Re: [R] cube root
Branimir K. Hackenberger wrote: This is some interesting: -8^(1/3) [1] -2 x=(-8:8) y=x^(1/3) y [1] NaN NaN NaN NaN NaN NaN NaN NaN 0.00 1.00 [11] 1.259921 1.442250 1.587401 1.709976 1.817121 1.912931 2.00 So, can anybody explain this?! (Why is x[1]^(1/3)=y[1]=NaN, but -8^(1/3)=-2?) Operator precedence. See R FAQ 7.33 Why are powers of negative numbers wrong? Berend -- View this message in context: http://r.789695.n4.nabble.com/cube-root-tp3455020p3455027.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] kpss
Dear All I will be glade if you help me to understand what difference is between trend Stationary and level Stationary at kpss.test in tseries package. best regards stat consult -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Counting elements in columns
Hi, try this; sum(sapply(a, function(x) (x==b))) -- View this message in context: http://r.789695.n4.nabble.com/Re-Counting-elements-in-columns-tp3454930p3454936.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] sweave options with variable
hello: Do any one know how to set sweave option by variable, for example I want set some of my selected code chunk with: eval=needRun= rather than eval=TRUE=, so I can change the action only in the head by change the variable needRun one times. I have tried use \Def and \newcommand, both do not work, so I suppose it is related with R/Sweave its self. thanks for any good suggestion. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cube root
Hey :) In addition, although (-2)^3 = -8, (-8)^(1/3) != -2, mathematically. A fractional power of a negative number is complex. And this can be obtained properly in R: (-8+0i)^(1/3) [1] 1+1.732051i Cheers, Tsjerk This is some interesting: -8^(1/3) [1] -2 x=(-8:8) y=x^(1/3) y [1] NaN NaN NaN NaN NaN NaN NaN NaN 0.00 1.00 [11] 1.259921 1.442250 1.587401 1.709976 1.817121 1.912931 2.00 So, can anybody explain this?! (Why is x[1]^(1/3)=y[1]=NaN, but -8^(1/3)=-2?) Operator precedence. See R FAQ 7.33 Why are powers of negative numbers wrong? Berend -- View this message in context: http://r.789695.n4.nabble.com/cube-root-tp3455020p3455027.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Tsjerk A. Wassenaar, Ph.D. post-doctoral researcher Molecular Dynamics Group * Groningen Institute for Biomolecular Research and Biotechnology * Zernike Institute for Advanced Materials University of Groningen The Netherlands __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] if else statements in data frame
Thanks alot B77S. That was a critical post. -- View this message in context: http://r.789695.n4.nabble.com/if-else-statements-in-data-frame-tp3454646p3455077.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] prb with data.table
I upgraded to R.2.13.0 and the problem disappeared. Thx -- View this message in context: http://r.789695.n4.nabble.com/prb-with-data-table-tp3454478p3455103.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cube root
Hey :) In addition, although (-2)^3 = -8, (-8)^(1/3) != -2, mathematically. A fractional power of a negative number is complex. _Not_ in addition. The citation of FAQ 7.33 did not address the correct issue since the sequence was enclosed in parens. Yours is the only correct answer. For further discussion about various approaches to fractional powers of negative numbers one can get further approaches by searching the archives. -- David. And this can be obtained properly in R: (-8+0i)^(1/3) [1] 1+1.732051i Cheers, Tsjerk This is some interesting: -8^(1/3) [1] -2 x=(-8:8) y=x^(1/3) y [1] NaN NaN NaN NaN NaN NaN NaN NaN 0.00 1.00 [11] 1.259921 1.442250 1.587401 1.709976 1.817121 1.912931 2.00 So, can anybody explain this?! (Why is x[1]^(1/3)=y[1]=NaN, but -8^(1/3)=-2?) Operator precedence. See R FAQ 7.33 Why are powers of negative numbers wrong? Berend Tsjerk Wassenaar wrote: Hey :) In addition, although (-2)^3 = -8, (-8)^(1/3) != -2, mathematically. A fractional power of a negative number is complex. And this can be obtained properly in R: (-8+0i)^(1/3) [1] 1+1.732051i Cheers, Tsjerk This is some interesting: -8^(1/3) [1] -2 x=(-8:8) y=x^(1/3) y [1] NaN NaN NaN NaN NaN NaN NaN NaN 0.00 1.00 [11] 1.259921 1.442250 1.587401 1.709976 1.817121 1.912931 2.00 So, can anybody explain this?! (Why is x[1]^(1/3)=y[1]=NaN, but -8^(1/3)=-2?) Operator precedence. See R FAQ 7.33 Why are powers of negative numbers wrong? Berend -- View this message in context: http://r.789695.n4.nabble.com/cube-root-tp3455020p3455027.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Tsjerk A. Wassenaar, Ph.D. post-doctoral researcher Molecular Dynamics Group * Groningen Institute for Biomolecular Research and Biotechnology * Zernike Institute for Advanced Materials University of Groningen The Netherlands __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Tsjerk Wassenaar wrote: Hey :) In addition, although (-2)^3 = -8, (-8)^(1/3) != -2, mathematically. A fractional power of a negative number is complex. And this can be obtained properly in R: (-8+0i)^(1/3) [1] 1+1.732051i Cheers, Tsjerk This is some interesting: -8^(1/3) [1] -2 x=(-8:8) y=x^(1/3) y [1] NaN NaN NaN NaN NaN NaN NaN NaN 0.00 1.00 [11] 1.259921 1.442250 1.587401 1.709976 1.817121 1.912931 2.00 So, can anybody explain this?! (Why is x[1]^(1/3)=y[1]=NaN, but -8^(1/3)=-2?) Operator precedence. See R FAQ 7.33 Why are powers of negative numbers wrong? Berend -- View this message in context: http://r.789695.n4.nabble.com/cube-root-tp3455020p3455027.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Tsjerk A. Wassenaar, Ph.D. post-doctoral researcher Molecular Dynamics Group * Groningen Institute for Biomolecular Research and Biotechnology * Zernike Institute for Advanced Materials University of Groningen The Netherlands __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://r.789695.n4.nabble.com/cube-root-tp3455020p3455118.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cube root
On 11-04-17 07:51, Branimir K. Hackenberger wrote: This is some interesting: -8^(1/3) [1] -2 x=(-8:8) y=x^(1/3) y [1] NaN NaN NaN NaN NaN NaN NaN NaN 0.00 1.00 [11] 1.259921 1.442250 1.587401 1.709976 1.817121 1.912931 2.00 So, can anybody explain this?! (Why is x[1]^(1/3)=y[1]=NaN, but -8^(1/3)=-2?) -8^(1/3) == (-8)^(1/3) # NA -8^(1/3) == -(8^(1/3)) # TRUE Thx!!! You're welcome. -- Sascha Vieweg, saschav...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to get the plot like this by R?
Thank you, John, for the blog website and the kind reminding. Thank Jinsong and Jinyan for your kindness. Xipei Wang, Ph.D. student Beijing, China At 2011-04-16 21:21:33£¬John Kane jrkrid...@yahoo.ca wrote: I posted my response to you to the mailing list. You replied only to me so other R-help readers will not see your response. You need to use Reply All to get your message to both the person and the list. This is something I forget regularly. I hope that Jim Lemon's code at http://www.r-statistics.com/2011/03/beeswarm-boxplot-and-plotting-it-with-r/ does what you want. If not, hopefully some of the other code or links will help. --- OnSat, 4/16/11, wangxipeiwangxi...@gmail.com wrote: From: wangxipei wangxi...@gmail.com Subject: Re:Re: [R] how to get the plot like this by R? To: John Kane jrkrid...@yahoo.ca Received: Saturday, April 16, 2011, 8:36 AM I am sorry for the fault. The figure is posted athttp://www.mediafire.com/i/?08b9f4um43vxvw8 (hope this works!). Thank John for the website. Xipei Wang, Ph.D. student -- Xipei Wang, Ph.D. student Department of Pharmaceutics, School of Pharmaceutical Sciences, Peking University Health Science Center, Beijing, China Tel: +86-0-152 1089 2231 Email:wangxi...@gmail.com ÌåÑéÍøÒ×ÓÊÏä2G³¬´ó¸½¼þ£¬ÇáËÉ·¢ÓÅÖÊ´óµçÓ°¡¢´óÕÕƬ£¬ÌáËÙ3±¶! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sweave options with variable
On 11-04-17 2:41 AM, Kai Ying wrote:
hello:
Do any one know how to set sweave option by variable, for example I want
set some of my selected code chunk with:
eval=needRun= rather thaneval=TRUE=, so I can change the action
only in the head by change the variable needRun one times.
