Re: [R] Potential minor GUI bug
Dear Any Thanks for your response. Maybe I did not explain the behavior well. I am aware that the Not Responding is a windows default. What I was trying to explain is that once the process that generated the Not Responding is finished and I can use R for othe computations the Not Responding caption will remain in the task bar icon but not in the caption on the main Gui form. Please see the attached screen caption for an example. Regards Francisco From: Liaw, Andy [EMAIL PROTECTED] To: 'Francisco J. Zagmutt' [EMAIL PROTECTED],R-help@stat.math.ethz.ch Subject: RE: [R] Potential minor GUI bug Date: Thu, 16 Jun 2005 17:26:03 -0400 I don't think that's a bug. Almost every Windows application can do that: when it's busy with computation, you'll see the not responding message. Andy From: Francisco J. Zagmutt Is this an interface bug? Using RGUI for windows I run into a Not Responding process (I smartly coded an infinite loop, yaiks!), I hit esc and the interpreter was stopped and I recovered the console functionality but the caption on the R icon in my windows taskbar (the individual icon shown for every software currently running in the session) was not updated so the caption still reads RGui (Not Responding). This behavior is repeated everytime I run into a Not responding process. Off course if I end the session and open a new session the icon caption goes back to the normal RGui. I am running R2.1.0 on Windows XP Pro V. 2002 SP2, Pentium M, 1.00 Gb Ram. Cheers Francisco __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Notice: This e-mail message, together with any attachments, contains information of Merck Co., Inc. (One Merck Drive, Whitehouse Station, New Jersey, USA 08889), and/or its affiliates (which may be known outside the United States as Merck Frosst, Merck Sharp Dohme or MSD and in Japan, as Banyu) that may be confidential, proprietary copyrighted and/or legally privileged. It is intended solely for the use of the individual or entity named on this message. If you are not the intended recipient, and have received this message in error, please notify us immediately by reply e-mail and then delete it from your system. -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Potential minor GUI bug
Now I understand. I get the same thing in SDI mode (R-2.1.0 on WinXPPro). No idea why... Andy From: Francisco J. Zagmutt Dear Any Thanks for your response. Maybe I did not explain the behavior well. I am aware that the Not Responding is a windows default. What I was trying to explain is that once the process that generated the Not Responding is finished and I can use R for othe computations the Not Responding caption will remain in the task bar icon but not in the caption on the main Gui form. Please see the attached screen caption for an example. Regards Francisco From: Liaw, Andy [EMAIL PROTECTED] To: 'Francisco J. Zagmutt' [EMAIL PROTECTED],R-help@stat.math.ethz.ch Subject: RE: [R] Potential minor GUI bug Date: Thu, 16 Jun 2005 17:26:03 -0400 I don't think that's a bug. Almost every Windows application can do that: when it's busy with computation, you'll see the not responding message. Andy From: Francisco J. Zagmutt Is this an interface bug? Using RGUI for windows I run into a Not Responding process (I smartly coded an infinite loop, yaiks!), I hit esc and the interpreter was stopped and I recovered the console functionality but the caption on the R icon in my windows taskbar (the individual icon shown for every software currently running in the session) was not updated so the caption still reads RGui (Not Responding). This behavior is repeated everytime I run into a Not responding process. Off course if I end the session and open a new session the icon caption goes back to the normal RGui. I am running R2.1.0 on Windows XP Pro V. 2002 SP2, Pentium M, 1.00 Gb Ram. Cheers Francisco __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html - - Notice: This e-mail message, together with any attachments, contains information of Merck Co., Inc. (One Merck Drive, Whitehouse Station, New Jersey, USA 08889), and/or its affiliates (which may be known outside the United States as Merck Frosst, Merck Sharp Dohme or MSD and in Japan, as Banyu) that may be confidential, proprietary copyrighted and/or legally privileged. It is intended solely for the use of the individual or entity named on this message. If you are not the intended recipient, and have received this message in error, please notify us immediately by reply e-mail and then delete it from your system. - - __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] CORRELATION MATRIX CONVERSION
How do I convert the output of cor(x) to a columnar format? Ex. from format below XYZ X 1.0 0.9 0.5 Y 0.9 1.0 0.1 Z 0.5 0.1 1.0 to format below X X 1.0 X Y 0.9 X Z 0.5 Y X 0.9 Y Y 1.0 Y Z 0.1 Z X 0.5 Z Y 0.1 Z Z 1.0 Thanks! Omer __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Potential minor GUI bug
Liaw, Andy wrote: Now I understand. I get the same thing in SDI mode (R-2.1.0 on WinXPPro). No idea why... I guess this is a Windows bug, because I have seen it in other applications as well. Hence I don't think we should waste our time here ... Uwe Ligges Andy From: Francisco J. Zagmutt Dear Any Thanks for your response. Maybe I did not explain the behavior well. I am aware that the Not Responding is a windows default. What I was trying to explain is that once the process that generated the Not Responding is finished and I can use R for othe computations the Not Responding caption will remain in the task bar icon but not in the caption on the main Gui form. Please see the attached screen caption for an example. Regards Francisco From: Liaw, Andy [EMAIL PROTECTED] To: 'Francisco J. Zagmutt' [EMAIL PROTECTED],R-help@stat.math.ethz.ch Subject: RE: [R] Potential minor GUI bug Date: Thu, 16 Jun 2005 17:26:03 -0400 I don't think that's a bug. Almost every Windows application can do that: when it's busy with computation, you'll see the not responding message. Andy From: Francisco J. Zagmutt Is this an interface bug? Using RGUI for windows I run into a Not Responding process (I smartly coded an infinite loop, yaiks!), I hit esc and the interpreter was stopped and I recovered the console functionality but the caption on the R icon in my windows taskbar (the individual icon shown for every software currently running in the session) was not updated so the caption still reads RGui (Not Responding). This behavior is repeated everytime I run into a Not responding process. Off course if I end the session and open a new session the icon caption goes back to the normal RGui. I am running R2.1.0 on Windows XP Pro V. 2002 SP2, Pentium M, 1.00 Gb Ram. Cheers Francisco __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html - - Notice: This e-mail message, together with any attachments, contains information of Merck Co., Inc. (One Merck Drive, Whitehouse Station, New Jersey, USA 08889), and/or its affiliates (which may be known outside the United States as Merck Frosst, Merck Sharp Dohme or MSD and in Japan, as Banyu) that may be confidential, proprietary copyrighted and/or legally privileged. It is intended solely for the use of the individual or entity named on this message. If you are not the intended recipient, and have received this message in error, please notify us immediately by reply e-mail and then delete it from your system. - - __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] CORRELATION MATRIX CONVERSION
Omer Bakkalbasi wrote: How do I convert the output of cor(x) to a columnar format? Ex. from format below XYZ X 1.0 0.9 0.5 Y 0.9 1.0 0.1 Z 0.5 0.1 1.0 to format below X X 1.0 X Y 0.9 X Z 0.5 Y X 0.9 Y Y 1.0 Y Z 0.1 Z X 0.5 Z Y 0.1 Z Z 1.0 See, e.g., ?reshape Uwe Ligges Thanks! Omer __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Potential minor GUI bug
