[R] text file imported incorrectly
Dear R-users, When I tried to import a text file (tab delimited) which has 2000+ rows with the following command (With the importData in S, it works though), x - read.table(textfile, sep= \t, skip=5, stringAsFactors=F) I received the following warning message: Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings,: number of items read is not a multiple of the number of columns. I checked the resulting data frame and found only about 1500 observations rather than 2000+ observations. Then, I used the command count.fields(textfile, sep=\t) and it showed that I have rows which have either 4 fields or 294 fields. (There are 294 variables altogether) When I tried to check those observations/rows which have only 4 fields indicated by count.fields, I realized that the problem is quite likely due to one of the variables I have. For this company variable, The problematic rows have names such as: BANK INT'L INDONESIA BEIJING CAP INT'L AIRP H BELLE INT'L HLDGS(CN) The other non-problematic rows have names like ANZ BANKING GROUP BABCOCK BROWN BEC WORLD which did not give problems. I believe the ' symbol is causing this variable for some of these rows to be read incorrectly. How do I read this field such that the names BANK INT'L INDONESIA BEIJING CAP INT'L AIRP H BELLE INT'L HLDGS(CN) etc can be interpreted as a single field and that all my rows will have 294 fields correctly interpreted by R. What will be the correct command to issue? Hope I am not unclear in my explanation of my problem. Hope to have your kind assistance! Best Regards, wy ** The information provided in this e-mail is confidential and is for the sole use of the recipient. It may not be disclosed, copied or distributed in any form without the express permission of Henderson Global Investors and to the extent that it is passed on, care must be taken to ensure that this is in a form which accurately reflects the information presented here. Whilst Henderson Global Investors believe that the information is correct at the date of this e-mail, no warranty or representation is given to this effect and no responsibility can be accepted by Henderson Global Investors to any end users for any action taken on the basis of this information. Henderson Global Investors is the name under which Henderson Global Investors Limited (registered no. 906355), Henderson Fund Management plc (registered no. 2607112), Henderson Investment Funds Limited (registered no. 2678531), Henderson Investment Management Limited (registered no. 1795354) Henderson Alternative Investment Advisor Limited (registered no. 962757) and Henderson Equity Partners Limited (registered no.2606646) (each incorporated and registered in England and Wales with registered office at 4 Broadgate, London EC2M 2DA and authorised and regulated by the Financial Services Authority) provide investment products and services. Henderson Secretarial Services Limited (incorporated and registered in England and Wales, registered no. 1471624, registered office 4 Broadgate, London EC2M 2DA) is the name under which company secretarial services are provided. All these companies are wholly owned subsidiaries of Henderson Group plc (incorporated and registered in England ! and Wales, registered no. 2072534, registered office 4 Broadgate, London EC2M 2DA). We may record telephone calls or email for our mutual protection and to improve customer service. ** [[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] Maximum likelihood estimation
Hi R-experts,
I'm new to R in mle. I tried to do the following but just couldn't get it
right. Hope someone can point out the mistakes. thanks a lot.
t - c(1:90)
y -
c(5,10,15,20,26,34,36,43,47,49,80,84,108,157,171,183,191,200,204,211,217,226,230,
234,236,240,243,252,254,259,263,264,268,271,277,290,309,324,331,346,367,375,381,
401,411,414,417,425,430,431,433,435,437,444,446,446,448,451,453,460,463,463,464,
464,465,465,465,466,467,467,467,468,469,469,469,469,470,472,472,473,473,473,473,
473,473,473,475,475,475,475)
fr - function(a, b, p, lambda){
l - 0.5*(lambda + b*p + (1-p)*(lambda-b))
l^2 lambda*b*p
delta - sqrt(abs(l^2 - b*p*lambda))
mt - a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
logl - sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
return(-logl)
}
optim(c(15,0.01,0.5,0.01),fr, method=L-BFGS-B,
lower = c(0.002, 0.002, 0.002, 0.002), upper = c(Inf, Inf, 0.999,
Inf),control=list(fnscale=-1))
--
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Re: [R] Binary Tree Testing in ape package (a bug?)
If I am right informed, 'is.binary.tree' cannot test for root polytomies. Consider this example: tree.hiv - read.tree(text=((rat,mouse,(human,chimp)), kangaroo);) is.binary.tree(tree.hiv)# will yield 'FALSE' For further questions you might be better advised to use the 'R-sig- phylo' mailing list. Christoph Heibl Systematic Botany Ludwig-Maximilians-Universität München Menzinger Str. 67 D-80638 München GERMANY phone: +49-(0)89-17861-251 e-mail:[EMAIL PROTECTED] http://www.christophheibl.de/ch-home.html On Sep 4, 2008, at 5:27 AM, Gundala Viswanath wrote: Dear all, I was testing the wonderful package APE. However upon testing a particular Newick's format tree - which I think to be a non-binary tree - it yields different result as expected. library(ape) tree.hiv - read.tree(text=(rat,mouse,(human,chimp));) is.binary.tree(tree.hiv) [1] TRUE Was that a bug in APE package? - 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-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] ggplot2: line plot with gaps in time axis
Brian, The easiest way is to create the entire timeseries and then set the missing values to NA. The NA values will lead to the gaps you want. 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 btcruiser Verzonden: woensdag 3 september 2008 20:33 Aan: r-help@r-project.org Onderwerp: [R] ggplot2: line plot with gaps in time axis Hello, I'm trying to plot data that has gaps in the timeline because my data only has the business day in it. When I do a line plot I get the data and then a blank area where a line goes the tail of the last data point to the head of the next data point. Is there a way I can do a line plot where the gaps are not plotted? Thanks in advance, Brian names(utildf) [1] Date_and_Time Utilization Direction utildf Date_and_Time Utilization Direction 12008-08-25 05:00:005.862601 Inbound 22008-08-25 05:05:00 10.025328 Inbound 32008-08-25 05:10:005.794900 Inbound 42008-08-25 05:15:009.862726 Inbound 52008-08-25 05:20:004.150328 Inbound 62008-08-25 05:25:005.559362 Inbound [...] # startDateTime and stopDateTime is user entered Start - as.numeric(as.POSIXct(startDateTime)) End - as.numeric(as.POSIXct(stopDateTime)) Period-as.numeric(seq.POSIXt(as.POSIXct(Start,origin=1970-1-1), as.POSIXct(End,origin=1970-1-1), by=DSTday)) Labels-as.Date(seq.POSIXt(as.POSIXct(Start,origin=1970-1-1), as.POSIXct(End,origin=1970-1-1), by=DSTday)) dt - qplot(as.numeric(utildf$Date_and_Time),Utilization,data=utildf,colour=Di rection,geom=segment) dt + scale_x_continuous(breaks=Period,label=Labels) -- View this message in context: http://www.nabble.com/ggplot2%3A-line-plot-with-gaps-in-time-axis-tp1929 5209p19295209.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. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document.%CRLF%The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document%CRLF% __ 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] text file imported incorrectly
Please do read the help page (as you were asked to do before posting). See the 'quote' argument. This is also covered in the 'R Data Import/Export Manual'. On Thu, 4 Sep 2008, Weiyang Lim wrote: Dear R-users, When I tried to import a text file (tab delimited) which has 2000+ rows with the following command (With the importData in S, it works though), x - read.table(textfile, sep= \t, skip=5, stringAsFactors=F) I received the following warning message: Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings,: number of items read is not a multiple of the number of columns. I checked the resulting data frame and found only about 1500 observations rather than 2000+ observations. Then, I used the command count.fields(textfile, sep=\t) and it showed that I have rows which have either 4 fields or 294 fields. (There are 294 variables altogether) When I tried to check those observations/rows which have only 4 fields indicated by count.fields, I realized that the problem is quite likely due to one of the variables I have. For this company variable, The problematic rows have names such as: BANK INT'L INDONESIA BEIJING CAP INT'L AIRP H BELLE INT'L HLDGS(CN) The other non-problematic rows have names like ANZ BANKING GROUP BABCOCK BROWN BEC WORLD which did not give problems. I believe the ' symbol is causing this variable for some of these rows to be read incorrectly. How do I read this field such that the names BANK INT'L INDONESIA BEIJING CAP INT'L AIRP H BELLE INT'L HLDGS(CN) etc can be interpreted as a single field and that all my rows will have 294 fields correctly interpreted by R. What will be the correct command to issue? Hope I am not unclear in my explanation of my problem. Hope to have your kind assistance! Best Regards, wy [[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] CV.Tree
See MASS (the book) for examples. BTW, R _is_ case sensitive. On Wed, 3 Sep 2008, Warren Schlechte wrote: I am using the tree package. One option is the cv.tree, which is supposed to run cross-validations on your tree models. Is there somewhere I can find some documentation on this function? I have the help file that comes with the library, but I need more, especially on what the output is. Thanks, Warren Schlechte HOH Fisheries Science Center 5103 Junction Hwy Mt. Home, TX 78058 Phone 830.866.3356 x.214 Fax830.866.3549 [[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] license for a university
2008/9/4 Ted Byers [EMAIL PROTECTED]: Erin, I trust you know what you risk when you assume. ;-) There IS a license, but it basically lets you copy or distribute it, or, in your case, install on as many machines as you wish. It is the GNU GENERAL PUBLIC LICENSE. Like most open source software I use, the Gnu license is in place primarly to ensure everyone can freely use it. Yes, there's no licence that covers usage - the GNU GPL covers copying, modification, and distribution only. Unlike certain statistical software's end-user licence agreement (EULA), someone who double-clicks an R icon and starts typing doesn't have to agree to anything. In practice, the person doing the classroom install of Other Stats Package will have clicked through the EULA and the users will never have seen the licence they are supposedly licensed under. Great. There's no EULA for R. Barry [practising the UK English practice of 'license' for a verb and 'licence' for a noun - I was advised this was good advice...] __ 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 update
Not sure if you can do it from within R, but if not surely you can just go to the website www.r-project.org Download and install the latest version and remove older versions if you wish. Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of giov Sent: Wednesday, September 03, 2008 4:16 PM To: r-help@r-project.org Subject: [R] R update Hi all, please, someone can explain me how update my R version? thank you! giov -- View this message in context: http://www.nabble.com/R-update-tp19291451p19291451.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] test if all predictors in a glm object are factors
On Wed, 3 Sep 2008, Marc Schwartz wrote:
on 09/03/2008 04:56 PM Michael Friendly wrote:
I'm trying to develop some graphic methods for glm objects, but they
only apply for models
where all predictors are discrete factors. How can I test for this in a
Is an ordered factor a 'discrete factor'? I suspect it is, so this needs
to be
is.discrete.glm - function(model)
all(attr(terms(model), dataClasses)[-1] %in% c(factor, ordered))
(removing redundant braces).
function, given the
glm model object?
That is, I want something that will serve as an equivalent of
is.discrete.glm() in the following
context:
myplot.glm -
function(model, ...) {
if (!inherits(model,glm)) stop(requires a glm object)
if (!is.discrete.glm(model)) stop(only factors are allowed)
...
}
A small example, for count data, a poisson glm:
GSS - data.frame(
expand.grid(sex=c(female, male), party=c(dem, indep, rep)),
count=c(279,165,73,47,225,191))
mod.glm - glm(count ~ sex + party, family = poisson, data = GSS)
So, the model terms are sex and party, both factors. Peeking inside
mod.glm, I
can find
mod.glm$xlevels
$sex
[1] female male
$party
[1] dem indep rep
and, in str(mod.glm$model) I see
str(mod.glm$model)
'data.frame': 6 obs. of 3 variables:
$ count: num 279 165 73 47 225 191
$ sex : Factor w/ 2 levels female,male: 1 2 1 2 1 2
$ party: Factor w/ 3 levels dem,indep,..: 1 1 2 2 3 3
- attr(*, terms)=Classes 'terms', 'formula' length 3 count ~ sex + party
so this is a keeper. Can someone help me improve on the following
is.discrete.glm() function.
It works for mod.glm, but isn't very general ;-)
is.discrete.glm - function(model) {
TRUE
}
Michael,
How about something like this:
is.discrete.glm - function(model) {
all(attr(terms(model), dataClasses)[-1] == factor)
}
Essentially, take the output of terms(model), check the 'dataClasses'
attribute, except for the first element, which is the DV.
is.discrete.glm(mod.glm)
[1] TRUE
HTH,
Marc Schwartz
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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] Coefficient of skewness
Hi, Is there a function in R to calculate the coefficient of skewness of some data? I had expected there to be one, but can find no information about it. Thanks for any pointers. Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] [[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] text file imported incorrectly
Thanks, Prof Ripley. I did read about the quote argument in the 'R Data Import/Export Manual'. But unfortunately, I did not really understand what it means and did not adjust that argument. But now after adjusting for it, it works for me. Thanks. Best Regards, wy -Original Message- From: Prof Brian Ripley [mailto:[EMAIL PROTECTED] Sent: 04 September 2008 15:43 To: Weiyang Lim Cc: r-help@r-project.org Subject: Re: [R] text file imported incorrectly Please do read the help page (as you were asked to do before posting). See the 'quote' argument. This is also covered in the 'R Data Import/Export Manual'. On Thu, 4 Sep 2008, Weiyang Lim wrote: Dear R-users, When I tried to import a text file (tab delimited) which has 2000+ rows with the following command (With the importData in S, it works though), x - read.table(textfile, sep= \t, skip=5, stringAsFactors=F) I received the following warning message: Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings,: number of items read is not a multiple of the number of columns. I checked the resulting data frame and found only about 1500 observations rather than 2000+ observations. Then, I used the command count.fields(textfile, sep=\t) and it showed that I have rows which have either 4 fields or 294 fields. (There are 294 variables altogether) When I tried to check those observations/rows which have only 4 fields indicated by count.fields, I realized that the problem is quite likely due to one of the variables I have. For this company variable, The problematic rows have names such as: BANK INT'L INDONESIA BEIJING CAP INT'L AIRP H BELLE INT'L HLDGS(CN) The other non-problematic rows have names like ANZ BANKING GROUP BABCOCK BROWN BEC WORLD which did not give problems. I believe the ' symbol is causing this variable for some of these rows to be read incorrectly. How do I read this field such that the names BANK INT'L INDONESIA BEIJING CAP INT'L AIRP H BELLE INT'L HLDGS(CN) etc can be interpreted as a single field and that all my rows will have 294 fields correctly interpreted by R. What will be the correct command to issue? Hope I am not unclear in my explanation of my problem. Hope to have your kind assistance! Best Regards, wy [[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 ** The information provided in this e-mail is confidential and is for the sole use of the recipient. It may not be disclosed, copied or distributed in any form without the express permission of Henderson Global Investors and to the extent that it is passed on, care must be taken to ensure that this is in a form which accurately reflects the information presented here. Whilst Henderson Global Investors believe that the information is correct at the date of this e-mail, no warranty or representation is given to this effect and no responsibility can be accepted by Henderson Global Investors to any end users for any action taken on the basis of this information. Henderson Global Investors is the name under which Henderson Global Investors Limited (registered no. 906355), Henderson Fund Management plc (registered no. 2607112), Henderson Investment Funds Limited (registered no. 2678531), Henderson Investment Management Limited (registered no. 1795354) Henderson Alternative Investment Advisor Limited (registered no. 962757) and Henderson Equity Partners Limited (registered no.2606646) (each incorporated and registered in England and Wales with registered office at 4 Broadgate, London EC2M 2DA and authorised and regulated by the Financial Services Authority) provide investment products and services. Henderson Secretarial Services Limited (incorporated and registered in England and Wales, registered no. 1471624, registered office 4 Broadgate, London EC2M 2DA) is the name under which company secretarial services are provided. All these companies are wholly owned subsidiaries of Henderson Group plc (incorporated and registered in England ! and Wales, registered no. 2072534, registered office 4 Broadgate, London EC2M 2DA). We may record telephone calls or email for our mutual protection and to improve customer service. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the
Re: [R] Coefficient of skewness
Dear Robin, RSiteSearch(skewness, restrict = functions) gave me 222 hits. There are several functions that calculate skewness. 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 Williams, Robin Verzonden: donderdag 4 september 2008 9:55 Aan: r-help@r-project.org Onderwerp: [R] Coefficient of skewness Hi, Is there a function in R to calculate the coefficient of skewness of some data? I had expected there to be one, but can find no information about it. Thanks for any pointers. Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] [[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. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document.%CRLF%The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document%CRLF% __ 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] restricted bootstrap
I see nothing here to do with the 'bootstrap', which is sampling with replacement. Do you know what you mean exactly by 'randomly sample'? In general the way to so this is to sample randomly (uniformly, whatever) and reject samples that do not meet your restriction. For some restrictions there are more efficient algorithms, but I don't understand yours. (What are the 'rows'? Do you want to sample rows in space or xy locations? How come 'dist' is not symmetric?) For some restrictions, an MCMC sampling scheme is needed, the hard-core spatial point process being a related example. On Wed, 3 Sep 2008, Grant Gillis wrote: Hello List, I am not sure that I have the correct terminology here (restricted bootstrap) which may be hampering my archive searches. I have quite a large spatially autocorrelated data set. I have xy coordinates and the corresponding pairwise distance matrix (metres) for each row. I would like to randomly sample some number of rows but restricting samples such that the distance between them is larger than the threshold of autocorrelation. I have been been unsuccessfully trying to link the 'sample' function to values in the distance matrix. My end goal is to randomly sample M thousand rows of data N thousand times calculating linear regression coefficients for each sample but am stuck on taking the initial sample. I believe I can figure out the rest. Example Question I would like to radomly sample 3 rows further but withe the restriction that they are greater than 100m apart example data: main data: y- c(1, 2, 9, 5, 6) x-c( 1, 3, 5, 7, 9) z-c(2, 4, 6, 8, 10) a-c(3, 9, 6, 4 ,4) maindata-cbind(y, x, z, a) y x x a [1,] 1 1 1 3 [2,] 2 3 3 9 [3,] 9 5 5 6 [4,] 5 7 7 4 [5,] 6 9 9 4 distance matrix: row1-c(0, 123, 567, 89) row2-c(98, 0, 345, 543) row3-c(765, 90, 0, 987) row4-c(654, 8, 99, 0) dist-rbind(row1, row2, row3, row4) [,1] [,2] [,3] [,4] row10 123 567 89 row2 980 345 543 row3 765 900 987 row4 6548 990 Thanks for all of the help in the past and now Cheers Grant [[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] Coefficient of skewness
It exists in many packages: e1071 was one of the earliest. Other packages include GLDEX, HyperbolicDist, QuantPsyc, TSA, agricolae, fUtilities, modeest, moments, npde, s20x It is also a common exercise in basic R programming courses. On Thu, 4 Sep 2008, Williams, Robin wrote: Hi, Is there a function in R to calculate the coefficient of skewness of some data? I had expected there to be one, but can find no information about it. Thanks for any pointers. Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] [[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.