I have tried use \Def and \newcommand, both do not work, so I suppose it
is related with R/Sweave its self.
thanks for any good suggestion.
You can't do that, but you can get a similar effect this way:
echo=FALSE=
needRun - TRUE
@
...
thecode,eval=FALSE=
someSlowFunction()
@
echo=FALSE=
if (needRun) {
thecode
}
@
Or you could use cacheSweave or weaver for caching, which may do what
you want, or write your own Sweave driver to do exactly what you want.
Duncan Murdoch
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Rsquared for anova
( did this msg make it through the lists as rich text? hotmail didn't seem to think it was plain text?) Anyway, having come in in the middle of this it isn't clear if your issues are with R or stats or both. Usually the hard core stats people punt the stats questions to other places but both can be addressed somewhat. In any case, exploratory work is a good way to learn both and I always like looking at new data. If you have one or a few dependent variable and many independent variable, it would probably help if you could visualize a surface with the response as a function of the input variables and then, maybe with the input of prior information or anecdotes, you have some idea what tests or analyses would make sense. just some thoughts for illustration only df-read.table(results_processedCP.txt,header=T) first it helps to make sure everything went ok and do quick checks, for example, str(df) unique(df$nh1) unique(df$nh2) unique(df$nh3) unique(df$randsize) unique(df$aweoghts) unique(df$aweights) now personally lots of binary variable confuse me and I can munge them all together since I expect I can later identify issues in following plots. So, with this data you can create a composite variable like this, ( now I have not checked any of this for accuracy and typos and other problems may render the results useless) x=df$nh1+2*df$nh2+4*df$nh3+2*df$randsize+32*df$aweights df2-cbind(df,x) str(df2) not sure if time was an input or output but you could see if there is any obvious trend or periodicity of time with your new made up variable, plot(df2$time,df2$x) Apparently x is a num rather than int, it can be changed for illustration but probably of no consequence, xi=as.integer(x) str(xi) and then you can add color based on this varaiable, min(xi) c=rainbow(56) cx=c[xi+1] str(cx) and make color coded scatter plots. Now, if you got lucky and guessed right you may see some patterns that you want to test, plot(df2$tos,df2$tws,col=cx) in this case, I get a cool red-yellow-green line along bottom ( very compelling linear fit question ) and scattered magenta( pink red? LOL ) and blue points everywhere with cluster near origin and nothing in top right quadrant. Also note a few blues lines above the red-green-yellow line but much shorter. And in fact, presumably you already knew this as it looks like it was designed in, if you just plot the red and green points the fit looks perfect for linear, good=which(df2$x20) plot(df2$tos[good],df2$tws[good],col=cx[good]) now if you look at results of fit of Good points vs all points, it isn't clear that anything like this would emerge from just looking at summaries of a linear fit, td=df2$tos[good] ti=df2$tws[good] lm(td~ti) lm(df2$tos~ df2$tws) summary(lm(td~ti)) summary(lm(df2$tos~ df2$tws)) Now of course tests need to be considered ahead of time or else it is easy to go shopping for the answer you want. Anything post hoc needs to be very complete and you should at least try to rationalize test results you don't happen to like ( assuming you are trying to understand the system from which the data was measured rather than justify some particular outcome). Date: Sun, 17 Apr 2011 11:34:14 +0200 From: dorien.herrem...@ua.ac.be To: dieter.me...@menne-biomed.de CC: r-help@r-project.org Subject: Re: [R] Rsquared for anova Thanks for your remarks. I've been reading about R for the last two days, but I don't really get when I should use lm or aov. I have attached the dataset, feel free to take a look at it. So far, running it with alle the combinations did not take too long and there seem to be some effects between the parameters. However, 2x2 combinations might suffice. Thanks for any help, or a pointer to some good documentation, Dorien On 16 April 2011 10:13, Dieter Menne dieter.me...@menne-biomed.de wrote: dorien wrote: fit - lm((tos~nh1*nh2*nh3*randsize*aweights*tt1*tt2*tt3*iters*length, data=expdata)) Error: unexpected ',' in fit - lm((tos~nh1*nh2*nh3*randsize*aweights*tt1*tt2*tt3*iters*length, Peter's point is the important one: too many interactions, and even with + instead of * you might be running into problems. But anyway: if you don't let us access /home/dorien/UA/meta-music/optimuse/optimuse1-build-desktop/results/results_processedCP you cannot expect a better answer which will depend on the structure of the data set. Dieter -- View this message in context: http://r.789695.n4.nabble.com/Rsquared-for-anova-tp3452399p3453719.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dorien Herremans *Department of Environment, Technology and Technology Management* Faculty of Applied
Re: [R] adding two colmns of a data frame
You mean like this? mydata - data.frame(aa=1:5, bb=1:5) mydata[,3] - mydata[,1] + mydata[,2] mydata --- On Sat, 4/16/11, ozgrerg ozgr...@gmail.com wrote: From: ozgrerg ozgr...@gmail.com Subject: [R] adding two colmns of a data frame To: r-help@r-project.org Received: Saturday, April 16, 2011, 4:22 PM It seems very simple but I could not find the way. I just want to add two columns of a data frame. I am doing the below operation: data$C=transform(data, data$A+data$B) However that return as a data frame. I will be glad if ypu help. Thanks a lot -- View this message in context: http://r.789695.n4.nabble.com/adding-two-colmns-of-a-data-frame-tp3454556p3454556.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sweave options with variable
thanks,
It can work but not as elegant as I expect because it add extra code to
every chunk of code.
I have tried the cache for this purpose it works in most case but have a lot
of other problems so I decided to control it by myself.
I guess that sweave running R code before tex, so tex variable do not work
and need find something run before sweave and set the variable,
but it seems currently no way do that. Write my own Sweave driver is too
hard for me hope in the future someone can refine the sweave driver.
best wishes
On Sun, Apr 17, 2011 at 7:30 AM, Duncan Murdoch murdoch.dun...@gmail.comwrote:
On 11-04-17 2:41 AM, Kai Ying wrote:
hello:
Do any one know how to set sweave option by variable, for example I
want
set some of my selected code chunk with:
eval=needRun= rather thaneval=TRUE=, so I can change the action
only in the head by change the variable needRun one times.
I have tried use \Def and \newcommand, both do not work, so I suppose
it
is related with R/Sweave its self.
thanks for any good suggestion.
You can't do that, but you can get a similar effect this way:
echo=FALSE=
needRun - TRUE
@
...
thecode,eval=FALSE=
someSlowFunction()
@
echo=FALSE=
if (needRun) {
thecode
}
@
Or you could use cacheSweave or weaver for caching, which may do what you
want, or write your own Sweave driver to do exactly what you want.
Duncan Murdoch
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] sweave options with variable
On Sun, Apr 17, 2011 at 11:04 AM, Kai Ying yi...@iastate.edu wrote: thanks, It can work but not as elegant as I expect because it add extra code to every chunk of code. I have tried the cache for this purpose it works in most case but have a lot of other problems so I decided to control it by myself. I guess that sweave running R code before tex, so tex variable do not work and need find something run before sweave and set the variable, but it seems currently no way do that. Write my own Sweave driver is too hard for me hope in the future someone can refine the sweave driver. best wishes You may be better off just directly emit the latex from R (not using Sweave at all). Another alternative is the brew package. -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sweave options with variable
[Forgot to cc: the list]
Hi.
On Sun, Apr 17, 2011 at 8:04 AM, Kai Ying yi...@iastate.edu wrote:
thanks,
It can work but not as elegant as I expect because it add extra code to
every chunk of code.
I have tried the cache for this purpose it works in most case but have a lot
of other problems so I decided to control it by myself.
I guess that sweave running R code before tex, so tex variable do not work
and need find something run before sweave and set the variable,
but it seems currently no way do that.
You can do exactly this by embedding your Sweave document with RSP markup, e.g.
% needRun - TRUE %
eval=%=needRun%=
someSlowFunction()
@
To compile this into a valid Sweave document/file, you have to append
filename extension *.rsp, e.g. report.Rnw.rsp. Then compile it as:
library(R.rsp);
rsp(report.Rnw.rsp, postprocess=FALSE);
which outputs a file report.Rnw containing:
eval=TRUE=
someSlowFunction()
@
If you also want to run Sweave on this and generate a PDF you can do
everything in one go (using the default postprocess=TRUE):
library(R.rsp);
rsp(report.Rnw.rsp);
RSP is a context-independent markup, i.e. it does not care what format
the underlying document has, as long as it is text based.