Dear Uwe I have not seen this behavior in other windows applications but I definitivelly agree with you that it is probably not worth spending time on this trivial issue. Thanks Francisco From: Uwe Ligges [EMAIL PROTECTED] To: Liaw, Andy [EMAIL PROTECTED] CC: 'Francisco J. Zagmutt' [EMAIL PROTECTED],R-help@stat.math.ethz.ch Subject: Re: [R] Potential minor GUI bug Date: Fri, 17 Jun 2005 08:21:05 +0200 Liaw, Andy wrote: Now I understand. I get the same thing in SDI mode (R-2.1.0 on WinXPPro). No idea why... I guess this is a Windows bug, because I have seen it in other applications as well. Hence I don't think we should waste our time here ... Uwe Ligges Andy From: Francisco J. Zagmutt Dear Any Thanks for your response. Maybe I did not explain the behavior well. I am aware that the Not Responding is a windows default. What I was trying to explain is that once the process that generated the Not Responding is finished and I can use R for othe computations the Not Responding caption will remain in the task bar icon but not in the caption on the main Gui form. Please see the attached screen caption for an example. Regards Francisco From: Liaw, Andy [EMAIL PROTECTED] To: 'Francisco J. Zagmutt' [EMAIL PROTECTED],R-help@stat.math.ethz.ch Subject: RE: [R] Potential minor GUI bug Date: Thu, 16 Jun 2005 17:26:03 -0400 I don't think that's a bug. Almost every Windows application can do that: when it's busy with computation, you'll see the not responding message. Andy From: Francisco J. Zagmutt Is this an interface bug? Using RGUI for windows I run into a Not Responding process (I smartly coded an infinite loop, yaiks!), I hit esc and the interpreter was stopped and I recovered the console functionality but the caption on the R icon in my windows taskbar (the individual icon shown for every software currently running in the session) was not updated so the caption still reads RGui (Not Responding). This behavior is repeated everytime I run into a Not responding process. Off course if I end the session and open a new session the icon caption goes back to the normal RGui. I am running R2.1.0 on Windows XP Pro V. 2002 SP2, Pentium M, 1.00 Gb Ram. Cheers Francisco __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html - - Notice: This e-mail message, together with any attachments, contains information of Merck Co., Inc. (One Merck Drive, Whitehouse Station, New Jersey, USA 08889), and/or its affiliates (which may be known outside the United States as Merck Frosst, Merck Sharp Dohme or MSD and in Japan, as Banyu) that may be confidential, proprietary copyrighted and/or legally privileged. It is intended solely for the use of the individual or entity named on this message. If you are not the intended recipient, and have received this message in error, please notify us immediately by reply e-mail and then delete it from your system. - - __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Potential minor GUI bug
Yes, it is a Windows bug: the frame is controlled by Windows and not by R. On Fri, 17 Jun 2005, Uwe Ligges wrote: Liaw, Andy wrote: Now I understand. I get the same thing in SDI mode (R-2.1.0 on WinXPPro). No idea why... I guess this is a Windows bug, because I have seen it in other applications as well. Hence I don't think we should waste our time here ... Uwe Ligges Andy From: Francisco J. Zagmutt Dear Any Thanks for your response. Maybe I did not explain the behavior well. I am aware that the Not Responding is a windows default. What I was trying to explain is that once the process that generated the Not Responding is finished and I can use R for othe computations the Not Responding caption will remain in the task bar icon but not in the caption on the main Gui form. Please see the attached screen caption for an example. Regards Francisco From: Liaw, Andy [EMAIL PROTECTED] To: 'Francisco J. Zagmutt' [EMAIL PROTECTED],R-help@stat.math.ethz.ch Subject: RE: [R] Potential minor GUI bug Date: Thu, 16 Jun 2005 17:26:03 -0400 I don't think that's a bug. Almost every Windows application can do that: when it's busy with computation, you'll see the not responding message. Andy From: Francisco J. Zagmutt Is this an interface bug? Using RGUI for windows I run into a Not Responding process (I smartly coded an infinite loop, yaiks!), I hit esc and the interpreter was stopped and I recovered the console functionality but the caption on the R icon in my windows taskbar (the individual icon shown for every software currently running in the session) was not updated so the caption still reads RGui (Not Responding). This behavior is repeated everytime I run into a Not responding process. Off course if I end the session and open a new session the icon caption goes back to the normal RGui. I am running R2.1.0 on Windows XP Pro V. 2002 SP2, Pentium M, 1.00 Gb Ram. Cheers Francisco __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html - - Notice: This e-mail message, together with any attachments, contains information of Merck Co., Inc. (One Merck Drive, Whitehouse Station, New Jersey, USA 08889), and/or its affiliates (which may be known outside the United States as Merck Frosst, Merck Sharp Dohme or MSD and in Japan, as Banyu) that may be confidential, proprietary copyrighted and/or legally privileged. It is intended solely for the use of the individual or entity named on this message. If you are not the intended recipient, and have received this message in error, please notify us immediately by reply e-mail and then delete it from your system. - - __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] lm and time series: simple question
On Thu, 16 Jun 2005 [EMAIL PROTECTED] wrote: This question is partly about R and partly out of my ignorance about time series. I want to regress one time series on another, taking into account the autocorrelation (in an AR1 model) within each series. I am interested in how the standard error changes when the acf is taken into account. This does not happen with least-squares fitting as done by lm. You can use arima or gls (in package nlme). Note that both assume a model for the residuals, not for the series themselves. You could also make a joint model of the two time series. That is probably not what you want. I've made both of my datasets into ts objects and used the basic lm function (with na.action=NULL) to no effect (i.e. the resulting standard error is the same as if they were not times series). I've also looked at binding the two series together with ts.union or ts.intersect, but then I am left with a single object, and don't understand how to regress one of the components of this onto the other. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] possible bug in merge with duplicate blank names in 'by' field.
What version of R is this (please do see the posting guide)?
In both 2.1.0 and 2.1.1 beta I get
all
Promoter ip.x ip.y ip
130 40 40
240 40 40
3a 10 NA NA
4c 20 20 20
5b NA 15 15
6d NA 30 30
so cannot reproduce your result. Are you sure that the `blanks' really are
empty and not some character that is printing as empty on your unstated
OS?
BTW ' ' is what is normally called `blank'.
BTW, these are not `names' but character strings: `names' has other
meanings in R.
On Thu, 16 Jun 2005, Frank Gibbons wrote:
Run this:
p - c('a', 'c', '', ''); a - c(10, 20, 30, 40); d1 -
data.frame(Promoter=p, ip=a) # Note duplicate empty names in p.
p - c('b', 'c', 'd', ''); a - c(15, 20, 30, 40); d2 -
data.frame(Promoter=p, ip=a)
all - merge(x=d1, y=d2, by=Promoter, all=T)
all - merge(x=all, y=d2, by=Promoter, all=T)
all
Data is this:
d1
Promoter ip
1a 10
2c 20
3 30
4 40
d2
Promoter ip
1b 15
2c 20
3d 30
4 40
Output looks like this:
Promoter ip.x ip.y ip
140 30 30
240 40 30
340 30 40
440 40 40
5b 15 NA NA
6c 20 20 20
7d 30 NA NA
8a NA 10 10
The weird thing about this is (in my view) that each instance of '' is
considered unique, so with each successive merge, all combinatorial
possibilities are explored, like a SQL outer join (Cartesian product). For
non-empty names, an inner join is performed.
Dealing with genomic data (10^4 datapoints), it's easy to have a couple of
blanks buried in the middle of things, and to combine several replicates
with successive merges. I couldn't understand how my three replicates of
6000 points, in which I expected substantial overlap in the labels, were
taking so long to merge and ultimately generating 57000 labels. The culprit
turned out to be a few hundred blanks buried in the middle.
Why does the empty (null) name merit special treatment? Perhaps I'm
missing something. I hesitate to submit this as a bug, since technically I
guess you could say that blank names, especially duplicates, are not
kosher. But on the other hand, this combinatorial behaviour seems to occur
only for blanks.
-Frank
PhD, Computational Biologist,
Harvard Medical School BCMP/SGM-322, 250 Longwood Ave, Boston MA 02115, USA.
Tel: 617-432-3555 Fax:
617-432-3557 http://llama.med.harvard.edu/~fgibbons
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] Analysing ordinal/nominal data
On Thu, 16 Jun 2005, Piotr Majdak wrote: I'm looking for a solution to analyse data, which consists of dichotomous responses (yes/no) for 2 multinomial ordinal variables. Please explain how you get a binary response for a `multinomial ordinal variables'? If you intend these variables to be explanatory variables, in what sense are they `multinomial'? I was trying glm() and got hierarhical models treating all variables as nominal, but I can't figure out how to tell glm() to use a model for ordinal data like this: log(Mij) = intercept + X + Y + Z + beta*(x-x')*(y-y') where beta is a regression factor for interaction between X and Y. What are Mij, X, x, x', Y, y, y' and Z? One normally fits a logistic regression to a binary response. Do you know a trick to code it in R or point me to some documentation? Probably no `trick' is required, but we need to start from a complete and accurate description of the model you want to fit. This could be a simple as R2-level factor U, V ordered factors glm(R ~ U + V + as.numeric(U)*as.numeric(V), family = binomial) -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] regressing each column of a matrix on all other columns
apply() is just a for() loop internally so why do you expect it to be
faster?
Some comments:
1) Here predict() is just extracting the fitted values.
2) Using lm.fit will be faster if fitted values is all you want.
3) You are actually regressing each column on all other columns plus an
intercept.
4) The as.matrix is wasteful.