[R] Any function to calculate returns
Hi, Sorry if this may be a simple question. Is there any function to calculate returns (percentage or non-percentage) directly like the function getReturns() in S Finmetrics? Thanks. Best Regards, wy ** The information provided in this e-mail is confidential and is for the sole use of the recipient. It may not be disclosed, copied or distributed in any form without the express permission of Henderson Global Investors and to the extent that it is passed on, care must be taken to ensure that this is in a form which accurately reflects the information presented here. Whilst Henderson Global Investors believe that the information is correct at the date of this e-mail, no warranty or representation is given to this effect and no responsibility can be accepted by Henderson Global Investors to any end users for any action taken on the basis of this information. Henderson Global Investors is the name under which Henderson Global Investors Limited (registered no. 906355), Henderson Fund Management plc (registered no. 2607112), Henderson Investment Funds Limited (registered no. 2678531), Henderson Investment Management Limited (registered no. 1795354) Henderson Alternative Investment Advisor Limited (registered no. 962757) and Henderson Equity Partners Limited (registered no.2606646) (each incorporated and registered in England and Wales with registered office at 4 Broadgate, London EC2M 2DA and authorised and regulated by the Financial Services Authority) provide investment products and services. Henderson Secretarial Services Limited (incorporated and registered in England and Wales, registered no. 1471624, registered office 4 Broadgate, London EC2M 2DA) is the name under which company secretarial services are provided. All these companies are wholly owned subsidiaries of Henderson Group plc (incorporated and registered in England ! and Wales, registered no. 2072534, registered office 4 Broadgate, London EC2M 2DA). We may record telephone calls or email for our mutual protection and to improve customer service. ** [[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] Free SQL Database with R
Thanks for that, looks like pretty much what I'm after. Kind Regards Chibisi On Wed, Sep 3, 2008 at 9:17 PM, Iain Gallagher [EMAIL PROTECTED] wrote: Hi Chibisi I'm sort of jumping into this thread but from your description of charts / plotting below I thought you might like to take a look at flot: http://code.google.com/p/flot/ From the page Flot is a pure Javascript plotting library for jQueryhttp://jquery.com/. It produces graphical *plots* of arbitrary datasets on-the-fly client-side. The focus is on *simple* usage (all settings are optional), * attractive* looks and *interactive* features like zooming. Although Flot is easy to use, it is also advanced enough to be suitable for Web 2.0 data mining/business intelligence purposes which is its original application. The plugin is targeting all newer browsers. If you find a problem, please report it. Drawing is done with the canvas tag introduced by Safari and now available on all major browsers, except Internet Explorer where the excanvas Javascript emulation helper is used. Maybe it could be useful to you for plots etc. Cheers Iain *Chibisi Chima-Okereke [EMAIL PROTECTED]* wrote: Dear Felix, Thanks for the reply, If you haven't already guessed I am new to web programming. The sort of webpage I want to build is one that presents quantitative information in graphs and charts, that people can interact with, e.g. select parts of charts to zoom into, highlight values, click buttons to do analysis on the data displayed, so yes some sort of interactive GUI. I initially thought of using flash as a front end but I don't know any actionscript, so learning that would to a suitable standard take alot of extra time, and I think it would be best if everything could be done in R as much as possible. If I used an RGUI I guess I would be using the playwith package? Do the consumers of the website need to have R to consume stuff displayed with an RGUI? The database itself would just be pretty static just being queried for information, unless some analysis was required in which case R would query the database do the analysis and write the analysis back to the database (I guessing that is the way it would be done), before it gets displayed on the web page. Kind Regards Chibisi On Wed, Sep 3, 2008 at 11:39 AM, drflxms wrote: Hello Chibisi, I am not shore whether I completely understand your needs: Do you want to build a webpage which relies on a content management system (cms)? Do you want to collect data (i.e. survey) which later on shall be analysed using R? Or shall it be a webpage with an interactive R GUI? What else? But personally I would prefer MySQL as backend for websites, as most professional (opensource) cms (i.e. typo3, wordpress etc.) are created with MySQL in mind. There is a good interface between R and MySQL called RMySQL as well. I use this on a daily basis, as all my data is stored in a local MySQL database (more flexible than always reading in text files - at least in my opinion). Hope this personal view might help a little bit. Cheers, Felix __ 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. [[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. [[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] cluster a distance(analogue)-object using agnes(cluster)
B == Birgitle [EMAIL PROTECTED]
on Tue, 2 Sep 2008 03:02:31 -0700 (PDT) writes:
B I try to perform a clustering using an existing dissimilarity matrix
that I
B calculated using distance (analogue)
B I tried two different things. One of them worked and one not and I don`t
B understand why.
B Here the code:
B not working example
B library(cluster)
B library(analogue)
B iris2 - as.data.frame(iris)
why that? After the above, iris2 is identical() to iris !
B str(iris2)
B 'data.frame':150 obs. of 5 variables:
B $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
B $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
B $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
B $ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
B $ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1 1 1
1
B 1 1 1 ...
B Test.Gower - distance(iris2, method =mixed)
why not just
daisy(iris2, metric = gower)
daisy() is in cluster which has been a recommended R package
forever.
So the solution (here, not in general!)
is to stay with package 'cluster' and use
daisy() before agnes().
Regards,
Martin Maechler, ETH Zurich
{same city! feel free to phone me..}
B Test.Gower.agnes-agnes(Test.Gower, diss=T)
B Fehler in agnes(Test.Gower, diss = T) :
B (list) Objekt kann nicht nach 'logical' umgewandelt werden
B Error in agnes(Test.Gower, diss=T).
B (list) object can`t be transformed to logical
B working example only numerics used:
B library(cluster)
B library(analogue)
B irisPart-subset(iris, select= Sepal.Length:Petal.Width)
B Dist.Gower - distance(irisPart, method =mixed)
B AgnesA - agnes(Dist.Gower, method=average, diss=TRUE)
B Would be great if somebody could help me.
B The dataset that I would like to use for the clustering also contains
B factors.
B and gives me the same Error message as in the not working example.
B Thanks in advance
B B.
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Re: [R] Maximum likelihood estimation
From ?optim
fn: A function to be minimized (or maximized), with first
argument the vector of parameters over which minimization is
to take place. It should return a scalar result.
I think you intended to optimize over c(a,b,p, lambda), so you need to
specify them as a single vector.
You may be making life unnecessarily hard for yourself: see function mle()
in package stats4.
Showing your code without a verbal description of what you are doing nor
the error message you got is less helpful than we need.
On Wed, 3 Sep 2008, toh wrote:
Hi R-experts,
I'm new to R in mle. I tried to do the following but just couldn't get it
right. Hope someone can point out the mistakes. thanks a lot.
t - c(1:90)
y -
c(5,10,15,20,26,34,36,43,47,49,80,84,108,157,171,183,191,200,204,211,217,226,230,
234,236,240,243,252,254,259,263,264,268,271,277,290,309,324,331,346,367,375,381,
401,411,414,417,425,430,431,433,435,437,444,446,446,448,451,453,460,463,463,464,
464,465,465,465,466,467,467,467,468,469,469,469,469,470,472,472,473,473,473,473,
473,473,473,475,475,475,475)
fr - function(a, b, p, lambda){
l - 0.5*(lambda + b*p + (1-p)*(lambda-b))
l^2 lambda*b*p
delta - sqrt(abs(l^2 - b*p*lambda))
mt - a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
logl - sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
return(-logl)
}
optim(c(15,0.01,0.5,0.01),fr, method=L-BFGS-B,
lower = c(0.002, 0.002, 0.002, 0.002), upper = c(Inf, Inf, 0.999,
Inf),control=list(fnscale=-1))
--
View this message in context:
http://www.nabble.com/Maximum-likelihood-estimation-tp19304249p19304249.html
Sent from the R help mailing list archive at Nabble.com.
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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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Re: [R] nls.control()
On Wed, 3 Sep 2008, Ranney, Steven wrote: and I'm trying to fit a simple Von Bertalanffy growth curve with program: VonB=nls(TL~Linf*(1-exp(-k*(age-t0))), data=box5.4, start=list(Linf=1000, k=0.1, t0=0.1), trace=TRUE) ... Everything works as it should right up until the confint(VonB) statement. When I ask for the confidence intervals, R goes through the process then stops and gives Error in prof$getProfile() : step factor 0.000488281 reduced below 'minFactor' of 0.000976562. However, when I use nls.control=(minFactor=0.01) OR nls.control(minFactor=0.01), In my experience, it never helps to reduce the control factors once this error comes up; it is a problem that by profiling you hit a boundary where convergence problems are serious. Suggestions: first, try with some smaller p-value, e.g. 0.8 or 0.7 instead of the default of 0.95; profiling will stay in safer region. It may be possible to use algorithm port in combination with lower/upper, but I am not sure if this is honoured by profile. Dieter I adjusted the nls() statement by adding the algorithm=port statement. Without changing the confint(vonB) request or including a lower/upper statement (I'm still not sure what they do), I was given 2.5% 97.5% Linf 462.8386NA k NA 0.6944959 t0 NA 0.7973458 It's not until I get down to confint(VonB, level=.31) that I get confidence interval values for all of the parameter estimates in the model. Any thoughts or ideas as to why? For the reason Dieter Menne has already told you. When profiling you are trying to fit models that cannot be fitted. I guess you need constraints on your parameters (as DM suggested), or perhaps better, to transform them. There are few things with nls profiling in R 2.7.x that are too conservative, so you might like to try R-devel (but I suspect you will still get into problems at 5%). Steven H. Ranney Graduate Research Assistant (Ph.D) USGS Montana Cooperative Fishery Research Unit Montana State University P.O. Box 173460 Bozeman, MT 59717-3460 phone: (406) 994-6643 fax: (406) 994-7479 [[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] nls convergence trouble
Hi, I agree with you that Excel is not the best tool for fittings, that's why I try to handle R. But I need to use this specific model (LgmAltFormula) and not a polynomial expression with the different parameters even if your method produced correct fitting. The parameters a and b are the Langmuir parameters that describe the adsorption of a compound onto activated carbon. I need to assess these parameters. Regards/Cordialement Benoit Boulinguiez -Message d'origine- De : Petr PIKAL [mailto:[EMAIL PROTECTED] Envoyé : mercredi 3 septembre 2008 17:58 À : Benoit Boulinguiez Cc : r-help@r-project.org Objet : Odp: [R] nls convergence trouble Hi Excel fit is not exceptionally good. Try fff-function(a,b) (V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 * C0 * + b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m * a * + b + (b * m * a)^2)^(1/2))/(2 * b * m) and with attached data frame plot(Qe,fff(364,0.0126)) abline(0,1) you clearly see linear relationship in smaller values but quite chaotic behaviour in bigger ones (or big deviation of experimental points from your model). So it is up to you if you want any fit (like from Excel) or only a good one (like from R). Seems to me that simple linear could be quite a good choice although there is some nelinearity. fit-lm(Qe~Ce+C0+V+m) summary(fit) Call: lm(formula = Qe ~ Ce + C0 + V + m) Residuals: Min 1Q Median 3Q Max -16.654 -8.653 2.426 9.971 11.912 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) -8.148e+02 1.330e+03 -0.613 0.549254 Ce -6.894e-02 4.982e-03 -13.839 6.02e-10 *** C0 3.284e-02 1.676e-03 19.589 4.26e-12 *** V2.153e+06 4.607e+05 4.674 0.000300 *** m -4.272e+04 1.218e+04 -3.509 0.003167 ** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Residual standard error: 10.87 on 15 degrees of freedom Multiple R-squared: 0.9903, Adjusted R-squared: 0.9877 F-statistic: 381.3 on 4 and 15 DF, p-value: 6.91e-15 plot(predict(fit), Qe) abline(0,1) Regards Petr [EMAIL PROTECTED] napsal dne 03.09.2008 16:01:36: Hi, Parameters assessment in R with nls doesn't work, though it works fine with MS Excel with the internal solver :( I use nls in R to determine two parameters (a,b) from experimental data. m VC0 CeQe 1 0.0911 0.0021740 3987.581 27.11637 94.51206 2 0.0911 0.0021740 3987.581 27.41915 94.50484 3 0.0911 0.0021740 3987.581 27.89362 94.49352 4 0.0906 0.0021740 5981.370 82.98477 189.37739 5 0.0906 0.0021740 5981.370 84.46435 189.34188 6 0.0906 0.0021740 5981.370 85.33213 189.32106 7 0.0911 0.0021740 7975.161 192.54276 233.30310 8 0.0911 0.0021740 7975.161 196.52891 233.20797 9 0.0911 0.0021740 7975.161 203.07467 233.05176 10 0.0906 0.0021872 9968.951 357.49157 328.29824 11 0.0906 0.0021872 9968.951 368.47609 328.03306 12 0.0906 0.0021872 9968.951 379.18904 327.77444 13 0.0904 0.0021740 13956.532 1382.61955 350.33391 14 0.0904 0.0021740 13956.532 1389.64915 350.16485 15 0.0904 0.0021740 13956.532 1411.87726 349.63030 16 0.0902 0.0021740 15950.322 2592.90486 367.38460 17 0.0902 0.0021740 15950.322 2606.34599 367.06064 18 0.0902 0.0021740 15950.322 2639.54301 366.26053 19 0.0906 0.0021872 17835.817 3894.12224 336.57036 20 0.0906 0.0021872 17835.817 3950.35273 335.21289 21 0.0906 0.0021872 17835.817 3972.29367 334.68320 the model LgmAltformula is Qe ~ (V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 * C0 * b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m * a * b + (b * m * a)^2)^(1/2))/(2 * b * m) the command in R is nls(formula=LgmAltFormula,data=bois.DATA,start=list(a=300,b=0.01),trace=TRUE ,control=nls.control(minFactor=0.9)) R has difficulties to converge and stops after the maximum of iterations 64650.47 : 2.945876e+02 3.837609e+08 64650.45 : 2.945876e+02 4.022722e+09 64650.45 : 2.945876e+02 1.695669e+09 64650.45 : 2.945876e+02 5.103971e+08 64650.44 : 2.945876e+02 8.497431e+08 64650.41 : 2.945876e+02 1.515243e+09 64650.36 : 2.945877e+02 5.482744e+09 64650.36 : 2.945877e+02 2.152294e+09 64650.36 : 2.945877e+02 7.953167e+08 64650.35 : 2.945877e+02 7.62e+07 Erreur dans nls(formula = LgmAltFormula, data = bois.DATA, start = list(a = 300, : le nombre d'itérations a dépassé le maximum de 50 The parameters a and b are estimated to be 364 and 0.0126 with Excel with the same data set. I tried with the algorithm=port with under and upper limits. One of the parameter reaches the limit and the regression stops. How can I succeed with R to make this regression? Regards/Cordialement - Benoit Boulinguiez Ph.D Ecole de Chimie de Rennes (ENSCR) Bureau 1.20 Equipe CIP UMR CNRS 6226 Sciences Chimiques de Rennes Campus de Beaulieu, 263 Avenue du Général Leclerc
[R] printing name of object inside lapply
Dear list members,
I am trying, within a lapply command, to print the name of the objects
in list or data frame. This is so that I can use odfWeave to print out a
report with a section for each object, including the object names.
I tried e.g.
a=b=c=1:5
lis=data.frame(a,b,c)
lapply(
lis, function (z) {
obj.nam - deparse(substitute(z))
cat(some other text,obj.nam,and so on,\n)
}
)
But instead of getting a b etc. I get X[[1L]] etc.
Any ideas?
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Re: [R] printing name of object inside lapply
On Thu, 4 Sep 2008, Steve Powell wrote:
Dear list members,
I am trying, within a lapply command, to print the name of the objects
in list or data frame. This is so that I can use odfWeave to print out a
report with a section for each object, including the object names.
I tried e.g.
a=b=c=1:5
lis=data.frame(a,b,c)
lapply(
lis, function (z) {
obj.nam - deparse(substitute(z))
cat(some other text,obj.nam,and so on,\n)
}
)
But instead of getting a b etc. I get X[[1L]] etc.
Any ideas?