Hope this helps
/Henrik
Write my own Sweave driver is too
hard for me hope in the future someone can refine the sweave driver.
best wishes
On Sun, Apr 17, 2011 at 7:30 AM, Duncan Murdoch
murdoch.dun...@gmail.comwrote:
On 11-04-17 2:41 AM, Kai Ying wrote:
hello:
Do any one know how to set sweave option by variable, for example I
want
set some of my selected code chunk with:
eval=needRun= rather thaneval=TRUE=, so I can change the action
only in the head by change the variable needRun one times.
I have tried use \Def and \newcommand, both do not work, so I suppose
it
is related with R/Sweave its self.
thanks for any good suggestion.
You can't do that, but you can get a similar effect this way:
echo=FALSE=
needRun - TRUE
@
...
thecode,eval=FALSE=
someSlowFunction()
@
echo=FALSE=
if (needRun) {
thecode
}
@
Or you could use cacheSweave or weaver for caching, which may do what you
want, or write your own Sweave driver to do exactly what you want.
Duncan Murdoch
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] startup script with library call to ggplot2 in R-2.13.0; was: (no subject)
1. Please do read the posting guide and perovide a sensible subject line. 2. Have you updated ggplot for R-2.13.0 before reporting the error? E.g. please try update.packages(checkBuilt=TRUE) and afterwards report again. Uwe Ligges On 16.04.2011 22:25, Steve Lianoglou wrote: Out of curiosity, do you have a problem calling `library(ggplot2)` from the R prompt (after R is finished spinning up)? I vaguely recall (~ 2 years ago) that I was trying to load some packages in my ~/.Rprofile since I thought I always wanted them in R anyway and had problems ... I've given up trying to do that, though :-) So -- doesn't answer your question, but if you have number of libraries you want to load up, but it's a pain to call individual `library()` calls, maybe you can define a .setup() function in your ~/.Rprofile that does all these library calls for you when you come to a point in your (interactive) work that you think you need them. -steve On Sat, Apr 16, 2011 at 2:47 PM, stephen sefickssef...@gmail.com wrote: I have just upgraded to R 2.13 and have library(ggplot2) in my .Rprofile (among other things). when i start R I get an error message. Has something in the start up scripts changed? Is there a better way to specify the library calls in .Rprofile? Thanks for all of the help in advance. Error: Loading required package: grid Loading required package: proto Error in rename(x, .base_to_ggplot) : could not find function setNames Error : unable to load R code in package 'ggplot2' Error: package/namespace load failed for 'ggplot2' [Previously saved workspace restored] Computer 1: R version 2.13.0 (2011-04-13) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=en_US.UTF-8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets grid methods [8] base other attached packages: [1] proto_0.3-9.1 reshape_0.8.4 plyr_1.5.1 Computer 2 R version 2.13.0 (2011-04-13) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=en_US.UTF-8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets grid methods [8] base other attached packages: [1] proto_0.3-9.1 reshape_0.8.4 plyr_1.5.1 -- Stephen Sefick | Auburn University | | Biological Sciences | | 331 Funchess Hall | | Auburn, Alabama | | 36849 | |___| | sas0...@auburn.edu | | http://www.auburn.edu/~sas0025 | |___| Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis A big computer, a complex algorithm and a long time does not equal science. -Robert Gentleman __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] boxplot with 2 treatments and 2 variables (with 2 corresponding y-axes)
On 15.04.2011 16:20, Ellis, David wrote: Hi, I am a complete newcomer to R and although I can plot standard box-plots I am struggling with this... I have two treatments - A B, and 2 variables 1 2. I want to compare boxplots of variable 1 with variable 2 for each treatment. I would also like them to all be on the same graphic. I would like treatment to be on the x axis (A and B) and above A and B have their corresponding boxplots for both variables.I would also like 2 seperate y-axes, one for each variable - Variable 1 has a continuous y axis, and variable 2 is a percentage (of variable 1 in-fact, not that this should make a difference). It would also be helpful if the boxplots were a different colour for each variable and that this corresponded somehow to the relevant axis. Or if someone can think of another way to show which treatment each boxplot is that would be gratefully recieved Hope this makes sense and isn't too much of a pain to answer! Not sure if it makes sense: If you have two boxplots beside each other, you tend to compare them directly. If they are unrelated (different variables with different scaling as in this case), I'd be careful not to confuse the reader of the plot Best, Uwe Ligges Thanks in advance Dave [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Boxplot with two or more Y vectors
On 13.04.2011 15:58, Håvard Wahl Kongsgård wrote: Hi, for a simple boxplot in R, in the formula is it possible to include two or more Y vectors directly. Or is that only possibility by aggregating the data first? Do you think about something like boxplot(cbind(y1, y2)) ? Uwe Ligges __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Rsquared for anova
On 2011-04-17 02:34, Dorien Herremans wrote: Thanks for your remarks. I've been reading about R for the last two days, but I don't really get when I should use lm or aov. I don't think that reading about R is the answer at this stage. It appears to me that you need to learn more about regression. There are many good introductory books. If you want to learn the R way at the same time, you could look at the books section on CRAN. Perhaps Peter Dalgaard's Intro to Stats with R or An R Companion to Applied Regression by J. Fox and S. Weisberg, or the books by Verzani or Heiberger/Holland. After that, you'll find that the R documentation is actually quite good. Most complaints about R's documentation seem to amount to complaints that it doesn't teach statistics. That's a good thing. About your data: I'm fairly sure that several, if not most, of your predictors should be factors. Peter Ehlers I have attached the dataset, feel free to take a look at it. So far, running it with alle the combinations did not take too long and there seem to be some effects between the parameters. However, 2x2 combinations might suffice. Thanks for any help, or a pointer to some good documentation, Dorien On 16 April 2011 10:13, Dieter Mennedieter.me...@menne-biomed.de wrote: dorien wrote: fit- lm((tos~nh1*nh2*nh3*randsize*aweights*tt1*tt2*tt3*iters*length, data=expdata)) Error: unexpected ',' in fit- lm((tos~nh1*nh2*nh3*randsize*aweights*tt1*tt2*tt3*iters*length, Peter's point is the important one: too many interactions, and even with + instead of * you might be running into problems. But anyway: if you don't let us access /home/dorien/UA/meta-music/optimuse/optimuse1-build-desktop/results/results_processedCP you cannot expect a better answer which will depend on the structure of the data set. Dieter -- View this message in context: http://r.789695.n4.nabble.com/Rsquared-for-anova-tp3452399p3453719.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Rsquared for anova
Thanks everyone. Yes Peter, I already added nh1=factor(nh1) to the 'routine'. Mostly, my collegues are helping me work out the results and they know more about regression, it has been a while for me... They just asked if I could also provide an R2, to see how good the model fits... therefore the question. I already have the P values for each factor. I tought it might be a simple command that I overlooked, such as summary(fit) :-) Mike, I will study what you propose first thing tomorrow morning when I am back at the office! Thanks a lot, Dorien On 17 April 2011 19:43, Peter Ehlers ehl...@ucalgary.ca wrote: On 2011-04-17 02:34, Dorien Herremans wrote: Thanks for your remarks. I've been reading about R for the last two days, but I don't really get when I should use lm or aov. I don't think that reading about R is the answer at this stage. It appears to me that you need to learn more about regression. There are many good introductory books. If you want to learn the R way at the same time, you could look at the books section on CRAN. Perhaps Peter Dalgaard's Intro to Stats with R or An R Companion to Applied Regression by J. Fox and S. Weisberg, or the books by Verzani or Heiberger/Holland. After that, you'll find that the R documentation is actually quite good. Most complaints about R's documentation seem to amount to complaints that it doesn't teach statistics. That's a good thing. About your data: I'm fairly sure that several, if not most, of your predictors should be factors. Peter Ehlers I have attached the dataset, feel free to take a look at it. So far, running it with alle the combinations did not take too long and there seem to be some effects between the parameters. However, 2x2 combinations might suffice. Thanks for any help, or a pointer to some good documentation, Dorien On 16 April 2011 10:13, Dieter Mennedieter.me...