So it would be faster to use
A1 - cbind(1, A)
for (i in 2:ncol(A)) B[,i-1] - lm.fit(A1[,-i], A1[,i])$fitted
You can do this reasonably efficiently in matrix algebra: one way is to
form the inverse of X^TX after removing column means and use the Goodnight
sweep operation on each column in turn.
On Thu, 16 Jun 2005, Stefan Mischke wrote:
DeaR list
I would like to predict the values of each column of a matrix A by
regressing it on all other columns of the same matrix A. I do this with
a for loop:
A - B - matrix(round(runif(10*3,1,10),0),10)
A
for (i in 1:length(A[1,]))B[,i] - as.matrix(predict(lm( A[,i] ~
A[,-i] )))
B
It works fine, but I need it to be faster. I've looked at *apply but
just can't seem to figure it out.
Maybe the solution could look somewhat like this:
mylm - function(y,ci) {
x - A[,-ci]
b - lm(y~x)
}
B - apply(A,2,mylm,ci=current_column_index(A))
Is there a way to pass the index of the current column in apply to my
function? Am I on the right path at all?
Thanks for your help.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] Computing generalized eigenvalues
On Thu, 16 Jun 2005, Joshua Gilbert wrote:
I need to compute generalized eigenvalues. The eigen function in base
doesn't do it and I can't find a package that does.
They are very rarely used in statistics, so this is not surprising.
I presume you mean solving Ax = lambda B x: if B is non-singular this
reduces to a conventional eigenproblem for B^{-1}A.
As I understand it, Lapack __can__ computer them
(http://www.netlib.org/lapack/lawn41/node111.html) and R can use
Lapack. If there is no function already, can I access Lapack from R
and use those routines directly?
Yes, you can: for real matrices the requisite routines are already
compiled into R. See DGGES or DGGEV.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] Plotting second axes outside xyplot
Paul,
thanks! An elegant solution.
Andrew
--
Andrew Robinson Ph: 208 885 7115
Department of Forest Resources Fa: 208 885 6226
University of Idaho E : [EMAIL PROTECTED]
PO Box 441133W : http://www.uidaho.edu/~andrewr
Moscow ID 83843 Or: http://www.biometrics.uidaho.edu
No statement above necessarily represents my employer's opinion.
- Original Message -
From: Paul Murrell [EMAIL PROTECTED]
Date: Thursday, June 16, 2005 7:07 pm
Subject: Re: [R] Plotting second axes outside xyplot
Hi
Andrew Robinson wrote:
Hi all,
I'm trying to find a way to get xyplot to produce a second set
of axes outside the right hand side of the graph. This is my
progress so far:
EE - equal.count(ethanol$E, number=9, overlap=1/4)
xyplot(NOx ~ C | EE, data = ethanol,
prepanel = function(x, y) prepanel.loess(x, y, span = 1),
xlab = Compression Ratio, ylab = NOx (micrograms/J),
panel = function(x, y) {
panel.grid(h=-1, v= 2)
panel.xyplot(x, y)
panel.loess(x,y, span=1)
panel.axis(side = right, at = c(1, 3),
labels = c(1, 3), outside = T)
},
aspect = xy)
Does anyone have any suggestions?
I suspect the output from panel.axis() is getting clipped. You
might
get it to work as follows ...
EE - equal.count(ethanol$E, number=9, overlap=1/4)
xyplot(NOx ~ C | EE, data = ethanol,
prepanel = function(x, y) {
prepanel.loess(x, y, span = 1)
},
xlab = Compression Ratio,
ylab = NOx (micrograms/J),
panel = function(x, y) {
panel.grid(h=-1, v= 2)
panel.xyplot(x, y)
panel.loess(x,y, span=1)
# don't call panel.axis in here
},
aspect = xy)
# return to the right-most panel WITH CLIPPING OFF
trellis.focus(panel, 9, 1, clip.off=TRUE)
# draw the extra axis
panel.axis(side = right, at = c(1, 3),
labels = c(1, 3), outside = T)
trellis.unfocus()
Paul
--
Dr Paul Murrell
Department of Statistics
The University of Auckland
Private Bag 92019
Auckland
New Zealand
64 9 3737599 x85392
[EMAIL PROTECTED]
http://www.stat.auckland.ac.nz/~paul/
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[R] glmmADMB: Mixed models for overdispersed and zero-inflated count data in R
Dear R-users, Earlier this year I posted a message to this list regarding negative binomial mixed models in R. It was suggested that the program I had written should be turned into an R-package. This has now been done, in collaboration with David Fournier and Anders Nielsen. The R-package glmmADMB provides the following GLMM framework: - Negative binomial or Poisson responses. - Zero-inflation (optionally), e.g. a mixture of a Poisson or negative binomial distribution and a point mass at zero. The computational method is based on the Laplace approximation for integrating out the random effects, together with the option of employing importance sampling at the posterior mode of the random effects to permit arbitrarily close approximation to the exact MLE. (However for these models differences appear to be very small.) Some of the generic convenience functions, such as predict(), fitted.values(), ... are still missing from this package, but will hopefully be added in later versions (contributions/suggestions are most welcome). Other response distributions than negative binomial or Poisson could easily be added. Download site: http://otter-rsch.com/admbre/examples/glmmadmb/glmmADMB.html The package is based on the software ADMB-RE, but the full unrestricted R-package is made freely available by Otter Research Ltd and does not require ADMB-RE to run with user supplied data. I you will find this useful, Hans Skaug -- Hans Skaug Department of Mathematics University of Bergen, Norway email: [EMAIL PROTECTED] ph. (+47) 55 58 48 61 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] About simulations
Hello I would like to generate covariance matrix with autoregressive structure. I saw some functions in nlme such as corAR1 for example but I don't know how to use it for my goal. Could someone help me or advise me another function? Thank you in advance Caroline __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] axis labels vertically
Hi, I have a plot and a custom axis labeling, e.g. x-c(...) plot(x,axes=FALSE) axis(2) axis(1,1:50,c(label1,...,label50)) now since the labels are quite long, only a few fit on the page. Can I rotate each label by 90 degree counterclockwise (so that they are vertical) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] About simulations
ARMAacf() will give you the acf for an autoregression, and toeplitz() wil turn this into a correlation matrix. Then just multiply by the desired variance. I am not sure what this has to do with your subject line, and ?arima.sim might be helpful for that. On Fri, 17 Jun 2005, Caroline TRUNTZER wrote: Hello I would like to generate covariance matrix with autoregressive structure. I saw some functions in nlme such as corAR1 for example but I don't know how to use it for my goal. Could someone help me or advise me another function? Thank you in advance Caroline -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Analysing ordinal/nominal data
Prof Brian Ripley wrote:
On Thu, 16 Jun 2005, Piotr Majdak wrote:
I'm looking for a solution to analyse data, which consists of
dichotomous responses (yes/no) for 2 multinomial ordinal variables.
Please explain how you get a binary response for a `multinomial ordinal
variables'? If you intend these variables to be explanatory variables,
in what sense are they `multinomial'?
My data are results from a pychoacoustical experiment:
- Response: 2 levels, frequencies for yes/no
- PR: factor, independent variable, 4 levels, ordinal
- ENV: factor, independent variable, 3 levels, ordinal
The hypothesis is that:
- PR is independent of ENV, given Response
- Response and ENV are conditionally dependent, given PR
- Response and PR are conditionally dependent, given ENV
The model:
fit=glm(count ~ PR+ENV+Response + PR:Response + ENV:Response,
data=table, family=poisson)
fits with p=0.04 only. My explanations are:
- I have a three-way interaction
- I must consider the ordinal information of PR and ENV
One normally fits a logistic
regression to a binary response.
Probably no `trick' is required, but we need to start from a complete
and accurate description of the model you want to fit.
I don't know how to include the interaction to the logit model.
I think the log-linear row effects model for the 3-dim. nominal-ordinal
table (Agresti 1984, p.89) would be the right one, but please correct me
if I'm on a wrong way:
log mijk = intercept + lambda^X_i + lambda^Y_j + lambda^Z_k +
tau^XY_i*(v_j-v') + tau^XZ_i*(w_k-w') +
beta^YZ*(v_j-v')*(w_k-w')
mijk: expected frequencies for cell with indicies i,j,k
v_j: scores of Y with {v_1 v_2 .. v_j}
w_k: scores of Z with {w_1 w_2 .. w_k}
lambda^X_i, lambda^Y_j, lambda^Z_k: estimated parameters for X, Y and Z
tau^XY_i, tau^XZ_j: association parameters for XY and XZ
beta^YZ: association parameter for ordinal factors Y and Z
In my case, X==Response, Y==PR and Z==ENV. I presume no interaction
between Y and Z and would like to test beta. My idea is: if beta is
significant, this model won't hold, the saturated model fits only and I
must calculate odds ratios and beta's from partial tables. But: how can
I tell glm() to use something like beta*PR*ENVm in the formula? And, am
I on the right way?