Use a for() loop on the names: lapply is overkill here. But you could use
lapply(names(lis), function (z) {
cat(some other text, z, and so on,\n)
## references to lis[[z]]
})
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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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Re: [R] nls convergence trouble
Well, the model you try to fit is almost independent of b try plot(Qe,fff1(364,0.0126)) points(Qe,fff1(364,1), col=2) So you can quite freely choose b without any substantial improvement to fit. I am not an expert in nonlinear fitting but seems to me that your data does not follow your model, so you need to change your model or change your data, depends on your preference. Regards Petr [EMAIL PROTECTED] napsal dne 04.09.2008 12:11:27: Hi, I agree with you that Excel is not the best tool for fittings, that's why I try to handle R. But I need to use this specific model (LgmAltFormula) and not a polynomial expression with the different parameters even if your method produced correct fitting. The parameters a and b are the Langmuir parameters that describe the adsorption of a compound onto activated carbon. I need to assess these parameters. Regards/Cordialement Benoit Boulinguiez -Message d'origine- De : Petr PIKAL [mailto:[EMAIL PROTECTED] Envoyé : mercredi 3 septembre 2008 17:58 À : Benoit Boulinguiez Cc : r-help@r-project.org Objet : Odp: [R] nls convergence trouble Hi Excel fit is not exceptionally good. Try fff-function(a,b) (V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 * C0 * + b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m * a * + b + (b * m * a)^2)^(1/2))/(2 * b * m) and with attached data frame plot(Qe,fff(364,0.0126)) abline(0,1) you clearly see linear relationship in smaller values but quite chaotic behaviour in bigger ones (or big deviation of experimental points from your model). So it is up to you if you want any fit (like from Excel) or only a good one (like from R). Seems to me that simple linear could be quite a good choice although there is some nelinearity. fit-lm(Qe~Ce+C0+V+m) summary(fit) Call: lm(formula = Qe ~ Ce + C0 + V + m) Residuals: Min 1Q Median 3Q Max -16.654 -8.653 2.426 9.971 11.912 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) -8.148e+02 1.330e+03 -0.613 0.549254 Ce -6.894e-02 4.982e-03 -13.839 6.02e-10 *** C0 3.284e-02 1.676e-03 19.589 4.26e-12 *** V2.153e+06 4.607e+05 4.674 0.000300 *** m -4.272e+04 1.218e+04 -3.509 0.003167 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 10.87 on 15 degrees of freedom Multiple R-squared: 0.9903, Adjusted R-squared: 0.9877 F-statistic: 381.3 on 4 and 15 DF, p-value: 6.91e-15 plot(predict(fit), Qe) abline(0,1) Regards Petr [EMAIL PROTECTED] napsal dne 03.09.2008 16:01:36: Hi, Parameters assessment in R with nls doesn't work, though it works fine with MS Excel with the internal solver :( I use nls in R to determine two parameters (a,b) from experimental data. m VC0 CeQe 1 0.0911 0.0021740 3987.581 27.11637 94.51206 2 0.0911 0.0021740 3987.581 27.41915 94.50484 3 0.0911 0.0021740 3987.581 27.89362 94.49352 4 0.0906 0.0021740 5981.370 82.98477 189.37739 5 0.0906 0.0021740 5981.370 84.46435 189.34188 6 0.0906 0.0021740 5981.370 85.33213 189.32106 7 0.0911 0.0021740 7975.161 192.54276 233.30310 8 0.0911 0.0021740 7975.161 196.52891 233.20797 9 0.0911 0.0021740 7975.161 203.07467 233.05176 10 0.0906 0.0021872 9968.951 357.49157 328.29824 11 0.0906 0.0021872 9968.951 368.47609 328.03306 12 0.0906 0.0021872 9968.951 379.18904 327.77444 13 0.0904 0.0021740 13956.532 1382.61955 350.33391 14 0.0904 0.0021740 13956.532 1389.64915 350.16485 15 0.0904 0.0021740 13956.532 1411.87726 349.63030 16 0.0902 0.0021740 15950.322 2592.90486 367.38460 17 0.0902 0.0021740 15950.322 2606.34599 367.06064 18 0.0902 0.0021740 15950.322 2639.54301 366.26053 19 0.0906 0.0021872 17835.817 3894.12224 336.57036 20 0.0906 0.0021872 17835.817 3950.35273 335.21289 21 0.0906 0.0021872 17835.817 3972.29367 334.68320 the model LgmAltformula is Qe ~ (V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 * C0 * b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m * a * b + (b * m * a)^2)^(1/2))/(2 * b * m) the command in R is nls(formula=LgmAltFormula,data=bois.DATA,start=list(a=300,b=0.01),trace=TRUE ,control=nls.control(minFactor=0.9)) R has difficulties to converge and stops after the maximum of iterations 64650.47 : 2.945876e+02 3.837609e+08 64650.45 : 2.945876e+02 4.022722e+09 64650.45 : 2.945876e+02 1.695669e+09 64650.45 : 2.945876e+02 5.103971e+08 64650.44 : 2.945876e+02 8.497431e+08 64650.41 : 2.945876e+02 1.515243e+09 64650.36 : 2.945877e+02 5.482744e+09 64650.36 : 2.945877e+02 2.152294e+09 64650.36 : 2.945877e+02 7.953167e+08 64650.35 : 2.945877e+02 7.62e+07 Erreur dans nls(formula = LgmAltFormula,
[R] read.table error
Dear all, I have a tab-delimited text (.txt) file which I'm trying to read into R. This file is of column format - there are in fact 3 columns and 259201 rows (including the column headers). I've been using the following commands, but receive an error each time which prevents the data from being read in: Jan - read.table(JanuaryAvBurntArea.txt, header=TRUE) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 6 elements I tried removing the 'header' argument, but receive a similar message: Jan - read.table(JanuaryAvBurntArea.txt) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 2 did not have 6 elements What's more confusing about this is that I know that none of the lines have 6 elements! They're not supposed to! Each row only has 3 values (one per column)! As a final resort I tried 'scan': - scan(JanuaryAvBurntArea.txt) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : scan() expected 'a real', got 'Latitude' ...which is obviously something to do with there being a header as the first row, but the 'scan' command doesn't seem to have an equivalent of 'header=TRUE' like read.table...? If anyone is able to shed some light on why I'm receiving these errors, and how I can get the data into R, then I'd be very grateful to hear them! I suspect I'm doing something very basic which is wrong! Many thanks, Steve _ Win New York holidays with Kellogg’s Live Search __ 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] read.table error
On Thu, 4 Sep 2008, Steve Murray wrote: Dear all, I have a tab-delimited text (.txt) file which I'm trying to read into R. This file is of column format - there are in fact 3 columns and 259201 rows (including the column headers). I've been using the following commands, but receive an error each time which prevents the data from being read in: How about telling R it is tab-delimited? Use read.delim or at least sep=\t. Also, see the footer to this message: we are not clairvoyant and cannot see the file unless you show us part of it. With a file of that size you should study the 'R Data Import/Export Manual' and take some steps to read it in efficiently (e.g. specify nrows and colClasses). Jan - read.table(JanuaryAvBurntArea.txt, header=TRUE) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 1 did not have 6 elements I tried removing the 'header' argument, but receive a similar message: Jan - read.table(JanuaryAvBurntArea.txt) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : line 2 did not have 6 elements What's more confusing about this is that I know that none of the lines have 6 elements! They're not supposed to! Each row only has 3 values (one per column)! I suspect there is whitespace in the 'values'. As a final resort I tried 'scan': - scan(JanuaryAvBurntArea.txt) Error in scan(file, what, nmax, sep, dec, quote, skip, nlines, na.strings, : scan() expected 'a real', got 'Latitude' ...which is obviously something to do with there being a header as the first row, but the 'scan' command doesn't seem to have an equivalent of 'header=TRUE' like read.table...? If anyone is able to shed some light on why I'm receiving these errors, and how I can get the data into R, then I'd be very grateful to hear them! I suspect I'm doing something very basic which is wrong! Many thanks, Steve _ Win New York holidays with Kellogg’s Live Search __ 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] restricted bootstrap
Hello Professor Ripely, Sorry for not being clear. I posted after a long day of struggling. Also my toy distance matrix should have been symmetrical. Simply put I have spatially autocorrelated data collected from many points. I would like to do a linear regression on these data. To deal with the autocrrelation I want to resample a subset of my data with replacement but I need to restrict subsets such that no two locations where data was collected are closer than Xm apart (further apart than the autocrrelation in the data). Thanks for having a look at this for me. I will look up the hard-core spatial point process. Grant 2008/9/4 Prof Brian Ripley [EMAIL PROTECTED] I see nothing here to do with the 'bootstrap', which is sampling with replacement. Do you know what you mean exactly by 'randomly sample'? In general the way to so this is to sample randomly (uniformly, whatever) and reject samples that do not meet your restriction. For some restrictions there are more efficient algorithms, but I don't understand yours. (What are the 'rows'? Do you want to sample rows in space or xy locations? How come 'dist' is not symmetric?) For some restrictions, an MCMC sampling scheme is needed, the hard-core spatial point process being a related example. On Wed, 3 Sep 2008, Grant Gillis wrote: Hello List, I am not sure that I have the correct terminology here (restricted bootstrap) which may be hampering my archive searches. I have quite a large spatially autocorrelated data set. I have xy coordinates and the corresponding pairwise distance matrix (metres) for each row. I would like to randomly sample some number of rows but restricting samples such that the distance between them is larger than the threshold of autocorrelation. I have been been unsuccessfully trying to link the 'sample' function to values in the distance matrix. My end goal is to randomly sample M thousand rows of data N thousand times calculating linear regression coefficients for each sample but am stuck on taking the initial sample. I believe I can figure out the rest. Example Question I would like to radomly sample 3 rows further but withe the restriction that they are greater than 100m apart example data: main data: y- c(1, 2, 9, 5, 6) x-c( 1, 3, 5, 7, 9) z-c(2, 4, 6, 8, 10) a-c(3, 9, 6, 4 ,4) maindata-cbind(y, x, z, a) y x x a [1,] 1 1 1 3 [2,] 2 3 3 9 [3,] 9 5 5 6 [4,] 5 7 7 4 [5,] 6 9 9 4 distance matrix: row1-c(0, 123, 567, 89) row2-c(98, 0, 345, 543) row3-c(765, 90, 0, 987) row4-c(654, 8, 99, 0) dist-rbind(row1, row2, row3, row4) [,1] [,2] [,3] [,4] row10 123 567 89 row2 980 345 543 row3 765 900 987 row4 6548 990 Thanks for all of the help in the past and now Cheers Grant [[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/http://www.stats.ox.ac.uk/%7Eripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 [[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] Projecting Survival Curve into the Future
Hello, I have a survivor curve that shows account cancellations during the past 3.5 months. Â Fortunately for our business, but unfortunately for my analysis, the survivor curve doesn't yet pass through 50%. Â Is there a safe way to extend the survivor curve and estimate at what time we'll hit 50%? We started a new program 3.5 months ago, and I believe that this set of accounts behaves differently than the rest of our company's accounts. Thanks very much, Alan -- Alan Cox Director, User Experience iContact, Corp. p 919.459.1038 f 919.287.2475 [[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] read.table error
Thanks Prof. Ripley! I knew it would be something simple - I'd missed the \t from the read.table command! I won't be doing that again...!! Thanks again, Steve __ 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] Correct for heteroscedasticity using car package
Dear all, Sorry if this is too obvious. I am trying to fit my multiple regression model using lm() Before starting model simplification using step() I checked whether the model presented heteroscedasticity with ncv.test() from the CAR package. It presents it. I want to correct for it, I used hccm() from the CAR package as well and got the Heteroscedasticity-Corrected Covariance Matrix. I am not sure what am I supposed to do with the matrix. I guess I should run my model again telling it to use that matrix but I don't really find the parameter in lm() to tell R so. I guess it should be somewhere in weights? I would really appracite if you could show me how I would do it or recommend a text on how to correct heteroscedasticity with R. Many thanks. Roman Carrasco. __ 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] Projecting Survival Curve into the Future
I have a survivor curve that shows account cancellations during the
past 3.5 months. Â Fortunately for our business, but unfortunately
for my analysis, the survivor curve doesn't yet pass through 50%.
 Is there a safe way to extend the survivor curve and estimate at
what time we'll hit 50%?
Without any example code it's hard to say, but take a look at
?predict.coxph and ?predict.survreg in the survival package.
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] nls convergence trouble
Hi Benoit,
another way of making Petr's point is by looking at the profile log likelihood
function
for b; that is, only estimating the a parameter for a grid of b values:
## Defining mean function for fixed b
lgma - function(b){
function(C0, m, V, a){ (V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 *
C0 *
b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m * a *
b + (b * m * a)^2)^(1/2))/(2 * b * m)}
}
## Defining profile log likelihood function
logLikb - function(b)
{
logLik(nls(Qe~(lgma(b))(C0, m, V, a),data = bois.DATA,start = list(a=300)))
}
logLikb2 - Vectorize(logLikb, b) # vectorising the function
## Plotting the profile function
plot(x-10^(seq(-4, 0, length.out=50)), logLikb2(x), type=l)
Essentially any b value from 0.2 upwards results in the same model fit.
Christian
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Re: [R] printing name of object inside lapply
Thanks Prof Ripley! How obvious in retrospect!
Prof Brian Ripley wrote:
On Thu, 4 Sep 2008, Steve Powell wrote:
Dear list members,
I am trying, within a lapply command, to print the name of the objects
in list or data frame. This is so that I can use odfWeave to print out a
report with a section for each object, including the object names.
I tried e.g.
a=b=c=1:5
lis=data.frame(a,b,c)
lapply(
lis, function (z) {
obj.nam - deparse(substitute(z))
cat(some other text,obj.nam,and so on,\n)
}
)
But instead of getting a b etc. I get X[[1L]] etc.
Any ideas?
Use a for() loop on the names: lapply is overkill here. But you could
use
lapply(names(lis), function (z) {
cat(some other text, z, and so on,\n)
## references to lis[[z]]
})
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and provide commented, minimal, self-contained, reproducible code.
[R] Adding a legend to R graph device with several plots (no to individual plots!)
Dear Users, I already posted this question: it either went unnoticed,
or it is to basic (if this is so, please sent me a hint).
I would like to know if there is a way to add a common
legend to an arrangement of plots. In the example below, I get four
plots in my device. each one has a density for 1995 and one for 2006.
I have found that using legend or smartlegend I can add a legend to
each plot, but I am looking for something in the spirit of mtext. That
is, putting the legend anywhere I want on the device. In my situation
I have:
par(mfrow=c(2,2), ann = FALSE)
p.names - c(1,2,3,4)
lapply(p.names, function(x) {
d1 - density(rnorm(100,0,1))
d2 - density(rnorm(100,.3,1.2))
plot(range(d1$x, d2$x), range(d1$y, d2$y), type = n, xlab = NULL,
ylab = NULL)
lines(d1, col = black, lty = 2)
lines(d2, col = black, lty = 1)
})
mtext(Learning R the hard way, side = 1, line=-1.5, outer=TRUE)
smartlegend(x= center, y=top, c(1995,2006), lty=2:1)
the last line puts the legend on the lower-right plot. I'd like to put
it in the middle - or top of the device. Any suggestion?
Thanks!
Nelson
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Re: [R] Projecting Survival Curve into the Future
You might consider a probit analysis using ln(Time) as the dose. At 09:24 AM 9/4/2008, [EMAIL PROTECTED] wrote: I have a survivor curve that shows account cancellations during the past 3.5 months. Â Fortunately for our business, but unfortunately for my analysis, the survivor curve doesn't yet pass through 50%. Â Is there a safe way to extend the survivor curve and estimate at what time we'll hit 50%? Without any example code it's hard to say, but take a look at ?predict.coxph and ?predict.survreg in the survival package. Regards, Richie. Mathematical Sciences Unit HSL Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: [EMAIL PROTECTED] Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ 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] Adding a legend to R graph device with several plots (no to individual plots!)
On 9/4/2008 9:35 AM, Nelson Villoria wrote:
Dear Users, I already posted this question: it either went unnoticed,
or it is to basic (if this is so, please sent me a hint).
I would like to know if there is a way to add a common
legend to an arrangement of plots. In the example below, I get four
plots in my device. each one has a density for 1995 and one for 2006.
I have found that using legend or smartlegend I can add a legend to
each plot, but I am looking for something in the spirit of mtext. That
is, putting the legend anywhere I want on the device. In my situation
I have:
par(mfrow=c(2,2), ann = FALSE)
p.names - c(1,2,3,4)
lapply(p.names, function(x) {
d1 - density(rnorm(100,0,1))
d2 - density(rnorm(100,.3,1.2))
plot(range(d1$x, d2$x), range(d1$y, d2$y), type = n, xlab = NULL,
ylab = NULL)
lines(d1, col = black, lty = 2)
lines(d2, col = black, lty = 1)
})
mtext(Learning R the hard way, side = 1, line=-1.5, outer=TRUE)
smartlegend(x= center, y=top, c(1995,2006), lty=2:1)
the last line puts the legend on the lower-right plot. I'd like to put
it in the middle - or top of the device. Any suggestion?
The likely reason you didn't get a reply is because you used
smartlegend(), which I believe comes from gplots. This means only
people who know gplots well will answer.
To do this using legend(), try this:
par(xpd=NA)
legend(locator(), c(1995,2006), lty=2:1)
This will allow you to place the legend manually whereever you like on
the page. If you want to code a particular position for re-use later,
run locator() on its own, and save the location. (You will probably
want to use grconvertX and grconvertY to change these into ndc
coordinates. For example, this looks okay to me:
par(xpd=NA)
legend(grconvertX(0.44, ndc, user),
grconvertY(0.55, ndc, user),
c(1995,2006), lty=2:1)
You might want to wrap this in a nice mlegend() function to save some
typing.