@menne-biomed.de wrote: dorien wrote: fit- lm((tos~nh1*nh2*nh3*randsize*aweights*tt1*tt2*tt3*iters*length, data=expdata)) Error: unexpected ',' in fit- lm((tos~nh1*nh2*nh3*randsize*aweights*tt1*tt2*tt3*iters*length, Peter's point is the important one: too many interactions, and even with + instead of * you might be running into problems. But anyway: if you don't let us access /home/dorien/UA/meta-music/optimuse/optimuse1-build-desktop/results/results_processedCP you cannot expect a better answer which will depend on the structure of the data set. Dieter -- View this message in context: http://r.789695.n4.nabble.com/Rsquared-for-anova-tp3452399p3453719.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dorien Herremans Department of Environment, Technology and Technology Management Faculty of Applied Economics University of Antwerp B.513 Prinsstraat 13 2000 Antwerp Belgium +32 3 265 41 25 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] multiple instance learning
Hi all, know somebody a package for R where are implemented multiple instance learning algos? I've just checked the forum, but couldnt find anythink... thanks in advance -- View this message in context: http://r.789695.n4.nabble.com/multiple-instance-learning-tp3455505p3455505.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] as.Date function error
Hi, I have some problems with as.Date function. After I applied as.Date for my data, 2010 changed to 2020 as below Where am I wrong? Thanks Wonjae x=c(11/16/2010,11/17/2010,11/18/2010,11/19/2010) x=as.Date(x,%m/%d/%y) x [1] 2020-11-16 2020-11-17 2020-11-18 2020-11-19 my R veision is 2.12.2. -- View this message in context: http://r.789695.n4.nabble.com/as-Date-function-error-tp3455466p3455466.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Multinomial Logit Model with lots of Dummy Variables
Hi Thanks to Jeremy for his response... I have been able to generate the factors and generate mlogit data using his code: mldata-mlogit.data(mydata, varying=NULL, choice=pitch_type_1, shape=wide) my mlogit data looks like: dependent_var,A variable,B Var,chid,alt FALSE,110,19,1,0 FALSE,110,19,1,1 FALSE,110,19,1,2 FALSE,110,19,1,3 FALSE,110,19,1,4 TRUE,110,19,1,5 FALSE,110,19,1,6 FALSE,110,19,1,7 FALSE,110,19,1,8 FALSE,110,19,2,0 FALSE,110,19,2,1 FALSE,110,19,2,2 FALSE,110,19,2,3 FALSE,110,19,2,4 FALSE,110,19,2,5 TRUE,110,19,2,6 FALSE,110,19,2,7 FALSE,110,19,2,8 TRUE,110,561,3,0 FALSE,110,561,3,1 FALSE,110,561,3,2 FALSE,110,561,3,3 FALSE,110,561,3,4 FALSE,110,561,3,5 FALSE,110,561,3,6 FALSE,110,561,3,7 FALSE,110,561,3,8 FALSE,110,149,4,0 FALSE,110,149,4,1 TRUE,110,149,4,2 ... The mldata contains 651431 rows. If I try to run this full data set I get the following error: mlogit.model- mlogit(dependent_var~0|A+B, data = mldata, reflevel=0) Error in model.matrix.default(formula, data) : allocMatrix: too many elements specified Calls: mlogit ... model.matrix.mFormula - model.matrix - model.matrix.default Execution halted Smaller datasets (595 mldata rows) and mlogit works fine and generates regression output. Is there a problem with mlogit and huge datasets? I suppose this is perhaps not the best way to assess this kind of data, but I am trying to replicate a previous analysis that was completed on a similar amount of similar data. -- View this message in context: http://r.789695.n4.nabble.com/Multinomial-Logit-Model-with-lots-of-Dummy-Variables-tp3439492p3455345.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cajolst
Well, it was a simple phrase problem. Maybe it is going to be useful for someone: summary(ca.jo(sjd,ecdet=const,type=eigen,K=2,spec=longrun,season=4)) summary(cajolst(sjd,season=4)) cajolst(sjd,season=4)@bp -- View this message in context: http://r.789695.n4.nabble.com/cajolst-tp3454306p3455628.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] reporting lme results
Hi, I have used lme4 and I have found a significant result when using anova to compare model 1 and model 2 (where I took out an interaction). The result looks like this: model.3: DIFFERENCE ~ (1 | MALE.ID) model.2: DIFFERENCE ~ MALE.SPECIES + (1 | MALE.ID) DfAICBIC logLik Chisq Chi Df Pr(Chisq) model.3 3 1379.7 1387.1 -686.86 model.2 4 1374.1 1384.0 -683.05 7.6235 1 0.005761 ** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 Just wondering how I would report this in a scientific paper? Thanks in advance, Tanya [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to retrieve a vector of a data.frame's variable attributes?
Hi, I have a data.frame with 100 variables and I have assigned a label, units and category attribute to each variable. I would like to reorder the variables in the data.frame by the category attributes but can't find a way. For example, the first variable is: attributes(hh$aez) $levels [1] coastal forest savannah $class [1] labelled factor $label [1] ecological zone 93 Levels: 10 quantiles of welfare ... year of the interview $units [1] class 24 Levels: '05 PPP USD / year cedis / year cedis /year class geo-1 ... years $category [1] geography 7 Levels: agriculture demography design expenditure geography ... welfare I have tried: hh - hh[, order(attr(hh, category))] hh - hh[, order(attr(hh[, 1:100], category))] hh - hh[, order(attr(dimnames(hh), category))] but all the right-hand side assignments above return NULL. Thanks very much for your help with this simple task! --Mel. ___ Melanie Bacou 3110 Wisconsin Ave, NW Washington DC, 20016 +1 (202) 492-7978 m...@mbacou.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] side by side histogram after splitting data by year
Hi everyone, I'm looking to produce a side-by-side histogram of the number of trips taken by jays with a particular number of acorns after accounting for year (year one and year two). I know this involves indexing first then creating a histogram but I'm not sure how I'd do this. I want to explore the possibilities that jays are altering their strategies in different years. Data are below. This is a common need for myself so any help would be greatly appreciated! Thanks, Jeff structure(list(year = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(one, two), class = factor), size = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
[R] Box plot with 5th and 95th percentiles instead of 1.5 * IQR: problems implementing an existing solution...
Hi all,
I'm just getting started with R and I would appreciate some help. I'm having
trouble creating a boxplot with whiskers at the 95th and 5th percentiles
instead of at 1.5 * IQR. I have read the relevant documentation, and checked
existing mails on this topic. I found a small modification that should work
: https://stat.ethz.ch/pipermail/r-help/2001-November/016817.html and tried
to implement it.
Basically, it says to replace boxplot.stats with:
myboxplot.stats - function (x, coef = NULL, do.conf = TRUE, do.out =
TRUE)
{
nna - !is.na(x)
n - sum(nna)
stats - quantile(x, c(.05,.25,.5,.75,.95), na.rm = TRUE)
iqr - diff(stats[c(2, 4)])
out - x stats[1] | x stats[5]
conf - if (do.conf)
stats[3] + c(-1.58, 1.58) * diff(stats[c(2, 4)])/sqrt(n)
list(stats = stats, n = n, conf = conf, out = x[out nna])
}
I entered the new function, and used fix(boxplot.default) to modify
boxplot.default so that it references myboxplot.stats instead of the
original boxplot.stats function.
If I now type boxplot.default, I can see that the code has been modified as
expected. However, I get the exact same result as before when I create a
boxplot - it shows the whiskers at 1.5 * IQR. You can test this out by
creating a boxplot from the iris dataset supplied with R using
boxplot(iris$Sepal.Length ~ iris$Species). You see that the boxplot is the
same before and after the fix. Does anybody know why this occurs, and how I
can get around this issue?
Thanks,
--
Regards,
Paul
=
Contact Details
=
Paul Raftery, BEng(Hons) (Mech), Fulbright Fellow, PhD
http://www.paulraftery.com/
http://www.paulraftery.com/
Postdoctoral Research Engineer
Informatics Research Unit for Sustainable Engineering (IRUSE)
http://www.iruse.ie/
Department of Civil Engineering,
National University of Ireland, Galway,
University Road,
Galway,
Ireland.