Thanks a lot for your response.
Piotr Majdak
Agresti 1984: Analysis of ordinal categorical data. John Wiley Sons Inc.
--
Piotr Majdak
Acoustics Research Institute
Austrian Academy of Sciences
Reichsratsstr. 17
A-1010 Vienna
AUSTRIA
phone: +43-1-4277-29511
fax: +43-1-4277-9296
email: [EMAIL PROTECTED]
WWW: http://www.kfs.oeaw.ac.at
__
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Re: [R] Computing generalized eigenvalues
Prof Brian Ripley [EMAIL PROTECTED] writes:
On Thu, 16 Jun 2005, Joshua Gilbert wrote:
I need to compute generalized eigenvalues. The eigen function in base
doesn't do it and I can't find a package that does.
They are very rarely used in statistics, so this is not surprising.
An aside, going a bit off-topic:
However, there's the related generalized singular value decomposition:
K = U Sigma inv(T)
L = V M inv(T)
U'U = V'V = I ; Sigma and M diagonal (Sigma^2 + M^2 = I by
convention) ; T regular
(Look at K'K and L'L to see the connection.)
This is used in Tikhonov regularization, which is penalized least
squares, which is a statistical issue (whether numerical analysts
realize it or not). Smoothing splines is a special case. Deconvolution
is another.
I presume you mean solving Ax = lambda B x: if B is non-singular this
reduces to a conventional eigenproblem for B^{-1}A.
(There are some complications if both A and B are singular. That's
why the GSVD has that peculiar-looking convention.)
--
O__ Peter Dalgaard ster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] Analysing ordinal/nominal data
As I suggested before, a binomial logistic model is appropriate here, not
a Poisson log-linear one. (They are equivalent, but the binomial version
is easier to interpret and less wasteful to fit.)
You have still not defined v' and w', nor the scores (are they estimated
or not). But the model I suggested is such a model with scores 1,2,...
On Fri, 17 Jun 2005, Piotr Majdak wrote:
Prof Brian Ripley wrote:
On Thu, 16 Jun 2005, Piotr Majdak wrote:
I'm looking for a solution to analyse data, which consists of
dichotomous responses (yes/no) for 2 multinomial ordinal variables.
Please explain how you get a binary response for a `multinomial ordinal
variables'? If you intend these variables to be explanatory variables, in
what sense are they `multinomial'?
My data are results from a pychoacoustical experiment:
- Response: 2 levels, frequencies for yes/no
- PR: factor, independent variable, 4 levels, ordinal
- ENV: factor, independent variable, 3 levels, ordinal
The hypothesis is that:
- PR is independent of ENV, given Response
- Response and ENV are conditionally dependent, given PR
- Response and PR are conditionally dependent, given ENV
The model:
fit=glm(count ~ PR+ENV+Response + PR:Response + ENV:Response, data=table,
family=poisson)
fits with p=0.04 only. My explanations are:
- I have a three-way interaction
- I must consider the ordinal information of PR and ENV
One normally fits a logistic regression to a binary response.
Probably no `trick' is required, but we need to start from a complete and
accurate description of the model you want to fit.
I don't know how to include the interaction to the logit model.
I think the log-linear row effects model for the 3-dim. nominal-ordinal table
(Agresti 1984, p.89) would be the right one, but please correct me if I'm on
a wrong way:
log mijk = intercept + lambda^X_i + lambda^Y_j + lambda^Z_k +
tau^XY_i*(v_j-v') + tau^XZ_i*(w_k-w') +
beta^YZ*(v_j-v')*(w_k-w')
mijk: expected frequencies for cell with indicies i,j,k
v_j: scores of Y with {v_1 v_2 .. v_j}
w_k: scores of Z with {w_1 w_2 .. w_k}
lambda^X_i, lambda^Y_j, lambda^Z_k: estimated parameters for X, Y and Z
tau^XY_i, tau^XZ_j: association parameters for XY and XZ
beta^YZ: association parameter for ordinal factors Y and Z
In my case, X==Response, Y==PR and Z==ENV. I presume no interaction between Y
and Z and would like to test beta. My idea is: if beta is significant, this
model won't hold, the saturated model fits only and I must calculate odds
ratios and beta's from partial tables. But: how can I tell glm() to use
something like beta*PR*ENVm in the formula? And, am I on the right way?
Thanks a lot for your response.
Piotr Majdak
Agresti 1984: Analysis of ordinal categorical data. John Wiley Sons Inc.
--
Piotr Majdak
Acoustics Research Institute
Austrian Academy of Sciences
Reichsratsstr. 17
A-1010 Vienna
AUSTRIA
phone: +43-1-4277-29511
fax: +43-1-4277-9296
email: [EMAIL PROTECTED]
WWW: http://www.kfs.oeaw.ac.at
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
__
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[R] Mixed model question
Hi, I am new to this list as a poster, but a reader for some time. I've using R for several weeks now, and I have a lot of questions about certain procedures. Here I go: I want to test if there are differences in the time spent by pollinators visiting flowers of a given plant species, according to a number of experimental manipulations made on those flowers. All experimental manipulations (factor with 5 levels) are replicated within plants (i.e. plant is my sampling unit). Further, I have two populations (two level factor), and a number of pollinator groups (again, two levels, the same in both populations). The response variables in the numbe of seconds invested in each probe. Further, I have plant floral display as a covariate, as it may influence visitation rates. I think I have to analyse this desing considering population, pollinator group and their interaction as fixed effects, and treatment nested within plant, and its interaction with population and pollinator group, as random factors. In SAS terminology, the model looks like this: proc mixed data=flwfunc.visitflower covtest method=reml; class site pollclass treatm plantid; model time = site|pollclass flwinflor / chisq; random treatm site*treatm pollclass*treatm / subject=plantid; lsmeans site pollclass site*pollclass; run; I've been successfully trying lm, but I think is not suitable for random effects. Thus, I've tried lme, but no success when defining the random part or trying to interpret the results... Any help will be welcome! -- -- Alfonso M. Sanchez-Lafuente Departamento de Biologia Vegetal y Ecologia Facultad de Biologia Universidad de Sevilla Avd. Reina Mercedes 9 E-41012, Sevilla, Spain email: [EMAIL PROTECTED] / [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Mixed model question
Hi, I am new to this list as a poster, but a reader for some time. I've using R for several weeks now, and I have a lot of questions about certain procedures. Here I go: I want to test if there are differences in the time spent by pollinators visiting flowers of a given plant species, according to a number of experimental manipulations made on those flowers. All experimental manipulations (factor with 5 levels) are replicated within plants (i.e. plant is my sampling unit). Further, I have two populations (two level factor), and a number of pollinator groups (again, two levels, the same in both populations). The response variables in the numbe of seconds invested in each probe. Further, I have plant floral display as a covariate, as it may influence visitation rates. I think I have to analyse this desing considering population, pollinator group and their interaction as fixed effects, and treatment nested within plant, and its interaction with population and pollinator group, as random factors. In SAS terminology, the model looks like this: proc mixed data=flwfunc.visitflower covtest method=reml; class site pollclass treatm plantid; model time = site|pollclass flwinflor / chisq; random treatm site*treatm pollclass*treatm / subject=plantid; lsmeans site pollclass site*pollclass; run; I've been successfully trying lm, but I think is not suitable for random effects. Thus, I've tried lme, but no success when defining the random part or trying to interpret the results... Any help will be welcome! -- -- Alfonso M. Sanchez-Lafuente Departamento de Biologia Vegetal y Ecologia Facultad de Biologia Universidad de Sevilla Avd. Reina Mercedes 9 E-41012, Sevilla, Spain email: [EMAIL PROTECTED] / [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] axis labels vertically
BoM DS wrote: Hi, I have a plot and a custom axis labeling, e.g. x-c(...) plot(x,axes=FALSE) axis(2) axis(1,1:50,c(label1,...,label50)) now since the labels are quite long, only a few fit on the page. Can I rotate each label by 90 degree counterclockwise (so that they are vertical) You should (re-)read ?axis, which points you to the las parameter: x - 1:50 plot(x, axes = FALSE) axis(1, x, paste(label, x), las = 2, cex.axis = 0.5) axis(2) box() --sundar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] CORRELATION MATRIX CONVERSION
Maybe like: dat X Y Z X 1.0 0.9 0.5 Y 0.9 1.0 0.1 Z 0.5 0.1 1.0 datrow - stack(as.data.frame(dat)) datrow$X=rownames(dat) datrow values ind X 11.0 X X 20.9 X Y 30.5 X Z 40.9 Y X 51.0 Y Y 60.1 Y Z 70.5 Z X 80.1 Z Y 91.0 Z Z Regards, Muhammad Subianto On this day 6/17/2005 8:14 AM, Omer Bakkalbasi wrote: How do I convert the output of cor(x) to a columnar format? Ex. from format below XYZ X 1.0 0.9 0.5 Y 0.9 1.0 0.1 Z 0.5 0.1 1.0 to format below X X 1.0 X Y 0.9 X Z 0.5 Y X 0.9 Y Y 1.0 Y Z 0.1 Z X 0.5 Z Y 0.1 Z Z 1.0 Thanks! Omer __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] axis labels vertically
BoM DS wrote: Hi, I have a plot and a custom axis labeling, e.g. x-c(...) plot(x,axes=FALSE) axis(2) axis(1,1:50,c(label1,...,label50)) now since the labels are quite long, only a few fit on the page. Can I rotate each label by 90 degree counterclockwise (so that they are vertical) See ?par and its argument las. Uwe Ligges __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Analysing ordinal/nominal data
Prof Brian Ripley wrote:
You have still not defined v' and w', nor the scores (are they estimated
or not). But the model I suggested is such a model with scores 1,2,...
