/corridor) in different colours i tried to include pch=19 to show points instead of labels in the syntax below but failed. biplot(model, main=, iter=0.1, cex=c(0.3,1.2), col=c(1,2)) i appreciate any hint. best regards, lukas °°° Lukas Indermaur, PhD student eawag / Swiss Federal Institute of Aquatic Science and Technology ECO - Department of Aquatic Ecology Überlandstrasse 133 CH-8600 Dübendorf Switzerland Phone: +41 (0) 71 220 38 25 Fax: +41 (0) 44 823 53 15 Email: [EMAIL PROTECTED] www.lukasindermaur.ch http://www.lukasindermaur.ch/ __ R-help@r-project.org 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.
[R] Flexible semivariogram in R (HELP)
Hey All, I am a PhD student in Forestry science and I wish use kriging to spatial LiDAR cloud points. The shape of my semivariogram is like hole (wave) model, but unfortunately I don't know the code of flexible model or hole model to fit my semivariogram in R. I Found it in google or in R help but I did't find nothing about this model in R. Thank you for Help Alessandro [[alternative HTML version deleted]] __ R-help@r-project.org 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] List names help
Sure, just set them to NULL. mylist - c(a=1, b=2, c=3) mylist a b c 1 2 3 names(mylist) - NULL mylist [1] 1 2 3 Sarah On Tue, Jul 29, 2008 at 1:45 PM, Rajasekaramya [EMAIL PROTECTED] wrote: hi , Is there anyways to delet the list names once created. i tried rm(names(mylist)) i didnt work kindly help me Ramya -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org 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] table questions
# is that what you want? table(cut(xy,seq(0,max(xy)+.4,by=.4))) # or this table(cut(xy,hist(xy)$breaks)) # not the same regards, PF +- | Patrizio Frederic | Research associate in Statistics, | Department of Economics, | University of Modena and Reggio Emilia, | Via Berengario 51, | 41100 Modena, Italy | | tel: +39 059 205 6727 | fax: +39 059 205 6947 | mail: [EMAIL PROTECTED] +- 2008/7/29 Edna Bell [EMAIL PROTECTED]: Hi again! Suppose I have the following: xy - round(rexp(20),1) xy [1] 0.1 3.4 1.6 0.4 1.0 1.4 0.2 0.3 1.6 0.2 0.0 0.1 0.1 1.0 2.0 0.9 2.5 0.1 1.5 0.4 table(xy) xy 0 0.1 0.2 0.3 0.4 0.9 1 1.4 1.5 1.6 2 2.5 3.4 1 4 2 1 2 1 2 1 1 2 1 1 1 Is there a way to set things up to have 0 - 0.4 0.5 - 0.9 etc. please? I know there is the cut functions, but breaks are required. If you don't have breaks, what should you do, please? Would using the breaks from the hist function work appropriately, please? thanks Edna Bell __ R-help@r-project.org 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. __ R-help@r-project.org 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] Graphics function question
Just a quick follow up to my own post here. In offline exchanges, Peter
noted a typo in the call to plot(), where I was missing the final ).
Also, I had the order of the unnamed arguments to segments() off.
Here is the corrected code, with some additional teaks:
plotdiffs - function(x, y, startage, endage, decline)
{
startvar - x - y
endvar - x
s1 - seq(47, 89, length = 10)
ymin - min(startvar, endvar, na.rm = TRUE)
ymax - max(startvar, endvar, na.rm = TRUE)
s2 - seq(ymin, ymax, length = 10)
label - gsub(^.+\\$, , deparse(substitute(x)))
plot(s1, s2, type = n, xlab = Age, ylab = label,
main = paste(Age, decline and, label))
segments(startage, startvar, endage, endvar, col = decline)
legend(topleft, legend = c(Stable, Decline), lty = 1,
col = c(1, 2))
}
Marc
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Re: [R] combining zoo series with an overlapping index?
My apologies it worked perfectly. On Tue, Jul 29, 2008 at 12:40 PM, stephen sefick [EMAIL PROTECTED] wrote: plot(replace(day, is.list(index(lin)), coredata(lin))) -- Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis [[alternative HTML version deleted]] __ R-help@r-project.org 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] ls() and memory question
It is memory. You should probably not have 40% of your RAM allocated to R so that if you do create some large objects, you will have room for them. How much physical memory do you have, can you put some numbers on how many objects you need, if you have a lot of them, have you considered using lists. On Tue, Jul 29, 2008 at 12:36 PM, Edna Bell [EMAIL PROTECTED] wrote: Hi again! I put in ls() to check the objects in my workspace. Is there a limit on how many objects I can have, please? Or does it depend on the memory, please? TIA, Edna Bell __ R-help@r-project.org 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org 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.