Landline: +353 91 49 3086
Mobile: +353 85 124 7947
Skype: praftery
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[R] converting data.frame into a numeric ( integer) form
Hi Everyone, I am relatively new to R and would appreciate your help on this problem that I encontered this morning. When running an ordinal IRT model using Ratings package on R, I keep getting this error message. ord.out-ordrating(UNORD, burnin = 1000, mcmc = 40,000, thin = 400, tune = 1, verbose = 1000, seed = NA) Error in as.vector(as.integer(Y)) : (list) object cannot be coerced to type 'integer' UNORD is my data.frame and it is in the form of delimited CSV file without either row or column name. According to the help page on Ordinal ratings, Y should be a Matrix of data to be analyzed. Entries must be integers from 1; : : : ;C or NA where C is the number of ordinal categories. Items are on the rows and subjects are on the columns. I am confused because my data.frame only contains numerica values of 1,2,3, and NA. I checked and my NAs were all numeric. Any help or advice you could provide on this issue would be greatly appreciated. Thank you very much. I would like to thank all the R experts navigating this forum in advance. Sincerely, Haillie -- View this message in context: http://r.789695.n4.nabble.com/converting-data-frame-into-a-numeric-integer-form-tp3455811p3455811.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] as.Date function error
as.Date(x,%m/%d/%Y) [1] 2010-11-16 2010-11-17 2010-11-18 2010-11-19 Notice the Y (four digit date). Dennis On Sun, Apr 17, 2011 at 6:34 AM, Wonjae Lee wjlee2...@naver.com wrote: Hi, I have some problems with as.Date function. After I applied as.Date for my data, 2010 changed to 2020 as below Where am I wrong? Thanks Wonjae x=c(11/16/2010,11/17/2010,11/18/2010,11/19/2010) x=as.Date(x,%m/%d/%y) x [1] 2020-11-16 2020-11-17 2020-11-18 2020-11-19 my R veision is 2.12.2. -- View this message in context: http://r.789695.n4.nabble.com/as-Date-function-error-tp3455466p3455466.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Color Key in RColorBrewer
Hello, I am using RcolorBrewer for plotting similarity scores, starting with pair-wise similarity matrix as input. similarity scores range is 0 to1, with 1 meaning 100% similar (see two maps below) http://r.789695.n4.nabble.com/file/n3455953/lingosimilarity.jpg http://r.789695.n4.nabble.com/file/n3455953/bingosimilarity.jpg I would like to have same color key in both the plots. My objective here is to compare two plots visually. By same color key, I mean, same range - like white for 0 - 0.2, grey for 0.2 - 0.4 and black for 0.8 - 1. Is it possible to make color scale constant in RColorBrewer? If so, please provide me pointers to command ! Thanks in Advance Chakri -- View this message in context: http://r.789695.n4.nabble.com/Color-Key-in-RColorBrewer-tp3455953p3455953.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] URL Scan
I am wondering why when I try to input data from the first site listed below into R using the scan() function, a different page is read in instead (the second site listed): http://data.visionappraisal.com/CanterburyCT/parcel.asp?pid=1242 http://www.visionappraisal.com/databases/ I am wondering if this is an issue with R or something in the source code of the web page that I am not familiar with. Since I can access the first site directly, I assume it is not within the source code. Any help would be appreciated. -- View this message in context: http://r.789695.n4.nabble.com/URL-Scan-tp3456084p3456084.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] (no subject)
Is there any news on this issue? I have the same problem but on a Mac. I have upgraded R and updated the built packages. The console output and sessionInfo are below. The problem is triggered by library(ggplot2) my .Rprofile If I do library(ggplot2) after the aborted start up ggplot2 is loaded properly, and I can manually do everything in my .Rprofile and my configuration is as originally intended. Thanks, Bryan Console Output: Loading required package: reshape Loading required package: plyr Attaching package: 'reshape' The following object(s) are masked from 'package:plyr': rename, round_any Loading required package: grid Loading required package: proto Error in rename(x, .base_to_ggplot) : could not find function setNames Error : unable to load R code in package 'ggplot2' Error: package/namespace load failed for 'ggplot2' [R.app GUI 1.40 (5751) x86_64-apple-darwin9.8.0] [History restored from /Users/bryanhanson/.Rhistory] and here is my session info after the aborted start up: R version 2.13.0 (2011-04-13) Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets grid methods base other attached packages: [1] proto_0.3-9.1 reshape_0.8.4 plyr_1.5.1 lattice_0.19-23 * Original Post from Stephen Sefick I have just upgraded to R 2.13 and have library(ggplot2) in my .Rprofile (among other things). when i start R I get an error message. Has something in the start up scripts changed? Is there a better way to specify the library calls in .Rprofile? Thanks for all of the help in advance. Error: Loading required package: grid Loading required package: proto Error in rename(x, .base_to_ggplot) : could not find function setNames Error : unable to load R code in package 'ggplot2' Error: package/namespace load failed for 'ggplot2' [Previously saved workspace restored] Computer 1: R version 2.13.0 (2011-04-13) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=en_US.UTF-8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets grid methods [8] base other attached packages: [1] proto_0.3-9.1 reshape_0.8.4 plyr_1.5.1 Computer 2 R version 2.13.0 (2011-04-13) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=en_US.UTF-8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets grid methods [8] base other attached packages: [1] proto_0.3-9.1 reshape_0.8.4 plyr_1.5.1 -- Stephen Sefick | Auburn University | | Biological Sciences | | 331 Funchess Hall | | Auburn, Alabama | | 36849 | |___| | sas0...@auburn.edu | | http://www.auburn.edu/~sas0025 | |___| -- View this message in context: http://r.789695.n4.nabble.com/no-subject-tp3454416p3456100.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 600 people's time series
Your questions are as basic that it smells like home/classwork. Try to
formulate precise questions and not questions like how do I solve this
general problem and you are much mor likely to get answers.
To get you started:
For questions on how to plot values from dataframes please have a look
either at an introduction to R (pdf), some basic textbooks or just
google: r-project plotting tutorial (the second link should get you
started)
For the regression I would loop through all ids and calculate the
regression for a subset of the dataframe like:
for (i:1:...)
{
invest = DF$invest[DF$id==i]
payout=
invest.norm=invest/mean(...)
l.model=lm(...)
}
You will have to find out the details yourself. We will happily help if
you post some precise (!) questions in case you get stuck!
Jannis
On 04/16/2011 07:15 PM, 苏江东Su Jiangdong wrote:
Hi there,
I have a data frame DF of over 600 people's short term trade data in time
order. Below is the super simplified structure of the data.
id invest payout
[1] 1 10 -1
[2] 1 33 33
[3] 1 20 -5
[4] 2 200 33
[5] 2 33-20
[6] 3 5 -5
[7] 3 5-5
id is each person's id. Each person have invested many times in the sampling
period, in temporal order.
What I want to check is the correlation between invest and payout.
1. How do I run the regression for each person, with the invest being
devided by the mean or medium of the person's invest?
2. How do I plot a graph with y axis being invest/mean(invest) and x axis
being payout, all 600 people's dots superimposed on one graph?
I tried to use
for (i in 1:(dim (DF)[1]-1))
{
if (DF[i,1]=DF[i+1,1]) id.lm- lm(invest ~ payput, data=DF)
}
But I don't know how to superimpose graphs onto each other.
Thanks a lot!
Su
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[R] R Reference Card for Data Mining
An R Reference Card for Data Mining is available for free download at http://www.rdatamining.com. It can be a quick reference card for you to use R for data mining applications. Regards -- Yanchang Zhao PhD Data Miner Email: yanchangz...@gmail.com RDataMining: http://www.rdatamining.com Twitter: http://www.twitter.com/RDataMining __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] as.Date function error
On Sun, 17 Apr 2011, Wonjae Lee wrote: Hi, I have some problems with as.Date function. After I applied as.Date for my data, 2010 changed to 2020 as below Where am I wrong? You used %y (two-digit year, here: 20 and then expanded to 2020) instead of %Y (four-digit year, here 2010). Using as.Date(x,%m/%d/%Y) yields the desired result. Z Thanks Wonjae x=c(11/16/2010,11/17/2010,11/18/2010,11/19/2010) x=as.Date(x,%m/%d/%y) x [1] 2020-11-16 2020-11-17 2020-11-18 2020-11-19 my R veision is 2.12.2. -- View this message in context: http://r.789695.n4.nabble.com/as-Date-function-error-tp3455466p3455466.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Box plot with 5th and 95th percentiles instead of 1.5 * IQR: problems implementing an existing solution...
Try
require(Hmisc)
?panel.bpplot
This implements extended box plots that can show a variety of quantiles.
Frank
Paul Raftery wrote:
Hi all,
I'm just getting started with R and I would appreciate some help. I'm
having
trouble creating a boxplot with whiskers at the 95th and 5th percentiles
instead of at 1.5 * IQR. I have read the relevant documentation, and
checked
existing mails on this topic. I found a small modification that should
work
: https://stat.ethz.ch/pipermail/r-help/2001-November/016817.html and
tried
to implement it.
Basically, it says to replace boxplot.stats with:
myboxplot.stats - function (x, coef = NULL, do.conf = TRUE, do.out =
TRUE)
{
nna - !is.na(x)
n - sum(nna)
stats - quantile(x, c(.05,.25,.5,.75,.95), na.rm = TRUE)
iqr - diff(stats[c(2, 4)])
out - x stats[1] | x stats[5]
conf - if (do.conf)
stats[3] + c(-1.58, 1.58) * diff(stats[c(2, 4)])/sqrt(n)
list(stats = stats, n = n, conf = conf, out = x[out nna])
}
I entered the new function, and used fix(boxplot.default) to modify
boxplot.default so that it references myboxplot.stats instead of the
original boxplot.stats function.
If I now type boxplot.default, I can see that the code has been modified
as
expected. However, I get the exact same result as before when I create a
boxplot - it shows the whiskers at 1.5 * IQR. You can test this out by
creating a boxplot from the iris dataset supplied with R using
boxplot(iris$Sepal.Length ~ iris$Species). You see that the boxplot is the
same before and after the fix. Does anybody know why this occurs, and how
I
can get around this issue?