Sorry for that, here it is:
scores v and w: integer scores, reflecting the ordering of columns/rows.
Agresti suggests to use 1,2,3,... as you did.
v' and w': mean of v and w, respectively
As I suggested before, a binomial logistic model is appropriate here,
not a Poisson log-linear one. (They are equivalent, but the binomial
version is easier to interpret and less wasteful to fit.)
After your suggestions I've read the logit-Chapter in Agresti (1984) and
tried to fit a logit model to my data:
log(m_ij2/m_ij1) = intercept + beta^PR_i(u_i-u') +
beta^ENV_j(v_j-v')
with:
mijk: expected frequencies for cell with indicies i,j,k
u_i: scores of PR with u={0,1,2,3}
v_j: scores of ENV with v={0,1,2}
beta^PR: association parameter for PR with responses
beta^ENV: association parameter for ENV with responses
I wrote in R:
count=c(250,274,285,241,279,299,247,246,255,280,355,362,
230,207,195,239,200,181,233,235,224,200,125,118)
PR=as.integer(gl(4,3,12))-1
ENV=as.integer(gl(3,1,12))-1
countM=matrix(count,12,2)
fit=glm(countM ~ as.integer(PR)+as.integer(ENV),
family=binomial(link=logit))
summary(fit)
R gives me:
Coefficients:
Estimate Std. Error z value Pr(|z|)
(Intercept) 0.184350.07737 2.383 0.017179 *
as.integer(PR) -0.105380.03085 -3.416 0.000635 ***
as.integer(ENV) -0.070750.04748 -1.490 0.136191
Looking at my data, I know that for PR=0 I have a very weak dependence
on ENV and for PR=3 very strong dependence on ENV. How can I get this
interpretation from the summary above? I'm new in R: is
fit$linearpredictors what I'm looking for?
--
Piotr Majdak
Institut fr Schallforschung
sterreichische Akademie der Wissenschaften
Reichsratsstr. 17
A-1010 Wien
Tel.: +43-1-4277-29511
Fax: +43-1-4277-9296
E-Mail: [EMAIL PROTECTED]
WWW: http://www.kfs.oeaw.ac.at
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Re: [R] axis labels vertically
2005/6/17, Sundar Dorai-Raj [EMAIL PROTECTED]: You should (re-)read ?axis, which points you to the las parameter: x - 1:50 plot(x, axes = FALSE) axis(1, x, paste(label, x), las = 2, cex.axis = 0.5) axis(2) box() thank you very much. This answer helps me in several ways: 1. it solves my particular problem 2. I wasn't aware that there is an online-help. (was using the tutorial on the web, which is nat at all detailed. 3. ?axis leaded me to ?par Thank you very much __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Error message from pamr
Hi, I got the fowllowing error message when I run pamr. Could you please advise me what does this error mean? Many thanks -- mydata - pamr.from.excel(datgrp4, 352, sample.labels=TRUE) Read 812768 items Read in 2307 genes Read in 350 samples Read in 350 sample labels Make sure these figures are correct!! mytrain - pamr.train(mydata) Error in if (from == to || length.out 2) by - 1 : missing value where logical needed - Rekindle the Rivalries. Sign up for Fantasy Football [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] mlbench xor
HI! I have 2D feature vectors and 2 classes. I want to solve this classification problem with SVM. I create an object XOR of the mlbench library. Then I replace the values of the XOR object with my values. Then I do a plot and I have my data plotted and coloured with the labels from the XOR object. So I replace the class labels in the xor object with my class labels. But when I do a plot with the new class labels every object is coloured in one colour. But half of the points should be in different colour(because different class label). As well when I apply svm on the xor object with th replaced data it finds just one class. someone any idea thx __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] wapply from gplots -- How do I get a local estimate of variance?
Dear list, I am trying to plot the local variance in a moving window on a dataset. The function that I am trying to use for this is wapply from gtools. However, from the lattice panel function code: snip cat(x) cat(y) wapply(x,y,method=range,width=1/10,fun=sd,na.rm=TRUE) - outvar /snip I get: snip 109 109 109 109 109 109 116 116 116 116 119 119 123 123 123 123 127 127 133 133 133 133 133 138 138 138 138 138 138 138 138138 138 138 142 142 142 142 142 142 142 147 147 147 147 147 158 158 158 158 158 158 158 158 158 158 158 158 158 20.234 28.431 10.762 45.613 15.028 10.775 18.516 13.371 18.516 13.371 20.61 31.08 9.338 210.77 11.927 39.438 -35.079 15.872 12.272 9.1122.665 7.355 22.057 -49.289 11.212 16.236 20.654 16.236 20.654 15.181 57.271 37.513 57.271 37.513 8.518 -108.517 -14.444 10.702 4.482 16.422 23.003 42.451 12.998 42.451 12.998 14.292 5.945 7.115 16.079 -13.172 10.62 16.079 -13.172 10.62 16.079 -13.172 10.62 20.477 22.656 Error in var(x, na.rm = na.rm) : 'x' is empty /snip What am I doing wrong with this? The variables that I think is supplied to the var function does not seem to be empty like the error message suggests. The gplots package is from the 2.0.7 version of the gregmisc bundle. /Fredrik Karlsson __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Fwd: lattice, panel.grid, and scales=list(tick.number=XXX)
I just realised that I had forgotten to copy this to the list.
-- Forwarded message --
From: Deepayan Sarkar [EMAIL PROTECTED]
Date: Jun 15, 2005 12:13 PM
Subject: Re: [R] lattice, panel.grid, and scales=list(tick.number=XXX)
To: Berton Gunter [EMAIL PROTECTED]
On 6/14/05, Berton Gunter [EMAIL PROTECTED] wrote:
If you look at the code of panel.grid, you'll see why it doesn't work -- it
does not use any of the scale parameters. Moreover the Help page for
panel.grid explicitly warns that the h,v=-1 specification may not work, so
no promises have been broken.
Right, the determination of tick mark locations can be arbitrarily
complicated, via the use of methods of the (unexported) generic
'formattedTicksAndLabels'. The only way I can think of to make
panel.grid aware of that is to store the results of that computation
somewhere. Note that these results could potentially be different for
every panel. While this is not impossible, it violates the principle
that a panel function only has access to the data for that panel (not
that this isn't ever violated, but not to this extent), and I would be
inclined to keep things as they are unless there are better reasons.
