Re: [R] bug in closing gzfile-opened connections?
Note that the use of read.table() does make a difference. If you did x - scan(gzfile(xxx.gz), list(,,)) you would leave an unused connection, and showConnections(all=TRUE) would show this. There is a finite pool of connections, and in general the correct way to use them is con - gzfile(xxx.gz) x - scan(con, list(,,)) close(con) read.table() is the exception, so I suspect it is other things that have been done in the session that have used up the pool of connections. On Tue, 3 Jul 2007, Duncan Murdoch wrote: On 03/07/2007 1:37 PM, David Reiss wrote: Hi, I am making multiple calls to gzfile() via read.table(), e.g. x - read.table( gzfile( xxx.gz ) ) After i do this many times (I haven't counted, but probably between 50 and 100 times) I get the error message: Error in open.connection(file, r) : unable to open connection In addition: Warning message: cannot open compressed file 'xxx.gz' however, I also find that: showConnections() description class mode text isopen can read can write so there are no (apparently) open connections. Calling closeAllConnections() does not fix the problem. I have to quit and re-start R. I am using R 2.5.0 on a Mac (OSX 10.4.9). Anyone know if this is a bug or a 'feature'? I see from the gzfile help that: In general functions using connections will open them if they are not open, but then close them again, so to leave a connection open call 'open' explicitly. You didn't give a reproducible example, so I couldn't say. When I create a gzipped version of a write.table output and run for(i in 1:1000) read.table(gzfile(f)) in R 2.5.0 I don't see a problem. This is on Windows, but I doubt that makes a difference. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] possible bug in ggplot2 v0.5.2???
On Tue, 3 Jul 2007, hadley wickham wrote: Hi Stephane, The problem is that the windows graphics device doesn't support transparent colours. You can get around this in two ways: It certainly does! Try col=transparent (and perhaps consult your dictionary). It was news to me that the windows() graphics device worked on Linux i586. What it does not support as yet is translucent colours, and that is a restriction imposed by Windows (translucency support was introduced for Windows XP, and we still try to support older versions of Windows, unlike the MacOS people). I have been working on a workaround, so translucency support is likely to be implemented in R 2.6.0 for users of XP or later. Given that neither of the two main screen devices and neither of the standard print devices support translucency, the subject line looks correct to me: the problem surely lies in the assumptions made in ggplot2. * export to a device that does support transparency (eg. pdf) * use a solid fill colour : + stat_smooth(method=lm, fill=grey50) Hadley On 7/3/07, Stephane Cruveiller [EMAIL PROTECTED] wrote: Dear R-Users, I recently gave a try to the nice package ggplot2. Everything went well until I tried to add a smoother (using lm method for instance). On the graphic device the regression line is displayed but not confidence intervals as it should be (at least on ggplot website). I tried to do the job on both MS winXP and Linux i586: same result. Did anyone encountered this problem? Did I miss something? My R version is 2.4.1. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] retrieving stats from bwplot
On 7/3/07, Héctor Villalobos [EMAIL PROTECTED] wrote: Hi all, I want to retrieve the stats from a 'bwplot' with one factor. I have read the help for 'panel' function and I'm aware of the option 'stats' which defaults to 'boxplot.stats' but I didn't understand it well and therefore I am unable to get what I need. I'm not sure what bwplot has to do with this. Perhaps this will help: foo - with(OrchardSprays, split(decrease, treatment)) str(foo) List of 8 $ A: num [1:8] 2 2 5 4 5 12 4 3 $ B: num [1:8] 8 6 4 10 7 4 8 14 $ C: num [1:8] 15 84 16 9 17 29 13 19 $ D: num [1:8] 57 36 22 51 28 27 20 39 $ E: num [1:8] 95 51 39 114 43 47 61 55 $ F: num [1:8] 90 69 87 20 71 44 57 114 $ G: num [1:8] 92 71 72 24 60 77 72 80 $ H: num [1:8] 69 127 72 130 81 76 81 86 boxplot.stats(foo$A) $stats [1] 2.0 2.5 4.0 5.0 5.0 $n [1] 8 $conf [1] 2.603464 5.396536 $out [1] 12 bxp.stats - lapply(foo, boxplot.stats) str(bxp.stats) List of 8 $ A:List of 4 ..$ stats: num [1:5] 2 2.5 4 5 5 ..$ n: int 8 ..$ conf : num [1:2] 2.60 5.40 ..$ out : num 12 $ B:List of 4 ..$ stats: num [1:5] 4 5 7.5 9 14 ..$ n: int 8 ..$ conf : num [1:2] 5.27 9.73 ..$ out : num(0) $ C:List of 4 ..$ stats: num [1:5] 9 14 16.5 24 29 ..$ n: int 8 ..$ conf : num [1:2] 10.9 22.1 ..$ out : num 84 $ D:List of 4 ..$ stats: num [1:5] 20 24.5 32 45 57 ..$ n: int 8 ..$ conf : num [1:2] 20.5 43.5 ..$ out : num(0) $ E:List of 4 ..$ stats: num [1:5] 39 45 53 78 114 ..$ n: int 8 ..$ conf : num [1:2] 34.6 71.4 ..$ out : num(0) $ F:List of 4 ..$ stats: num [1:5] 20 50.5 70 88.5 114 ..$ n: int 8 ..$ conf : num [1:2] 48.8 91.2 ..$ out : num(0) $ G:List of 4 ..$ stats: num [1:5] 60 65.5 72 78.5 92 ..$ n: int 8 ..$ conf : num [1:2] 64.7 79.3 ..$ out : num 24 $ H:List of 4 ..$ stats: num [1:5] 69 74 81 106 130 ..$ n: int 8 ..$ conf : num [1:2] 62.8 99.2 ..$ out : num(0) If you want combinations defined by more than one factor, you could use something like with(OrchardSprays, split(decrease, interaction(treatment, colpos))) (although this is a bad example, since there is only one observation per combination) -Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] read items
Hello, I write us because I wanna know if it's possible to don't display read item when I execute a scan(textConnection()) thanks _ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] probabilty plot
Hi all, I am a freshman of R,but I am interested in it! Those days,I am learning pages on NIST,with url http://www.itl.nist.gov/div898/handbook/eda/section3/probplot.htm, I am meeting a problem about probability plot and I don't know how to plot a data set with R. Could somebody tell me the answer,and a example is the best! I will look forward to your answer. Thank you very much. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] possible bug in ggplot2 v0.5.2???
On 7/4/07, Prof Brian Ripley [EMAIL PROTECTED] wrote: On Tue, 3 Jul 2007, hadley wickham wrote: Hi Stephane, The problem is that the windows graphics device doesn't support transparent colours. You can get around this in two ways: It certainly does! Try col=transparent (and perhaps consult your dictionary). It was news to me that the windows() graphics device worked on Linux i586. Well my dictionary defines transparent as allowing light to pass through so that objects behind can be distinctly seen which I believe applies here (ie. stained glass windows and blue points with alpha 0.5 are both transparent). What does your dictionary say? What it does not support as yet is translucent colours, and that is a restriction imposed by Windows (translucency support was introduced for Windows XP, and we still try to support older versions of Windows, unlike the MacOS people). I have been working on a workaround, so translucency support is likely to be implemented in R 2.6.0 for users of XP or later. I am confused by your implication that windows (prior to XP) does not support translucency. Perhaps it is not supported at the operating system level, but it has certainly been available at the application level for a very long time. Given that neither of the two main screen devices and neither of the standard print devices support translucency, the subject line looks correct to me: the problem surely lies in the assumptions made in ggplot2. The features of the windows and X11 devices clearly lag behind the quartz and pdf devices. I can program for the lowest common denominator or I can use modern features that support the tasks I am working on. I choose the later, and it is certainly your prerogative to declare that a bug in me. Hadley __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to install R 2.5 with Synaptic in Ubuntu?
Hi, I can't get this approach to work. When I first added the repsitory line to /etc/apt/sources.list synaptic complained that it was a malformed line. I fixed this by adding main to end of the entry making it: deb http://my.favorite.cran.mirror/bin/linux/ubuntu feisty main However after this it still complains that it can't find packages.gz It appears to be looking in http://my.favorite.cran.mirror/bin/linux/ubuntu/distsfeisty which isn't the directory structure of the cran repository, but I can see anyway to modify this behaviour. Every other Ubuntu repositoy I have looked at contains the dists directory. Any suggestions for modifying this behaviour are gratefully recieved. Many thanks Mike Smith quot;Stefan Großequot; wrote: I'm using Ubuntu dapper, which only have R of Version 2.2.1. Would anybody tell me how to install the latest version of R with from the CRAN Ubuntu readme- works for synaptic as well: * UBUNTU R packages for Ubuntu on i386 are available. The plans are to support at least the latest Ubuntu release and the latest LTS release. Currently (April 2007), these are Feisty Fawn (7.04) and Dapper Drake (6.06), respectively. Since Feisty was released very shortly before R 2.5.0, binary packages *for this release of R* are also available for Edgy Eft (6.10). To obtain the latest R packages, add an entry like deb http://my.favorite.cran.mirror/bin/linux/ubuntu feisty/ or deb http://my.favorite.cran.mirror/bin/linux/ubuntu edgy/ or deb http://my.favorite.cran.mirror/bin/linux/ubuntu dapper/ in your /etc/apt/sources.list file. See http://cran.r-project.org/mirrors.html for the list of CRAN mirrors. To install the complete R system, use sudo apt-get update sudo apt-get install r-base Users who need to compile packages should also install the r-base-dev package: sudo apt-get install r-base-dev The R packages for Ubuntu should otherwise behave like the Debian ones. For more information, see the README file in http://cran.R-project.org/bin/linux/debian/ * SECURE APT The Ubuntu archives on CRAN are signed with the key of Vincent Goulet [EMAIL PROTECTED] with key ID E2A11821. You can fetch this with gpg --keyserver subkeys.pgp.net --recv-key E2A11821 and then you feed the key to apt-key with gpg -a --export E2A11821 | sudo apt-key add - Some people have reported difficulties using this approach. The issue was usually related to a firewall blocking port 11371. An alternative approach is to search for the key at http://keyserver.noreply.org/ and copy the key to a plain text file, say key.txt. Then, feed the key to apt-key with sudo apt-key add key.txt * ACKNOWLEDGEMENT The Debian R packages are maintained by Dirk Eddelbuettel and Doug Bates. The Ubuntu packages are compiled for i386 by Vincent Goulet. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/How-to-install-R-2.5-with-Synaptic-in-Ubuntu--tf3998481.html#a11426235 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Adding data to existing plot with new=TRUE does not appear to work
Dear all, I am trying to shove a number of cmdscale() results into a single plot (k=1 so I'm trying to get multiple columns in the plot). From ?par I learned that I can/should set new=TRUE in either par() or the plot function itself. However with the following reduced code, I get only a plot with a column of data points with x==2. plot(1,10, xlim=range(0,3), ylim=range(0,10), type='n') aa - rep(1,10) bb - 1:10 plot(aa,bb, xlim=range(0,3), ylim=range(0,10), new=TRUE) aa - rep(2,10) plot(aa,bb, xlim=range(0,3), ylim=range(0,10), new=TRUE) Also, when I insert a op - par(new=TRUE) either before or immediately after the first plot statement (the type='n' one) in the above code fragment, the resulting graph still only shows one column of data. Have I misinterpreted the instructions or the functionality of new=TRUE? Thank you, Paul Lemmens __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to plot a monthplot from a ts object where all individual years are shown (e.g. as lines) and can be compared with a average or median year?
Dear R help, I'm working with regular 8-daily time-series from 2000 up till now and would like to be able to compare years with each other. E.g. by creating a monthplot via the result of the stl() method it looks ok but I was wondering whether there exist other methods to plot the different years as lines on top of each other such that years can be compared with each other (temporal change detection)? This is the type of time-series we are working with: forest - ts(data, frequency=46, start=c(2000,8), end=c(2006,46)) I tried it as follows but this is not very clear. year0 - window(forest, start=2000,end=c(2000,46)) year1 - window(forest, start=2001,end=c(2001,46)) year2 - window(forest, start=2002,end=c(2002,46)) year3 - window(forest, start=2003,end=c(2003,46)) year4 - window(forest, start=2004,end=c(2004,46)) year5 - window(forest, start=2005,end=c(2005,46)) year6 - window(forest, start=2006,end=c(2006,46)) plot(1:46,years[,1], col=2) lines(1:46,years[,2], col=3) lines(1:46,years[,3], col=4) lines(1:46,years[,4], col=5) lines(1:46,years[,5], col=6) lines(1:46,years[,6], col=7) lines(1:46,years[,7], col=8) Are there other options to be able to compare years with each other in order to detect change (e.g., per month)? Thanks a lot, J R 2.5, Win XP [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fine tunning rgenoud
I think fine tuning the function might be in order.