Thanks,
--
Regards,
Paul
=
Contact Details
=
Paul Raftery, BEng(Hons) (Mech), Fulbright Fellow, PhD
http://www.paulraftery.com/
lt;http://www.paulraftery.com/gt;
Postdoctoral Research Engineer
Informatics Research Unit for Sustainable Engineering (IRUSE)
http://www.iruse.ie/
Department of Civil Engineering,
National University of Ireland, Galway,
University Road,
Galway,
Ireland.
Landline: +353 91 49 3086
Mobile: +353 85 124 7947
Skype: praftery
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to retrieve a vector of a data.frame's variable attributes?
On Apr 17, 2011, at 4:21 PM, Bacou, Melanie wrote: Hi, I have a data.frame with 100 variables and I have assigned a label, units and category attribute to each variable. I would like to reorder the variables in the data.frame by the category attributes but can't find a way. Something like lapply(hh, attr, which=category) might return something potentially useful. You obviously have a test case, but have failed to offer it up. Possibly using order() around that might get all the like category variables together. For example, the first variable is: attributes(hh$aez) $levels [1] coastal forest savannah $class [1] labelled factor $label [1] ecological zone 93 Levels: 10 quantiles of welfare ... year of the interview $units [1] class 24 Levels: '05 PPP USD / year cedis / year cedis /year class geo-1 ... years $category [1] geography 7 Levels: agriculture demography design expenditure geography ... welfare I have tried: hh - hh[, order(attr(hh, category)) ] Did you look at what order(attr(hh, category)) returns. Since you assigned the attribute to individual columns (which are arranged as a list, you cannot expect the whole object to return anything useable when queried with attr(). hh - hh[, order(attr(hh[, 1:100], category))] (It would be the same since hh == hh[,1:100] ) hh - hh[, order(attr(dimnames(hh), category))] dimnames would _not_ have any attributes. And attr can only work on one object at a a time anyway, but all the right-hand side assignments above return NULL. Thanks very much for your help with this simple task! --Mel. ___ Melanie Bacou David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] URL Scan
On Sun, Apr 17, 2011 at 9:40 PM, jmsc michaelfp...@gmail.com wrote: I am wondering why when I try to input data from the first site listed below into R using the scan() function, a different page is read in instead (the second site listed): http://data.visionappraisal.com/CanterburyCT/parcel.asp?pid=1242 http://www.visionappraisal.com/databases/ I am wondering if this is an issue with R or something in the source code of the web page that I am not familiar with. Since I can access the first site directly, I assume it is not within the source code. Any help would be appreciated. I can't access the first URL directly - even from my web browser without R being involved at all. Is that pid a parcel ID that you need to be logged in to see? Or not a valid parcel id anymore? If you want to access a web site from R that needs a login/password then you need to send the appropriate login form info from R and keep the cookie session info that gets returned. Web sessions from R and from a web browser are independent. Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] different colors in a segplot centers (package latticeExtra)
Hi, how can I change colors in the centers of my segplot? I'm not interested in coloring the lines (Standard error limits) but the centers (means)? here's my code: segplot(reorder(factor(Species), MeanBiom) ~ MinSEBiom + MaxSEBiom , data=dfRatioAftBefBiom, draw.bands = FALSE, centers = MeanBiom, col=as.numeric (Commercial.Value)) thanks a lot, barbara [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] side by side histogram after splitting data by year
Will one of these do it for you: str(x) 'data.frame': 550 obs. of 5 variables: $ year: Factor w/ 2 levels one,two: 1 1 1 1 1 1 1 1 1 1 ... $ size: Factor w/ 2 levels large,small: 2 2 2 2 2 2 2 2 2 2 ... $ distance: num 30.9 121.5 46.1 46.1 46.1 ... $ taken : int 10 2 12 1 4 1 10 3 5 5 ... $ mass: num 13.88 2.78 16.65 1.39 5.55 ... require(lattice) histogram(~taken|year, x) histogram(~taken|year*size, x) On Sun, Apr 17, 2011 at 10:51 AM, Stratford, Jeffrey jeffrey.stratf...@wilkes.edu wrote: Hi everyone, I'm looking to produce a side-by-side histogram of the number of trips taken by jays with a particular number of acorns after accounting for year (year one and year two). I know this involves indexing first then creating a histogram but I'm not sure how I'd do this. I want to explore the possibilities that jays are altering their strategies in different years. Data are below. This is a common need for myself so any help would be greatly appreciated! Thanks, Jeff structure(list(year = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c(one, two), class = factor), size = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
Re: [R] Dump the source code of data frame
If you want to save it so that you can read it back later, then look at 'save' and 'load'. On Wed, Apr 13, 2011 at 3:09 AM, C.H. chainsawti...@gmail.com wrote: Dear R experts, I remember a similar function existed and have been mentioned in R-help before. I tried my best to search but I really can't find it out. suppose I have an data frame like this: somedata - data.frame(age.min = 1, age.max = 1.5, male = TRUE, l = -1.013, m=16.133, s=0.07656) In order to back up the data and I don't want to use write.table(), I would like to back up the source code of the data frame. When I apply that function (let's call it dumpdf() ), the function will reproduce the source code that creates the data.frame. For example: dumpdf(somedata) somedata - data.frame(age.min = 1, age.max = 1.5, male = TRUE, l = -1.013, m=16.133, s=0.07656) Is there any function similar to the dumpdf() above? Thank you so much! Regards, CH -- CH Chan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] URL Scan
The site does not require a login/password. Another way to access the first site would be to go to the second site, click Connecticut, click Canterbury, CT, enter the online database, click search under Query by Location with nothing in the search fields, and click the first property. Viewing the frame source on this page redirects to the second site. Also, could you direct me to or give me some instructions on scanning from sites that do require a login/password? Thanks. On Sun, Apr 17, 2011 at 6:33 PM, Barry Rowlingson [via R] ml-node+3456231-1127256797-230...@n4.nabble.com wrote: On Sun, Apr 17, 2011 at 9:40 PM, jmsc [hidden email]http://user/SendEmail.jtp?type=nodenode=3456231i=0by-user=t wrote: I am wondering why when I try to input data from the first site listed below into R using the scan() function, a different page is read in instead (the second site listed): http://data.visionappraisal.com/CanterburyCT/parcel.asp?pid=1242 http://www.visionappraisal.com/databases/ I am wondering if this is an issue with R or something in the source code of the web page that I am not familiar with. Since I can access the first site directly, I assume it is not within the source code. Any help would be appreciated. I can't access the first URL directly - even from my web browser without R being involved at all. Is that pid a parcel ID that you need to be logged in to see? Or not a valid parcel id anymore? If you want to access a web site from R that needs a login/password then you need to send the appropriate login form info from R and keep the cookie session info that gets returned. Web sessions from R and from a web browser are independent. Barry __ [hidden email]http://user/SendEmail.jtp?type=nodenode=3456231i=1by-user=tmailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/URL-Scan-tp3456084p3456231.html To unsubscribe from URL Scan, click herehttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=3456084code=bWljaGFlbGZwYWdlQGdtYWlsLmNvbXwzNDU2MDg0fC04NTEyNDQyOTE=. -- View this message in context: http://r.789695.n4.nabble.com/URL-Scan-tp3456084p3456257.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] NEW ONLINE R COURSE: Fundamentals of Using R
EARLY REGISTRATION ENDS APRIL 22 The non-profit organization *Information Institute* ( http://www.information-institute.org) and faculty from Virginia Commonwealth University (VCU) are offering a live, interactive, synchronous online course entitled Fundamentals of Using R. The early registration cost (through April 22) for the 14-hour, 5 week course is $195 (student); $250 (faculty); and $295 (practitioner). All of the live class sessions that registered participants attend are recorded and provided to those participants so they can repeatedly review the live sessions at their convenience. This course is designed for people who are curious about R, new to R, and who would like to obtain a basic knowledge of R capabilities that apply to any ultimate intended use of R. This course instructs about the R environment: using workspaces; importing and exporting data; R data types, data structures, and objects; writing simple and complex scripts; essential R functions and packages; writing your own R functions and packages; programming with R; publication-style 2-D and 3-D graphics capabilities using special R packages, and many other essential topics. The course is designed to serve as a useful knowledge foundation regardless of ones ultimate use of R (for example, statistical and data analyses, programming, writing scripts or creating packages, using graphics, etc.). There is more information at the URLs below. The informational (and registration) site for the AM version (AM by US Eastern Time) is here: https://www.regonline.com/R-fund-may-AM This course runs on five consecutive Fridays from May 27 to June 24, from 11AM until 2PM ET (GMT-4). The informational (and registration) site for the PM version (PM by US Eastern Time) is here: https://www.regonline.com/R-fund-may-PM This course runs on the same five consecutive Fridays (May 27-June 24), but in the evenings from 6PM until 9PM ET (GMT-4). Please email ghub...@vcu.edu with any questions or for more information. Geoff Hubona Information Systems Department Virginia Commonwealth University [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] as.Date error
Thank you for replying the as.Date error question. I have one more question as below. I used cbind command, and data x changed, 2010-11-16 to 14929, 2010-11-17 to 14930. What happened to them? What should I do to see -mm-dd format data? x=c(11/16/2010,11/17/2010,11/18/2010,11/19/2010) x=as.Date(x,%m/%d/%Y) x [1] 2010-11-16 2010-11-17 2010-11-18 2010-11-19 y=c(1753.75,15077,1759.35,15078) cbind(x,y) xy [1,] 14929 1753.75 [2,] 14930 15077.00 [3,] 14931 1759.35 [4,] 14932 15078.00 Thanks Wonjae -- View this message in context: http://r.789695.n4.nabble.com/as-Date-error-tp3456279p3456279.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Incremental ReadLines
Hi again,
Changing my code by defining vectors outside the loop and combining them
afterwards helped a lot so now the code does not slow down anymore and I was
able to parse the file in less than 2 hours. Not fantastic but it works.