In this particular case, the tick computations are simple, so one
could do (using the recently added 'current.panel.limits', only
available if you update lattice to the latest version):
bwplot(voice.part ~ height, data=singer,
scales = list(tick.number = 10),
panel = function(...) {
ref.line - trellis.par.get(reference.line)
panel.abline(v = pretty(current.panel.limits()$xlim, n = 10),
col = ref.line$col,
lty = ref.line$lty,
lwd = ref.line$lwd)
panel.bwplot(...)
})
[...]
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of M. K.
Sent: Tuesday, June 14, 2005 3:37 PM
To: R-help mailing list
Subject: [R] lattice, panel.grid, and scales=list(tick.number=XXX)
I have a Lattice plot in which I want to adjust the number of tick
marks used, and I want to have the drawn grid reflect that change.
Here is what I'm doing:
bwplot(var1 ~ var2, data=df, scales=list(tick.number=10),
panel=function(...) {
panel.grid(h=0,v=-1,...);
panel.stripplot(col=gray40, pch=|, cex=2, ...);
panel.bwplot(...);
})
Unfortunately this doesn't quite work. Although the bwplot's tick
marks are indeed increased as requested, the panel.grid produces the
same (3 line) grid as before, seemingly unaware of the changed # of
ticks.
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] mlbench xor
Thorstensen Nicolas wrote: HI! I have 2D feature vectors and 2 classes. I want to solve this classification problem with SVM. I create an object XOR of the mlbench library. Then I replace the values of the XOR object with my values. Then I do a plot and I have my data plotted and coloured with the labels from the XOR object. So I replace the class labels in the xor object with my class labels. But when I do a plot with the new class labels every object is coloured in one colour. But half of the points should be in different colour(because different class label). As well when I apply svm on the xor object with th replaced data it finds just one class. someone any idea No, almost impossible without having a reproducible example Uwe Ligges thx __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Error message from pamr
luk wrote: Hi, I got the fowllowing error message when I run pamr. Could you please advise me what does this error mean? Many thanks -- mydata - pamr.from.excel(datgrp4, 352, sample.labels=TRUE) Read 812768 items Read in 2307 genes Read in 350 samples Read in 350 sample labels Make sure these figures are correct!! mytrain - pamr.train(mydata) Error in if (from == to || length.out 2) by - 1 : missing value where logical needed There is no reproducible example given, hence we can only guess. Might be a bug in package pamr or user error, if the first, you want to contact the package maintainer rather than R help. Uwe Ligges - Rekindle the Rivalries. Sign up for Fantasy Football [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] mlbench xor
On Fri, 17 Jun 2005 13:28:21 +0200, Thorstensen Nicolas (TN) wrote: HI! I have 2D feature vectors and 2 classes. I want to solve this classification problem with SVM. I create an object XOR of the mlbench library. Then I replace the values of the XOR object with my values. Then I do a plot and I have my data plotted and coloured with the labels from the XOR object. So I replace the class labels in the xor object with my class labels. But when I do a plot with the new class labels every object is coloured in one colour. But half of the points should be in different colour(because different class label). As well when I apply svm on the xor object with th replaced data it finds just one class. It is not the purpose of the r-help mailing list that undergraduate students ask questions about their classwork. We have a lab session scheduled for next Wednesday where you can ask questions like this in person. My apologies to the list for the noise that has been generated. Best, Fritz Leisch -- --- Friedrich Leisch Institut fr Statistik Tel: (+43 1) 58801 10715 Technische Universitt WienFax: (+43 1) 58801 10798 Wiedner Hauptstrae 8-10/1071 A-1040 Wien, Austria http://www.ci.tuwien.ac.at/~leisch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Rmpi installation over MPICH
This is a rather obscure question, I realize. I have written to the package author but have not heard back as of yet. I have read the README in the package, as well, but it didn't give me enough detail to diagnose the problems that I am having. (I did edit out the LAM-MPI check in zzz.R.in.) I am working to install Rmpi on top of MPICH on a beowulf cluster. Below is the output of the install command (which seems to go OK and then the start of the R session, which fails. I was hoping someone with some experience installing Rmpi on MPICH could give me some help with doing so. Any help would be greatly appreciated Sean [EMAIL PROTECTED] Rmpi]$ R CMD INSTALL --configure-args=--with-mpi=/opt/lib32/usr/lib --library='/home/sdavis/R/library' ../Rmpi_0.4-9.tar.gz * Installing *source* package 'Rmpi' ... Try to find mpi.h ... checking for gcc... gcc checking for C compiler default output... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ANSI C... none needed checking how to run the C preprocessor... gcc -E checking for egrep... grep -E checking for ANSI C header files... yes checking for sys/types.h... yes checking for sys/stat.h... yes checking for stdlib.h... yes checking for string.h... yes checking for memory.h... yes checking for strings.h... yes checking for inttypes.h... yes checking for stdint.h... yes checking for unistd.h... yes checking mpi.h usability... yes checking mpi.h presence... yes checking for mpi.h... yes Try to find libmpi ... checking for main in -lmpi... yes Try to find liblam ... checking for main in -llam... no liblam not found. Probably not LAM-MPI checking for openpty in -lutil... yes checking for main in -lpthread... yes configure: creating ./config.status config.status: creating src/Makevars config.status: creating R/zzz.R ** libs gcc -I/usr/lib/R/include -DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -DSTDC_HEADERS=1 -DHAVE_SYS_TYPES_H=1 -DHAVE_SYS_STAT_H=1 -DHAVE_STDLIB_H=1 -DHAVE_STRING_H=1 -DHAVE_MEMORY_H=1 -DHAVE_STRINGS_H=1 -DHAVE_INTTYPES_H=1 -DHAVE_STDINT_H=1 -DHAVE_UNISTD_H=1-I/usr/local/include -fPIC -O2 -g -pipe -march=i386 -mcpu=i686 -c conversion.c -o conversion.o gcc -I/usr/lib/R/include -DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -DSTDC_HEADERS=1 -DHAVE_SYS_TYPES_H=1 -DHAVE_SYS_STAT_H=1 -DHAVE_STDLIB_H=1 -DHAVE_STRING_H=1 -DHAVE_MEMORY_H=1 -DHAVE_STRINGS_H=1 -DHAVE_INTTYPES_H=1 -DHAVE_STDINT_H=1 -DHAVE_UNISTD_H=1-I/usr/local/include -fPIC -O2 -g -pipe -march=i386 -mcpu=i686 -c internal.c -o internal.o gcc -I/usr/lib/R/include -DPACKAGE_NAME=\\ -DPACKAGE_TARNAME=\\ -DPACKAGE_VERSION=\\ -DPACKAGE_STRING=\\ -DPACKAGE_BUGREPORT=\\ -DSTDC_HEADERS=1 -DHAVE_SYS_TYPES_H=1 -DHAVE_SYS_STAT_H=1 -DHAVE_STDLIB_H=1 -DHAVE_STRING_H=1 -DHAVE_MEMORY_H=1 -DHAVE_STRINGS_H=1 -DHAVE_INTTYPES_H=1 -DHAVE_STDINT_H=1 -DHAVE_UNISTD_H=1-I/usr/local/include -fPIC -O2 -g -pipe -march=i386 -mcpu=i686 -c Rmpi.c -o Rmpi.o gcc -shared -L/usr/local/lib -o Rmpi.so conversion.o internal.o Rmpi.o -lmpi -lutil -lpthread ** R ** demo ** inst ** preparing package for lazy loading ** help Building/Updating help pages for package 'Rmpi' Formats: text html latex example hosts texthtmllatex internal texthtmllatex mpi.abort texthtmllatex mpi.barrier texthtmllatex mpi.bcast texthtmllatex mpi.bcast.Robjtexthtmllatex mpi.bcast.cmd texthtmllatex mpi.comm texthtmllatex example mpi.comm.disconnect texthtmllatex mpi.comm.free texthtmllatex mpi.comm.intertexthtmllatex mpi.comm.set.errhandler texthtmllatex mpi.comm.spawntexthtmllatex mpi.const texthtmllatex mpi.exit texthtmllatex mpi.finalize texthtmllatex mpi.gathertexthtmllatex mpi.get.count texthtmllatex mpi.get.processor.nametexthtmllatex mpi.get.sourcetag texthtmllatex mpi.info texthtmllatex mpi.init.sprngtexthtmllatex mpi.intercomm.merge texthtmllatex mpi.parallel.sim texthtmllatex
Re: [R] CORRELATION MATRIX CONVERSION
Something like:
dat - data.frame(x=runif(10), y=runif(10), z=runif(10))
m - cor(dat)
m
x y z
x 1.000 0.1183305 0.1096394
y 0.1183305 1.000 -0.2819285
z 0.1096394 -0.2819285 1.000
mat2col - function(m) {
+ m2 - matrix(m, ncol=1)
+ rownames(m2) - outer(rownames(m), colnames(m), paste)
+ m2
+ }
mat2col(m)
[,1]
x x 1.000
y x 0.1183305
z x 0.1096394
x y 0.1183305
y y 1.000
z y -0.2819285
x z 0.1096394
y z -0.2819285
z z 1.000
Andy
From: Omer Bakkalbasi
How do I convert the output of cor(x) to a columnar format?