The function has just a single penalty for not meeting
the constraints no matter how close it is to meeting
them. A better approach is to have a penalty that
depends on the amount by which all of the constraints
are breached.
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and A Guide for the Unwilling S User)
Paul Smith wrote:
Dear All,
I am trying to solve the following maximization problem, but I cannot
have rgenoud giving me a reliable solution.
Any ideas?
Thanks in advance,
Paul
library(rgenoud)
v - 0.90
O1 - 10
O2 - 20
O0 - v*O1+(1-v)*O2
myfunc - function(x) {
U0 - x[1]
U1 - x[2]
U2 - x[3]
q0 - x[4]
q1 - x[5]
q2 - x[6]
p - x[7]
if (U0 0)
return(-1e+200)
else if (U1 0)
return(-1e+200)
else if (U2 0)
return(-1e+200)
else if ((U0-(U1+(O1-O0)*q1)) 0)
return(-1e+200)
else if ((U0-(U2+(O2-O0)*q2)) 0)
return(-1e+200)
else if ((U1-(U0+(O0-O1)*q0)) 0)
return(-1e+200)
else if ((U1-(U2+(O2-O1)*q2)) 0)
return(-1e+200)
else if((U2-(U0+(O0-O2)*q0)) 0)
return(-1e+200)
else if((U2-(U1+(O1-O2)*q1)) 0)
return(-1e+200)
else if(p 0)
return(-1e+200)
else if(p 1)
return(-1e+200)
else if(q0 0)
return(-1e+200)
else if(q1 0)
return(-1e+200)
else if(q2 0)
return(-1e+200)
else
return(p*(sqrt(q0)-(O0*q0+U0))+(1-p)*(v*(sqrt(q1)-(O1*q1+U1))+(1-v)*(sqrt(q2)-(O2*q2+U2
}
genoud(myfunc,nvars=7,max=T,pop.size=6000,starting.values=runif(7),wait.generations=150,max.generations=300,boundary.enforcement=2)
__
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and provide commented, minimal, self-contained, reproducible code.
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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] How to install R 2.5 with Synaptic in Ubuntu?
to end of the entry making it: deb http://my.favorite.cran.mirror/bin/linux/ubuntu feisty main However after this it still complains that it can't find packages.gz Just a guess: have you replaced the my.favorite.cran.mirror by a mirror which is close to you? If you're in UK it would be for example deb http://www.stats.bris.ac.uk/R/bin/linux/ubuntu feisty main ;o) Stefan It appears to be looking in http://my.favorite.cran.mirror/bin/linux/ubuntu/distsfeisty which isn't the directory structure of the cran repository, but I can see anyway to modify this behaviour. Every other Ubuntu repositoy I have looked at contains the dists directory. Any suggestions for modifying this behaviour are gratefully recieved. Many thanks Mike Smith -=-=- ... The simple truth is that interstellar distances will not fit into the human imagination - (The Hitchhiker's Guide to the Galaxy) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Adding data to existing plot with new=TRUE does not appear to work
Hi if you change your code plot(1,10, xlim=range(0,3), ylim=range(0,10), type='n') aa - rep(1,10) bb - 1:10 plot(aa,bb, xlim=range(0,3), ylim=range(0,10), new=TRUE) aa - rep(2,10) par(new=T) plot(aa,bb, xlim=range(0,3), ylim=range(0,10), new=TRUE) you will get both columns plotted. However you can get similar result with using points plot(1,10, xlim=range(0,3), ylim=range(0,10), type='n') aa - rep(1,10) bb - 1:10 points(aa,bb) aa - rep(2,10) points(aa,bb) Regards Petr [EMAIL PROTECTED] [EMAIL PROTECTED] napsal dne 04.07.2007 09:48:15: Dear all, I am trying to shove a number of cmdscale() results into a single plot (k=1 so I'm trying to get multiple columns in the plot). From ?par I learned that I can/should set new=TRUE in either par() or the plot function itself. However with the following reduced code, I get only a plot with a column of data points with x==2. plot(1,10, xlim=range(0,3), ylim=range(0,10), type='n') aa - rep(1,10) bb - 1:10 plot(aa,bb, xlim=range(0,3), ylim=range(0,10), new=TRUE) aa - rep(2,10) plot(aa,bb, xlim=range(0,3), ylim=range(0,10), new=TRUE) Also, when I insert a op - par(new=TRUE) either before or immediately after the first plot statement (the type='n' one) in the above code fragment, the resulting graph still only shows one column of data. Have I misinterpreted the instructions or the functionality of new=TRUE? Thank you, Paul Lemmens __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to install R 2.5 with Synaptic in Ubuntu?
Hi, Thanks for the suggestion and I wish the solution was that obvious, but I have changed it to really point at my favourite mirror. Using your example Synaptic reports the following error when I try to update the repositories: http://www.stats.bris.ac.uk/R/bin/linux/ubuntu/dists/feisty/main/binary-i386/Packages.gz: 404 Not Found This is understandable since that location doesn't exist, but it makes me think that the directory structure of the R mirrors is not compatible with Ubuntu and Synaptic, since it automatically seeks /dists/feisty/ rather than just /feisty/ as it is on the CRAN mirrors. Thanks again Mike Smith Stefan Grosse-2 wrote: to end of the entry making it: deb http://my.favorite.cran.mirror/bin/linux/ubuntu feisty main However after this it still complains that it can't find packages.gz Just a guess: have you replaced the my.favorite.cran.mirror by a mirror which is close to you? If you're in UK it would be for example deb http://www.stats.bris.ac.uk/R/bin/linux/ubuntu feisty main ;o) Stefan It appears to be looking in http://my.favorite.cran.mirror/bin/linux/ubuntu/distsfeisty which isn't the directory structure of the cran repository, but I can see anyway to modify this behaviour. Every other Ubuntu repositoy I have looked at contains the dists directory. Any suggestions for modifying this behaviour are gratefully recieved. Many thanks Mike Smith -=-=- ... The simple truth is that interstellar distances will not fit into the human imagination - (The Hitchhiker's Guide to the Galaxy) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/How-to-install-R-2.5-with-Synaptic-in-Ubuntu--tf3998481.html#a11427837 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] QP for solving Support Vector Regression
The ksvm object is probably what you need to use. [quote] Dear R users, I'm trying to run the Support Vector Regression by a general quadratic programming function like ipop ( ) in kernlab or solve.QP ( ) in quadprog packages. Since they are general, their application in Support Vector Regression can lead to misunderstanding, particularly when constructing matrices. Even their examples are general and applied in Support Vector Classification. Could anybody please introduce an example code for regression case. [\quote] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Fine tunning rgenoud
On 7/4/07, RAVI VARADHAN [EMAIL PROTECTED] wrote:
Here is another approach: I wrote an R function that would generate interior
points as starting values for constrOptim. This might work better than the
LP approach, since the LP approach gives you a starting value that is on the
boundary of the feasible region, i.e a vertex of the polyhedron, whereas this
new approach gives you points on the interior. You can generate as many
points as you wish, but the approach is brute-force and is very inefficient -
it takes on the order of a 1000 tries to find one feasible point.
Thanks again, Ravi. Actually, the LP approach also works here. Let
g(X) = k be the constraints. Then, by solving a LP problem with the
constraints
g(X) = (k+0.2)
returns an interior starting value for constrOptim. I am aware that
the new set of constraints may correspond to an impossible linear
system, but it works in many cases.
Paul
- Original Message -
From: Paul Smith [EMAIL PROTECTED]
Date: Tuesday, July 3, 2007 7:32 pm
Subject: Re: [R] Fine tunning rgenoud
To: R-help r-help@stat.math.ethz.ch
On 7/4/07, Ravi Varadhan [EMAIL PROTECTED] wrote:
It should be easy enough to check that your solution is valid (i.e.
a local
minimum): first, check to see if the solution satisfies all the
constraints; secondly, check to see if it is an interior point
(i.e. none of
the constraints become equality); and finally, if the solution is an
interior point, check to see whether the gradient there is close to
zero.
Note that if the solution is one of the vertices of the polyhedron,
then the
gradient may not be zero.
I am having bad luck: all constraints are satisfied, but the solution
given by constrOptim is not interior; the gradient is not equal to
zero.
Paul
-Original Message-
From: [EMAIL PROTECTED]
[ On Behalf Of Paul Smith
Sent: Tuesday, July 03, 2007 5:10 PM
To: R-help
Subject: Re: [R] Fine tunning rgenoud
On 7/3/07, Ravi Varadhan [EMAIL PROTECTED] wrote:
You had indicated in your previous email that you are having trouble
finding
a feasible starting value for constrOptim(). So, you basically
need to
solve a system of linear inequalities to obtain a starting point.
Have
you
considered using linear programming? Either simplex() in the boot
package
or solveLP() in linprog would work. It seems to me that you
could use
any
linear objective function in solveLP to obtain a feasible
starting point.
This is not the most efficient solution, but it might be worth a
try.
I am aware of other methods for generating n-tuples that satisfy
linear
inequality constraints, but AFAIK those are not available in R.
Thanks, Ravi. I had already conceived the solution that you suggest,
actually using lpSolve. I am able to get a solution for my problem
with constrOptim, but I am not enough confident that the solution is
right. That is why I am trying to get a solution with rgenoud, but
unsuccessfully until now.
Paul
-Original Message-
From: [EMAIL PROTECTED]
[ On Behalf Of Paul Smith
Sent: Tuesday, July 03, 2007 4:10 PM
To: R-help
Subject: [R] Fine tunning rgenoud
Dear All,
I am trying to solve the following maximization problem, but I cannot
have rgenoud giving me a reliable solution.
Any ideas?
Thanks in advance,
Paul
library(rgenoud)
v - 0.90
O1 - 10
O2 - 20
O0 - v*O1+(1-v)*O2
myfunc - function(x) {
U0 - x[1]
U1 - x[2]
U2 - x[3]
q0 - x[4]
q1 - x[5]
q2 - x[6]
p - x[7]
if (U0 0)
return(-1e+200)
else if (U1 0)
return(-1e+200)
else if (U2 0)
return(-1e+200)
else if ((U0-(U1+(O1-O0)*q1)) 0)
return(-1e+200)
else if ((U0-(U2+(O2-O0)*q2)) 0)
return(-1e+200)
else if ((U1-(U0+(O0-O1)*q0)) 0)
return(-1e+200)
else if ((U1-(U2+(O2-O1)*q2)) 0)
return(-1e+200)
else if((U2-(U0+(O0-O2)*q0)) 0)
return(-1e+200)
else if((U2-(U1+(O1-O2)*q1)) 0)
return(-1e+200)
else if(p 0)
return(-1e+200)
else if(p 1)
return(-1e+200)
else if(q0 0)
return(-1e+200)
else if(q1 0)
return(-1e+200)
else if(q2 0)
return(-1e+200)
else
return(p*(sqrt(q0)-(O0*q0+U0))+(1-p)*(v*(sqrt(q1)-(O1*q1+U1))+(1-v)*(sqrt(q2
)-(O2*q2+U2
}
genoud(myfunc,nvars=7,max=T,pop.size=6000,starting.values=runif(7),wait.gene
rations=150,max.generations=300,boundary.enforcement=2)
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Re: [R] sequences
Hi all, thank you very much. livia wrote: Hi, I would like to generate a series in the following form (0.8^1, 0.8^2, ..., 0.8^600) Could anyone tell me how can I achieve that? I am really new to R. -- View this message in context: http://www.nabble.com/sequences-tf4019146.html#a11428415 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice: shifting strips to left of axes
[EMAIL PROTECTED] wrote:
myYlabGrob -
function(..., main.ylab = ) ## ...is lab1, lab2, etc
{
## you can add arguments to textGrob for more control
## in the next line
labs - lapply(list(...), textGrob, rot=90)
main.ylab - textGrob(main.ylab, rot = 90)
nlabs - length(labs)
lab.heights -
lapply(labs,
function(lab) unit(1, grobheight,
data=list(lab)))
unit1 - unit(1.2, grobheight, data = list(main.ylab))
unit2 - do.call(max, lab.heights)
lab.layout -
grid.layout(ncol = 2, nrow = nlabs,
heights = unit(1, null),
widths = unit.c(unit1, unit2),
respect = TRUE)
lab.gf - frameGrob(layout=lab.layout)
for (i in seq_len(nlabs))
{
lab.gf - placeGrob(lab.gf, labs[[i]], row = i, col = 2)
}
lab.gf - placeGrob(lab.gf, main.ylab, col = 1)
lab.gf
}
Wow. I don't think I would have been able to come up with that on my
own. Thank you!