Duncan Murdoch
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Re: [R] Projecting Survival Curve into the Future
On Thu, 4 Sep 2008, [EMAIL PROTECTED] wrote: I have a survivor curve that shows account cancellations during the past 3.5 months. Â Fortunately for our business, but unfortunately for my analysis, the survivor curve doesn't yet pass through 50%. Â Is there a safe way to extend the survivor curve and estimate at what time we'll hit 50%? Without any example code it's hard to say, but take a look at ?predict.coxph and ?predict.survreg in the survival package. You will not be able to do this with coxph: there will be no events to estimate the baseline hazard from. Whether using a parametric accelerated life model (survreg) is 'safe' depends on what you are prepared to assume. I'd say it was pretty dubious unless you have theoretical reasons to suppose that e.g. an exponential is a good approximation and it fits the data you do have. Regards, Richie. -- 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] Installing rgl
Duncan Murdoch wrote: On 03/09/2008 4:34 PM, Kevin E. Thorpe wrote: Hello. I'm having trouble installing rgl. I have a theory as to the problem. First, the error message and session info. install.packages(rgl) trying URL 'http://probability.ca/cran/src/contrib/rgl_0.81.tar.gz' Content type 'application/x-gzip' length 1636939 bytes (1.6 Mb) opened URL == downloaded 1.6 Mb * Installing *source* package 'rgl' ... checking for gcc... gcc -std=gnu99 checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc -std=gnu99 accepts -g... yes checking for gcc -std=gnu99 option to accept ISO C89... none needed checking how to run the C preprocessor... gcc -std=gnu99 -E checking for gcc... (cached) gcc -std=gnu99 checking whether we are using the GNU C compiler... (cached) yes checking whether gcc -std=gnu99 accepts -g... (cached) yes checking for gcc -std=gnu99 option to accept ISO C89... (cached) none needed checking for libpng-config... yes configure: using libpng-config configure: using libpng dynamic linkage checking for X... libraries , headers checking GL/gl.h usability... yes checking GL/gl.h presence... yes checking for GL/gl.h... yes checking GL/glu.h usability... yes checking GL/glu.h presence... yes checking for GL/glu.h... yes checking for glEnd in -lGL... no configure: error: missing required library GL ERROR: configuration failed for package 'rgl' ** Removing '/usr/local/lib/R/library/rgl' The downloaded packages are in /tmp/RtmpZAeU9W/downloaded_packages Updating HTML index of packages in '.Library' Warning message: In install.packages(rgl) : installation of package 'rgl' had non-zero exit status sessionInfo() R version 2.7.2 Patched (2008-09-02 r46486) i686-pc-linux-gnu locale: C attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.7.2 So, the error appears to be related to libGL (or libGLU?). I am running OpenSUSE 10.2. I have the OpenGL packages installed. I have the MESA and MESA-devel packages installed which provide libGL.so.1 and libGLU.so.1. I also have the fglrx (proprietary ATI driver package installed) which also provides libGL.so.1 (but no libGLU.so.1). I suspect that an incompatible libGL is being found (both versions appear to be on the system in different directories). How do I confirm if this is the case and force a particular lib file during package installation? rgl is now hosted on R-forge, at https://r-forge.r-project.org/projects/rgl/. You could post questions there, and be one of the first to do so :-). But the general idea is probably to run the configure script manually, and look at the output. So you would download and untar rgl (or just check it out using svn), then in the main directory, run ./configure. I think this will produce a config.log file that you can check for all the gory details. Duncan Murdoch Thanks Duncan. I followed your advice and ran the configure script manually. I looked at config.log which showed me where to look. I discovered that /usr/lib/libGL.so.1 was not there. I think I know how that happened, but I don't want to talk about it. :-) I re-installed my MESA packages and everything worked. Kevin -- Kevin E. Thorpe Biostatistician/Trialist, Knowledge Translation Program Assistant Professor, Dalla Lana School of Public Health University of Toronto email: [EMAIL PROTECTED] Tel: 416.864.5776 Fax: 416.864.6057 __ 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 reduce stress value in isoMDS?
no I mean maybe use a higher dimension solution- the other thing you may wish to check out is package vegan which has an mds function and very good viginettes, and is my favorite package for doing ordination analysis (metaMDS uses isoMDS). hope this helps On Wed, Sep 3, 2008 at 9:03 PM, 陈武 [EMAIL PROTECTED] wrote: No, the two dimensions are the same. It's a dissimilarity matrix in the txt file. data[x][x]=0 data[x][y]=data[y][x] and 0=data[x][y]=1. 2008/9/3, stephen sefick [EMAIL PROTECTED]: different dimensions? On Wed, Sep 3, 2008 at 8:13 AM, 陈武 [EMAIL PROTECTED] wrote: haha...wrong code again, it's isoMDS not sammon in the 5th line. Thanks for Victor Lemes Landeiro's and Brian D. Ripley's advice. ÔÚ08-9-3£¬Prof Brian Ripley [EMAIL PROTECTED] дµÀ£º There is still no call to isoMDS in your code, and nothing we can reproduce. It all depends on the dissimilarity matrix you have not given us: maybe there is no good 2D representation of it. Looks like you need to ask a local expert about what you are doing, for this is a statistical and not an R question. On Wed, 3 Sep 2008, ³ÂÎä wrote: Sorry, wrong code. The right one here: library(MASS) cl-read.table(e:/data.txt,header=T,sep=,) row.names(cl)-colnames(cl) cm-as.matrix(cl) loc-sammon(cm) jpeg(filename=e:/plot.gif,width = 480, height = 480, units = px, pointsize = 12, quality = 75, bg = white, res = NA, restoreConsole = TRUE) plot(loc$points,type=p) text(loc$points,rownames(cl),cex=1,pos=1,offset=1) dev.off() And e:/data.txt contains a 40*40 dissimilarity matrix. Thanks for you advices! 2008/9/3, [EMAIL PROTECTED]: I apply isoMDS to my data, but the result turns out to be bad as the stress value stays around 31! Yeah, 31 ,not 3.1... I don't know if I ignore something before recall isoMDS. My code as follow: m - read.table(e:/tsdata.txt,header=T,sep=,) article_number - ts(m, start = 2004,end=2008, frequency = 1 ,names=colnames(m)) jpeg(filename=e:/tsmap.gif,width = 480, height = 480, units = px, pointsize = 12, quality = 75, bg = white, res = NA, restoreConsole = TRUE) plot(article_number, plot.type=single, lty=c(1,1,1,1,1),col=c(1,2,3,4,5),las=1) max-range(m) x-c(2004,2004,2004,2004,2004) y-c(max[2]*0.96,max[2]*0.9199,max[2]*0.88,max[2]*0.84,max[2]*0.7999) points(x,y,col=c(1,2,3,4,5),pch=15) text(x,y,colnames(m),pos=4,offset=0.4) dev.off() A 40*40 matrix in e:/tsdata.txt. How should I do to improve the effect? Thank you! [[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 [[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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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 -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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 http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Correct for heteroscedasticity using car package
Dear Roman, You can use the coefficient-covariance matrix returned by hccm() for calculating corrected standard errors for the coefficients. Alternatively, if you know the pattern of heteroscedasticity [as you probably do if you used ncv.test()], you could try to correct for it by a transformation of the response variable or by weighted-least-squares estimation. I hope this helps, John -- John Fox, Professor Department of Sociology McMaster University Hamilton, Ontario, Canada web: socserv.mcmaster.ca/jfox -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Carrasco-Torrecilla, Roman R Sent: September-04-08 9:03 AM To: r-help@r-project.org Subject: [R] Correct for heteroscedasticity using car package Dear all, Sorry if this is too obvious. I am trying to fit my multiple regression model using lm() Before starting model simplification using step() I checked whether the model presented heteroscedasticity with ncv.test() from the CAR package. It presents it. I want to correct for it, I used hccm() from the CAR package as well and got the Heteroscedasticity-Corrected Covariance Matrix. I am not sure what am I supposed to do with the matrix. I guess I should run my model again telling it to use that matrix but I don't really find the parameter in lm() to tell R so. I guess it should be somewhere in weights? I would really appracite if you could show me how I would do it or recommend a text on how to correct heteroscedasticity with R. Many thanks. Roman Carrasco. __ 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.
[R] A ternary graph's scales
Hi all, I am drawing a ternary graph. Everything is fine with both ternaryplot (package vcd) and triangle.plot (package ade4), but I want to present scales in neither percents nor from 0 to 1 (this is actually the only option I found in both functions). I want the scales to be in a natural scale (from 0 to k, k being the number of objects). Is it at all possible? (Descriptions of both functions do not mention this possibility.) Thanks, Marcin -- Build up your weaknesses until they become your strong points -- Knute Rockne __ 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] Correct for heteroscedasticity using car package
On Thu, 4 Sep 2008, Carrasco-Torrecilla, Roman R wrote: Dear all, Sorry if this is too obvious. I am trying to fit my multiple regression model using lm() Before starting model simplification using step() I checked whether the model presented heteroscedasticity with ncv.test() from the CAR package. It presents it. I want to correct for it, I used hccm() from the CAR package as well and got the Heteroscedasticity-Corrected Covariance Matrix. I am not sure what am I supposed to do with the matrix. I guess I should run my model again telling it to use that matrix but I don't really find the parameter in lm() to tell R so. I guess it should be somewhere in weights? If you have a reasonable approximation of the pattern of heteroskedasticity, you can supply it in the weights argument to lm() and perform WLS. hccm() on the other hand does not assume a particular pattern of heteroskedasticity (with the obvious advantages and disadvantages). You can easily employ it for inference based on Wald statistics. The car package provides linear.hypothesis() for this and the package lmtest provides functions coeftest() and waldtest(). I would really appracite if you could show me how I would do it or recommend a text on how to correct heteroscedasticity with R. The sandwich package which provides more flexible implementations of the estimators underlying hccm() as well as other estimators has vignette(sandwich, package = sandwich) with some background information and hands-on examples. hth, Z __ 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] 3 Courses*** R/Splus Fundamentals and Programming Techniques: September 2008 at 3 locations by XLSolutions Corp
rXLSolutions Corporation (www.xlsolutions-corp.com) is proud to announce our*** R/Splus Fundamentals and Programming Techniques***course at 3 USA locations for September 2008. (1) R/Splus Fundamentals and Programming Techniques http://www.xlsolutions-corp.com/Rfund.htm * San Francisco ** September 22-23, 2008 * Princeton, NJ ** September 25-26, 2008 * Washington, DC ** September 29-30, 2008 Looking for R/Splus Advanced Programming? Please email sue Turner: [EMAIL PROTECTED] Ask for group discount and reserve your seat Now - Earlybird Rates. Payment due after the class! Email Sue Turner: [EMAIL PROTECTED] (1) R/Splus Fundamentals and Programming Techniques http://www.xlsolutions-corp.com/Rfund.htm (2) R/Splus Advanced Programming http://www.xlsolutions-corp.com/Radv.htm Email us for group discounts. Email Sue Turner: [EMAIL PROTECTED] Phone: 206-686-1578 Please let us know if you and your colleagues are interested in this class to take advantage of group discount. Register now to secure your seat! Cheers, Elvis Miller, PhD Manager Training. XLSolutions Corporation 206 686 1578 www.xlsolutions-corp.com [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] nls convergence trouble
# 1. Supplying the derivatives results in convergence:
lgmg - function(a, b, C0, m, V) {
+ e - expression((V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 *
+ C0 * b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m *
+ a * b + (b * m * a)^2))/(2 * b * m))
+ val - eval(e)
+ attr(val, gradient) - cbind(a = eval(D(e, a)), b = eval(D(e, b)))
+ val
+ }
nls(Qe ~ lgmg(a, b, C0, m, V), bois.DATA, start = c(a = 300, b = 1))
Nonlinear regression model
model: Qe ~ lgmg(a, b, C0, m, V)
data: bois.DATA
a b
337.74912 0.03864
residual sum-of-squares: 15473
Number of iterations to convergence: 9
Achieved convergence tolerance: 3.16e-06
# 2. As mentioned before squaring both sides results in convergence:
# even without derivatives
# though admittedly that is a slightly different objective.
lgm - function(a, b, C0, m, V) {
+ (V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 *
+ C0 * b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m *
+ a * b + (b * m * a)^2))/(2 * b * m)
+ }
nls(Qe^2 ~ lgm(a, b, C0, m, V)^2, bois.DATA, start = c(a = 300, b = 1))
Nonlinear regression model
model: Qe^2 ~ lgm(a, b, C0, m, V)^2
data: bois.DATA
ab
225.6474 0.3568
residual sum-of-squares: 9.98e+10
Number of iterations to convergence: 16
Achieved convergence tolerance: 6.096e-06
# 3. Also using the reciprocal it converges
# without derivatives
nls(Qe ~ lgm(a, 1/b, C0, m, V), bois.DATA, start = c(a = 300, b = 1))
Nonlinear regression model
model: Qe ~ lgm(a, 1/b, C0, m, V)
data: bois.DATA
a b
337.75 25.88
residual sum-of-squares: 15473
Number of iterations to convergence: 12
Achieved convergence tolerance: 1.722e-06
transform(as.list(coef(.Last.value)), b = 1/b)
a b
1 337.7492 0.03863738
On Wed, Sep 3, 2008 at 10:36 AM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
Try squaring both sides of the formula.
On Wed, Sep 3, 2008 at 10:01 AM, Benoit Boulinguiez
[EMAIL PROTECTED] wrote:
Hi,
Parameters assessment in R with nls doesn't work, though it works fine with
MS Excel with the internal solver :(
I use nls in R to determine two parameters (a,b) from experimental data.
m VC0 CeQe
1 0.0911 0.0021740 3987.581 27.11637 94.51206
2 0.0911 0.0021740 3987.581 27.41915 94.50484
3 0.0911 0.0021740 3987.581 27.89362 94.49352
4 0.0906 0.0021740 5981.370 82.98477 189.37739
5 0.0906 0.0021740 5981.370 84.46435 189.34188
6 0.0906 0.0021740 5981.370 85.33213 189.32106
7 0.0911 0.0021740 7975.161 192.54276 233.30310
8 0.0911 0.0021740 7975.161 196.52891 233.20797
9 0.0911 0.0021740 7975.161 203.07467 233.05176
10 0.0906 0.0021872 9968.951 357.49157 328.29824
11 0.0906 0.0021872 9968.951 368.47609 328.03306
12 0.0906 0.0021872 9968.951 379.18904 327.77444
13 0.0904 0.0021740 13956.532 1382.61955 350.33391
14 0.0904 0.0021740 13956.532 1389.64915 350.16485
15 0.0904 0.0021740 13956.532 1411.87726 349.63030
16 0.0902 0.0021740 15950.322 2592.90486 367.38460
17 0.0902 0.0021740 15950.322 2606.34599 367.06064
18 0.0902 0.0021740 15950.322 2639.54301 366.26053
19 0.0906 0.0021872 17835.817 3894.12224 336.57036
20 0.0906 0.0021872 17835.817 3950.35273 335.21289
21 0.0906 0.0021872 17835.817 3972.29367 334.68320
the model LgmAltformula is
Qe ~ (V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 * C0 *
b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m * a *
b + (b * m * a)^2)^(1/2))/(2 * b * m)
the command in R is
nls(formula=LgmAltFormula,data=bois.DATA,start=list(a=300,b=0.01),trace=TRUE
,control=nls.control(minFactor=0.9))
R has difficulties to converge and stops after the maximum of iterations
64650.47 : 2.945876e+02 3.837609e+08
64650.45 : 2.945876e+02 4.022722e+09
64650.45 : 2.945876e+02 1.695669e+09
64650.45 : 2.945876e+02 5.103971e+08
64650.44 : 2.945876e+02 8.497431e+08
64650.41 : 2.945876e+02 1.515243e+09
64650.36 : 2.945877e+02 5.482744e+09
64650.36 : 2.945877e+02 2.152294e+09
64650.36 : 2.945877e+02 7.953167e+08
64650.35 : 2.945877e+02 7.62e+07
Erreur dans nls(formula = LgmAltFormula, data = bois.DATA, start = list(a =
300, :
le nombre d'itérations a dépassé le maximum de 50
The parameters a and b are estimated to be 364 and 0.0126 with Excel
with the same data set.
I tried with the algorithm=port with under and upper limits. One of the
parameter reaches the limit and the regression stops.
How can I succeed with R to make this regression?
Regards/Cordialement
-
Benoit Boulinguiez
Ph.D
Ecole de Chimie de Rennes (ENSCR) Bureau 1.20
Equipe CIP UMR CNRS 6226 Sciences Chimiques de Rennes
Campus de Beaulieu, 263 Avenue du Général Leclerc
35700 Rennes, France
Tel 33 (0)2 23 23 80 83
Fax 33 (0)2 23 23 81 20
http://www.ensc-rennes.fr/
[[alternative HTML version deleted]]
__
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[R] pass data to log-likelihood function
Hi there,
When I do bootstrap on a maximum likelihood estimation, I try the
following code, however, I get error:
Error in minuslogl(alpha = 0, beta = 0) : object x not found
It seems that mle() only get data from workspace, other than the
boot.fun().
My question is how to pass the data to mle() in my case.
I really appreciated to any suggestions.