I will William's the last suggestion of how to parse it without looping
through for next time I have to parse a large file.
Many thanks for your help!
Frederik
On Thu, Apr 14, 2011 at 4:58 PM, William Dunlap wdun...@tibco.com wrote:
[see below]
From: Frederik Lang [mailto:frederikl...@gmail.com]
Sent: Thursday, April 14, 2011 12:56 PM
To: William Dunlap
Cc: r-help@r-project.org
Subject: Re: [R] Incremental ReadLines
Hi Bill,
Thank you so much for your suggestions. I will try and alter my
code.
Regarding the even shorter solution outside the loop it looks
good but my problem is that not all observations have the same variables
so that three different observations might look like this:
Id: 1
Var1: false
Var2: 6
Var3: 8
Id: 2
missing
Id: 3
Var1: true
3 4 5
Var2: 7
Var3: 3
Doing it without looping through I thought my data had to quite
systematic, which it is not. I might be wrong though.
Doing the simple preallocation that I describe should speed it up
a lot with very little effort. It is more work to manipulate the
columns one at a time instead of using data.frame subscripting and
it may not be worth it if you have lots of columns.
If you have a lot of this sort of file and feel that it will be worth
the programming time to do something fancier, here is some code that
reads lines of the form
cat(lines, sep=\n)
Id: First
Var1: false
Var2: 6
Var3: 8
Id: Second
Id: Last
Var1: true
Var3: 8
and produces a matrix with the Id's along the rows and the Var's
along the columns:
f(lines)
Var1Var2 Var3
First false 6 8
Second NA NA NA
Last true NA 8
The function f is:
f - function (lines)
{
# keep only lines with colons
lines - grep(value = TRUE, ^.+:, lines)
lines - gsub(^[[:space:]]+|[[:space:]]+$, , lines)
isIdLine - grepl(^Id:, lines)
group - cumsum(isIdLine)
rownames - sub(^Id:[[:space:]]*, , lines[isIdLine])
lines - lines[!isIdLine]
group - group[!isIdLine]
varname - sub([[:space:]]*:.*$, , lines)
value - sub(.*:[[:space:]]*, , lines)
colnames - unique(varname)
col - match(varname, colnames)
retval - array(NA_character_, c(length(rownames),
length(colnames)),
dimnames = list(rownames, colnames))
retval[cbind(group, col)] - value
retval
}
The main trick is the matrix subscript given to retval on the
penultimate line.
Thanks again,
Frederik
On Thu, Apr 14, 2011 at 12:56 PM, William Dunlap
wdun...@tibco.com wrote:
I have two suggestions to speed up your code, if you
must use a loop.
First, don't grow your output dataset at each iteration.
Instead of
cases - 0
output - numeric(cases)
while(length(line - readLines(input, n=1))==1) {
cases - cases + 1
output[cases] - as.numeric(line)
}
preallocate the output vector to be about the size of
its eventual length (slightly bigger is better),
replacing
output - numeric(0)
with the likes of
output - numeric(50)
and when you are done with the loop trim down the length
if it is too big
if (cases length(output)) length(output) - cases
Growing your dataset in a loop can cause quadratic or
worse
growth in time with problem size and the above sort of
code should make the time grow linearly with problem
size.
Second, don't do data.frame subscripting inside your
loop.
Instead of
data - data.frame(Id=numeric(cases))
while(...) {
data[cases, 1] - newValue
}
do
Id - numeric(cases)
while(...) {
Id[cases] - newValue
}
data - data.frame(Id = Id)
This is just the general principal that you don't want
to
repeat the same operation over and over in a loop.
dataFrame[i,j] first extracts column j then extracts
element
i from that column. Since the column is the same every
iteration
you may as well extract the column outside of the loop.
Avoiding the loop altogether is the fastest. E.g., the
code
you showed does the same
[R] Repeating a function
Hello all,
I currently have this function:
drift -function(p0=0.4,N=40,ngen=55){
p = p0
for( i in 1:ngen){
p = rbinom(1,2*N,p)/(2*N)
}
return( p )
}
I want to repeat it 1000 times, then do some analysis on the results. I've
tried using the rep() function, but that only gives me repeats of the first
value of p. How can I get 1000 values of p for different iterations of the
function?
Thanks in advance
--
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Re: [R] Repeating a function
Hi,
try this;
output - list()
times - 1000
drift -function(p0=0.4,N=40,ngen=55){
p = p0
for( i in 1:ngen){
p = rbinom(1,2*N,p)/(2*N)
}
return( p )
}
for(i in 1:times){
result - drift(0.4, 40, 55)
output - c(output, list(result))
}
Then, you can use the output with unlist() to do analysis. For example
mean(unlist(output))
Helin.
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[R] Deleting the last value of a vector
Hey guys, I've search a few threads about deleting a value from a vector, but no one has addressed this question so far. I want to delete the last value from a string of values I have: r = [ 1, 2, 3, 4, 5 ], and i want to make r2 = to [ 1, 2, 3, 4] So that r2 is just like r, except that it missing the final value. Thanks, -- View this message in context: http://r.789695.n4.nabble.com/Deleting-the-last-value-of-a-vector-tp3456363p3456363.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Deleting the last value of a vector
Hi,
Try this,
snip = function(x, n=1) {
rand = sample(1:3, 1)
print(paste(using algorithm #, rand))
switch(rand,
'1' = head(x, length(x) - n),
'2' = x[ seq(1, length(x) - n) ],
'3' = x[ - seq(length(x), by=-1, length=n) ])
}
snip(1:5)
HTH, but please do read the posting guide.
baptiste
On 18 April 2011 12:51, empyrean ctr...@ucdavis.edu wrote:
Hey guys,
I've search a few threads about deleting a value from a vector, but no one
has addressed this question so far.
I want to delete the last value from a string of values
I have:
r = [ 1, 2, 3, 4, 5 ], and i want to make r2 = to [ 1, 2, 3, 4]
So that r2 is just like r, except that it missing the final value.