Ex. from format below
XYZ
X 1.0 0.9 0.5
Y 0.9 1.0 0.1
Z 0.5 0.1 1.0
to format below
X X 1.0
X Y 0.9
X Z 0.5
Y X 0.9
Y Y 1.0
Y Z 0.1
Z X 0.5
Z Y 0.1
Z Z 1.0
Thanks!
Omer
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Re: [R] CORRELATION MATRIX CONVERSION
Excellent! This is the most flexible and intuitive option. Thanks!
Omer
Cell: (914) 671-7447
-Original Message-
From: Liaw, Andy [mailto:[EMAIL PROTECTED]
Sent: Friday, June 17, 2005 8:40 AM
To: '[EMAIL PROTECTED]'; r-help@stat.math.ethz.ch
Subject: RE: [R] CORRELATION MATRIX CONVERSION
Something like:
dat - data.frame(x=runif(10), y=runif(10), z=runif(10))
m - cor(dat)
m
x y z
x 1.000 0.1183305 0.1096394
y 0.1183305 1.000 -0.2819285
z 0.1096394 -0.2819285 1.000
mat2col - function(m) {
+ m2 - matrix(m, ncol=1)
+ rownames(m2) - outer(rownames(m), colnames(m), paste)
+ m2
+ }
mat2col(m)
[,1]
x x 1.000
y x 0.1183305
z x 0.1096394
x y 0.1183305
y y 1.000
z y -0.2819285
x z 0.1096394
y z -0.2819285
z z 1.000
Andy
From: Omer Bakkalbasi
How do I convert the output of cor(x) to a columnar format?
Ex. from format below
XYZ
X 1.0 0.9 0.5
Y 0.9 1.0 0.1
Z 0.5 0.1 1.0
to format below
X X 1.0
X Y 0.9
X Z 0.5
Y X 0.9
Y Y 1.0
Y Z 0.1
Z X 0.5
Z Y 0.1
Z Z 1.0
Thanks!
Omer
__
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http://www.R-project.org/posting-guide.html
--
Notice: This e-mail message, together with any attachments,...{{dropped}}
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Re: [R] possible bug in merge with duplicate blank names in 'by' field.
Thanks for your quick responses, Gabor and Brian. I'm currently running R version 1.9.1 on Linux. Actually, I have just tested this on R v.2.1.0 running under Windows XP, and indeed, as you both indicate, the problem does not exist on that version for that OS. So, at an appropriate time I'll upgrade my Linux installation to the most recent version (1.9.1 is a year old, I guess). -Frank At 03:26 AM 6/17/2005, Prof Brian Ripley wrote: What version of R is this (please do see the posting guide)? In both 2.1.0 and 2.1.1 beta I get all Promoter ip.x ip.y ip 130 40 40 240 40 40 3a 10 NA NA 4c 20 20 20 5b NA 15 15 6d NA 30 30 so cannot reproduce your result. Are you sure that the `blanks' really are empty and not some character that is printing as empty on your unstated OS? BTW ' ' is what is normally called `blank'. BTW, these are not `names' but character strings: `names' has other meanings in R. PhD, Computational Biologist, Harvard Medical School BCMP/SGM-322, 250 Longwood Ave, Boston MA 02115, USA. Tel: 617-432-3555 Fax: 617-432-3557 http://llama.med.harvard.edu/~fgibbons __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Trying to build R sources on Windows
I am trying without success to build the R-2.1.0 sources on Windows 2003 server. I have MinGW/bin in front of cygwin/bin in the Windows path. However, I try to build, I get failures trying to include headers which are not part of MinGW. I am definitely using the MinGW compilers, so why are the sources trying to access headers which are not included? Examples would be: argz.h included from l10nflist.c alloca.h from errors.c langinfo.h from main.c Any clues about what I am doing wrong? Bill Northcott __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] reading csv-data
Hi! I have had this problem for a long time. I have tried to study the manuals and search the mailing lists, but I can not solve this. I think there has to be one simple solution to this, but I just can not find it. I have saved the data in excel (csv-format). Then I read the data in R e.g. data - read.csv2(example.csv, header=TRUE) I look the data and it looks ok. E.g data Mean1 1 4.4332 2 8.5113 3 35.1624 4 9.1693 52.974 6 65.1578 7 43.2241 8 3.1278 9 5.3364 10 3.9767 However, this Mean1 is categorical when it should be real numbers. Mean1 [1] 4.4332 8.5113 35.1624 9.1693 2.974 65.1578 43.2241 3.1278 5.3364 Levels: 2.974 3.1278 35.1624 4.4332 43.2241 5.3364 65.1578 8.5113 9.1693 Why R does not understand that this should be real numbers? What am I doing wrong here? Thanks for your help. Regards, Johanna Sundvik __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] [R-pkgs] New CRAN package sp: classes and methods for spatial data
We're happy to announce the CRAN release of sp, an R package which has new-style classes and methods for spatial data, version 0.7-9. Spatial data types that sp implements are: points, grids, lines, and polygons (i.e., rings) optionally with holes. Methods include + the usual print, summary, plot, [, [[, $, ... + coercion between types (e.g. points and grids, matrices, data.frames) + coordinates(x), which returns the spatial coordinates of x + bbox(x), returns the coordinates bounding box of x + overlay, to query the value of e.g. points in polygons or grid (essentially does a point-in-polygon or point-in-raster cell) + spsample, for random sampling methods over a spatial domain. An additional package (spproj) provides coordinate reference system transformation (projection and re-projection) using the PROJ.4 library [2]. Others (will) provide interfaces to GRASS 6 and gdal. A good deal of work has also gone into providing plotting methods using base, grid and lattice graphics, through the spplot function, a front-end to lattice plots for spatial data (see gallery [1]). The home page of these packages is found at http://r-spatial.sourceforge.net/ See also the Task View on Spatial Data Analysis, linked from CRAN. The reason why we wrote this package is that we think R is an excellent environment to deal with spatial data, but that it lacks a uniform way to deal with spatial data. Compared to the handling of dates and times, which can utilize base classes or those provided in the chron package, spatial data handling is much more fragmented. As a consequence: - various packages make their own assumptions about how spatial data are organized - spatial data organized for a certain package cannot easily be used for another package - few (or no) packages address the full range of spatial data types (points, grids, lines, polygons) and their interaction - generic spatial functionality (e.g. I/O to GIS, plotting, projection) is scattered and often limited in functionality. It also means that many different package authors have to use time writing similar data handling code, rather than concentrating on analytical functions. If the sp package achieves its goals, data I/O will become many-to-one, and data access for analysis one-to-many, providing a shared data object layer for which shared methods can be written. Classes and methods for spatial data are only useful when the spatial packages support them. The sp development team includes maintainers of a number of spatial R packages, on which we will work, and we hope that over time other spatial package maintainers will also provide support for the classes provided by this package. Although we are working towards a fixed set of classes at this moment of the development of sp we cannot fully guarantee that the exact representation of the sp classes will not change in the future. Therefore, we advise users to keep the scripts with instructions of building the data into sp classes; it is most likely that we will not change the building functions and methods. Work in progress currently involves: - support of sp classes by several spatial statistics packages - automatic determination of hole polygons from shapefiles - plotting filled polygons with holes - gdal read/write support - GRASS 6.0 support The development of this package is a joint effort of Virgilio Gomez-Rubio, Barry Rowlinson, Roger Bivand and Edzer Pebesma, and followed from discussions held at a pre-DSC2003 workshop [3], announcements on R-sig-geo [4], and a meeting held last November in Lancaster [5]. A beta release of sp was announced a while ago on R-sig-geo [6]. With best regards, -- Roger Bivand and Edzer Pebesma [1] http://r-spatial.sourceforge.net/ [2] http://www.remotesensing.org/proj/ [3] http://spatial.nhh.no/meetings/vienna/index.html [4] e.g. https://stat.ethz.ch/pipermail/r-sig-geo/2003-October/28.html [5] http://elearning.maths.lancs.ac.uk:8080/RSpatial/ [6] https://stat.ethz.ch/pipermail/r-sig-geo/2005-April/000378.html ___ R-packages mailing list [EMAIL PROTECTED] https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] drop elements of vector by class