Best wishes,
Jinsong
#---code start here---
x - c(32, 16, 8, 4, 2, 1)
r - c(20, 12, 10, 8, 6, 0)
n - c(20, 20, 20, 20, 20, 20)
mydata - data.frame(x = x, r = r, n = n)
rm(x, r, n) #if not rmed, it will affect the final result.
ll - function(alpha, beta) { #how to pass the data to this function?
x - log10(x)
P - pnorm(alpha + beta * x)
P - pmax(pmin(P,1),0)
-(sum(r * log(P)) + sum((n - r)* log(1-P)))
}
boot.fun - function(data, index) {
boot.data - data[index, ]
# it seems that the following three line dose nothing with the mle()
x - boot.data$x
r - boot.data$r
n - boot.data$n
fit - mle(ll, start = list(alpha = 0, beta = 0), method = BFGS)
boot.coef - coef(fit)
stats - -boot.coef[1] / boot.coef[2]
}
library(stats4)
library(boot)
myboot - boot(mydata, boot.fun, R = 199)
# give the error message:
# Error in minuslogl(alpha = 0, beta = 0) : object x not found
#---code end here---
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] pass data to log-likelihood function
On 9/4/2008 10:54 AM, Jinsong Zhao wrote:
Hi there,
When I do bootstrap on a maximum likelihood estimation, I try the
following code, however, I get error:
Error in minuslogl(alpha = 0, beta = 0) : object x not found
It seems that mle() only get data from workspace, other than the
boot.fun().
This is the way R does scoping. Your ll function was defined in the
global environment, so that's where it will look for x, n and r. If you
want it to look at the local variables in boot.fun, then you should
define it in boot.fun.
Duncan Murdoch
My question is how to pass the data to mle() in my case.
I really appreciated to any suggestions.
Best wishes,
Jinsong
#---code start here---
x - c(32, 16, 8, 4, 2, 1)
r - c(20, 12, 10, 8, 6, 0)
n - c(20, 20, 20, 20, 20, 20)
mydata - data.frame(x = x, r = r, n = n)
rm(x, r, n) #if not rmed, it will affect the final result.
ll - function(alpha, beta) { #how to pass the data to this function?
x - log10(x)
P - pnorm(alpha + beta * x)
P - pmax(pmin(P,1),0)
-(sum(r * log(P)) + sum((n - r)* log(1-P)))
}
boot.fun - function(data, index) {
boot.data - data[index, ]
# it seems that the following three line dose nothing with the mle()
x - boot.data$x
r - boot.data$r
n - boot.data$n
fit - mle(ll, start = list(alpha = 0, beta = 0), method = BFGS)
boot.coef - coef(fit)
stats - -boot.coef[1] / boot.coef[2]
}
library(stats4)
library(boot)
myboot - boot(mydata, boot.fun, R = 199)
# give the error message:
# Error in minuslogl(alpha = 0, beta = 0) : object x not found
#---code end here---
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[R] Stepwise
Hi, Is there any facility in R to perform a stepwise process on a model, which will remove any highly-correlated explanatory variables? I am told there is in SPSS. I have a large number of variables (some correlated), which I would like to just chuck in to a model and perform stepwise and see what comes out the other end, to give me an idea perhaps as to which variables I should focus on. Thanks for any help / suggestions. Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] [[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 on jarque test
Hi all, I used the function jarque.test (in the moments package) on my data set and I obtained something like this: Jarque-Bera Normality Test data: x JB = 4.8381, p-value = 0.089 alternative hypothesis: greater or Jarque-Bera Normality Test data: x JB = 2.6018, p-value = 0.2723 alternative hypothesis: greater I cannot understand this. Please, someone can help me? thank you giov -- View this message in context: http://www.nabble.com/help-on-jarque-test-tp19312270p19312270.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] for 1:length and removing rows
Hello everyone,
I have a problem using a for-loop to go through a matrix. I want to remove all
rows with a sum of 0.
What I do is basically:
for(i in 1:length(data))
{
if(sum(add(data[i,]) == 0)
{
data - data[-i,]
}
}
I get a error message: Error in `[.data.frame`(data, i) : undefined columns
selected, because the length is reduced when removing in the if-clause.
How do I make this?
Best wishes,
Markus
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[R] saving an object
Hi there, I have a dataset stored in an object which has very huge volume of rows.I want to reuse it for comparing with other datasets.I dont want it to reload every time i run the script.Is there a way of saving a particular loaded object in the workspace and reusing it. Kindly help me. Ramya Thulasingam -- View this message in context: http://www.nabble.com/saving-an-object-tp19313276p19313276.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] saving an object
?save I assume you can connect to it from within the script, presumably by supplying the path of the object to the appropriate argument in your script. I'm no expert though. HTH, Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Rajasekaramya Sent: Thursday, September 04, 2008 4:36 PM To: r-help@r-project.org Subject: [R] saving an object Hi there, I have a dataset stored in an object which has very huge volume of rows.I want to reuse it for comparing with other datasets.I dont want it to reload every time i run the script.Is there a way of saving a particular loaded object in the workspace and reusing it. Kindly help me. Ramya Thulasingam -- View this message in context: http://www.nabble.com/saving-an-object-tp19313276p19313276.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] Stepwise
Robin Williams wrote Is there any facility in R to perform a stepwise process on a model, which will remove any highly-correlated explanatory variables? I am told there is in SPSS. I have a large number of variables (some correlated), which I would like to just chuck in to a model and perform stepwise and see what comes out the other end, to give me an idea perhaps as to which variables I should focus on. Thanks for any help / suggestions. Stepwise is a bad method of selecting variables. Far better methods are LASSO and LAR (least angle regression), available in the LARS package and the LASSO2 package. However, while both these methods are good, neither is a substitute for substantive knowledge. Also, the key thing is not so much whether variables are correlated, but whether they are co-linear, which is different. If you have a great many variables, then you can have a high degree of colinearity even with no high pairwise correlations. I've not done this in R, but RSiteSearch(collinearity, restrict = 'functions') yields 34 hits. HTH Peter [[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] for 1:length and removing rows
On 04-Sep-08 15:33:06, Markus Mühlbacher wrote:
Hello everyone,
I have a problem using a for-loop to go through a matrix. I want to
remove all rows with a sum of 0.
What I do is basically:
for(i in 1:length(data))
{
if(sum(add(data[i,]) == 0)
{
data - data[-i,]
}
}
I get a error message: Error in `[.data.frame`(data, i) : undefined
columns selected, because the length is reduced when removing in the
if-clause.
How do I make this?
The function rowSums() gives you the row-sums of a matrix, which
you can test for ==0, and carry on from there. Example:
M-matrix(c(1,2,3,0,-1,-2,-3,1,-2,3,0,1,2,-1),ncol=2)
M
# [,1] [,2]
# [1,]11
# [2,]2 -2
# [3,]33
# [4,]00
# [5,] -11
# [6,] -22
# [7,] -3 -1
ix-which(rowSums(M)==0)
M[-ix,]
# [,1] [,2]
# [1,]11
# [2,]33
# [3,] -3 -1
Hoping this helps!
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 04-Sep-08 Time: 16:49:46
-- XFMail --
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[R] table and colnames
I have a table statement that returns the following: [10.839,10.841] (10.841,10.843] (10.843,10.846] (10.846,10.848] (10.848,10.85] 0 0 0 0 1 (10.85,10.852] (10.852,10.854] (10.854,10.857] (10.857,10.859] (10.859,10.861] 0 0 0 0 0 What I want to do is get the upper bound value into a vector for each of the 10-buckets. (e.g. v-c(10.841,10.843...) I'm totally clueless on how to proceed. Any help appreciated. TIA, Joe [[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] A ternary graph's scales
There is also the tri function in the cwhtool package, triax.plot in the plotrix package, and triplot in the TeachingDemos package (I think it is between this plot and progress bars as to what functionality is reproduced in the most packages). One of the others may do what you want, or be modifiable to do what you want. The triplot (TeachingDemos) does not put scales on by default, but does had an 'add' argument that could possibly be used to add scales after the fact, or the code is fairly short and simple to see what the transformation is to add scale manually. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Marcin Kozak Sent: Thursday, September 04, 2008 8:34 AM To: r-help@r-project.org Subject: [R] A ternary graph's scales Hi all, I am drawing a ternary graph. Everything is fine with both ternaryplot (package vcd) and triangle.plot (package ade4), but I want to present scales in neither percents nor from 0 to 1 (this is actually the only option I found in both functions). I want the scales to be in a natural scale (from 0 to k, k being the number of objects). Is it at all possible? (Descriptions of both functions do not mention this possibility.) Thanks, Marcin -- Build up your weaknesses until they become your strong points -- Knute Rockne __ 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] table and colnames
Try this: v - as.numeric(gsub(]|\\), , unlist(lapply(strsplit(names(tb), ,), [, 2 Where tb is your table On Thu, Sep 4, 2008 at 10:20 AM, Trubisz, Joseph [EMAIL PROTECTED] wrote: I have a table statement that returns the following: [10.839,10.841] (10.841,10.843] (10.843,10.846] (10.846,10.848] (10.848,10.85] 0 0 0 0 1 (10.85,10.852] (10.852,10.854] (10.854,10.857] (10.857,10.859] (10.859,10.861] 0 0 0 0 0 What I want to do is get the upper bound value into a vector for each of the 10-buckets. (e.g. v-c(10.841,10.843...) I'm totally clueless on how to proceed. Any help appreciated. TIA, Joe [[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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] table and colnames
Here is one way: y - levels(cut(x,5)) y [1] (0.0124,0.209] (0.209,0.405] (0.405,0.601] (0.601,0.797] (0.797,0.993] sapply(y, function(x) sub(.*,(.*)], \\1, x)) (0.0124,0.209] (0.209,0.405] (0.405,0.601] (0.601,0.797] (0.797,0.993] 0.2090.4050.6010.7970.993 as.numeric(sapply(y, function(x) sub(.*,(.*)], \\1, x))) [1] 0.209 0.405 0.601 0.797 0.993 On Thu, Sep 4, 2008 at 9:20 AM, Trubisz, Joseph [EMAIL PROTECTED] wrote: I have a table statement that returns the following: [10.839,10.841] (10.841,10.843] (10.843,10.846] (10.846,10.848] (10.848,10.85] 0 0 0 0 1 (10.85,10.852] (10.852,10.854] (10.854,10.857] (10.857,10.859] (10.859,10.861] 0 0 0 0 0 What I want to do is get the upper bound value into a vector for each of the 10-buckets. (e.g. v-c(10.841,10.843...) I'm totally clueless on how to proceed. Any help appreciated. TIA, Joe [[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 that 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] A ternary graph's scales
Thank you, Greg. After all I noticed that in ternaryplot it can be easily made by setting a 'scale' parameter. Thanks, Marcin On Thu, Sep 4, 2008 at 6:00 PM, Greg Snow [EMAIL PROTECTED] wrote: There is also the tri function in the cwhtool package, triax.plot in the plotrix package, and triplot in the TeachingDemos package (I think it is between this plot and progress bars as to what functionality is reproduced in the most packages). One of the others may do what you want, or be modifiable to do what you want. The triplot (TeachingDemos) does not put scales on by default, but does had an 'add' argument that could possibly be used to add scales after the fact, or the code is fairly short and simple to see what the transformation is to add scale manually. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Marcin Kozak Sent: Thursday, September 04, 2008 8:34 AM To: r-help@r-project.org Subject: [R] A ternary graph's scales Hi all, I am drawing a ternary graph. Everything is fine with both ternaryplot (package vcd) and triangle.plot (package ade4), but I want to present scales in neither percents nor from 0 to 1 (this is actually the only option I found in both functions). I want the scales to be in a natural scale (from 0 to k, k being the number of objects). Is it at all possible? (Descriptions of both functions do not mention this possibility.) Thanks, Marcin -- Build up your weaknesses until they become your strong points -- Knute Rockne __ 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.
[R] Plotting using 'if' statements
Dear all, I have a dataset of four columns, and I wish to plot (as a scatter graph) the values of the third column where the values are greater than zero, and the fourth column. I tried doing this via the plot command itself, but got into a bit of a mess (resulting in errors!). My dataframe is called 'January': plot(January[3(0):4]) Error: unexpected '' in plot(January[3( After a few variations on this, I thought I'd try making a new object which includes all values from the third column of January0 (to plot in a separate step) as follows: JanFilter - January[3]0 No error here. However, when I display the 'values' of JanFilter, it shows that instead of keeping the numerical values, the above operation simply displays the results of the logical test: head(JanFilter) Value [1,] FALSE [2,]TRUE [3,]TRUE [4,]TRUE [5,]TRUE [6,]TRUE This is obviously no good for plotting the numerical values on axes! So my question is, how do I perform 'if' statements in order to filter out various parts of a dataset, for plotting on a graph. Many thanks, Steve __ 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 and colnames
See http://gsubfn.googlecode.com for more on the gsubfn package. Below we use strapply to pick out every string of digits and decimal points converting each to numeric returning a list of pairs and rbinding that list into a matrix. x - c([10.839,10.841], (10.841,10.843], (10.843,10.846], (10.846,10.848], (10.848,10.85]) library(gsubfn) strapply(x, [0-9.]+, as.numeric, simplify = rbind) [,1] [,2] [1,] 10.839 10.841 [2,] 10.841 10.843 [3,] 10.843 10.846 [4,] 10.846 10.848 [5,] 10.848 10.850 On Thu, Sep 4, 2008 at 9:20 AM, Trubisz, Joseph [EMAIL PROTECTED] wrote: I have a table statement that returns the following: [10.839,10.841] (10.841,10.843] (10.843,10.846] (10.846,10.848] (10.848,10.85] 0 0 0 0 1 (10.85,10.852] (10.852,10.854] (10.854,10.857] (10.857,10.859] (10.859,10.861] 0 0 0 0 0 What I want to do is get the upper bound value into a vector for each of the 10-buckets. (e.g. v-c(10.841,10.843...) I'm totally clueless on how to proceed. Any help appreciated. TIA, Joe [[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] Plotting using 'if' statements
Try this: plot(January[January[,3] 0, 3:4]) On Thu, Sep 4, 2008 at 1:28 PM, Steve Murray [EMAIL PROTECTED] wrote: Dear all, I have a dataset of four columns, and I wish to plot (as a scatter graph) the values of the third column where the values are greater than zero, and the fourth column. I tried doing this via the plot command itself, but got into a bit of a mess (resulting in errors!). My dataframe is called 'January': plot(January[3(0):4]) Error: unexpected '' in plot(January[3( After a few variations on this, I thought I'd try making a new object which includes all values from the third column of January0 (to plot in a separate step) as follows: JanFilter - January[3]0 No error here. However, when I display the 'values' of JanFilter, it shows that instead of keeping the numerical values, the above operation simply displays the results of the logical test: head(JanFilter) Value [1,] FALSE [2,]TRUE [3,]TRUE [4,]TRUE [5,]TRUE [6,]TRUE This is obviously no good for plotting the numerical values on axes! So my question is, how do I perform 'if' statements in order to filter out various parts of a dataset, for plotting on a graph. Many thanks, Steve __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] saving an object
Alternatively you can use saveObject() and loadObject() of R.utils - that will not hardwire the name of the loaded object avoiding name conflicts, e.g. library(R.utils); foo - 1:10; saveObject(foo, file=foo.RData); bar - loadObject(foo.RData); /HB On Thu, Sep 4, 2008 at 8:44 AM, Williams, Robin [EMAIL PROTECTED] wrote: ?save I assume you can connect to it from within the script, presumably by supplying the path of the object to the appropriate argument in your script. I'm no expert though. HTH, Robin Williams Met Office summer intern - Health Forecasting [EMAIL PROTECTED] -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Rajasekaramya Sent: Thursday, September 04, 2008 4:36 PM To: r-help@r-project.org Subject: [R] saving an object Hi there, I have a dataset stored in an object which has very huge volume of rows.I want to reuse it for comparing with other datasets.I dont want it to reload every time i run the script.Is there a way of saving a particular loaded object in the workspace and reusing it. Kindly help me. Ramya Thulasingam -- View this message in context: http://www.nabble.com/saving-an-object-tp19313276p19313276.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. __ 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] Plotting using 'if' statements
Ah that's great, thank you very much. As a follow-on, in the same format, how would I plot where column 3 is greater than 0 *but also less than 2*? Once again, any help is much appreciated. Thanks, Steve Date: Thu, 4 Sep 2008 13:39:12 -0300 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] Plotting using 'if' statements CC: r-help@r-project.org Try this: plot(January[January[,3] 0, 3:4]) On Thu, Sep 4, 2008 at 1:28 PM, Steve Murray wrote: Dear all, I have a dataset of four columns, and I wish to plot (as a scatter graph) the values of the third column where the values are greater than zero, and the fourth column. I tried doing this via the plot command itself, but got into a bit of a mess (resulting in errors!). My dataframe is called 'January': plot(January[3(0):4]) Error: unexpected '' in plot(January[3( After a few variations on this, I thought I'd try making a new object which includes all values from the third column of January0 (to plot in a separate step) as follows: JanFilter - January[3]0 No error here. However, when I display the 'values' of JanFilter, it shows that instead of keeping the numerical values, the above operation simply displays the results of the logical test: head(JanFilter) Value [1,] FALSE [2,]TRUE [3,]TRUE [4,]TRUE [5,]TRUE [6,]TRUE [[elided Hotmail spam]] So my question is, how do I perform 'if' statements in order to filter out various parts of a dataset, for plotting on a graph. Many thanks, Steve __ R-help@r-project.org mailing list PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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] Plotting using 'if' statements
Use '': plot(January[January[,3] 0 January[,3] 2, 3:4]) On Thu, Sep 4, 2008 at 1:47 PM, Steve Murray [EMAIL PROTECTED] wrote: Ah that's great, thank you very much. As a follow-on, in the same format, how would I plot where column 3 is greater than 0 *but also less than 2*? Once again, any help is much appreciated. Thanks, Steve Date: Thu, 4 Sep 2008 13:39:12 -0300 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] Subject: Re: [R] Plotting using 'if' statements CC: r-help@r-project.org Try this: plot(January[January[,3] 0, 3:4]) On Thu, Sep 4, 2008 at 1:28 PM, Steve Murray wrote: Dear all, I have a dataset of four columns, and I wish to plot (as a scatter graph) the values of the third column where the values are greater than zero, and the fourth column. I tried doing this via the plot command itself, but got into a bit of a mess (resulting in errors!). My dataframe is called 'January': plot(January[3(0):4]) Error: unexpected '' in plot(January[3( After a few variations on this, I thought I'd try making a new object which includes all values from the third column of January0 (to plot in a separate step) as follows: JanFilter - January[3]0 No error here. However, when I display the 'values' of JanFilter, it shows that instead of keeping the numerical values, the above operation simply displays the results of the logical test: head(JanFilter) Value [1,] FALSE [2,]TRUE [3,]TRUE [4,]TRUE [5,]TRUE [6,]TRUE This is obviously no good for plotting the numerical values on axes! So my question is, how do I perform 'if' statements in order to filter out various parts of a dataset, for plotting on a graph. Many thanks, Steve __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O _ Make a mini you and download it into Windows Live Messenger http://clk.atdmt.com/UKM/go/111354029/direct/01/ -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] R base 2.7.2 packaged for Mandriva 2008.1 x86_64: anyone interested?