--
Michael Hoffman
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Re: [R] Adding data to existing plot with new=TRUE does not appear to work
Hi Petr, On 7/4/07, Petr PIKAL [EMAIL PROTECTED] wrote: par(new=T) plot(aa,bb, xlim=range(0,3), ylim=range(0,10), new=TRUE) So I need to activate the par(new=T) really just ahead of time when I need it, not as sort of a general clause at the beginning of my script? However you can get similar result with using points Yes I new that, but I wanted to try and go without an if() for deciding between the first and consecutive columns. Thnx for helping out! Paul Lemmens __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] The R Book by M. J. Crawley
G'day Uwe, On Tue, 03 Jul 2007 14:33:05 +0200 Uwe Ligges [EMAIL PROTECTED] wrote: Pietrzykowski, Matthew (GE, Research) wrote: I saw the new book, The R Book, by Michael J. Crawley and wanted to know what R users thoughts of it. The author seems to be an expert in (almost?) all available statistical programming languages I would have thought that honour would go to Brian S. Everitt. :-) M.J. Crawley only seems to write books using S-PLus and R. His book on Statistical Computing (using S-Plus) is about 750 pages and his book Statistics: An introduction using R is about 320 pages. I do not know and have not seen The R Book yet, so I cannot comment on it. The statistical material presented in the other two books is pretty sound and well explained. M.J. Crawley definitely has some strong opinions on how certain data should be analysed and how statistics should be used. The S-Plus code or R code he uses can, on occasions, be somewhat improved. Cheers, Berwin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Calling C Code from R
Hi R Users,
Thanks in advance.
I am using R-2.5.1 on Windows XP.
I am trying to call C code (testCX1.C) from R. testCX1.c calls another C code
(funcC1.c) and returning a value to testCX1.c. I like to have this value in R.
My steps are below:
1. R CMD SHLIB testCX1.c funcC1.c (at command propmt)
2. It creates testCX1.dll with warning (but testCX1.dll works):
testCX1.c:38: warning: implicit declaration of function 'func'
How to get rid off this error ?
What is the best way to call funcC1.c from testCX1.c?
I have provided the codes below:
Once again thank you very much for the time you have given.
Regards,
Debabrata Midya (Deb)
Statistician
NSW Department of Commerce
Sydney, Australia
testCX1.C
--
/*
*
testCX1.c
*
*/
#include R.h
#include Rdefines.h
#include Rmath.h
SEXP testC1(SEXP a)
{
int i, nr;
double *xa, *xw, tmp;
SEXP w;
PROTECT(a = coerceVector(a, REALSXP));
nr = length(a);
printf( Length : %d \n, nr);
PROTECT(w = allocVector(REALSXP, 1));
xa = REAL(a);
xw = REAL(w);
tmp = 0.0;
for (i = 0; i nr; i++)
{
tmp += xa[i];
}
// tmp = 0.0;
xw[0] = func(xa);
UNPROTECT(2);
return(w);
}
funcC1.c
/*
*
funcC1.c
*
*/
#include R.h
#include Rdefines.h
#include Rmath.h
SEXP func(SEXP a)
{
int i, nr = 3;
double *xa, *xw, tmp;
SEXP w;
PROTECT(a = coerceVector(a, REALSXP));
PROTECT(w = allocVector(REALSXP, 1));
xa = REAL(a);
xw = REAL(w);
tmp = 0.0;
for (i = 0; i nr; i++)
{
tmp += xa[i] * xa[i];
}
xw[0] = tmp;
UNPROTECT(2);
return(w);
}
R script
---
dyn.load(testCX1.dll)
xL - 1:5
xL - as.double(as.vector(t(xL)))
.Call(testC1,xL)
[1] 55
-
[[alternative HTML version deleted]]
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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] Calling C Code from R
On Wed, Jul 04, 2007 at 04:39:18AM -0700, Deb Midya wrote: Hi R Users, Thanks in advance. I am using R-2.5.1 on Windows XP. I am trying to call C code (testCX1.C) from R. testCX1.c calls another C code (funcC1.c) and returning a value to testCX1.c. I like to have this value in R. My steps are below: 1. R CMD SHLIB testCX1.c funcC1.c (at command propmt) 2. It creates testCX1.dll with warning (but testCX1.dll works): testCX1.c:38: warning: implicit declaration of function 'func' How to get rid off this error ? By adding the prototype of 'func' to testCX1.c: SEXP func(SEXP a); Probably it is simplest to collect all prototypes in a single header file and include that from all .c files. What is the best way to call funcC1.c from testCX1.c? See .C and .Call and in particular the 'Writing R Extensions' manual, 5 System and foreign language interfaces. Gabor [...] -- Csardi Gabor [EMAIL PROTECTED]MTA RMKI, ELTE TTK __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to install R 2.5 with Synaptic in Ubuntu?
msmith schrieb: Hi, Thanks for the suggestion and I wish the solution was that obvious, but I have changed it to really point at my favourite mirror. Using your example Synaptic reports the following error when I try to update the repositories: http://www.stats.bris.ac.uk/R/bin/linux/ubuntu/dists/feisty/main/binary-i386/Packages.gz: 404 Not Found This is understandable since that location doesn't exist, but it makes me think that the directory structure of the R mirrors is not compatible with Ubuntu and Synaptic, since it automatically seeks /dists/feisty/ rather than just /feisty/ as it is on the CRAN mirrors. Hm. I have Fedora 7 so I cannot really check what my entry would look like. I recently installed Kubuntu on a friends notebook so there the readme from http://www.stats.bris.ac.uk/R/bin/linux/ubuntu/README.html did actually work. I just had a second look at your mail so could it be that there is a missing slash after feisty? If that does not work try another mirror like deb http://stat.ethz.ch/CRAN/bin/linux/ubuntu/ feisty/ there is also no main stated there in the readme... sorry I copied that from your mail so thats obviously not following the readme... I hope this works now. Stefan -=-=- ... Satisfaction does not come with achievement, but with effort. Full effort is full victory. (M.Gandhi) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to install R 2.5 with Synaptic in Ubuntu?
Stefan Grosse schrieb: deb http://www.stats.bris.ac.uk/R/bin/linux/ubuntu feisty main Sorry, I copied a mistake there, it should be: deb http://www.stats.bris.ac.uk/R/bin/linux/ubuntu feisty/ Stefan -=-=- ... The simple truth is that interstellar distances will not fit into the human imagination - (The Hitchhiker's Guide to the Galaxy) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] probabilty plot
Is this what you mean ? --- mydata - c(1,2,3,4,5,7,5,4,3) plot(mydata) --- --- along zeng [EMAIL PROTECTED] wrote: Hi all, I am a freshman of R,but I am interested in it! Those days,I am learning pages on NIST,with url http://www.itl.nist.gov/div898/handbook/eda/section3/probplot.htm, I am meeting a problem about probability plot and I don't know how to plot a data set with R. Could somebody tell me the answer,and a example is the best! I will look forward to your answer. Thank you very much. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calling C Code from R
Gabor, Thank you very much for such a quick response. As I am new to this area, will you please explain where can I put SEXP func(SEXP a); in my program. Once again, thank you very much for your quick response. Regards, Deb Gabor Csardi [EMAIL PROTECTED] wrote: On Wed, Jul 04, 2007 at 04:39:18AM -0700, Deb Midya wrote: Hi R Users, Thanks in advance. I am using R-2.5.1 on Windows XP. I am trying to call C code (testCX1.C) from R. testCX1.c calls another C code (funcC1.c) and returning a value to testCX1.c. I like to have this value in R. My steps are below: 1. R CMD SHLIB testCX1.c funcC1.c (at command propmt) 2. It creates testCX1.dll with warning (but testCX1.dll works): testCX1.c:38: warning: implicit declaration of function 'func' How to get rid off this error ? By adding the prototype of 'func' to testCX1.c: SEXP func(SEXP a); Probably it is simplest to collect all prototypes in a single header file and include that from all .c files. What is the best way to call funcC1.c from testCX1.c? See .C and .Call and in particular the 'Writing R Extensions' manual, 5 System and foreign language interfaces. Gabor [...] -- Csardi Gabor MTA RMKI, ELTE TTK - Boardwalk for $500? In 2007? Ha! [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calling C Code from R
On Wed, Jul 04, 2007 at 05:15:15AM -0700, Deb Midya wrote: Gabor, Thank you very much for such a quick response. As I am new to this area, will you please explain where can I put SEXP func(SEXP a); in my program. Deb, anywhere before calling it. (Well outside a function definition.) Typically after the #include lines. Or put all these prototypes into a header file called myfuncs.h and add line #include myfuncs.h just after the other #include lines at the beginning of the file. Gabor Once again, thank you very much for your quick response. Regards, Deb -- Csardi Gabor [EMAIL PROTECTED]MTA RMKI, ELTE TTK __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Loop and cbind
Hi, I would like to apply the following function for i between 1 and 12, and
then construct a list of the return series.
for (i in 1:12){
ewma[i] - emaTA(calm[[i]]^2,0.03)
standard[i]- calm[[i]]/sqrt(ewma[i])
standard - cbind(standard[i])
}
But it does not work. Could anyone give me some advice how can I achieve
this? Many thanks
--
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http://www.nabble.com/Loop-and-cbind-tf4024291.html#a11430500
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[R] Problem/bug with smooth.spline and all.knots=T
Dear list, if I do smooth.spline(tmpSec, tmpT, all.knots=T) with the attached data, I get this error-message: Error in smooth.spline(tmpSec, tmpT, all.knots = T) : smoothing parameter value too small If I do smooth.spline(tmpSec[-single arbitrary number], tmpT[-single arbitrary number], all.knots=T) it works! I just don't see it. It works for hundrets other datasets, but not for this one. Would be glad if anyone could help! Hubertus __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loop and cbind
Hi
In what way does it not work?
My guess is that you have not declared your values outside the for loop.
As they are local they will be lost on exit.
You need to declare them before:
ewma-vector(length=12)
standard-vector(length=12)
for ... {
}
John Seers
---
Hi, I would like to apply the following function for i between 1 and 12,
and then construct a list of the return series.
for (i in 1:12){
ewma[i] - emaTA(calm[[i]]^2,0.03)
standard[i]- calm[[i]]/sqrt(ewma[i])
standard - cbind(standard[i])
}
But it does not work. Could anyone give me some advice how can I achieve
this? Many thanks
--
View this message in context:
http://www.nabble.com/Loop-and-cbind-tf4024291.html#a11430500
Sent from the R help mailing list archive at Nabble.com.
__
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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.
__
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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] Loop and cbind
A more elegant way to do this is
standard - sapply(calm, function(calmi){calmi / sqrt(emaTA(calmi ^ 2,
0.03))})
Cheers,
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
Do not put your faith in what statistics say until you have carefully
considered what they do not say. ~William W. Watt
A statistical analysis, properly conducted, is a delicate dissection of
uncertainties, a surgery of suppositions. ~M.J.Moroney
-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Namens john seers (IFR)
Verzonden: woensdag 4 juli 2007 15:19
Aan: livia; r-help@stat.math.ethz.ch
Onderwerp: Re: [R] Loop and cbind
Hi
In what way does it not work?
My guess is that you have not declared your values outside
the for loop.
As they are local they will be lost on exit.
You need to declare them before:
ewma-vector(length=12)
standard-vector(length=12)
for ... {
}
John Seers
---
Hi, I would like to apply the following function for i
between 1 and 12, and then construct a list of the return series.
for (i in 1:12){
ewma[i] - emaTA(calm[[i]]^2,0.03)
standard[i]- calm[[i]]/sqrt(ewma[i])
standard - cbind(standard[i])
}
But it does not work. Could anyone give me some advice how
can I achieve this? Many thanks
--
View this message in context:
http://www.nabble.com/Loop-and-cbind-tf4024291.html#a11430500
Sent from the R help mailing list archive at Nabble.com.
__
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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.