[R] R command history -- can it be like Matlab's?
Hi all, In R GUI window, if you use up and down key, you will be able to recall the previous and next command that has been used and stored in the command history cache. But there is one inconvenience: In Matlab, you can type the first a few characters of the command that you previously used, and then press “up key, you will be able to get to that previous command very fast, no matter how long ago it has been used... In R GUI, there is no such functionality. Any thoughts on making R work my productivity? -- Also where do I find good debugger for R? Thanks! -- View this message in context: http://www.nabble.com/R-command-historycan-it-be-like-Matlab%27s--tp18719530p18719530.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] Coarsening the Resolution of a Dataset
I assume that you are doing this on one column of the matrix which should only have 2160 entries in it. can you send the actual code you are using. I tried it with 10,000 samples and it works fine. So I need to understand the data structure you are using. Also the infinite recursion sounds strange; do you have function like 'cut' or 'c' redefined? So it would help if you could supply a reproducible example. On Tue, Jul 29, 2008 at 10:09 AM, Steve Murray [EMAIL PROTECTED] wrote: Unfortunately, when I get to the 'myCuts' line, I receive the following error: Error: evaluation nested too deeply: infinite recursion / options(expressions=)? ...and I also receive warnings about memory allocation being reached (even though I've already used memory.limit() to maximise the memory) - this is a fairly sizeable dataset afterall, 2160 rows by 4320 columns. Therefore I was wondering if there are any alternative ways of coarsening a dataset? Or are there any packages/commands built for this sort of thing? Any advice would be much appreciated! Thanks again, Steve _ Find the best and worst places on the planet http://clk.atdmt.com/UKM/go/101719807/direct/01/ -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@r-project.org 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 command history -- can it be like Matlab's?
A patch to do this was posted on 2007-09-29 by Glenn Davis. Some people not addicted to Matlab find the behaviour very inconvenient and prefer the getline/readline behaviour (triggered by ^R/^S) of Rterm and R on Unixen. On Tue, 29 Jul 2008, losemind wrote: Hi all, In R GUI window, if you use up and down key, you will be able to recall the previous and next command that has been used and stored in the command history cache. But there is one inconvenience: Or an inconvenience of Matlab not shared by R, depending on your preferences. In Matlab, you can type the first a few characters of the command that you previously used, and then press “up key, you will be able to get to that previous command very fast, no matter how long ago it has been used... In R GUI, there is no such functionality. Any thoughts on making R work my productivity? -- Also where do I find good debugger for R? Thanks! -- View this message in context: http://www.nabble.com/R-command-historycan-it-be-like-Matlab%27s--tp18719530p18719530.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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. -- 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@r-project.org 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] try question
Edna Bell [EMAIL PROTECTED] wrote: Hi still yet again! I have the following code: try(log(rnorm(25)),silent=TRUE) [1] -0.26396185 NaN NaN -0.13078069 -2.44997193 -2.15603971 NaN 0.94917495 0.07244544 NaN [11] -1.06341127 -0.42293099 -0.53769569 0.95134763 0.93403340 NaN -0.10502078 NaN 0.30283262 NaN [21] -0.11696872 -3.84122332 NaN NaN -0.12808690 Warning message: In log(rnorm(25)) : NaNs produced I thought that putting the silent = TRUE would suppress the warnings, please. What should I do instead, please? It's not a great idea to take logs of negative numbers. Better than supressing the resulting messages, you might try something like this: a - rnorm(25) b - log(ifelse(a 0, NA, a)) which gives this: a [1] -0.04816269 -0.50745059 0.15229031 0.54735811 [5] -0.29896853 1.81854119 0.19462259 -0.48984075 [9] 0.63489288 1.47432484 1.15295160 0.75842227 [13] 0.07918115 1.04596643 1.31722543 -0.03614219 [17] 0.44072181 0.25358843 -0.49626405 -2.10954780 [21] -0.85815654 1.38983430 0.66592947 0.64700068 [25] -1.17829527 b [1] NA NA -1.8819 -0.60265201 [5] NA 0.59803463 -1.63669302 NA [9] -0.45429898 0.38820015 0.14232526 -0.27651496 [13] -2.53601706 0.04494127 0.27552758 NA [17] -0.81934141 -1.37204269 NA NA [21] NA 0.32918453 -0.40657152 -0.43540794 [25] NA See the help page for ifelse() for more information. -- Mike Prager, NOAA, Beaufort, NC * Opinions expressed are personal and not represented otherwise. * Any use of tradenames does not constitute a NOAA endorsement. __ R-help@r-project.org 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.