I have packaged R base 2.7.2 for Mandriva 2008.1 x86_64 who should I send it to so that it can be made available to everybody? It is my first attempt and it works well on my computer, but it will need some testing. Best, -- Corrado Topi Global Climate Change Biodiversity Indicators Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-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.
[R] Interpolation Function f(y)
Hello friends!!! I have a list of values called y. The list is y=c(221.0, 212.0, 206.0, 202.7, 198.4, 195.1, 192.2, 189.7, 187.6, 185.8); y is f(x) and x=c(10,20,30,40,50,60,70,80,90,100). I only have x and y. I don´t know f(x). I would like interpolate f(x) to obtain other values as f(15), f(25), f(35) and if it was possible obtain f(110), f(120). is there any function that allow me obtain this values?? Thank you very much, Luismi -- View this message in context: http://www.nabble.com/Interpolation-Function-f%28y%29-tp19314985p19314985.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] Interpolation Function f(y)
I think that you can use the splinefun function: f - splinefun(x, y) f(15) On Thu, Sep 4, 2008 at 1:52 PM, ermimi [EMAIL PROTECTED] wrote: Hello friends!!! I have a list of values called y. The list is y=c(221.0, 212.0, 206.0, 202.7, 198.4, 195.1, 192.2, 189.7, 187.6, 185.8); y is f(x) and x=c(10,20,30,40,50,60,70,80,90,100). I only have x and y. I don´t know f(x). I would like interpolate f(x) to obtain other values as f(15), f(25), f(35) and if it was possible obtain f(110), f(120). is there any function that allow me obtain this values?? Thank you very much, Luismi -- View this message in context: http://www.nabble.com/Interpolation-Function-f%28y%29-tp19314985p19314985.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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] Interpolation Function f(y)
Thank you very much, You have helped me to resolve the problem. Thank you!! A greetings, Luismi Henrique Dallazuanna wrote: I think that you can use the splinefun function: f - splinefun(x, y) f(15) On Thu, Sep 4, 2008 at 1:52 PM, ermimi [EMAIL PROTECTED] wrote: Hello friends!!! I have a list of values called y. The list is y=c(221.0, 212.0, 206.0, 202.7, 198.4, 195.1, 192.2, 189.7, 187.6, 185.8); y is f(x) and x=c(10,20,30,40,50,60,70,80,90,100). I only have x and y. I don´t know f(x). I would like interpolate f(x) to obtain other values as f(15), f(25), f(35) and if it was possible obtain f(110), f(120). is there any function that allow me obtain this values?? Thank you very much, Luismi -- View this message in context: http://www.nabble.com/Interpolation-Function-f%28y%29-tp19314985p19314985.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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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. -- View this message in context: http://www.nabble.com/Interpolation-Function-f%28y%29-tp19314985p19315377.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] Hazard Rate Confidence Intervals
Hello, Is it possible to create confidence intervals for hazard rates? Â I'm creating two muhaz objects: haz1 - muhaz(NumDaysCustomer[cRV==true],status[cRV==true]) haz2 - muhaz(NumDaysCustomer[cRV==false],status[cRV==false]) and plotting them. Â There are many, many more observations in the cohort cRV==false than ==true. Â And, there are many more observations with lifetimes in the middle of the range than at the ends. Â I suspect that this is a common phenomenon. When I plot the two hazard rate curves, haz1 looks very different than haz2, but I'd like to see if it's different enough. Â How can I go about creating confidence intervals and plotting them? Thanks, Alan -- Alan Cox Director, User Experience iContact, Corp. p 919.459.1038 f 919.287.2475 [[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] help on jarque test
giov biowoman at libero.it writes: Hi all, I used the function jarque.test (in the moments package) on my data set and I obtained something like this: Jarque-Bera Normality Test data: x JB = 4.8381, p-value = 0.089 alternative hypothesis: greater or Jarque-Bera Normality Test data: x JB = 2.6018, p-value = 0.2723 alternative hypothesis: greater I cannot understand this. Please, someone can help me? You need to give us more information. What don't you understand? Depending on your precise question, you may need to look up information about the Jarque-Bera test (a reference is given in the help file), or ask a local consultant more about the statistical issues involved, or perhaps read more about statistical hypothesis testing in general ... ? Generally speaking, neither of the examples you show above gives strong indication of non-normality. In the first case, there is a 9% chance under the null hypothesis that the test statistic would be as greater than or greater than that observed in the data; in the second case that probability is 27%. 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.
[R] Error with update.packages()
R-helpers: I just updated from R 7.0 to R 7.2.2 today. I am using MAC OS X version 10.5.4 on a Macbook to run R. sessionInfo() R version 2.7.2 (2008-08-25) i386-apple-darwin8.11.1 locale: en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.7.2 When I tried to update my packages, I got the following messages: update.packages() Amelia : Version 1.1-30 installed in /Library/Frameworks/R.framework/ Resources/library Version 1.1-33 available at http://www.ibiblio.org/pub/languages/R/ CRAN Update (y/N/c)? y empiricalBayes : Version 2.0 installed in /Library/Frameworks/R.framework/Resources/ library Version 2.1 available at http://www.ibiblio.org/pub/languages/R/CRAN Update (y/N/c)? y Matrix : Version 0.999375-9 installed in /Library/Frameworks/R.framework/ Resources/library Version 0.999375-11 available at http://www.ibiblio.org/pub/languages/R/CRAN Update (y/N/c)? y ...snip, all seems well also installing the dependencies sandwich, strucchange, vcd, rrcov, spam trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/sandwich_2.1-0.tgz' Content type 'application/x-gzip' length 678518 bytes (662 Kb) opened URL == downloaded 662 Kb trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/strucchange_1.3-3.tgz' Content type 'application/x-gzip' length 777150 bytes (758 Kb) opened URL == downloaded 758 Kb ...snip, the errors begin trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/rrcov_0.4-06.tgz' Content type 'application/x-gzip' length 365965 bytes (357 Kb) opened URL == downloaded 304 Kb trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/spam_0.15-0.tgz' Error in download.file(url, destfile, method, mode = wb, ...) : cannot open URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/spam_0.15-0.tgz' In addition: Warning messages: 1: In download.file(url, destfile, method, mode = wb, ...) : downloaded length 312279 != reported length 365965 2: In download.file(url, destfile, method, mode = wb, ...) : unable to connect to 'mirrors.ibiblio.org' on port 80. Warning in download.packages(p0, destdir = tmpd, available = available, : download of package 'spam' failed trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/Amelia_1.1-33.tgz' Error in download.file(url, destfile, method, mode = wb, ...) : cannot open URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/Amelia_1.1-33.tgz' In addition: Warning message: In download.file(url, destfile, method, mode = wb, ...) : unable to connect to 'mirrors.ibiblio.org' on port 80. Warning in download.packages(p0, destdir = tmpd, available = available, : download of package 'Amelia' failed ...snip, the errors continue In addition: Warning message: In download.file(url, destfile, method, mode = wb, ...) : unable to connect to 'mirrors.ibiblio.org' on port 80. Warning in download.packages(p0, destdir = tmpd, available = available, : download of package 'zoo' failed gzip: stdin: unexpected end of file tar: Unexpected EOF in archive tar: Unexpected EOF in archive tar: Error is not recoverable: exiting now Error in sprintf(gettext(fmt, domain = domain), ...) : argument is missing, with no default The first part of the procedure seemed to work, then came page after page of error messages. Can someone tell me what is going on here and how I can fix the problem? I found a similar report on R-help that was never resolved and also involved a MAC (http://tolstoy.newcastle.edu.au/R/e4/help/08/06/15556.html ). Could this be a MAC issue? Thanks very much for your time. Brant Inman [[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] Stepwise
Peter Flom peterf at brainscope.com writes: Robin Williams wrote Is there any facility in R to perform a stepwise process on a model, which will remove any highly-correlated explanatory variables? I am told there is in SPSS. I have a large number of variables (some correlated), which I would like to just chuck in to a model and perform stepwise and see what comes out the other end, to give me an idea perhaps as to which variables I should focus on. Thanks for any help / suggestions. Stepwise is a bad method of selecting variables. Far better methods are LASSO and LAR (least angle regression), available in the LARS package and the LASSO2 package. However, while both these methods are good, neither is a substitute for substantive knowledge. Also, the key thing is not so much whether variables are correlated, but whether they are co-linear, which is different. If you have a great many variables, then you can have a high degree of colinearity even with no high pairwise correlations. I've not done this in R, but RSiteSearch(collinearity, restrict = 'functions') yields 34 hits. HTH Peter Another suggestion would be to do PCA on the predictor variables. And to read Frank Harrell's book on _Regression modeling strategies_. cheers 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.
[R] combning rows
I do have a txt file where each row is a record and the first element of each record is an id for an individual. I am looking into combing the records into one row if the id is the same and save as a txt file. Notice that the rows may not have the same length in the result file that I want to create . Can I do this using R. I appreciate your help. Thanks -- View this message in context: http://www.nabble.com/combning-rows-tp19316079p19316079.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] Error in table lookup
I am trying to lookup a value in 1 of 10 loaded two column-data sets (Bins)
by displaying the value of the second column based on the value of the
first. For instance in
Bin1_Acres Bin1_parprobBin1_TAZ
[1,] 0.0004420.978 356
[2,] 0.0004530.954 356
[3,] 0.0005830.925 366
[4,] 0.0006350.893 403
[5,] 0.0007560.854 358
[6,] 0.0007740.815 530
[7,] 0.0008130.773 405
[8,] 0.0009700.724 576
[9,] 0.0010220.672 569
[10,] 0.0010660.618 620
I would like to display the column value on Bin1_parprob based on the
closest match to Bin1_acres. So if the value i am referenceing in column 1
(Bin1_Acres) is .00 the value outputted would be .672. I keep
getting a numeric(0) error with the code i am using(see below). I think the
issue is i usually dont have an exact match and i am using a = sign which
may be causing the problem. I need the closest match and it needs to be
larger, not smaller if not exact match is found( which is very common).
#Test value for Vacant acres in TAZ
TAZDetermine = 24
#Test value for and development size
Dev_Size= 3.5
#Determines Bin number based on vacant acres in TAZ
BinSize=function(Dev_Size,BinNumer){
if (TAZDetermine=3.999)
BinNumber=1
if (TAZDetermine=4:6.999)
(BinNumber=2)
if (TAZDetermine =10:16.999)
(BinNumber=3)
if (TAZDetermine=17:27.999)
(BinNumber=4)
if (TAZDetermine=28:49.999)
(BinNumber=5)
if (TAZDetermine=50:90.999)
(BinNumber=6)
if (TAZDetermine=91:150.999)
(BinNumber=7)
if (TAZDetermine=151:340.999)
(BinNumber=8)
if (TAZDetermine=341:650.999)
(BinNumber=9)
if (TAZDetermine=651:3000)
(BinNumber=10)
BinNumber
}
#so in this case Bin 4 is selected
#Based on previously selected bin, display second column
value(Bin1_parprob). Selected value in column 1 may be slightly larger but
closest match is desirable.
if (BinNumber==1)
(Loc_Prop=Bin1Main.data[Bin1Main.data$Bin1_Acres =Dev_Size,1])
if (BinNumber==2)
(Loc_Prop=Bin2Main.data[Bin2Main.data$Bin2_Acres=Dev_Size,1])
if (BinNumber==3)
(Loc_Prop=Bin3Main.data[Bin3Main.data$Bin3_Acres=Dev_Size,1])
if (BinNumber==4)
(Loc_Prop=Bin4Main.data[Bin4Main.data$Bin4_Acres=Dev_Size,1])
if (BinNumber==5)
(Loc_Prop=Bin5Main.data[Bin5Main.data$Bin5_Acres=Dev_Size,1])
if (BinNumber==6)
(Loc_Prop=Bin6Main.data[Bin6Main.data$Bin6_Acres=Dev_Size,1])
if (BinNumber==7)
(Loc_Prop=Bin7Main.data[Bin7Main.data$Bin7_Acres=Dev_Size,1])
if (BinNumber==8)
(Loc_Prop=Bin8Main.data[Bin8Main.data$Bin8_Acres=Dev_Size,1])
if (BinNumber==9)
(Loc_Prop=Bin9Main.data[Bin9Main.data$Bin9_Acres=Dev_Size,1])
if (BinNumber==10)
(Loc_Prop=Bin10Main.data[Bin10Main.data$Bin10_Acres=Dev_Size,1])
I hope this question is clear. I have tried a number of different lines of
code but get the same error (Numeric (0))
Cheers,
JR
--
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__
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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] Stepwise
Also consider the redun function in the Hmisc package, which does not use the response variable but uses flexible nonlinear additive models to predict each predictor variable from all the others, using a stepwise procedure in a formal redundancy analysis. Frank Ben Bolker wrote: Peter Flom peterf at brainscope.com writes: Robin Williams wrote Is there any facility in R to perform a stepwise process on a model, which will remove any highly-correlated explanatory variables? I am told there is in SPSS. I have a large number of variables (some correlated), which I would like to just chuck in to a model and perform stepwise and see what comes out the other end, to give me an idea perhaps as to which variables I should focus on. Thanks for any help / suggestions. Stepwise is a bad method of selecting variables. Far better methods are LASSO and LAR (least angle regression), available in the LARS package and the LASSO2 package. However, while both these methods are good, neither is a substitute for substantive knowledge. Also, the key thing is not so much whether variables are correlated, but whether they are co-linear, which is different. If you have a great many variables, then you can have a high degree of colinearity even with no high pairwise correlations. I've not done this in R, but RSiteSearch(collinearity, restrict = 'functions') yields 34 hits. HTH Peter Another suggestion would be to do PCA on the predictor variables. And to read Frank Harrell's book on _Regression modeling strategies_. cheers Ben Bolker Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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] Error in table lookup
Your IF statement (if (x = 3:6)) does not work as you probably expect
it since it does not test for the range. You probably want to use
cut to convert a value to a bucket in a range:
x - 3.5
cut(x, breaks=seq(-3,5,.2), labels=FALSE)
[1] 33
seq(-3,5,.2)
[1] -3.0 -2.8 -2.6 -2.4 -2.2 -2.0 -1.8 -1.6 -1.4 -1.2 -1.0 -0.8 -0.6
-0.4 -0.2 0.0 0.2
[18] 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0 2.2 2.4 2.6 2.8
3.0 3.2 3.4 3.6
[35] 3.8 4.0 4.2 4.4 4.6 4.8 5.0
# it is in the 33rd group
On Thu, Sep 4, 2008 at 1:52 PM, PDXRugger [EMAIL PROTECTED] wrote:
I am trying to lookup a value in 1 of 10 loaded two column-data sets (Bins)
by displaying the value of the second column based on the value of the
first. For instance in
Bin1_Acres Bin1_parprobBin1_TAZ
[1,] 0.0004420.978 356
[2,] 0.0004530.954 356
[3,] 0.0005830.925 366
[4,] 0.0006350.893 403
[5,] 0.0007560.854 358
[6,] 0.0007740.815 530
[7,] 0.0008130.773 405
[8,] 0.0009700.724 576
[9,] 0.0010220.672 569
[10,] 0.0010660.618 620
I would like to display the column value on Bin1_parprob based on the
closest match to Bin1_acres. So if the value i am referenceing in column 1
(Bin1_Acres) is .00 the value outputted would be .672. I keep
getting a numeric(0) error with the code i am using(see below). I think the
issue is i usually dont have an exact match and i am using a = sign which
may be causing the problem. I need the closest match and it needs to be
larger, not smaller if not exact match is found( which is very common).