__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
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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] Fine tunning rgenoud
On 7/4/07, RAVI VARADHAN [EMAIL PROTECTED] wrote: My point is that it might be better to try multiple (feasible) starting values for constrOptim to ensure that you have a good local minimum, since it appears that constrOptim converges to a boundary solution where the gradient is non-zero. That is why my code could be useful. Thanks, Ravi. I have used your function, which works pretty fine. However, constrOptim returns solutions markedly different, depending on the starting values. That is true that I am expecting a solution in the boundary, but should not constrOptim find boundary solutions correctly? The set of solution that I got is below. Paul 2.67682495728743e-080.676401684216637 5.18627076390355e-09 0.00206463986063195 0.871859686128364.32039325909089e-11 0.9996234 3.71711020733097e-080.539853580957444 1.82592937615235e-08 0.00206941041763503 0.933052503934472.08076621230984e-11 0.9995774 1.55648443014316e-080.356047772992972 8.61341165816411e-09 0.00207149128044574 0.939531540703735 2.55211186629222e-12 0.424 2.20685747493755e-070.575689534431218 5.30976753476747e-08 0.00210500604605837 0.588947341576757 3.1310360048386e-10 0.9998789 1.92961662926727e-080.773588030510204 1.04841835042200e-08 0.00206723852358352 0.816755014708394 3.89478290348532e-11 0.9997794 0.0002798240512890820.0003109923855228861.01467522935252e-06 3.11645639181419e-050.00249801538651552 3.0978819115532e-05 7.11821104872585e-06 2.81901448690893e-070.381718731525906 4.72860507882539e-08 0.00206807672109157 0.769178513763055 1.39278079797628e-09 0.9967123 5.58938545019597e-050.00171253668169328 4.54005998518212e-09 0.00165663757292733 0.00247994862102590 6.20992250482468e-06 0.419169641865998 1.03300938985890e-080.438357835603591 6.89854079723234e-09 0.00206693286138396 0.977554885433201 1.17209206267609e-10 0.996921 7.63336821363444e-050.00177141538041517 1.88050423143828e-10 0.00169507950991094 0.00249739505142207 8.4814984916537e-06 0.470929220605509 9.16005846107533e-090.682179815036755 1.63255733785783e-09 0.00206922107327189 0.919323193130209 5.71436138398897e-11 0.999629 1.40968913167328e-080.343606628343661 1.33227447885302e-08 0.00206789984370423 0.343671264496824 1.11679312116211e-11 0.822 4.76054734844857e-090.593022549313178 2.28102966623129e-09 0.00206625165098398 0.947562121256448 8.9437610753173e-11 0.992 1.96950784184139e-070.579488113726155 1.61915231214025e-07 0.00208000350528798 1.008913405950401.22248906754713e-10 0.9996493 8.1448937742933e-09 0.441088618716555 4.54846390087941e-09 0.00207634940425852 0.446155700100820 4.81439647816238e-12 0.39 4.82439218405912e-080.557771049256698 3.53737879481732e-08 0.0020663035737319 0.588137767965923 2.6568947800491e-11 0.9988615 2.43086751126363e-080.522927598354163 2.26886829089137e-08 0.00206533531066324 0.611696593543814 4.51226610050184e-11 0.087 3.05498959434100e-080.465522202845817 1.09246302124670e-08 0.00207004066920179 0.465583376966915 3.24213847202457e-11 0.9997366 1.88687179088788e-070.783614197203923 4.51346471059839e-08 0.00222403775221293 0.786422171740329 8.17865794171933e-10 0.9986103 1.0154423824979e-08 0.30265579883 9.06923080122203e-09 0.00206615353968094 0.359722316646974 8.27866320956902e-12 0.998461 8.91008717665837e-080.0020661526864997 3.08619455858999e-09 0.00206579199039568 0.00275523149199496 9.55650084108725e-09 0.985185595958656 1.25320647920029e-070.635217955401437 7.44627883600107e-08 0.00206656250455391 0.855937507707323 3.70326032870889e-10 0.9998375 2.57618374406559e-080.636499151952225 1.09822023878715e-08 0.00206677354204888 0.772636071860102 8.99370944431481e-11 0.9978744 1.09474196877990e-080.501469973722704 1.19992915868609e-10 0.00206117941606503 0.501594064757161 1.34320044786225e-11 0.9991232 5.24203710193977e-050.0001279983401441093.33258623630601e-09 7.55779680724378e-050.00248898574263025 5.82411313482383e-06 0.0221497278110802 3.80217498132259e-070.576645687031891.01755510162620e-08 0.00207232950382402 0.944031557945531 5.30703662426069e-10 0.9995957 1.45159816281038e-09
Re: [R] Fine tunning rgenoud
On 7/4/07, Paul Smith [EMAIL PROTECTED] wrote: On 7/4/07, RAVI VARADHAN [EMAIL PROTECTED] wrote: My point is that it might be better to try multiple (feasible) starting values for constrOptim to ensure that you have a good local minimum, since it appears that constrOptim converges to a boundary solution where the gradient is non-zero. That is why my code could be useful. Thanks, Ravi. I have used your function, which works pretty fine. However, constrOptim returns solutions markedly different, depending on the starting values. That is true that I am expecting a solution in the boundary, but should not constrOptim find boundary solutions correctly? The set of solution that I got is below. Unless, there are many local optimal solutions... Paul 2.67682495728743e-080.676401684216637 5.18627076390355e-09 0.00206463986063195 0.871859686128364.32039325909089e-11 0.9996234 3.71711020733097e-080.539853580957444 1.82592937615235e-08 0.00206941041763503 0.933052503934472.08076621230984e-11 0.9995774 1.55648443014316e-080.356047772992972 8.61341165816411e-09 0.00207149128044574 0.939531540703735 2.55211186629222e-12 0.424 2.20685747493755e-070.575689534431218 5.30976753476747e-08 0.00210500604605837 0.588947341576757 3.1310360048386e-10 0.9998789 1.92961662926727e-080.773588030510204 1.04841835042200e-08 0.00206723852358352 0.816755014708394 3.89478290348532e-11 0.9997794 0.0002798240512890820.0003109923855228861.01467522935252e-06 3.11645639181419e-050.00249801538651552 3.0978819115532e-05 7.11821104872585e-06 2.81901448690893e-070.381718731525906 4.72860507882539e-08 0.00206807672109157 0.769178513763055 1.39278079797628e-09 0.9967123 5.58938545019597e-050.00171253668169328 4.54005998518212e-09 0.00165663757292733 0.00247994862102590 6.20992250482468e-06 0.419169641865998 1.03300938985890e-080.438357835603591 6.89854079723234e-09 0.00206693286138396 0.977554885433201 1.17209206267609e-10 0.996921 7.63336821363444e-050.00177141538041517 1.88050423143828e-10 0.00169507950991094 0.00249739505142207 8.4814984916537e-06 0.470929220605509 9.16005846107533e-090.682179815036755 1.63255733785783e-09 0.00206922107327189 0.919323193130209 5.71436138398897e-11 0.999629 1.40968913167328e-080.343606628343661 1.33227447885302e-08 0.00206789984370423 0.343671264496824 1.11679312116211e-11 0.822 4.76054734844857e-090.593022549313178 2.28102966623129e-09 0.00206625165098398 0.947562121256448 8.9437610753173e-11 0.992 1.96950784184139e-070.579488113726155 1.61915231214025e-07 0.00208000350528798 1.008913405950401.22248906754713e-10 0.9996493 8.1448937742933e-09 0.441088618716555 4.54846390087941e-09 0.00207634940425852 0.446155700100820 4.81439647816238e-12 0.39 4.82439218405912e-080.557771049256698 3.53737879481732e-08 0.0020663035737319 0.588137767965923 2.6568947800491e-11 0.9988615 2.43086751126363e-080.522927598354163 2.26886829089137e-08 0.00206533531066324 0.611696593543814 4.51226610050184e-11 0.087 3.05498959434100e-080.465522202845817 1.09246302124670e-08 0.00207004066920179 0.465583376966915 3.24213847202457e-11 0.9997366 1.88687179088788e-070.783614197203923 4.51346471059839e-08 0.00222403775221293 0.786422171740329 8.17865794171933e-10 0.9986103 1.0154423824979e-08 0.30265579883 9.06923080122203e-09 0.00206615353968094 0.359722316646974 8.27866320956902e-12 0.998461 8.91008717665837e-080.0020661526864997 3.08619455858999e-09 0.00206579199039568 0.00275523149199496 9.55650084108725e-09 0.985185595958656 1.25320647920029e-070.635217955401437 7.44627883600107e-08 0.00206656250455391 0.855937507707323 3.70326032870889e-10 0.9998375 2.57618374406559e-080.636499151952225 1.09822023878715e-08 0.00206677354204888 0.772636071860102 8.99370944431481e-11 0.9978744 1.09474196877990e-080.501469973722704 1.19992915868609e-10 0.00206117941606503 0.501594064757161 1.34320044786225e-11 0.9991232 5.24203710193977e-050.0001279983401441093.33258623630601e-09 7.55779680724378e-050.00248898574263025 5.82411313482383e-06 0.0221497278110802
[R] copula estimation wih time series marginals
I am using R 2.5.1 for windows and my purpose is to estimate a clayton copula . Since I have two time series marginals, I found that the most appropriate model was an ARMA(1,0)+GARCH(1,1) model for both with sstd as conditional distribution. Can anyone give me some tips about the code to estimate the copula? Thanks in advance Gaetano Rossi -- Scegli infostrada: ADSL gratis per tutta lestate e telefoni senza canone Telecom http://click.libero.it/infostrada __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Please help with legend command
Hi R-ers: I'm drawing a plot and have used different line types (lty) for different race/ethnicity groups. I want a legend that explains what line types correspond to the different race/ethnicity groups. I used the following code: legend( 1992 , 42 , c(Hispanic , non-Hispanic white (NHW) , non-Hispanic black , AI/AN , Asian ) , lty=1:5 ,cex = .6 , bty='n' ) Guess what? The legend box was so narrow that the line types that show up in that legend box look essentially the same, because they are short. I.e, although a line type might be a long dash followed by a short dash, only the long dash shows up in the box. The consequence of this is that the race/ethnic group that corresponds to the line type that is only a long dash cannot be distinguished from the legend. How do I stretch that legend box out so as to allow lty to draw longer line segments? Please reply to: [EMAIL PROTECTED] Many thanks! Phil Smith Centers for Disease Control and Prevention __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A More efficient method?
Keith Alan Chamberlain Keith.Chamberlain at Colorado.EDU writes:
Cat=c('a','a','a','b','b','b','a','a','b')# Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
ifelse(Cat == a, -1, 1)
[1] -1 -1 -1 1 1 1 -1 -1 1
HTH
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[R] A More efficient method?
Dear Rhelpers,
Is there a faster way than below to set a vector based on values from
another vector? I'd like to call a pre-existing function for this, but one
which can also handle an arbitrarily large number of categories. Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
Sincerely,
KeithC.
Psych Undergrad, CU Boulder (US)
RE McNair Scholar
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Re: [R] A More efficient method?
Cat - c('a','a','a','b','b','b','a','a','b')
C1 - ifelse(Cat == 'a', -1, 1)
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
Do not put your faith in what statistics say until you have carefully
considered what they do not say. ~William W. Watt
A statistical analysis, properly conducted, is a delicate dissection of
uncertainties, a surgery of suppositions. ~M.J.Moroney
-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Namens Keith Alan
Chamberlain
Verzonden: woensdag 4 juli 2007 15:45
Aan: r-help@stat.math.ethz.ch
Onderwerp: [R] A More efficient method?
Dear Rhelpers,
Is there a faster way than below to set a vector based on
values from another vector? I'd like to call a pre-existing
function for this, but one which can also handle an
arbitrarily large number of categories. Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b')# Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
Sincerely,
KeithC.
Psych Undergrad, CU Boulder (US)
RE McNair Scholar
__
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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.
__
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Re: [R] how to solve a min problem
S is an array 1-dimensional, for example 1 X 10, and mean(S) is the mean of these 10 elements. So, I want to do: minimize mean(S) with 0 b_func(S) 800. That is, there are some boundaries on S according the b_funct The function apply an iterative convergent criterion: f_1=g(S), f_2=g(f_1), f_3=g(f_2), ecc The function stops when f_n - f_n-1 =0.1e-09 and g(S) is a non-linear function of S and the convergence is mathematically assured. Is it possible to use 'optimize'? thanks domenico 2007/7/3, Spencer Graves [EMAIL PROTECTED]: Do you mean minimize mu with 0 b_func(S+mu) 800? For this kind of problem, I'd first want to know the nature of b_func. Without knowing more, I might try to plot b_func(S+mu) vs. mu, then maybe use 'optimize'. If this is not what you mean, please be more specific: I'm confused. Hope this helps. Spencer Graves domenico pestalozzi wrote: I know it's possible to solve max e min problems by using these functions: nlm, optimize, optim but I don't know how to use them (...if possible...) to solve this problem. I have a personal function called b_func(S) where S is an input array (1 X n) and I'd like: minimize mean(S) with 0 b_funct 800. I know that the solution exists, but It's possible to calculate it in R? The b_func is non linear and it calculates a particular value using S as input and applying a convergent iterative algorithm. thanks domenico [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A More efficient method?