Thanks,
--
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__
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Deleting the last value of a vector
A easier solution: r - c(1, 2, 3, 4, 5) r2-r[1:length(r)-1] On Mon, Apr 18, 2011 at 10:51 AM, empyrean ctr...@ucdavis.edu wrote: Hey guys, I've search a few threads about deleting a value from a vector, but no one has addressed this question so far. I want to delete the last value from a string of values I have: r = [ 1, 2, 3, 4, 5 ], and i want to make r2 = to [ 1, 2, 3, 4] So that r2 is just like r, except that it missing the final value. Thanks, -- View this message in context: http://r.789695.n4.nabble.com/Deleting-the-last-value-of-a-vector-tp3456363p3456363.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Daisy Englert Duursma Department of Biological Sciences Room E8C156 Macquarie University, North Ryde, NSW 210 Australia Tel +61 2 9850 9256 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Deleting the last value of a vector
Or perhaps even more parsimoniously (by a couple of characters) - r - c(1, 2, 3, 4, 5) r2-r[-length(r)] Min-Han On Sun, Apr 17, 2011 at 10:23 PM, Daisy Englert Duursma daisy.duur...@gmail.com wrote: A easier solution: r - c(1, 2, 3, 4, 5) r2-r[1:length(r)-1] On Mon, Apr 18, 2011 at 10:51 AM, empyrean ctr...@ucdavis.edu wrote: Hey guys, I've search a few threads about deleting a value from a vector, but no one has addressed this question so far. I want to delete the last value from a string of values I have: r = [ 1, 2, 3, 4, 5 ], and i want to make r2 = to [ 1, 2, 3, 4] So that r2 is just like r, except that it missing the final value. Thanks, -- View this message in context: http://r.789695.n4.nabble.com/Deleting-the-last-value-of-a-vector-tp3456363p3456363.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Daisy Englert Duursma Department of Biological Sciences Room E8C156 Macquarie University, North Ryde, NSW 210 Australia Tel +61 2 9850 9256 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Repeating a function
Hi:
The rdist functions are vectorized, so the loop is unnecessary:
drift - rbinom(55, 80, 0.4)
or within a function,
drift - function(p0 = 0.4, N = 40, ngen = 55) rbinom(ngen, 2 * N, p0)
To repeat it 1000 times, you have at least two options:
(1) [1000 columns]
simmat - replicate(1000, drift())
(2) [1000 rows]
simmat - matrix(drift(0.4, 40, 1000 * 55), nrow = 1000)
HTH,
Dennis
On Sun, Apr 17, 2011 at 4:48 PM, Sclera kn0wza...@gmail.com wrote:
Hello all,
I currently have this function:
drift -function(p0=0.4,N=40,ngen=55){
p = p0
for( i in 1:ngen){
p = rbinom(1,2*N,p)/(2*N)
}
return( p )
}
I want to repeat it 1000 times, then do some analysis on the results. I've
tried using the rep() function, but that only gives me repeats of the first
value of p. How can I get 1000 values of p for different iterations of the
function?
Thanks in advance
--
View this message in context:
http://r.789695.n4.nabble.com/Repeating-a-function-tp3456295p3456295.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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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] cairo device and locator on windows
In several tries, I am finding the locator and identify functions on the cairo device on Windows, with R-2.13.0, do not seem to work correctly. Here is my experience with locator where I click the four corners of the device window. First, with the windows() device, the results are consistent with what I see on screen windows() plot(1,1) locator() $x [1] 0.6075000 1.3981252 1.3653126 0.6090625 $y [1] 0.6117440 0.6239533 1.3931395 1.3983721 Second, with the Cairo device, the results are not consistent with what I see on screen. library(cairoDevice) Cairo() plot(1,1) locator() $x [1] 0.6715602 1.7007429 1.7071620 0.6672808 $y [1] 0.2775717 0.2945700 1.3144688 1.3290388 Any guidance on making Cairo work would be appreciated. Rich [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Print out data frames into neat images
Hi Group, I often need to print out data frames with results of analysis into a neat little image to copy and paste into documents. I need apply formatting like bold, currency signs, number formats, header shading etc. I currently output the data into csv and format using good old excel. Any suggestions if there are packages to help with such activities (to some level of basic functionality). I know that formatting can get quite adhoc but just wondering if there is anyway to output reasonably standard output. (I'm borrowing this idea from javascript plugins that can do such activities in the HTML5 world). Thanks, Santosh __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] covariance matrix: a erro and simple mixed model question, but id not know answer sorry
Dear list I need your help: Execuse me for my limited R knowledge. #example data set set.seed (134) lm=c(1:4) block = c(rep(lm,6)) gen - c(rep(1, 4), rep(2, 4), rep(3, 4), rep(4, 4),rep(5, 4),rep(6, 4)) X1 = c( rnorm (4, 10, 4), rnorm (4, 12, 6), rnorm (4, 10, 7),rnorm (4, 5, 2), rnorm (4, 8, 4), rnorm (4,7, 2)) X2 = X1 + rnorm(length(X1), 0,3) yvar - c(X1, X2) X - c(rep( 1, length(X1)), rep( 2, length(X2))) # dummy x variable dataf - data.frame(as.factor(block), as.factor(gen), as.factor(X), yvar ) My objective to estimate variance-covariance between two variables X1 and X2. Means that I need to fit something like unstructure (UN) covariance structure. Question 1: I got the following error require(lme4); fm1Gen - lmer(yvar ~ X + gen +(1|block), data= dataf) # Question 1: should I consider X fixed or random Error in model.frame.default(data = dataf, formula = yvar ~ X + gen + : variable lengths differ (found for 'gen') A tried nlme too. require(nlme) fm2Gen - lme(yvar ~ X + gen, random= ~ 1|block, data= dataf) Error in model.frame.default(formula = ~yvar + X + gen + block, data = list( : variable lengths differ (found for 'gen') # similar error Question 2: How can get I covariance matrix between X1 and X2 either using lme4 or lmer. X1X2 X1 Var (X1) Cov(X1,X2) X2 Cov(X1, X2) Var(X2) Should I put gen in the model to do this? Should I specify something in * correlation* = Thank you for your time Maya [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Using jitter function with differing variable lengths
Hi, I am trying to make a scatter plot with 4 different categories using the jitter function. My code returns a variable length error and will not plot because my four categories have different numbers of samples. When I delete samples from my spreadsheet so that each of the categories has the same number of variables, it plots just fine. Is there any way to get around this and plot all of my samples using jitter? Thanks for the help! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to retrieve a vector of a data.frame's variable attributes?
David, Thanks very much. That was the right method. --Mel. -Original Message- From: David Winsemius [mailto:dwinsem...@comcast.net] Sent: Sunday, April 17, 2011 6:00 PM To: Bacou, Melanie Cc: r-help@r-project.org Subject: Re: [R] How to retrieve a vector of a data.frame's variable attributes? On Apr 17, 2011, at 4:21 PM, Bacou, Melanie wrote: Hi, I have a data.frame with 100 variables and I have assigned a label, units and category attribute to each variable. I would like to reorder the variables in the data.frame by the category attributes but can't find a way. Something like lapply(hh, attr, which=category) might return something potentially useful. You obviously have a test case, but have failed to offer it up. Possibly using order() around that might get all the like category variables together. For example, the first variable is: attributes(hh$aez) $levels [1] coastal forest savannah $class [1] labelled factor $label [1] ecological zone 93 Levels: 10 quantiles of welfare ... year of the interview $units [1] class 24 Levels: '05 PPP USD / year cedis / year cedis /year class geo-1 ... years $category [1] geography 7 Levels: agriculture demography design expenditure geography ... welfare I have tried: hh - hh[, order(attr(hh, category)) ] Did you look at what order(attr(hh, category)) returns. Since you assigned the attribute to individual columns (which are arranged as a list, you cannot expect the whole object to return anything useable when queried with attr(). hh - hh[, order(attr(hh[, 1:100], category))] (It would be the same since hh == hh[,1:100] ) hh - hh[, order(attr(dimnames(hh), category))] dimnames would _not_ have any attributes. And attr can only work on one object at a a time anyway, but all the right-hand side assignments above return NULL. Thanks very much for your help with this simple task! --Mel. ___ Melanie Bacou David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] error message while running IRT model
I have been struggling with the same problem for the past few hours and am desperately in need of some help. The code I am running is as follows: ## opening a data set on the desktop setwd(C:/Users/haillie/Desktop) UN2010- read.csv(UN2010.csv,header=TRUE) ##calling libraries library(reshape) library(car) ## these are the variables I want to recode colname - names(UN2010) ## this applies the recode functionto all the variable I want to recode UN2010[colname] - lapply(UN2010[colname], function(x) recode(x, recodes = 8=NA, as.factor.result = FALSE, as.numeric.result = TRUE)) UN2010frame-data.matrix(lapply(UN2010,as.numeric)) ord.out1-ordrating(UN2010, beta.constraint=1, tune=.035, ma=1, mb=-5, vinva=1, vinvb=0.05, gamma.start=c(-300, 0, 1.5, 3.0, 4.5, 300), thin=20, burnin=2, mcmc=10, verbose=1000) ## However, when i try to run the above ordrating code, an error message comes up NA/NaN/Inf in foreign function call (arg 24). Because I interpreted this as that this model has no ability to perform na.omit, I did na.omit myself using the code UN2010-na.omit(UN2010). However, I encoutered another problem then. I recieved the below error messages this time.. Error in dimnames(x) - dn : length of 'dimnames' [1] not equal to array extent In addition: Warning messages: 1: In min(Y, na.rm = TRUE) : no non-missing arguments to min; returning Inf 2: In max(Y, na.rm = TRUE) : no non-missing arguments to max; returning -Inf 3: In max(Y, na.rm = TRUE) : no non-missing arguments to max; returning -Inf ## I have tried everything I could but nothing has worked so far. Any advice or help would be deeply appreciated. Thank you very much. Haillie -- View this message in context: http://r.789695.n4.nabble.com/error-message-while-running-IRT-model-tp3456603p3456603.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