#Test value for Vacant acres in TAZ
TAZDetermine = 24
#Test value for and development size
Dev_Size= 3.5
#Determines Bin number based on vacant acres in TAZ
BinSize=function(Dev_Size,BinNumer){
if (TAZDetermine=3.999)
BinNumber=1
if (TAZDetermine=4:6.999)
(BinNumber=2)
if (TAZDetermine =10:16.999)
(BinNumber=3)
if (TAZDetermine=17:27.999)
(BinNumber=4)
if (TAZDetermine=28:49.999)
(BinNumber=5)
if (TAZDetermine=50:90.999)
(BinNumber=6)
if (TAZDetermine=91:150.999)
(BinNumber=7)
if (TAZDetermine=151:340.999)
(BinNumber=8)
if (TAZDetermine=341:650.999)
(BinNumber=9)
if (TAZDetermine=651:3000)
(BinNumber=10)
BinNumber
}
#so in this case Bin 4 is selected
#Based on previously selected bin, display second column
value(Bin1_parprob). Selected value in column 1 may be slightly larger but
closest match is desirable.
if (BinNumber==1)
(Loc_Prop=Bin1Main.data[Bin1Main.data$Bin1_Acres =Dev_Size,1])
if (BinNumber==2)
(Loc_Prop=Bin2Main.data[Bin2Main.data$Bin2_Acres=Dev_Size,1])
if (BinNumber==3)
(Loc_Prop=Bin3Main.data[Bin3Main.data$Bin3_Acres=Dev_Size,1])
if (BinNumber==4)
(Loc_Prop=Bin4Main.data[Bin4Main.data$Bin4_Acres=Dev_Size,1])
if (BinNumber==5)
(Loc_Prop=Bin5Main.data[Bin5Main.data$Bin5_Acres=Dev_Size,1])
if (BinNumber==6)
(Loc_Prop=Bin6Main.data[Bin6Main.data$Bin6_Acres=Dev_Size,1])
if (BinNumber==7)
(Loc_Prop=Bin7Main.data[Bin7Main.data$Bin7_Acres=Dev_Size,1])
if (BinNumber==8)
(Loc_Prop=Bin8Main.data[Bin8Main.data$Bin8_Acres=Dev_Size,1])
if (BinNumber==9)
(Loc_Prop=Bin9Main.data[Bin9Main.data$Bin9_Acres=Dev_Size,1])
if (BinNumber==10)
(Loc_Prop=Bin10Main.data[Bin10Main.data$Bin10_Acres=Dev_Size,1])
I hope this question is clear. I have tried a number of different lines of
code but get the same error (Numeric (0))
Cheers,
JR
--
View this message in context:
http://www.nabble.com/Error-in-table-lookup-tp19316307p19316307.html
Sent from the R help mailing list archive at Nabble.com.
__
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.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
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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] combning rows
Read your data in, parse off the ID, split the records (using 'split') and then combine the like values. Since you did not read the posting guide, it is hard to give explicit help. On Thu, Sep 4, 2008 at 1:42 PM, kayj [EMAIL PROTECTED] wrote: I do have a txt file where each row is a record and the first element of each record is an id for an individual. I am looking into combing the records into one row if the id is the same and save as a txt file. Notice that the rows may not have the same length in the result file that I want to create . Can I do this using R. I appreciate your help. Thanks -- View this message in context: http://www.nabble.com/combning-rows-tp19316079p19316079.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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that 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] Possible R graphics devices
Hello all, I've been working on a new R Graphics device that targets Adobe's Shockwave Flash format (SWF for short). It uses http://www.libming.org/ on the backend. Here are some example outputs so far: http://160.129.129.41/~hornerj/plots/ Once you click on the above, choose a directory like 'smooth' and then click on test.html to see a side-by-side comparison of the swf file with the png equivalent. Fonts aren't working yet (but they will), and libming doesn't support dashed lines either, so there are some areas upon which to improve. My goal is to support limited interactivity in the form of popup labels for key data points. Maybe someone has some other ideas? Also, there's a fascinating new project that aims to create interactive graphics within the HTML canvas tag using the Processing.js javascript library. It's quite eye-opening: http://ejohn.org/blog/processingjs I believe an R graphics device could target javascript as well! Thoughts? Jeff -- http://biostat.mc.vanderbilt.edu/JeffreyHorner __ 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] Plotting the complex fft in 3D?
Hello, I'm new to R (using it since about two weeks), but absolutely a fan of it from the beginning on. :-) Best tool for working with data I found. :-) I tried using the fft() and other funcitons for analysing time series. What I would be glad to have, would be a convenient way to display the complex result of a fft in a way, that real and imaginary parts each use an axis for themselves, and the index of the resulting values use the third axe. When displaying this as a 3D-2D picture, it also would be nice, to change the view, like it can be done with persp(). Is there already a package or script for preparing the data of an fft to be displayed in this way? Ciao, Oliver __ 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-SIG-Mac] Error with update.packages()
Try another mirror: this is not the first time we've seen problems at www.ibiblio.org. On Thu, 4 Sep 2008, Brant Inman wrote: R-helpers: I just updated from R 7.0 to R 7.2.2 today. I am using MAC OS X version 10.5.4 on a Macbook to run R. sessionInfo() R version 2.7.2 (2008-08-25) i386-apple-darwin8.11.1 locale: en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.7.2 When I tried to update my packages, I got the following messages: update.packages() Amelia : Version 1.1-30 installed in /Library/Frameworks/R.framework/ Resources/library Version 1.1-33 available at http://www.ibiblio.org/pub/languages/R/ CRAN Update (y/N/c)? y empiricalBayes : Version 2.0 installed in /Library/Frameworks/R.framework/Resources/ library Version 2.1 available at http://www.ibiblio.org/pub/languages/R/CRAN Update (y/N/c)? y Matrix : Version 0.999375-9 installed in /Library/Frameworks/R.framework/ Resources/library Version 0.999375-11 available at http://www.ibiblio.org/pub/languages/R/CRAN Update (y/N/c)? y ...snip, all seems well also installing the dependencies ?sandwich?, ?strucchange?, ?vcd?, ?rrcov?, ?spam? trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/sandwich_2.1-0.tgz' Content type 'application/x-gzip' length 678518 bytes (662 Kb) opened URL == downloaded 662 Kb trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/strucchange_1.3-3.tgz' Content type 'application/x-gzip' length 777150 bytes (758 Kb) opened URL == downloaded 758 Kb ...snip, the errors begin trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/rrcov_0.4-06.tgz' Content type 'application/x-gzip' length 365965 bytes (357 Kb) opened URL == downloaded 304 Kb trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/spam_0.15-0.tgz' Error in download.file(url, destfile, method, mode = wb, ...) : cannot open URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/spam_0.15-0.tgz' In addition: Warning messages: 1: In download.file(url, destfile, method, mode = wb, ...) : downloaded length 312279 != reported length 365965 2: In download.file(url, destfile, method, mode = wb, ...) : unable to connect to 'mirrors.ibiblio.org' on port 80. Warning in download.packages(p0, destdir = tmpd, available = available, : download of package 'spam' failed trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/Amelia_1.1-33.tgz' Error in download.file(url, destfile, method, mode = wb, ...) : cannot open URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/Amelia_1.1-33.tgz' In addition: Warning message: In download.file(url, destfile, method, mode = wb, ...) : unable to connect to 'mirrors.ibiblio.org' on port 80. Warning in download.packages(p0, destdir = tmpd, available = available, : download of package 'Amelia' failed ...snip, the errors continue In addition: Warning message: In download.file(url, destfile, method, mode = wb, ...) : unable to connect to 'mirrors.ibiblio.org' on port 80. Warning in download.packages(p0, destdir = tmpd, available = available, : download of package 'zoo' failed gzip: stdin: unexpected end of file tar: Unexpected EOF in archive tar: Unexpected EOF in archive tar: Error is not recoverable: exiting now Error in sprintf(gettext(fmt, domain = domain), ...) : argument is missing, with no default The first part of the procedure seemed to work, then came page after page of error messages. Can someone tell me what is going on here and how I can fix the problem? I found a similar report on R-help that was never resolved and also involved a MAC (http://tolstoy.newcastle.edu.au/R/e4/help/08/06/15556.html ). Could this be a MAC issue? Thanks very much for your time. Brant Inman [[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.
Re: [R] (with subject)
Dirk, Gabor, thanks for your advice, I have now tried to study these vignettes, but I must say, as an aging Gynecologist, I am facing an enormous learning curve :-)-O. This works nicely: rawData - data.frame(x = c(2008-03-01, 2008-03-21, 2008-03-23, 2008-04-08, 2008-04-20, 2008-05-10, 2008-06-20), y = c(4, 6, 8, 5, 7, 2 ,1)) rawData[,x] - as.Date(rawData[,x]) with(rawData, plot(x, y, main=Some text, type='l')) How do I make the X-Axis start on 2008-01-01, end on 2008-12-31, with tick marks for every month (ie month.abb[1:12] or something similar)? A pointer is enough, weekend is coming up and I am off :-)-O I also need to add 2 lines statements, but I reckon I can figure that out by myself :-)-O greetings, el -- Dr. Eberhard W. Lisse \/ Obstetrician Gynaecologist (Saar) [EMAIL PROTECTED] el108-ARIN / * | Telephone: +264 81 124 6733 (cell) PO Box 8421 \ / Please send DNS/NA-NiC related e-mail Bachbrecht, Namibia ;/ to [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] Plotting the complex fft in 3D?
Oliver Bandel wrote: Hello, I'm new to R (using it since about two weeks), but absolutely a fan of it from the beginning on. :-) Best tool for working with data I found. :-) I tried using the fft() and other funcitons for analysing time series. What I would be glad to have, would be a convenient way to display the complex result of a fft in a way, that real and imaginary parts each use an axis for themselves, and the index of the resulting values use the third axe. When displaying this as a 3D-2D picture, it also would be nice, to change the view, like it can be done with persp(). Is there already a package or script for preparing the data of an fft to be displayed in this way? I don't find this very enlightening, but here you go: x - rnorm(1000) f - fft(x) library(rgl) plot3d(1:length(f), Re(f), Im(f)) Duncan Murdoch __ 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] Plotting using 'if' statements
Thanks again Henrique - that's really useful to know! _ Discover Bird's Eye View now with Multimap from Live Search __ 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] adding NA to subset
Hi All- I have a data set (spdco2) spdco2 [,1] [,2] [,3] [1,]1 5.4 382.4212 [2,]2 5.1 383.0315 [3,]3 4.8 383.9520 [4,]4 4.7 384.4376 [5,]5 4.7 384.5929 [6,]6 4.4 384.8864 [7,]7 4.1 385.2156 [8,]8 3.8 385.2919 [9,]9 3.7 385.5925 [10,] 10 3.9 385.6801 I am subsetting it to output when [,2] is = 4.7 . x-subset(spdco2,spdco2[,2]=4.7) This works, but I would like to add 'NA' to the data that it does exclude based on my subset criteria. I have searched through the archives and two R books, but I cannot figure out how to add the NA. Any suggestions would be appreciated- thanks! sherri x-subset(spdco2,spdco2[,2]=4.7) __ 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-SIG-Mac] Error with update.packages()
Thank you for your suggestion. I did as you suggested and switched from USA (NC) to USA (IA). Again, the packages seemed to download OK but this time a single error message appeared: -- ...snip... == downloaded 168 Kb trying URL 'http://streaming.stat.iastate.edu/CRAN/bin/macosx/universal/contrib/2.7/zoo_1.5-4.tgz' Content type 'application/x-gzip' length 749912 bytes (732 Kb) opened URL == downloaded 732 Kb Error: cannot remove prior installation of package ‘fSeries’ -- This error appears to have stopped the package updating, since when I type update.packages() again, the whole list of packages is generated, implying that they have not been updated. Any other suggestions? Would uninstalling R and then reinstalling it help? Brant - On Sep 4, 2008, at 2:59 PM, Prof Brian Ripley wrote: Try another mirror: this is not the first time we've seen problems at www.ibiblio.org. On Thu, 4 Sep 2008, Brant Inman wrote: R-helpers: I just updated from R 7.0 to R 7.2.2 today. I am using MAC OS X version 10.5.4 on a Macbook to run R. sessionInfo() R version 2.7.2 (2008-08-25) i386-apple-darwin8.11.1 locale: en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.7.2 When I tried to update my packages, I got the following messages: update.packages() Amelia : Version 1.1-30 installed in /Library/Frameworks/R.framework/ Resources/library Version 1.1-33 available at http://www.ibiblio.org/pub/languages/R/ CRAN Update (y/N/c)? y empiricalBayes : Version 2.0 installed in /Library/Frameworks/R.framework/Resources/ library Version 2.1 available at http://www.ibiblio.org/pub/languages/R/CRAN Update (y/N/c)? y Matrix : Version 0.999375-9 installed in /Library/Frameworks/R.framework/ Resources/library Version 0.999375-11 available at http://www.ibiblio.org/pub/languages/R/CRAN Update (y/N/c)? y ...snip, all seems well also installing the dependencies ?sandwich?, ?strucchange?, ?vcd?, ?rrcov?, ?spam? trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/sandwich_2.1-0.tgz' Content type 'application/x-gzip' length 678518 bytes (662 Kb) opened URL == downloaded 662 Kb trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/strucchange_1.3-3.tgz' Content type 'application/x-gzip' length 777150 bytes (758 Kb) opened URL == downloaded 758 Kb ...snip, the errors begin trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/rrcov_0.4-06.tgz' Content type 'application/x-gzip' length 365965 bytes (357 Kb) opened URL == downloaded 304 Kb trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/spam_0.15-0.tgz' Error in download.file(url, destfile, method, mode = wb, ...) : cannot open URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/spam_0.15-0.tgz' In addition: Warning messages: 1: In download.file(url, destfile, method, mode = wb, ...) : downloaded length 312279 != reported length 365965 2: In download.file(url, destfile, method, mode = wb, ...) : unable to connect to 'mirrors.ibiblio.org' on port 80. Warning in download.packages(p0, destdir = tmpd, available = available, : download of package 'spam' failed trying URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/Amelia_1.1-33.tgz' Error in download.file(url, destfile, method, mode = wb, ...) : cannot open URL 'http://www.ibiblio.org/pub/languages/R/CRAN/bin/macosx/universal/contrib/2.7/Amelia_1.1-33.tgz' In addition: Warning message: In download.file(url, destfile, method, mode = wb, ...) : unable to connect to 'mirrors.ibiblio.org' on port 80. Warning in download.packages(p0, destdir = tmpd, available = available, : download of package 'Amelia' failed ...snip, the errors continue In addition: Warning message: In download.file(url, destfile, method, mode = wb, ...) : unable to connect to 'mirrors.ibiblio.org' on port 80. Warning in download.packages(p0, destdir = tmpd, available = available, : download of package 'zoo' failed gzip: stdin: unexpected end of file tar: Unexpected EOF in archive tar: Unexpected EOF in archive tar: Error is not recoverable: exiting now Error in sprintf(gettext(fmt, domain = domain), ...) : argument is missing, with no default The first part of the procedure seemed to work, then came page after page of error messages. Can someone tell me what is going on here
Re: [R] adding NA to subset
Dear Sherri, Perhaps: spdco2=matrix(scan(),ncol=3,byrow=TRUE) 1 5.4 382.4212 2 5.1 383.0315 3 4.8 383.9520 4 4.7 384.4376 5 4.7 384.5929 6 4.4 384.8864 7 4.1 385.2156 8 3.8 385.2919 9 3.7 385.5925 10 3.9 385.6801 spdco2[,2]-ifelse(spdco2[,2]=4.7,spdco2[,2],NA) spdco2 HTH, Jorge On Thu, Sep 4, 2008 at 3:21 PM, Sherri Heck [EMAIL PROTECTED] wrote: Hi All- I have a data set (spdco2) spdco2 [,1] [,2] [,3] [1,]1 5.4 382.4212 [2,]2 5.1 383.0315 [3,]3 4.8 383.9520 [4,]4 4.7 384.4376 [5,]5 4.7 384.5929 [6,]6 4.4 384.8864 [7,]7 4.1 385.2156 [8,]8 3.8 385.2919 [9,]9 3.7 385.5925 [10,] 10 3.9 385.6801 I am subsetting it to output when [,2] is = 4.7 . x-subset(spdco2,spdco2[,2]=4.7) This works, but I would like to add 'NA' to the data that it does exclude based on my subset criteria. I have searched through the archives and two R books, but I cannot figure out how to add the NA. Any suggestions would be appreciated- thanks! sherri x-subset(spdco2,spdco2[,2]=4.7) __ 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. [[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] adding NA to subset
Sherri Heck wrote: Hi All- I have a data set (spdco2) spdco2 [,1] [,2] [,3] [1,]1 5.4 382.4212 [2,]2 5.1 383.0315 [3,]3 4.8 383.9520 [4,]4 4.7 384.4376 [5,]5 4.7 384.5929 [6,]6 4.4 384.8864 [7,]7 4.1 385.2156 [8,]8 3.8 385.2919 [9,]9 3.7 385.5925 [10,] 10 3.9 385.6801 I am subsetting it to output when [,2] is = 4.7 . x-subset(spdco2,spdco2[,2]=4.7) This works, but I would like to add 'NA' to the data that it does exclude based on my subset criteria. I have searched through the archives and two R books, but I cannot figure out how to add the NA. Any suggestions would be appreciated- Use is.na(): x - subset(spdco2, spdco2[,2] = 4.7 | is.na(spdco2[,2])) Make sure that the inclusion or exclusion of the row can be computed: TRUE | NA is TRUE, but FALSE | NA is NA. Duncan Murdoch Duncan Murdoch __ 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] adding NA to subset