C1 - rep(-1, length(Cat))
C1[Cat == b]] - 1
b
On Jul 4, 2007, at 9:44 AM, Keith Alan Chamberlain wrote:
Dear Rhelpers,
Is there a faster way than below to set a vector based on values from
another vector? I'd like to call a pre-existing function for this,
but one
which can also handle an arbitrarily large number of categories.
Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b')# Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
Sincerely,
KeithC.
Psych Undergrad, CU Boulder (US)
RE McNair Scholar
__
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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.
__
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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] A More efficient method?
Cat=c('a','a','a','b','b','b','a','a','b')# Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
how about:
Cat-c('a','a','a','b','b','b','a','a','b')
c1- -2*(Cat==a)+1
-=-=-
... Time is an illusion, lunchtime doubly so. (Ford Prefect)
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Re: [R] A More efficient method?
On 04-Jul-07 13:44:44, Keith Alan Chamberlain wrote:
Dear Rhelpers,
Is there a faster way than below to set a vector based on values
from another vector? I'd like to call a pre-existing function for
this, but one which can also handle an arbitrarily large number
of categories. Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b')# Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
Cat=c('a','a','a','b','b','b','a','a','b')
Cat==b
[1] FALSE FALSE FALSE TRUE TRUE TRUE FALSE FALSE TRUE
(Cat==b) - 0.5
[1] -0.5 -0.5 -0.5 0.5 0.5 0.5 -0.5 -0.5 0.5
2*((Cat==b) - 0.5)
[1] -1 -1 -1 1 1 1 -1 -1 1
to give one example of a way to do it. But you don't say why you
really want to do this. You may really want factors. And what do
you want to see if there is an arbitrarily large number of
categories?
For instance:
factor(Cat,labels=c(-1,1))
[1] -1 -1 -1 1 1 1 -1 -1 1
but this is not a vector, but a factor object. To get the vector,
you need to convert Cat to an integer:
as.integer(factor(Cat))
[1] 1 1 1 2 2 2 1 1 2
where (unless you've specified otherwise in factor()) the values
will correspond to the elements of Cat in natural order, in this
case first a (- 1), then b (- 2).
E.g.
Cat2-c(a,a,c,b,a,b)
as.integer(factor(Cat2))
[1] 1 1 3 2 1 2
so, with C2-as.integer(factor(Cat2)), you get a vector of distinct
integers 91,2,3) for the distinct levels (a,b,c) of Cat2.
If you want integer values for these levels, you can write a function
to change them.
Hoping this helps to beark the ice!
Ted.
E-Mail: (Ted Harding) [EMAIL PROTECTED]
Fax-to-email: +44 (0)870 094 0861
Date: 04-Jul-07 Time: 16:44:20
-- XFMail --
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Re: [R] A More efficient method?
Here are two ways. The second way is more than 10x faster.
set.seed(1)
C - sample(c(a, b), 10, replace = TRUE)
system.time(s1 - ifelse(C == a, 1, -1))
user system elapsed
0.370.010.38
system.time(s2 - 2 * (C == a) - 1)
user system elapsed
0.020.000.02
identical(s1, s2)
[1] TRUE
On 7/4/07, Keith Alan Chamberlain [EMAIL PROTECTED] wrote:
Dear Rhelpers,
Is there a faster way than below to set a vector based on values from
another vector? I'd like to call a pre-existing function for this, but one
which can also handle an arbitrarily large number of categories. Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
Sincerely,
KeithC.
Psych Undergrad, CU Boulder (US)
RE McNair Scholar
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] A More efficient method?
or
Cat - c('a','a','a','b','b','b','a','a','b')
C1 - (Cat=='a')*1
ONKELINX,
Thierry
Thierry.ONKELINX To
@inbo.be Keith Alan Chamberlain
Sent by: [EMAIL PROTECTED],
[EMAIL PROTECTED] r-help@stat.math.ethz.ch
at.math.ethz.chcc
Subject
04/07/2007 17:17 Re: [R] A More efficient method?
Cat - c('a','a','a','b','b','b','a','a','b')
C1 - ifelse(Cat == 'a', -1, 1)
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
Do not put your faith in what statistics say until you have carefully
considered what they do not say. ~William W. Watt
A statistical analysis, properly conducted, is a delicate dissection of
uncertainties, a surgery of suppositions. ~M.J.Moroney
-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Namens Keith Alan
Chamberlain
Verzonden: woensdag 4 juli 2007 15:45
Aan: r-help@stat.math.ethz.ch
Onderwerp: [R] A More efficient method?
Dear Rhelpers,
Is there a faster way than below to set a vector based on
values from another vector? I'd like to call a pre-existing
function for this, but one which can also handle an
arbitrarily large number of categories. Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b') # Categorical
variable
C1=vector(length=length(Cat))# New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
Sincerely,
KeithC.
Psych Undergrad, CU Boulder (US)
RE McNair Scholar
__
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Re: [R] A More efficient method?
Gabor Grothendieck wrote: set.seed(1) C - sample(c(a, b), 10, replace = TRUE) system.time(s1 - ifelse(C == a, 1, -1)) user system elapsed 0.370.010.38 system.time(s2 - 2 * (C == a) - 1) user system elapsed 0.020.000.02 system.time(s1 - ifelse(C == a, 1, -1)) user system elapsed 0.040.010.08 system.time(s2 - 2 * (C == a) - 1) user system elapsed 0 0 0 I am just wondering: how comes the time does add up to 0.05 while elapsed states 0.08 on my system? (Vista+R2.5.1) Stefan -=-=- ... Time is an illusion, lunchtime doubly so. (Ford Prefect) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A More efficient method?
#Given
Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable
#and defining
coding-array(c(-1,1), dimnames=list(unique(Cat) ))
#(ie an array of values corresponding to your character array levels, and with
names set to those levels)
coding[Cat]
#does what you want.
Keith Alan Chamberlain [EMAIL PROTECTED] 04/07/2007 14:44:44
Dear Rhelpers,
Is there a faster way than below to set a vector based on values from
another vector? I'd like to call a pre-existing function for this, but one
which can also handle an arbitrarily large number of categories. Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
Sincerely,
KeithC.
Psych Undergrad, CU Boulder (US)
RE McNair Scholar
__
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and provide commented, minimal, self-contained, reproducible code.
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[R] working with matrix
I am new in R and I want to solve this problem; I have a matrix X (with n-rows and p-colums) my problem is to obtain the products of the vectors of rows and print out only the maximum value and identify the row that gives the maximum value. Thanks Oyeyemi, G.M - Don't pick lemons. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] working with matrix
which.max(apply(mat, 1, prod)) Gabor On Wed, Jul 04, 2007 at 09:18:41AM -0700, Yemi Oyeyemi wrote: I am new in R and I want to solve this problem; I have a matrix X (with n-rows and p-colums) my problem is to obtain the products of the vectors of rows and print out only the maximum value and identify the row that gives the maximum value. Thanks Oyeyemi, G.M - Don't pick lemons. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Csardi Gabor [EMAIL PROTECTED]MTA RMKI, ELTE TTK __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A More efficient method?
Dear Ted, You are correct in that factors are probably what I had in mind since I would be using them as predictors in a regression. I didn't know the syntax to get R to do the arithmetic. Many thanks to everyone who replied! Sincerely, KeithC. Psych Undergrad, CU Boulder (US) RE McNair Scholar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A More efficient method?
In thinking about this a bit more I have found a slightly faster one still.
See s3. Also I have added s0, the original solution, to the timings.
set.seed(1)
C - sample(c(a, b), 100, replace = TRUE)
system.time({
+ s0 - vector(length = length(C))
+ for(i in seq_along(C)) s0[i] - if (C[i] == a) 1 else -1
+ s0
+ })
user system elapsed
21.750.02 25.99
system.time(s1 - ifelse(C == a, 1, -1))
user system elapsed
2.320.172.54
system.time(s2 - 2 * (C == a) - 1)
user system elapsed
0.290.020.32
system.time({tmp - C == a; tmp - !tmp})
user system elapsed
0.210.000.21
identical(s0, s1)
[1] TRUE
identical(s0, s2)
[1] TRUE
identical(s0, s3)
[1] TRUE
On 7/4/07, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are two ways. The second way is more than 10x faster.
set.seed(1)
C - sample(c(a, b), 10, replace = TRUE)
system.time(s1 - ifelse(C == a, 1, -1))
user system elapsed
0.370.010.38
system.time(s2 - 2 * (C == a) - 1)
user system elapsed
0.020.000.02
identical(s1, s2)
[1] TRUE
On 7/4/07, Keith Alan Chamberlain [EMAIL PROTECTED] wrote:
Dear Rhelpers,
Is there a faster way than below to set a vector based on values from
another vector? I'd like to call a pre-existing function for this, but one
which can also handle an arbitrarily large number of categories. Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
Sincerely,
KeithC.
Psych Undergrad, CU Boulder (US)
RE McNair Scholar
__
R-help@stat.math.ethz.ch 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.
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Re: [R] working with matrix
Yemi,
Try
which.max(apply(X,1,prod))
(or possibly abs(apply(X,1,prod)) if you're only interested in unsigned product
max.
S
Yemi Oyeyemi [EMAIL PROTECTED] 04/07/2007 17:18:41
I am new in R and I want to solve this problem;
I have a matrix X (with n-rows and p-colums) my problem is to obtain the
products of the vectors of rows and print out only the maximum value and
identify the row that gives the maximum value. Thanks
Oyeyemi, G.M
-
Don't pick lemons.
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] A More efficient method?
This was in error since s3 was not set. The as.numeric in the calculation
of s3 can be omitted if its ok to have an integer rather than numeric result
and in that case its still faster yet.
set.seed(1)
C - sample(c(a, b), 100, replace = TRUE)
system.time({
+ s0 - vector(length = length(C))
+ for(i in seq_along(C)) s0[i] - if (C[i] == a) 1 else -1
+ s0
+ })
user system elapsed
21.320.02 26.10
system.time(s1 - ifelse(C == a, 1, -1))
user system elapsed
2.370.262.64
system.time(s2 - 2 * (C == a) - 1)
user system elapsed
0.320.020.35
system.time({tmp - C == a; s3 - as.numeric(tmp - !tmp)})
user system elapsed
0.280.020.31
identical(s0, s1)
[1] TRUE
identical(s0, s2)
[1] TRUE
identical(s0, s3)
[1] TRUE
On 7/4/07, Gabor Grothendieck [EMAIL PROTECTED] wrote:
In thinking about this a bit more I have found a slightly faster one still.
See s3. Also I have added s0, the original solution, to the timings.
set.seed(1)
C - sample(c(a, b), 100, replace = TRUE)
system.time({
+ s0 - vector(length = length(C))
+ for(i in seq_along(C)) s0[i] - if (C[i] == a) 1 else -1
+ s0
+ })
user system elapsed
21.750.02 25.99
system.time(s1 - ifelse(C == a, 1, -1))
user system elapsed
2.320.172.54
system.time(s2 - 2 * (C == a) - 1)
user system elapsed
0.290.020.32
system.time({tmp - C == a; tmp - !tmp})
user system elapsed
0.210.000.21
identical(s0, s1)
[1] TRUE
identical(s0, s2)
[1] TRUE
identical(s0, s3)
[1] TRUE
On 7/4/07, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are two ways. The second way is more than 10x faster.
set.seed(1)
C - sample(c(a, b), 10, replace = TRUE)
system.time(s1 - ifelse(C == a, 1, -1))
user system elapsed
0.370.010.38
system.time(s2 - 2 * (C == a) - 1)
user system elapsed
0.020.000.02
identical(s1, s2)
[1] TRUE
On 7/4/07, Keith Alan Chamberlain [EMAIL PROTECTED] wrote:
Dear Rhelpers,
Is there a faster way than below to set a vector based on values from
another vector? I'd like to call a pre-existing function for this, but one
which can also handle an arbitrarily large number of categories. Any
ideas?
Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
Sincerely,
KeithC.
Psych Undergrad, CU Boulder (US)
RE McNair Scholar
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] About dataset
Hi R hep group member, I want to know that how I can call my data in R for princomp function. I want to calculate PCA of 200 descriptors of 4000 molecule(I am using Linux). How I can call this in R. Thanking you. -- Nitish Kumar Mishra Junior Research Fellow BIC, IMTECH, Chandigarh, India E-Mail Address: [EMAIL PROTECTED] [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Working with matrix
Dear, Thanks for your prompt assistant. The problem I posted is just a bit of my problem. The whole problem is that; 1. I simulated a multivariate data set X (n by p) 2. Then draw sample without replacement from X (k less than n) 3. Compute the eigen values of the variance-covariance matrix of the sample drawn 4. Store the eigen values in matrix eigenvals 5. Store the samples(that is the row identifier) drawn in rowvals 6. Bind the two matrices using rbind. The problem is; 7. Now I want the print out of the maximum values of the eigenvalues product and samples that give such maximum product. Thanks Oyeyemi Gafar M. - Fussy? Opinionated? Impossible to please? Perfect. Join Yahoo!'s user panel and lay it on us. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Long-tail model in R ... anyone?