[R] optimize simultaneously two binomials inequalities using nlm
Dear R users,
I´m trying to optimize simultaneously two binomials inequalities used to
acceptance sampling, which are nonlinear solution, so there is no simple
direct solution. Please, let me explain shortly the the problem and the
question as following.
The objective is to obtain the smallest value of 'n' (sample size)
satisfying both inequalities:
(1-alpha) = pbinom(c, n, p1) pbinom(c, n, p2) = beta
Where p1 (AQL) and p2 (LTPD) are probabilities parameters (Consumer and
Producer), the alpha and beta are conumer and producer risks, and finally,
the 'n' represents the sample size and the 'c' an acceptance number or
maximum number of defects (nonconforming) in sample size.
Considering that the 'n' and 'c' values are integer variables, it is
commonly not possible to derive an OC curve including the both (p1,1-alpha)
and (p2,beta) points. Some adjacency compromise is commonly required,
achieved by searching a more precise OC curve with respect to one of the
points.
I´m using Mathematica 6 but it is a Trial, so I would like use R intead (or
better, I need it)!
To exemplify, In Mathematica I call the function using NMinimize passing
the restriction and parameters:
/* function name findOpt and parameters... */
restriction = (1 - alpha) = CDF[BinomialDistribution[sample_n, p1], c]
betha = CDF[BinomialDistribution[sample_n, p2], c]
0 alpha alphamax 0 betha bethamax
1 sample_n = lot_Size 0 = c lot_size
p1 p2 p2max ;
fcost = sample_n/lot_Size;
result = NMinimize[{fcost, restriction}, {sample_n, c, alpha, betha, p2max},
Method - NelderMead, AccuracyGoal - 10];
/* Calling the function findOpt */
findOpt[p1=0.005, lot_size=1000, alphamax=0.05, bethamax =0.05, p2max =
0.04]
/* and I got the return of values of; minimal n, c, alpha, betha and
the p2 or (LTPD) were computed */ {0.514573, {alpha$74 - 0.0218683,
sample_n$74 - 155.231, betha$74 - 0.05,
c$74 - 2, p2$74 - 0.04}}
Now, using R, I would define the pbinom(c, n, prob) given only the minimum
and maximum values to n and c and limits to p1 and p2 probabilities
(Consumer and Producer), similar on the example above.
I found some examples to optimize equations in R and some tips, but I not be
able to define the sintaxe to use with that functions. Among the functions
that could be used to resolve the problem presented, I found the function
optim() that it is used for unconstrained optimization and the nlm() which
is used for solving nonlinear unconstrained minimization problems. May I
wrong, but the nlm() function would be appropriate to solve this problem, is
it right?
Can I get a pointer to solve this problem using the nlm() function or where
could I get some tips/example to help me, please?
Thank you very much.
EToktar
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[R] finding a faster way to do an iterative computation
useR's, I am trying trying to find out if there is a faster way to do a certain computation. I have successfully used FOR loops and the apply function to do this, but it can take some time to fully compute, but I was wondering if anyone may know of a different function or way to do this: x [1] 1 2 3 4 5 xk [1] 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 I want to do: abs(x-xk[i]) where i = 1 to length(xk)=13. It should result in a 13 by 5 matrix or data frame. Does anyone have a quicker solution than FOR loops or apply()? Much appreciation and thanks, dxc13 -- View this message in context: http://www.nabble.com/finding-a-faster-way-to-do-an-iterative-computation-tp18718233p18718233.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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] finding a faster way to do an iterative computation
matrix(rep(x, each=13) - xk, nrow=13) -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of dxc13 Sent: Tuesday, July 29, 2008 2:13 PM To: r-help@r-project.org Subject: [R] finding a faster way to do an iterative computation useR's, I am trying trying to find out if there is a faster way to do a certain computation. I have successfully used FOR loops and the apply function to do this, but it can take some time to fully compute, but I was wondering if anyone may know of a different function or way to do this: x [1] 1 2 3 4 5 xk [1] 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 I want to do: abs(x-xk[i]) where i = 1 to length(xk)=13. It should result in a 13 by 5 matrix or data frame. Does anyone have a quicker solution than FOR loops or apply()? Much appreciation and thanks, dxc13 -- View this message in context: http://www.nabble.com/finding-a-faster-way-to-do-an-iterative- computation-tp18718233p18718233.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org 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. __ R-help@r-project.org 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] finding a faster way to do an iterative computation
On 30/07/2008, at 6:12 AM, dxc13 wrote:
useR's,
I am trying trying to find out if there is a faster way to do a
certain
computation. I have successfully used FOR loops and the apply
function to
do this, but it can take some time to fully compute, but I was
wondering if
anyone may know of a different function or way to do this:
x
[1] 1 2 3 4 5
xk
[1] 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
I want to do: abs(x-xk[i]) where i = 1 to length(xk)=13. It should
result
in a 13 by 5 matrix or data frame. Does anyone have a quicker
solution
than FOR loops or apply()?