Try this: is.na(spdco2[,2]) - which(spdco2[,2] 4.7) spdco2 On Thu, Sep 4, 2008 at 4:21 PM, Sherri Heck [EMAIL PROTECTED] wrote: Hi All- I have a data set (spdco2) spdco2 [,1] [,2] [,3] [1,]1 5.4 382.4212 [2,]2 5.1 383.0315 [3,]3 4.8 383.9520 [4,]4 4.7 384.4376 [5,]5 4.7 384.5929 [6,]6 4.4 384.8864 [7,]7 4.1 385.2156 [8,]8 3.8 385.2919 [9,]9 3.7 385.5925 [10,] 10 3.9 385.6801 I am subsetting it to output when [,2] is = 4.7 . x-subset(spdco2,spdco2[,2]=4.7) This works, but I would like to add 'NA' to the data that it does exclude based on my subset criteria. I have searched through the archives and two R books, but I cannot figure out how to add the NA. Any suggestions would be appreciated- thanks! sherri x-subset(spdco2,spdco2[,2]=4.7) __ 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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[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] adding NA to subset
Fantastic! Each suggestion worked beautifully. Thanks so much. Henrique Dallazuanna wrote: Try this: is.na http://is.na(spdco2[,2]) - which(spdco2[,2] 4.7) spdco2 On Thu, Sep 4, 2008 at 4:21 PM, Sherri Heck [EMAIL PROTECTED] mailto:[EMAIL PROTECTED] wrote: Hi All- I have a data set (spdco2) spdco2 [,1] [,2] [,3] [1,]1 5.4 382.4212 [2,]2 5.1 383.0315 [3,]3 4.8 383.9520 [4,]4 4.7 384.4376 [5,]5 4.7 384.5929 [6,]6 4.4 384.8864 [7,]7 4.1 385.2156 [8,]8 3.8 385.2919 [9,]9 3.7 385.5925 [10,] 10 3.9 385.6801 I am subsetting it to output when [,2] is = 4.7 . x-subset(spdco2,spdco2[,2]=4.7) This works, but I would like to add 'NA' to the data that it does exclude based on my subset criteria. I have searched through the archives and two R books, but I cannot figure out how to add the NA. Any suggestions would be appreciated- thanks! sherri x-subset(spdco2,spdco2[,2]=4.7) __ R-help@r-project.org mailto: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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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] (with subject)
Is this what you want? library(ggplot2) rawData - data.frame(Date = c(2008-03-01, 2008-03-21, 2008-03-23, 2008-04-08, 2008-04-20, 2008-05-10, 2008-06-20), y = c(4, 6,8, 5, 7, 2 ,1)) rawData$Date - as.Date(rawData$Date) qplot(Date,y,data=rawData,geom=line,xlim=c(as.Date(2008-1-1),as.Date(2008-12-1))) --- On Thu, 9/4/08, Dr Eberhard W Lisse [EMAIL PROTECTED] wrote: From: Dr Eberhard W Lisse [EMAIL PROTECTED] Subject: Re: [R] (with subject) To: r-help@r-project.org Cc: Dr Eberhard W Lisse [EMAIL PROTECTED] Date: Thursday, September 4, 2008, 12:15 PM Dirk, Gabor, thanks for your advice, I have now tried to study these vignettes, but I must say, as an aging Gynecologist, I am facing an enormous learning curve :-)-O. This works nicely: rawData - data.frame(x = c(2008-03-01, 2008-03-21, 2008-03-23, 2008-04-08, 2008-04-20, 2008-05-10, 2008-06-20), y = c(4, 6, 8, 5, 7, 2 ,1)) rawData[,x] - as.Date(rawData[,x]) with(rawData, plot(x, y, main=Some text, type='l')) How do I make the X-Axis start on 2008-01-01, end on 2008-12-31, with tick marks for every month (ie month.abb[1:12] or something similar)? A pointer is enough, weekend is coming up and I am off :-)-O I also need to add 2 lines statements, but I reckon I can figure that out by myself :-)-O greetings, el -- Dr. Eberhard W. Lisse \/ Obstetrician Gynaecologist (Saar) [EMAIL PROTECTED] el108-ARIN / * | Telephone: +264 81 124 6733 (cell) PO Box 8421 \ / Please send DNS/NA-NiC related e-mail Bachbrecht, Namibia ;/ to [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-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] showing the image
Hi everyone, I have a matrix containing color values of an image same as following, [1,] #44 #44 #434343 #404040 #3D3D3D #3D3D3D #3E3E3E [2,] #414141 #414141 #414141 #404040 #3F3F3F #3F3F3F #3F3F3F [3,] #3E3E3E #3E3E3E #3F3F3F #404040 #404040 #404040 #404040 [4,] #3E3E3E #3D3D3D #3E3E3E #404040 #414141 #414141 #404040 [5,] #3E3E3E #3E3E3E #3F3F3F #3F3F3F #3F3F3F #3F3F3F #3E3E3E [6,] #3E3E3E #3F3F3F #3E3E3E #3D3D3D #3B3B3B #3C3C3C #3B3B3B I wonder if there is any function that helps me to show the image. Thanks, Rostam [[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] Possible R graphics devices
I think being able to say lable your points with a third column of data would be cool (which I am sure is already do-able), and also to be able to hover over a point and find out the value. On a huge wishlist that I don't have a clue how to implement- a zoom window function close to kaliedagraphs would make me very happy. On Thu, Sep 4, 2008 at 2:41 PM, Jeffrey Horner [EMAIL PROTECTED] wrote: Hello all, I've been working on a new R Graphics device that targets Adobe's Shockwave Flash format (SWF for short). It uses http://www.libming.org/ on the backend. Here are some example outputs so far: http://160.129.129.41/~hornerj/plots/ Once you click on the above, choose a directory like 'smooth' and then click on test.html to see a side-by-side comparison of the swf file with the png equivalent. Fonts aren't working yet (but they will), and libming doesn't support dashed lines either, so there are some areas upon which to improve. My goal is to support limited interactivity in the form of popup labels for key data points. Maybe someone has some other ideas? Also, there's a fascinating new project that aims to create interactive graphics within the HTML canvas tag using the Processing.js javascript library. It's quite eye-opening: http://ejohn.org/blog/processingjs I believe an R graphics device could target javascript as well! Thoughts? Jeff -- http://biostat.mc.vanderbilt.edu/JeffreyHorner __ 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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy 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 http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plotting the complex fft in 3D?
Zitat von Duncan Murdoch [EMAIL PROTECTED]: Oliver Bandel wrote: Hello, I'm new to R (using it since about two weeks), but absolutely a fan of it from the beginning on. :-) Best tool for working with data I found. :-) I tried using the fft() and other funcitons for analysing time series. What I would be glad to have, would be a convenient way to display the complex result of a fft in a way, that real and imaginary parts each use an axis for themselves, and the index of the resulting values use the third axe. When displaying this as a 3D-2D picture, it also would be nice, to change the view, like it can be done with persp(). Is there already a package or script for preparing the data of an fft to be displayed in this way? I don't find this very enlightening, but here you go: x - rnorm(1000) f - fft(x) library(rgl) plot3d(1:length(f), Re(f), Im(f)) [...] Ok, this is a starting point. :-) It would be enlightening, if you have a timeseries that is not noise only, and if the plot would not use dots. So, using a timeseries that is derived from some data could be very interesting. But rnorm creates noise only, and not a deterministic signal. So the resulting fft values look quite boring ;-) Instead of dots, a line from the index-axis at Re=0, Im=0 to the value of the fft at that index should be drawn. BTW: how to change the perspective? I did not found an angle-parameter for the plot3d()-function. Ciao, Oliver __ 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] isoMDS and dist
I am starting with a matrix in which rows are vegetation plots and columns are various characteristics including ID# and elevation. I removed elevation and ID columns to avoid having those characteristics influence the distances between points which I calculated using the dist command. The resulting distance file was then used in isoMDS. What I want to know is whether I can reattach the ID and elevation onto the point location information given from isoMDS? or is it possible that the order given in the isoMDS output is not the same as the order of the original data used to calculate distances? Thanks, Carrie Pucko University of Vermont __ 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] using complete.cases() with nested factors
Hello, This maybe a newbie question. I have a dataframe that looks like the sample at the bottom of the email. I have monthly precipitation data from several sites over several years. For each site, I need to extract years that have a complete series of 12 monthly precipitation values, while excluding that year for sites with incomplete data. I can't figure out how to do this gracefully (i.e. without a silly for loop). Any help will be appreciate, thanks! Andrew SiteIDyearmonthprecip(mm) 6700901941jan2998 6700901941feb1299 6700901941mar1007 6700901941apr354 6700901941may88 6700901941jun156 6700901941jul8 6700901941aug4 6700901941sep8 6700901941oct58 6700901941nov397 6700901941dec248 6700901942janNA 6700901942feb380 6700901942mar797 6700901942apr142 6700901942may43 6700901942jun14 6700901942jul70 6700901942aug51 6700901942sep0 6700901942oct10 6700901942nov235 6700901942dec405 -- Andrew [[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] Plotting the complex fft in 3D?
On 04/09/2008 4:44 PM, Oliver Bandel wrote: Zitat von Duncan Murdoch [EMAIL PROTECTED]: Oliver Bandel wrote: Hello, I'm new to R (using it since about two weeks), but absolutely a fan of it from the beginning on. :-) Best tool for working with data I found. :-) I tried using the fft() and other funcitons for analysing time series. What I would be glad to have, would be a convenient way to display the complex result of a fft in a way, that real and imaginary parts each use an axis for themselves, and the index of the resulting values use the third axe. When displaying this as a 3D-2D picture, it also would be nice, to change the view, like it can be done with persp(). Is there already a package or script for preparing the data of an fft to be displayed in this way? I don't find this very enlightening, but here you go: x - rnorm(1000) f - fft(x) library(rgl) plot3d(1:length(f), Re(f), Im(f)) [...] Ok, this is a starting point. :-) It would be enlightening, if you have a timeseries that is not noise only, and if the plot would not use dots. So, using a timeseries that is derived from some data could be very interesting. But rnorm creates noise only, and not a deterministic signal. So the resulting fft values look quite boring ;-) Instead of dots, a line from the index-axis at Re=0, Im=0 to the value of the fft at that index should be drawn. plot3d doesn't support that directly, but you could plot with type='n', then use segments3d to add the lines. BTW: how to change the perspective? I did not found an angle-parameter for the plot3d()-function. Just grab it with your mouse and drag. Alternatively, play3d(spin3d()) will spin it, or par3d(userMatrix=rotationMatrix(...)) for a fixed setting. Duncan Murdoch __ 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] Building a time series.
I have a need to build a time series and there are a couple of aspects about the time series object that are confusing me. First it seems that ts.union is not doing what I would expect. For example: x0 - rep(0,10) x1 - rep(1,10) xt0 - ts(x0, frequency=10) xt1 - ts(x1, frequency=10) st2 - ts.union(xt0, xt1) xt2 Time Series: Start = c(1, 1) End = c(1, 10) Frequency = 10 xt0 xt1 1.0 0 1 1.1 0 1 1.2 0 1 1.3 0 1 1.4 0 1 1.5 0 1 1.6 0 1 1.7 0 1 1.8 0 1 1.9 0 1 spectrum(xt2) Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In xy.coords(x, y, xlabel, ylabel, log = log) : 10 y values = 0 omitted from logarithmic plot 2: In min(x) : no non-missing arguments to min; returning Inf 3: In max(x) : no non-missing arguments to max; returning -Inf 4: In xy.coords(x, y, xlabel, ylabel, log) : 5 y values = 0 omitted from logarithmic plot I would expect that ts.union would concatenate the time series. I would expect xt2 from above to be a time series from 1:20. If I do xt2 - c(xt0, xt1) xt2 [1] 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 This seems to get rid of the time series nature of each of the objects. I can do: xt2 - ts(c(xt0,xt1), frequency=10) xt2 Time Series: Start = c(1, 1) End = c(2, 10) Frequency = 10 [1] 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 But whst am I to interpret the start and end (c(1,1), and c(2,10)) to be? Is this the best way to concatenate two time series? 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.
Re: [R] using complete.cases() with nested factors
Andrew Barr wabarr at gmail.com writes:
This maybe a newbie question. I have a dataframe
that looks like the sample
at the bottom of the email. I have monthly
precipitation data from several
sites over several years. For each site,
I need to extract years that have
a complete series of 12 monthly precipitation
values, while excluding that
year for sites with incomplete data.
I can't figure out how to do this
gracefully (i.e. without a silly for loop).
Any help will be appreciate,
thanks!
SiteIDyearmonthprecip(mm)
6700901941jan2998
6700901941feb1299
6700901941mar1007
6700901941apr354
6700901941may88
6700901941jun156
6700901941jul8
6700901941aug4
6700901941sep8
6700901941oct58
6700901941nov397
6700901941dec248
6700901942janNA
6700901942feb380
6700901942mar797
6700901942apr142
6700901942may43
6700901942jun14
6700901942jul70
6700901942aug51
6700901942sep0
6700901942oct10
6700901942nov235
6700901942dec405
There are likely more elegant solutions but this seems to work.
If the data frame is in a variable named dd
lapply(unique(dd$year), function(x) {s - subset(dd, year == x)
if (nrow(s) == 12) s})
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] showing the image
Here is one approach: tmp - scan(what='') 1: #44 #44 #434343 #404040 #3D3D3D #3D3D3D #3E3E3E 8: #414141 #414141 #414141 #404040 #3F3F3F #3F3F3F #3F3F3F 15: #3E3E3E #3E3E3E #3F3F3F #404040 #404040 #404040 #404040 22: #3E3E3E #3D3D3D #3E3E3E #404040 #414141 #414141 #404040 29: #3E3E3E #3E3E3E #3F3F3F #3F3F3F #3F3F3F #3F3F3F #3E3E3E 36: #3E3E3E #3F3F3F #3E3E3E #3D3D3D #3B3B3B #3C3C3C #3B3B3B 43: Read 42 items tmp - factor(tmp) tmp2 - matrix(as.numeric(tmp), nrow=6, byrow=TRUE) image(tmp2, col=levels(tmp)) But the image is pretty gray. Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of rostam shahname Sent: Thursday, September 04, 2008 2:28 PM To: r-help@r-project.org Subject: [R] showing the image Hi everyone, I have a matrix containing color values of an image same as following, [1,] #44 #44 #434343 #404040 #3D3D3D #3D3D3D #3E3E3E [2,] #414141 #414141 #414141 #404040 #3F3F3F #3F3F3F #3F3F3F [3,] #3E3E3E #3E3E3E #3F3F3F #404040 #404040 #404040 #404040 [4,] #3E3E3E #3D3D3D #3E3E3E #404040 #414141 #414141 #404040 [5,] #3E3E3E #3E3E3E #3F3F3F #3F3F3F #3F3F3F #3F3F3F #3E3E3E [6,] #3E3E3E #3F3F3F #3E3E3E #3D3D3D #3B3B3B #3C3C3C #3B3B3B I wonder if there is any function that helps me to show the image. Thanks, Rostam [[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] Plotting the complex fft in 3D?
Zitat von Duncan Murdoch [EMAIL PROTECTED]: On 04/09/2008 4:44 PM, Oliver Bandel wrote: Zitat von Duncan Murdoch [EMAIL PROTECTED]: Oliver Bandel wrote: Hello, [...] plot3d doesn't support that directly, but you could plot with type='n', then use segments3d to add the lines. BTW: how to change the perspective? I did not found an angle-parameter for the plot3d()-function. Just grab it with your mouse and drag. Wow, coool! :-) Well, rgl I think gl stands for OpenGl. Fine. :-) Alternatively, play3d(spin3d()) will spin it, or par3d(userMatrix=rotationMatrix(...)) for a fixed setting. Ok, some thinsg to play with. Thank you. Ciao, Oliver __ 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] Building a time series.
On Thu, 4 Sep 2008, [EMAIL PROTECTED] wrote: I have a need to build a time series and there are a couple of aspects about the time series object that are confusing me. First it seems that ts.union is not doing what I would expect. For example: x0 - rep(0,10) x1 - rep(1,10) xt0 - ts(x0, frequency=10) xt1 - ts(x1, frequency=10) st2 - ts.union(xt0, xt1) xt2 Time Series: Start = c(1, 1) End = c(1, 10) Frequency = 10 xt0 xt1 1.0 0 1 1.1 0 1 1.2 0 1 1.3 0 1 1.4 0 1 1.5 0 1 1.6 0 1 1.7 0 1 1.8 0 1 1.9 0 1 spectrum(xt2) Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In xy.coords(x, y, xlabel, ylabel, log = log) : 10 y values = 0 omitted from logarithmic plot 2: In min(x) : no non-missing arguments to min; returning Inf 3: In max(x) : no non-missing arguments to max; returning -Inf 4: In xy.coords(x, y, xlabel, ylabel, log) : 5 y values = 0 omitted from logarithmic plot I would expect that ts.union would concatenate the time series. You expect wrong. ts.union() is more like cbind() for all combined time indexes, ts.intersect() for the intersection of time indexes. I would expect xt2 from above to be a time series from 1:20. If I do This would never make sense. You have defined both xt0 and xt1 to be time series starting at 1(1) and ending at 1(10). Why should xt1 be shifted to start at 2(1) and end at 2(10)? xt2 - c(xt0, xt1) xt2 [1] 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 This seems to get rid of the time series nature of each of the objects. I can do: xt2 - ts(c(xt0,xt1), frequency=10) xt2 Time Series: Start = c(1, 1) End = c(2, 10) Frequency = 10 [1] 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 But whst am I to interpret the start and end (c(1,1), and c(2,10)) to be? Is this the best way to concatenate two time series? Package zoo has a c() command for zoo series. For your example this complains R c(as.zoo(xt0), as.zoo(xt1)) Error in rbind.zoo(...) : indexes overlap But if a concatenation makes sense, this works R xt1 - ts(x1, start = 2, frequency=10) R c(as.zoo(xt0), as.zoo(xt1)) 1(1) 1(2) 1(3) 1(4) 1(5) 1(6) 1(7) 1(8) 1(9) 1(10) 2(1) 2(2) 2(3) 0 0 0 0 0 0 0 0 0 0 1 1 1 2(4) 2(5) 2(6) 2(7) 2(8) 2(9) 2(10) 1 1 1 1 1 1 1 R as.ts(c(as.zoo(xt0), as.zoo(xt1))) Time Series: Start = c(1, 1) End = c(2, 10) Frequency = 10 [1] 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 hth, Z 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.