Dear all,
first I would like to tell you that I've been using R for two days... (so,
you can predict my knowledge of the language!).
Yet, I managed to implement some stuff related with the Long-Tail model [1].
I did some tests with the data in table 1 (from [1]), and plotted figure 2
(from [1]). (See R code and CSV file at the end of the email)
Now, I'm stuck in the nonlinear regression model of F(x). I got a nice error:
Error in nls(~F(r, N50, beta, alfa), data = dataset, start = list(N50 =
N50, : singular gradient
And, yes, I've been looking for how to solve this (via this mailing list +
some google), and I could not come across to a proper solution. That's why
I am asking the experts to help me! :-)
So, any help would be much appreciated...
Cheers, Oscar
[1] http://www.firstmonday.org/issues/issue12_5/kilkki/
PS: R code and CVS file
FILE: data.R (data taken from [1] Table 1, columns 1 and 2)
--8=---
rank,cum_value
10, 17396510
32, 31194809
96, 53447300
420,100379331
1187, 152238166
24234, 432238757
91242, 581332371
294180, 650880870
1242185,665227287
--=8---
R CODE:
#
# F(x). The long-tail model
# Reference: http://www.firstmonday.org/issues/issue12_5/kilkki/
# Params:
# x : Rank (either an integer or a list)
# N50 : the number of objects that cover half of the whole volume
# beta: total volume
# alfa: the factor that defines the form of the function
F - function (x, N50, beta=1.0, alfa=0.49)
{
xx - as.numeric(x) # as.numeric() prevents overflow
Fx = beta / ( (N50/xx)^alfa + 1 )
Fx
}
# Read CSV file (rank, cum_value)
lt - read.csv(file=data.R,head=TRUE,sep=,)
r - lt$rank
v - lt$cum_value
pcnt - v/v[length(v)] *100 # get cumulative percentage
plot(r, pcnt, log=x, type='l', xlab='Ranking', ylab='Cumulative
percentatge of sales', main=Books Popularity, sub=The long-tail
effect, col='blue')
# Set some default values to be used by F(x)...
alfa = 0.49
beta = 1.38
N50 = 30714
# Start using F(x). Results are in 'f' ...
f - c(0) # oops! is this the best initialization for 'f'?
for (i in 1:24234) f[i] - F(i, N50, beta, alfa)*100
# Plot some estimated values from F(x) (N50, beta, and alfa values come
from the paper. See ref. [1])
plot(f, log=x, type='l', xlab='Ranking', ylab='Cumulative percentatge of
sales', main=Books Popularity, sub=Plotting first values of F(x) and
some real points)
points(r, pcnt, col=blue) # adding the real points
# Create a dataset to be used by nls()
dataset - data.frame(r, pcnt)
# Verifying that F(x) works fine... (comparing with the real values
contained in the dataset)
dataset
F(10, N50, beta, alfa) * 100
F(32, N50, beta, alfa) * 100
F(96, N50, beta, alfa) * 100
F(420, N50, beta, alfa) * 100
F(1187, N50, beta, alfa) * 100
F(24234, N50, beta, alfa) * 100
F(91242, N50, beta, alfa) * 100
F(294180, N50, beta, alfa) * 100
F(1242185, N50, beta, alfa) * 100
#dataset - data.frame(pcnt) # which dataset should I use? Should I
include the ranks in it?
nls( ~ F(r, N50, beta, alfa), data = dataset, start = list(N50=N50,
beta=beta, alfa=alfa), trace = TRUE )
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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] A More efficient method?
[Keith Alan Chamberlain]
Is there a faster way than below to set a vector based on values
from another vector? I'd like to call a pre-existing function for
this, but one which can also handle an arbitrarily large number of
categories. Any ideas?
Cat=c('a','a','a','b','b','b','a','a','b') # Categorical variable
C1=vector(length=length(Cat)) # New vector for numeric values
# Cycle through each column and set C1 to corresponding value of Cat.
for(i in 1:length(C1)){
if(Cat[i]=='a') C1[i]=-1 else C1[i]=1
}
C1
[1] -1 -1 -1 1 1 1 -1 -1 1
Cat
[1] a a a b b b a a b
For handling an arbitrarily large number of categories, one may go
through a recoding vector, like this for the example above:
Cat - c('a', 'a', 'a', 'b', 'b', 'b', 'a', 'a', 'b')
C1 - c(a=-1, b=1)[Cat]
C1
a a a b b b a a b
-1 -1 -1 1 1 1 -1 -1 1
--
François Pinard http://pinard.progiciels-bpi.ca
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Re: [R] A More efficient method?
User and System are a measure of the CPU time that was consumed. Elapsed time is the wall clock and even though they are both measured in seconds, they are not really the same units. The reason for the difference is any idle time that they system may have waiting for I/O to complete which does not consume CPU time for your process, but does consume Elasped time. For some instances of CPU intensive code (with no I/O of competing tasks), the User + System ~= Elapsed. Also you have to take into account the granularity of the clock when looking at numbers like 0.04. So serious comparisons of timing, you want runs of at least 10s of seconds or more. On 7/4/07, Stefan Grosse [EMAIL PROTECTED] wrote: Gabor Grothendieck wrote: set.seed(1) C - sample(c(a, b), 10, replace = TRUE) system.time(s1 - ifelse(C == a, 1, -1)) user system elapsed 0.370.010.38 system.time(s2 - 2 * (C == a) - 1) user system elapsed 0.020.000.02 system.time(s1 - ifelse(C == a, 1, -1)) user system elapsed 0.040.010.08 system.time(s2 - 2 * (C == a) - 1) user system elapsed 0 0 0 I am just wondering: how comes the time does add up to 0.05 while elapsed states 0.08 on my system? (Vista+R2.5.1) Stefan -=-=- ... Time is an illusion, lunchtime doubly so. (Ford Prefect) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to solve a min problem
Whether you can use optim or not depends on the nature of the constraints on S. If you have simple box constraints, you can use the L-BFGS-B method in optim. If not, optim may not be directly applicable, unless you can somehow transform your problem into an unconstrained minimization problem. Ravi. - Original Message - From: domenico pestalozzi [EMAIL PROTECTED] Date: Wednesday, July 4, 2007 11:26 am Subject: Re: [R] how to solve a min problem To: R-help r-help@stat.math.ethz.ch S is an array 1-dimensional, for example 1 X 10, and mean(S) is the mean of these 10 elements. So, I want to do: minimize mean(S) with 0 b_func(S) 800. That is, there are some boundaries on S according the b_funct The function apply an iterative convergent criterion: f_1=g(S), f_2=g(f_1), f_3=g(f_2), ecc The function stops when f_n - f_n-1 =0.1e-09 and g(S) is a non-linear function of S and the convergence is mathematically assured. Is it possible to use 'optimize'? thanks domenico 2007/7/3, Spencer Graves [EMAIL PROTECTED]: Do you mean minimize mu with 0 b_func(S+mu) 800? For this kind of problem, I'd first want to know the nature of b_func. Without knowing more, I might try to plot b_func(S+mu) vs. mu, then maybe use 'optimize'. If this is not what you mean, please be more specific: I'm confused. Hope this helps. Spencer Graves domenico pestalozzi wrote: I know it's possible to solve max e min problems by using these functions: nlm, optimize, optim but I don't know how to use them (...if possible...) to solve this problem. I have a personal function called b_func(S) where S is an input array (1 X n) and I'd like: minimize mean(S) with 0 b_funct 800. I know that the solution exists, but It's possible to calculate it in R? The b_func is non linear and it calculates a particular value using S as input and applying a convergent iterative algorithm. thanks domenico [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list PLEASE do read the posting guide and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list PLEASE do read the posting guide and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] A More efficient method?
One other thing, in a multiprocessor configuration, if your application is making use of the additional CPUs, then User + System Elapsed In some cases. On 7/4/07, jim holtman [EMAIL PROTECTED] wrote: User and System are a measure of the CPU time that was consumed. Elapsed time is the wall clock and even though they are both measured in seconds, they are not really the same units. The reason for the difference is any idle time that they system may have waiting for I/O to complete which does not consume CPU time for your process, but does consume Elasped time. For some instances of CPU intensive code (with no I/O of competing tasks), the User + System ~= Elapsed. Also you have to take into account the granularity of the clock when looking at numbers like 0.04. So serious comparisons of timing, you want runs of at least 10s of seconds or more. On 7/4/07, Stefan Grosse [EMAIL PROTECTED] wrote: Gabor Grothendieck wrote: set.seed(1) C - sample(c(a, b), 10, replace = TRUE) system.time(s1 - ifelse(C == a, 1, -1)) user system elapsed 0.370.010.38 system.time(s2 - 2 * (C == a) - 1) user system elapsed 0.020.000.02 system.time(s1 - ifelse(C == a, 1, -1)) user system elapsed 0.040.010.08 system.time(s2 - 2 * (C == a) - 1) user system elapsed 0 0 0 I am just wondering: how comes the time does add up to 0.05 while elapsed states 0.08 on my system? (Vista+R2.5.1) Stefan -=-=- ... Time is an illusion, lunchtime doubly so. (Ford Prefect) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to solve a min problem
If the constraints on S are linear inequalities, then linear programming methods would work. See function solveLP in package linprog or simplex in boot or package lpSolve. Ravi. - Original Message - From: domenico pestalozzi [EMAIL PROTECTED] Date: Wednesday, July 4, 2007 11:26 am Subject: Re: [R] how to solve a min problem To: R-help r-help@stat.math.ethz.ch S is an array 1-dimensional, for example 1 X 10, and mean(S) is the mean of these 10 elements. So, I want to do: minimize mean(S) with 0 b_func(S) 800. That is, there are some boundaries on S according the b_funct The function apply an iterative convergent criterion: f_1=g(S), f_2=g(f_1), f_3=g(f_2), ecc The function stops when f_n - f_n-1 =0.1e-09 and g(S) is a non-linear function of S and the convergence is mathematically assured. Is it possible to use 'optimize'? thanks domenico 2007/7/3, Spencer Graves [EMAIL PROTECTED]: Do you mean minimize mu with 0 b_func(S+mu) 800? For this kind of problem, I'd first want to know the nature of b_func. Without knowing more, I might try to plot b_func(S+mu) vs. mu, then maybe use 'optimize'. If this is not what you mean, please be more specific: I'm confused. Hope this helps. Spencer Graves domenico pestalozzi wrote: I know it's possible to solve max e min problems by using these functions: nlm, optimize, optim but I don't know how to use them (...if possible...) to solve this problem. I have a personal function called b_func(S) where S is an input array (1 X n) and I'd like: minimize mean(S) with 0 b_funct 800. I know that the solution exists, but It's possible to calculate it in R? The b_func is non linear and it calculates a particular value using S as input and applying a convergent iterative algorithm. thanks domenico [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list PLEASE do read the posting guide and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list PLEASE do read the posting guide and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] read items
?scan Help page says to set quiet=TRUE On 7/4/07, elyakhlifi mustapha [EMAIL PROTECTED] wrote: Hello, I write us because I wanna know if it's possible to don't display read item when I execute a scan(textConnection()) thanks _ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Kmeans performance difference
Hi All, A question from a newbie using R 2-5-0 on windows XP. Why is it that kmeans clustering with apparently the exact same parameters behaves so differently between the two following examples : cl1 - kmeans(subset(pointsUXO1, select = c(2:4)), 10) Takes about 2 seconds to deliver a result cl1 - clust(subset(pointsUXO1, select = c(2:4)), k=10, method=kmeansHartigan) Dies after about 10 minutes and fills up RAM : *** running kmeansHartigan cluster algorithm... *** calculating validity measure... Erreur : impossible d'allouer un vecteur de taille 922.9 Mo De plus : Warning messages: 1: Reached total allocation of 1023Mb: see help(memory.size) 2: Reached total allocation of 1023Mb: see help(memory.size) 3: Reached total allocation of 1023Mb: see help(memory.size) 4: Reached total allocation of 1023Mb: see help(memory.size) If I understand correctly, both methods should give the sameish results (modulo the initial random locations) since the default in kmeans is Hartigan-Wong. My data frame is 3 columns X 1 lines. It must be that kmeans is more a core R function whereas clust id from the clustTool package, but isn't clustTool simply wrapping the core kmeans method ? Why such a difference ? TIA, Yves Moisan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Lookups in R
Hey all; I'm a beginner++ user of R, trying to use it to do some processing
of data sets of over 1M rows, and running into a snafu. imagine that my
input is a huge table of transactions, each linked to a specif user id. as
I run through the transactions, I need to update a separate table for the
users, but I am finding that the traditional ways of doing a table lookup
are way too slow to support this kind of operation.
i.e:
for(i in 1:100) {
userid = transactions$userid[i];
amt = transactions$amounts[i];
users[users$id == userid,'amt'] += amt;
}
I assume this is a linear lookup through the users table (in which there are
10's of thousands of rows), when really what I need is O(constant time), or
at worst O(log(# users)).
is there any way to manage a list of ID's (be they numeric, string, etc) and
have them efficiently mapped to some other table index?