outer(xk,x,function(a,b){abs(a-b)})
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Re: [R] R command history -- can it be like Matlab's?
On Tue, Jul 29, 2008 at 2:45 PM, Prof Brian Ripley [EMAIL PROTECTED] wrote: A patch to do this was posted on 2007-09-29 by Glenn Davis. Some people not addicted to Matlab find the behaviour very inconvenient and prefer the getline/readline behaviour (triggered by ^R/^S) of Rterm and R on Unixen. On unixen you can redefine your up/down arrows in your .inputrc: \e[A: history-search-backward \e[B: history-search-forward which I find really useful, but it definitely takes a few weeks of getting used to. I suspect there maybe an equivalent for Rterm. Hadley -- http://had.co.nz/ __ R-help@r-project.org 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 command history -- can it be like Matlab's?
On 30/07/2008, at 9:16 AM, hadley wickham wrote:
On Tue, Jul 29, 2008 at 2:45 PM, Prof Brian Ripley
[EMAIL PROTECTED] wrote:
A patch to do this was posted on 2007-09-29 by Glenn Davis. Some
people not
addicted to Matlab find the behaviour very inconvenient and prefer
the
getline/readline behaviour (triggered by ^R/^S) of Rterm and R on
Unixen.
On unixen you can redefine your up/down arrows in your .inputrc:
\e[A: history-search-backward
\e[B: history-search-forward
which I find really useful, but it definitely takes a few weeks of
getting used to. I suspect there maybe an equivalent for Rterm.
Those of us who (sensibly! :-) ) use vi have:
set editing-mode vi
in our .inputrc files.
Then esc puts you into vi editing mode whence ``/'' searches
``forwards'' and ``?'' searches ``backwards'' --- with the convention
that the most recent command is the ``top'' of the file.
And then when you've found the command that you want, you can edit it
(with vi syntax) before pressing return to re-issue the command.
No home should be without one.
cheers,
Rolf Turner
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Re: [R] R command history -- can it be like Matlab's?
On Tue, Jul 29, 2008 at 4:39 PM, Rolf Turner [EMAIL PROTECTED] wrote: On 30/07/2008, at 9:16 AM, hadley wickham wrote: On Tue, Jul 29, 2008 at 2:45 PM, Prof Brian Ripley [EMAIL PROTECTED] wrote: A patch to do this was posted on 2007-09-29 by Glenn Davis. Some people not addicted to Matlab find the behaviour very inconvenient and prefer the getline/readline behaviour (triggered by ^R/^S) of Rterm and R on Unixen. On unixen you can redefine your up/down arrows in your .inputrc: \e[A: history-search-backward \e[B: history-search-forward which I find really useful, but it definitely takes a few weeks of getting used to. I suspect there maybe an equivalent for Rterm. Those of us who (sensibly! :-) ) use vi have: set editing-mode vi in our .inputrc files. Then esc puts you into vi editing mode whence ``/'' searches ``forwards'' and ``?'' searches ``backwards'' --- with the convention that the most recent command is the ``top'' of the file. And then when you've found the command that you want, you can edit it (with vi syntax) before pressing return to re-issue the command. And you might also want to beef up the size of your history file (and automatically remove duplicates). I have the following in my bash_profile (and I assume similar functionality exists for other shells): export HISTSIZE=1000 export HISTFILESIZE=5000 export HISTCONTROL=erasedups Hadley -- http://had.co.nz/ __ R-help@r-project.org 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.