I see the CRAN package for SQLite hashes, but that seems to be going a bit
too far.
thanks,
Mike
Intern, Oyster Card Group, Transport for London
(feel free to email back to this address, I'm posting through NAbble so I
hope it works).
--
View this message in context:
http://www.nabble.com/Lookups-in-R-tf4026062.html#a11435994
Sent from the R help mailing list archive at Nabble.com.
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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] Long-tail model in R ... anyone?
I think you simply had your nls() syntax wrong. Works here: ## first a neat trick to read the data from embedded text fmdata - read.csv(textConnection( + rank,cum_value 10, 17396510 32, 31194809 96, 53447300 420,100379331 1187, 152238166 24234, 432238757 91242, 581332371 294180, 650880870 1242185,665227287)) ## then compute cumulative share fmdata[,cumshare] - fmdata[,cum_value] / fmdata[nrow(fmdata),cum_value] ## then check the data, just in case summary(fmdata) rank cum_valuecumshare Min. : 10 Min. : 17396510 Min. :0.02615 1st Qu.: 96 1st Qu.: 53447300 1st Qu.:0.08034 Median : 1187 Median :152238166 Median :0.22885 Mean : 183732 Mean :298259489 Mean :0.44836 3rd Qu.: 91242 3rd Qu.:581332371 3rd Qu.:0.87389 Max. :1242185 Max. :665227287 Max. :1.0 ## finally estimate the model, using only the first seven rows of data ## using the parametric form from the paper and some wild guesses as ## starting values: fit - nls(cumshare ~ Beta / ((N50 / rank)^Alpha + 1), data=fmdata[1:7,], start=list(Alpha=1, Beta=1, N50=1e4)) summary(fit) Formula: cumshare ~ Beta/((N50/rank)^Alpha + 1) Parameters: Estimate Std. Error t value Pr(|t|) Alpha 4.829e-01 5.374e-03 89.86 9.20e-08 *** Beta 1.429e+00 2.745e-02 52.07 8.14e-07 *** N50 3.560e+04 3.045e+03 11.69 0.000306 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 0.002193 on 4 degrees of freedom Number of iterations to convergence: 8 Achieved convergence tolerance: 1.297e-06 which is reasonably close to the quoted N50 = 30714, α = 0.49, and β = 1.38. You can probably play a little with the nls options to see what effect this has. That said, seven observations for three parameters in non-linear model may be a little hazardous. One indication is that the estimated parameters values are not too stable once you add the eights and nineth row of data. Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lookups in R
mfrumin wrote:
Hey all; I'm a beginner++ user of R, trying to use it to do some processing
of data sets of over 1M rows, and running into a snafu. imagine that my
input is a huge table of transactions, each linked to a specif user id. as
I run through the transactions, I need to update a separate table for the
users, but I am finding that the traditional ways of doing a table lookup
are way too slow to support this kind of operation.
i.e:
for(i in 1:100) {
userid = transactions$userid[i];
amt = transactions$amounts[i];
users[users$id == userid,'amt'] += amt;
}
I assume this is a linear lookup through the users table (in which there are
10's of thousands of rows), when really what I need is O(constant time), or
at worst O(log(# users)).
is there any way to manage a list of ID's (be they numeric, string, etc) and
have them efficiently mapped to some other table index?
I see the CRAN package for SQLite hashes, but that seems to be going a bit
too far.
Sometimes you need a bit of lateral thinking. I suspect that you could
do it like this:
tbl - with(transactions, tapply(amount, userid, sum))
users$amt - users$amt + tbl[users$id]
one catch is that there could be users with no transactions, in which
case you may need to replace userid by factor(userid, levels=users$id).
None of this is tested, of course.
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] New function combinations using tree structures?
Hi everybody, I'm still interesting the possibility to use R for genetic programming (see https://stat.ethz.ch/pipermail/r-help/2007-April/128782.html) and I'd like to know, how to express for instance this kind of functions (x^2+3x+1 etc, see the picture) using some kind of tree structures and what are needed to make similar recombination operations with them as seen in the picture: http://users.utu.fi/attenka/Kuva1.png I know the basic principles of mutation and mating etc, as far as genetic algorithms are concerned, but now I'd like to know the practical issues considering R and mating of these functions using trees. All suggestions are warmly appreciated. With best wishes, Atte __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to install R 2.5 with Synaptic in Ubuntu?
mike, try installing directly using apt-get instead of Synaptic. in my /etc/apt/sources.list i added the line: deb http://cran.R-project.org/bin/linux/ubuntu/ dapper/ and then i did: bash$ sudo apt-get install r-base r-doc-info r-doc-pdf r-doc-html r-mathlib r-base-html r-base-latex r-base-dev r-gnome recently to update to R 2.5.1 on my version of Ubuntu (6.06). cheers, thomas. Message: 98 Date: Wed, 4 Jul 2007 02:34:37 -0700 (PDT) From: msmith [EMAIL PROTECTED] Subject: Re: [R] How to install R 2.5 with Synaptic in Ubuntu? To: r-help@stat.math.ethz.ch Message-ID: [EMAIL PROTECTED] Content-Type: text/plain; charset=us-ascii Hi, Thanks for the suggestion and I wish the solution was that obvious, but I have changed it to really point at my favourite mirror. Using your example Synaptic reports the following error when I try to update the repositories: http://www.stats.bris.ac.uk/R/bin/linux/ubuntu/dists/feisty/main/binary-i386/Packages.gz: 404 Not Found This is understandable since that location doesn't exist, but it makes me think that the directory structure of the R mirrors is not compatible with Ubuntu and Synaptic, since it automatically seeks /dists/feisty/ rather than just /feisty/ as it is on the CRAN mirrors. Thanks again Mike Smith Stefan Grosse-2 wrote: to end of the entry making it: deb http://my.favorite.cran.mirror/bin/linux/ubuntu feisty main However after this it still complains that it can't find packages.gz Just a guess: have you replaced the my.favorite.cran.mirror by a mirror which is close to you? If you're in UK it would be for example deb http://www.stats.bris.ac.uk/R/bin/linux/ubuntu feisty main ;o) Stefan It appears to be looking in http://my.favorite.cran.mirror/bin/linux/ubuntu/distsfeisty which isn't the directory structure of the cran repository, but I can see anyway to modify this behaviour. Every other Ubuntu repositoy I have looked at contains the dists directory. Any suggestions for modifying this behaviour are gratefully recieved. Many thanks Mike Smith -=-=- ... The simple truth is that interstellar distances will not fit into the human imagination - (The Hitchhiker's Guide to the Galaxy) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/How-to-install-R-2.5-with-Synaptic-in-Ubuntu--tf3998481.html#a11427837 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lookups in R
Michael,
A hash provides constant-time access, though the resulting perl-esque
data structures (a hash of lists, e.g.) are not convenient for other
manipulations
n_accts - 10^3
n_trans - 10^4
t - list()
t$amt - runif(n_trans)
t$acct - as.character(round(runif(n_trans, 1, n_accts)))
uhash - new.env(hash=TRUE, parent=emptyenv(), size=n_accts)
## keys, presumably account ids
for (acct in as.character(1:n_accts)) uhash[[acct]] - list(amt=0, n=0)
system.time(for (i in seq_along(t$amt)) {
+ acct - t$acct[i]
+ x - uhash[[acct]]
+ uhash[[acct]] - list(amt=x$amt + t$amt[i], n=x$n + 1)
+ })
user system elapsed
0.264 0.000 0.262
udf - data.frame(amt=0, n=rep(0L, n_accts),
+ row.names=as.character(1:n_accts))
system.time(for (i in seq_along(t$amt)) {
+ idx - row.names(udf)==t$acct[i]
+ udf[idx, ] - c(udf[idx,amt], udf[idx, n]) + c(t$amt[i], 1)
+ })
user system elapsed
18.398 0.000 18.394
Peter Dalgaard [EMAIL PROTECTED] writes:
mfrumin wrote:
Hey all; I'm a beginner++ user of R, trying to use it to do some processing
of data sets of over 1M rows, and running into a snafu. imagine that my
input is a huge table of transactions, each linked to a specif user id. as
I run through the transactions, I need to update a separate table for the
users, but I am finding that the traditional ways of doing a table lookup
are way too slow to support this kind of operation.
i.e:
for(i in 1:100) {
userid = transactions$userid[i];
amt = transactions$amounts[i];
users[users$id == userid,'amt'] += amt;
}
I assume this is a linear lookup through the users table (in which there are
10's of thousands of rows), when really what I need is O(constant time), or
at worst O(log(# users)).
is there any way to manage a list of ID's (be they numeric, string, etc) and
have them efficiently mapped to some other table index?
I see the CRAN package for SQLite hashes, but that seems to be going a bit
too far.
Sometimes you need a bit of lateral thinking. I suspect that you could
do it like this:
tbl - with(transactions, tapply(amount, userid, sum))
users$amt - users$amt + tbl[users$id]
one catch is that there could be users with no transactions, in which
case you may need to replace userid by factor(userid, levels=users$id).
None of this is tested, of course.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Martin Morgan
Bioconductor / Computational Biology
http://bioconductor.org
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lookups in R
i wish it were that simple. unfortunately the logic i have to do on
each transaction is substantially more complicated, and involves
referencing the existing values of the user table through a number of
conditions.
any other thoughts on how to get better-than-linear performance time?
is there a recommended binary searching/sorting (i.e. BTree) module that
I could use to maintain my own index?
thanks,
mike
Peter Dalgaard wrote:
mfrumin wrote:
Hey all; I'm a beginner++ user of R, trying to use it to do some
processing
of data sets of over 1M rows, and running into a snafu. imagine that my
input is a huge table of transactions, each linked to a specif user
id. as
I run through the transactions, I need to update a separate table for
the
users, but I am finding that the traditional ways of doing a table
lookup
are way too slow to support this kind of operation.
i.e:
for(i in 1:100) {
userid = transactions$userid[i];
amt = transactions$amounts[i];
users[users$id == userid,'amt'] += amt;
}
I assume this is a linear lookup through the users table (in which
there are
10's of thousands of rows), when really what I need is O(constant
time), or
at worst O(log(# users)).
is there any way to manage a list of ID's (be they numeric, string,
etc) and
have them efficiently mapped to some other table index?
I see the CRAN package for SQLite hashes, but that seems to be going
a bit
too far.
Sometimes you need a bit of lateral thinking. I suspect that you could
do it like this:
tbl - with(transactions, tapply(amount, userid, sum))
users$amt - users$amt + tbl[users$id]
one catch is that there could be users with no transactions, in which
case you may need to replace userid by factor(userid,
levels=users$id). None of this is tested, of course.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lookups in R
__ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Empirical copula in R
I just got 203 hits from RSiteSearch(copula) and 60 from RSiteSearch(copula, fun). Most if not all of the first 24 or the hits in the latter referred to the 'copula' package. Have you reviewed these? The 25th hit in the latter referred to an 'fgac' package for 'Generalized Archimedean Copula', with a Brazilian author and maintainer, who presumably is not related to the 'copula' author and maintainer at U. Iowa. Also, have you tried contacting the official maintainers of these packages? You can get an email address for them from help(package=copula) and help (package=fgac). Hope this helps. Spencer Graves GWeiss wrote: Hi, I would like to implement the empirical copula in R, does anyone know if it is included in a package? I know it is not in the Copula package. This one only includes a gof-test based on the empirical copula process. Thanks for your help! Gregor __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Newbie creating package with compiled code
Hi R Gurus! I'm trying to create a test package using the package.skeleton function. I wanted to add some compiled code too. In the src library, I put together a baby subroutine, compiled it and created a test.dll When I use the R cmd build, it works fine. But I get into trouble with the R CMD check section. /home/Desktop/R-2.5.1/bin # ./R CMD check mypkg * checking for working latex ... OK * using log directory '/home/Desktop/R-2.5.1/bin/mypkg.Rcheck' * using R version 2.5.1 (2007-06-27) * checking for file 'mypkg/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'mypkg' version '1.0' * checking package dependencies ... OK * checking if this is a source package ... OK * checking whether package 'mypkg' can be installed ... OK * checking package directory ... OK * checking for portable file names ... OK * checking for sufficient/correct file permissions ... OK * checking DESCRIPTION meta-information ... OK * checking top-level files ... OK * checking index information ... OK * checking package subdirectories ... OK * checking R files for non-ASCII characters ... OK * checking R files for syntax errors ... OK * checking whether the package can be loaded ... ERROR Error in .find.package(package, lib.loc, verbose = verbose) : there are no packages called 'mypkg', 'stats', 'graphics', 'grDevices', 'utils', 'datasets', 'methods', 'base' Error in library(mypkg) : .First.lib failed for 'mypkg' Execution halted It looks like this package has a loading problem: see the messages for details. Here is the mypkg.R file sss - /home/hodgesse/Desktop/R-2.5.1 .First.lib - function(lib=sss,pkg=mypkg) library.dynam(mypkg.so,pkg=mypkg,lib=sss) f - function(x,y) x+y g -function(x,y) x-y Thanks for any help Edna mailto: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Newbie creating package with compiled code
On 04/07/2007 6:43 PM, Edna Bell wrote: Hi R Gurus! I'm trying to create a test package using the package.skeleton function. I wanted to add some compiled code too. In the src library, I put together a baby subroutine, compiled it and created a test.dll When I use the R cmd build, it works fine. But I get into trouble with the R CMD check section. /home/Desktop/R-2.5.1/bin # ./R CMD check mypkg * checking for working latex ... OK * using log directory '/home/Desktop/R-2.5.1/bin/mypkg.Rcheck' * using R version 2.5.1 (2007-06-27) * checking for file 'mypkg/DESCRIPTION' ... OK * checking extension type ... Package * this is package 'mypkg' version '1.0' * checking package dependencies ... OK * checking if this is a source package ... OK * checking whether package 'mypkg' can be installed ... OK * checking package directory ... OK * checking for portable file names ... OK * checking for sufficient/correct file permissions ... OK * checking DESCRIPTION meta-information ... OK * checking top-level files ... OK * checking index information ... OK * checking package subdirectories ... OK * checking R files for non-ASCII characters ... OK * checking R files for syntax errors ... OK * checking whether the package can be loaded ... ERROR Error in .find.package(package, lib.loc, verbose = verbose) : there are no packages called 'mypkg', 'stats', 'graphics', 'grDevices', 'utils', 'datasets', 'methods', 'base' Error in library(mypkg) : .First.lib failed for 'mypkg' Execution halted It looks like this package has a loading problem: see the messages for details. Here is the mypkg.R file sss - /home/hodgesse/Desktop/R-2.5.1 .First.lib - function(lib=sss,pkg=mypkg) library.dynam(mypkg.so,pkg=mypkg,lib=sss) That's a very strange .First.lib. I think you'll have more success with a simpler one: .First.lib - function(libname, pkgname) library.dynam(mypkg, package=pkgname, lib.loc=libname) (and sss is not needed at all). Duncan Murdoch f - function(x,y) x+y g -function(x,y) x-y Thanks for any help Edna mailto: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lookups in R
Michael Frumin wrote:
i wish it were that simple. unfortunately the logic i have to do on
each transaction is substantially more complicated, and involves
referencing the existing values of the user table through a number of
conditions.
any other thoughts on how to get better-than-linear performance time?
is there a recommended binary searching/sorting (i.e. BTree) module that
I could use to maintain my own index?
The point remains: To do anything efficient in R, you need to get rid of
that for loop and use something vectorized. Notice that you can expand
values from the user table into the transaction table by indexing with
transactions$userid, or you can use a merge operation.
thanks,
mike
Peter Dalgaard wrote:
mfrumin wrote:
Hey all; I'm a beginner++ user of R, trying to use it to do some
processing
of data sets of over 1M rows, and running into a snafu. imagine that my
input is a huge table of transactions, each linked to a specif user
id. as
I run through the transactions, I need to update a separate table for
the
users, but I am finding that the traditional ways of doing a table
lookup
are way too slow to support this kind of operation.
i.e:
for(i in 1:100) {
userid = transactions$userid[i];
amt = transactions$amounts[i];
users[users$id == userid,'amt'] += amt;
}
I assume this is a linear lookup through the users table (in which
there are
10's of thousands of rows), when really what I need is O(constant
time), or
at worst O(log(# users)).
is there any way to manage a list of ID's (be they numeric, string,
etc) and
have them efficiently mapped to some other table index?
I see the CRAN package for SQLite hashes, but that seems to be going
a bit
too far.
Sometimes you need a bit of lateral thinking. I suspect that you could
do it like this:
tbl - with(transactions, tapply(amount, userid, sum))
users$amt - users$amt + tbl[users$id]
one catch is that there could be users with no transactions, in which
case you may need to replace userid by factor(userid,
levels=users$id). None of this is tested, of course.
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[R] nls() lower/upper bound specification
Dear all, In optim() all parameters of a function to be adjusted is stored in a single vector, with lower/upper bounds can be specified by a vector of the same length. In nls(), is it true that if I want to specify lower/upper bounds, functions must be re-written so that each parameter is contained in a single-valued vector? ## data input x - 1:10 y - 3*x+4*x^2+rnorm(10,250) ## this one does not work f - function(x) function(beta) beta[1]+ beta[2]*x+beta[3]*x^2 out - nls(y~f(x)(beta),data=data.frame(x,y), alg=port, start=list(beta=1:3), lower=list(beta=rep(0,3))) (However, this works if I do not specify a lower bound) ## this one works g - function(x) function(beta1,beta2,beta3) beta1+ beta2*x+beta3*x^2 out - nls(y~g(x)(beta1,beta2,beta3),data=data.frame(x,y), alg=port, start=list(beta1=1,beta2=1,beta3=1), lower=list(beta1=1,beta2=1,beta3=1)) Thanks in advance! Stephen __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lookups in R
On 7/4/07, Martin Morgan [EMAIL PROTECTED] wrote:
Michael,
A hash provides constant-time access, though the resulting perl-esque
data structures (a hash of lists, e.g.) are not convenient for other
manipulations
n_accts - 10^3
n_trans - 10^4
t - list()
t$amt - runif(n_trans)
t$acct - as.character(round(runif(n_trans, 1, n_accts)))
uhash - new.env(hash=TRUE, parent=emptyenv(), size=n_accts)
## keys, presumably account ids
for (acct in as.character(1:n_accts)) uhash[[acct]] - list(amt=0, n=0)
system.time(for (i in seq_along(t$amt)) {
+ acct - t$acct[i]
+ x - uhash[[acct]]
+ uhash[[acct]] - list(amt=x$amt + t$amt[i], n=x$n + 1)
+ })
user system elapsed
0.264 0.000 0.262
udf - data.frame(amt=0, n=rep(0L, n_accts),
+ row.names=as.character(1:n_accts))
system.time(for (i in seq_along(t$amt)) {
+ idx - row.names(udf)==t$acct[i]
+ udf[idx, ] - c(udf[idx,amt], udf[idx, n]) + c(t$amt[i], 1)
+ })
user system elapsed
18.398 0.000 18.394
I don't think that's a fair comparison--- much of the overhead comes
from the use of data frames and the creation of the indexing vector. I
get
n_accts - 10^3
n_trans - 10^4
t - list()
t$amt - runif(n_trans)
t$acct - as.character(round(runif(n_trans, 1, n_accts)))
uhash - new.env(hash=TRUE, parent=emptyenv(), size=n_accts)
for (acct in as.character(1:n_accts)) uhash[[acct]] - list(amt=0, n=0)
system.time(for (i in seq_along(t$amt)) {
+ acct - t$acct[i]
+ x - uhash[[acct]]
+ uhash[[acct]] - list(amt=x$amt + t$amt[i], n=x$n + 1)
+ }, gcFirst = TRUE)
user system elapsed
0.508 0.008 0.517
udf - matrix(0, nrow = n_accts, ncol = 2)
rownames(udf) - as.character(1:n_accts)
colnames(udf) - c(amt, n)
system.time(for (i in seq_along(t$amt)) {
+ idx - t$acct[i]
+ udf[idx, ] - udf[idx, ] + c(t$amt[i], 1)
+ }, gcFirst = TRUE)
user system elapsed
1.872 0.008 1.883
The loop is still going to be the problem for realistic examples.
-Deepayan
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Re: [R] questions on lme function
Ana You are estimating a random coefficient model on 5 individuals (mean and variance). Are you sure this is wise? Ross Darnell - Original Message - From: Ana Conesa [EMAIL PROTECTED] Date: Thursday, July 5, 2007 1:21 am Subject: [R] questions on lme function Dear list, I am using the lme funcion to fit a mixed model for the time response of a number of physiological variables. The random variable would be the individual on which physiological variables are measured at different time points. I have 4 time points, 5 individuals and 3 replicates per condition (time/individual), and I would like to fit a quadratic model on time. The model I am using is mm - lme(myvar ~ time + time2, random= ~ time|individual, data=clinical) being time2 = time*time I have a number of questions 1) I am not very sure the random effect is correctly modeled. Would I need to include the time2 variable aswell? 2) I would like to extract the F statistics of the model, and I do not find a function for this. Is this possible? 3) depending of the variable I take, I frequently obtain a convergence error as a result of the lme funcion. Any ideas on what to do to improve convergence? Thank you Ana Conesa, PhD Centro de Investigacion Principe Felipe Avda. Autopista Saler 16 46013 Valencia http://bioinfo.cipf.es __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.htmland provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] (Statistics question) - Nonlinear regression and simultaneous equation
Hi,I have a fundamental questions that I'm a bit confused. If any guru from this circle could help me out, I would really appreciate.I have a system of equations in which some of the endogs appear on right hand sides of some equations. To solve this, one needs a technique like 2SLS or FIML to circumvent inconsistency of the estimated coefficients. My question is that if I apply the nonlinear regression like SVM regression. Do I still need to worry about endogeneity? Meaning, what I only need to care is the 1st step of 2SLS. That would mean that I only need to carry out the SVM regression on all the exogs. Am I missing anything here? Thank you so much.Regards,- adschai [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Question about framework to weighting different classes in SVM
Hi gurus, I have a doubt about multiclass classification SVM. The population in my data includes a couple of class labels that have relatively small proportion of the entire population compared to other classes. I would like SVM to pay more attention to these classes. However, the question I am having here is that is there any systematic/theoretic framework to determine the weights for each class? My second question is directly related to R. I would like to use the class.weights attribute in svm function. However, I'm quite confused a bit about how to use it from the description I got from ?svm. Below is the quote. 'a named vector of weights for the different classes, used for asymetric class sizes. Not all factor levels have to be supplied (default weight: 1). All components have to be named.' Is the name of the vector has to match the levels in my factor used as target labels for my classification? Any simple example would be really appreciated. Thank you! - adschai __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] speed up crr function in cmprsk package
I am trying to use the crr function in the cmprsk package to analyze a large patient dataset (45000 +), The model has 100 + covariates and 5 competing risks. I am finding that R seems to get bogged down and even if I let it run for several hours I don't get anything back. Am I expecting too much, or are there ways to speed up the process? Any help is appreciated. Best, Spencer [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Question for svm function in e1071
Hi, Sorry that I have many questions today. I am using svm function on about 180,000 points of training set. It takes very long time to run. However, I would like it to spit out something to make sure that the run is not dead in between. Would you please suggest anyway to do so? And is there anyway to speed up the performance of this svm function? Thank you. - adschai __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] about stableFit() and hypFit() of fBasics package
Dear R users, I'm trying to fit stable distribution and hyperbolic distribution to my data using stableFit(), and hypFit() of fBasics. However, there are some problems This is the result == stableFit(lm, alpha = 1, beta = 0, gamma = 1, delta = 0, doplot = TRUE, trace = FALSE, title = NULL, description = NULL) Title: Stable Parameter Estimation Call: .qStableFit(x = x, doplot = doplot, title = title, description = description) Model: Student-t Distribution Estimated Parameter(s): alpha beta gamma delta NANANANA == first, this is stable distribution, but in Model, it's always Student-t Distribution. Second, everytime I run stableFit(), the result of Estimated Parameter(s) is NA. I can't really find what's wrong in my code. In the case of hyperbolic distribution, this is the result == Model: Hyperbolic Distribution Estimated Parameter(s): alpha beta deltamu 63.201132 1.991194 11.165716 2.921906 There were 41 warnings (use warnings() to see them) Warning messages: 1: NA/Inf replaced by maximum positive value 2: NA/Inf replaced by maximum positive value ... == First, I don't what the warning messages mean and why they appeared. Many thanks in advance Junhee [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