i'm trying to build a little summary table for the contents of a data frame. t-sapply(macro, data.class) c-sapply(macro, length) m-sapply(macro, mean, na.rm=T, digits=2) cbind(type=t, n=c , mean=m) I want to drop the variables that are factors so I can include -max- and -min- in my table. -macro- contacts the data--how do I drop the variables according to their data.class thanks, michael foster E. Michael Foster (W) 814-865-1923 (Fax) 814-863- After 7/1 Work: Professor of Maternal and Child Health School of Public Health University of North Carolina, Chapel Hill Rosenau Hall, CB# 7445 Chapel Hill, NC 27599-7445 Home: 309 Old Larkspur Way Chapel Hill, NC 27516 UNC School of Public Health--voted #2 SPH by . US News . Association of Hand-to-Fin Carp Fisherman . Iron Chef (not to be confused with the Iron Dukes!) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] reading csv-data
On Fri, 17 Jun 2005, Johanna Sundvik wrote: However, this Mean1 is categorical when it should be real numbers. Mean1 [1] 4.4332 8.5113 35.1624 9.1693 2.974 65.1578 43.2241 3.1278 5.3364 Levels: 2.974 3.1278 35.1624 4.4332 43.2241 5.3364 65.1578 8.5113 9.1693 Why R does not understand that this should be real numbers? What am I doing wrong here? Thanks for your help. Your files must have some entries that are not numbers, such as . or something. R then can't tell that the field is supposed to be numeric. This may happen with missing data, in which case the na.strings= argument can be used to tell R how missing data are specified. You can convert the data to numeric as described in FAQ 7.10 -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to calculate random matrices from the multivariate normal distribution
Hi R users, I am trying to calculate MonteCarlo from the multivariate normal distribution. I am utilizing the parameters vector (how mean) and covariance matrix (or 1/hessian) how input. Can anyone provide guidance on how I could do this? Thank you. Juan Carlos Quiroz Instituto de Fomento Pesquero Blanco 839 Valparaiso, CHILE. Casilla 8V Office: 56+032-322497 Email: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Fit values for NA's in linear regression
Hi, To obtain estimates for some missing values in my data I fitted a linear regression and then used the command fitted(model) to get the fitted values from the model, but R doesn't return any values for the NA's. I can calculate the fitted values from the estimates obtained from the summary of model, but that's not very handy. Is there a way to include the missing values in the analysis and get fitted values for the NA's? I tried to use with the function na.action, but that didn't work. thanks Luca __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Analysing ordinal/nominal data
Hi Brian (and the list of course!), I still have problems analysing data in R, because I don't know how to tell glm() use row-effect model, please. The models are well defined by Agresti, but can't get the link from the theory to the implementations in R. Different names, definitions and no explanation of results in the documentation of glm() or summary() make it hard to get in... But I found another solution for my data: I just use the saturated log-linear model for categorial data and refer to the estimated parameter of the three-way-interaction only. That's enough statistics for my paper ;-). Thanks for your help, Piotr Majdak Prof Brian Ripley wrote: As I suggested before, a binomial logistic model is appropriate here, not a Poisson log-linear one. (They are equivalent, but the binomial version is easier to interpret and less wasteful to fit.) You have still not defined v' and w', nor the scores (are they estimated or not). But the model I suggested is such a model with scores 1,2,... -- Piotr Majdak Acoustics Research Institute Austrian Academy of Sciences Reichsratsstr. 17 A-1010 Vienna AUSTRIA phone: +43-1-4277-29511 fax: +43-1-4277-9296 email: [EMAIL PROTECTED] WWW: http://www.kfs.oeaw.ac.at __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] reading csv-data
Thomas Lumley wrote: On Fri, 17 Jun 2005, Johanna Sundvik wrote: However, this Mean1 is categorical when it should be real numbers. Mean1 [1] 4.4332 8.5113 35.1624 9.1693 2.974 65.1578 43.2241 3.1278 5.3364 Levels: 2.974 3.1278 35.1624 4.4332 43.2241 5.3364 65.1578 8.5113 9.1693 Why R does not understand that this should be real numbers? What am I doing wrong here? Thanks for your help. Your files must have some entries that are not numbers, such as . or something. R then can't tell that the field is supposed to be numeric. This may happen with missing data, in which case the na.strings= argument can be used to tell R how missing data are specified. You can convert the data to numeric as described in FAQ 7.10 -thomas I think the problem can be that you use read.csv2(), which expect a comma (,) as decimal-indicator (as is common in Scandinavia), and a semi-colon (;) as separator between columns. Either you should try read.csv(), or you can try read.csv2(example.csv, dec=., header=TRUE) Have a look at ?read.csv (read.csv2 is in the same help-text). Ivar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] reading csv-data
In my experience, this has always been due to the presence of non-numeric values in the input. In the example you show, it is not obvious that there is any. I would start by first inspecting the input file very carefully, using a text editor outside of R. Since your example appears to have only one column of data, you could try reading it with the scan() function. This might produce additional information that would help you identify any non-numeric data. Using count.fields() on the data file might reveal something. If Mean1 is an element of data, then simply typing Mean1 at the prompt should produce a not found message. Yet Mean1 was found. Have you omitted something in your email, or is there another object named Mean1? -Don At 4:50 PM +0300 6/17/05, Johanna Sundvik wrote: Hi! I have had this problem for a long time. I have tried to study the manuals and search the mailing lists, but I can not solve this. I think there has to be one simple solution to this, but I just can not find it. I have saved the data in excel (csv-format). Then I read the data in R e.g. data - read.csv2(example.csv, header=TRUE) I look the data and it looks ok. E.g data Mean1 1 4.4332 2 8.5113 3 35.1624 4 9.1693 52.974 6 65.1578 7 43.2241 8 3.1278 9 5.3364 10 3.9767 However, this Mean1 is categorical when it should be real numbers. Mean1 [1] 4.4332 8.5113 35.1624 9.1693 2.974 65.1578 43.2241 3.1278 5.3364 Levels: 2.974 3.1278 35.1624 4.4332 43.2241 5.3364 65.1578 8.5113 9.1693 Why R does not understand that this should be real numbers? What am I doing wrong here? Thanks for your help. Regards, Johanna Sundvik __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- -- Don MacQueen Environmental Protection Department Lawrence Livermore National Laboratory Livermore, CA, USA __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to calculate random matrices from the multivariate normal distribution
On 6/17/2005 11:03 AM, Juan Carlos Quiroz Espinosa wrote: Hi R users, I am trying to calculate MonteCarlo from the multivariate normal distribution. I am utilizing the parameters vector (how mean) and covariance matrix (or 1/hessian) how input. Can anyone provide guidance on how I could do this? You probably want the mvrnorm function from the MASS package. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] drop elements of vector by class
try this # to get only the factors macro.f - macro[sapply(macro, is.factor)] to drop the factors macro.nf - macro[sapply(macro, !is.factor)] # to get only numerics macro.n - macro[sapply(macro, is.numeric)] and so on. I hope it helps. Best, Dimitris Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.be/biostat/ http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm - Original Message - From: E. Michael Foster [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Sent: Friday, June 17, 2005 4:57 PM Subject: [R] drop elements of vector by class i'm trying to build a little summary table for the contents of a data frame. t-sapply(macro, data.class) c-sapply(macro, length) m-sapply(macro, mean, na.rm=T, digits=2) cbind(type=t, n=c , mean=m) I want to drop the variables that are factors so I can include -max- and -min- in my table. -macro- contacts the data--how do I drop the variables according to their data.class thanks, michael foster E. Michael Foster (W) 814-865-1923 (Fax) 814-863- After 7/1 Work: Professor of Maternal and Child Health School of Public Health University of North Carolina, Chapel Hill Rosenau Hall, CB# 7445 Chapel Hill, NC 27599-7445 Home: 309 Old Larkspur Way Chapel Hill, NC 27516 UNC School of Public Health--voted #2 SPH by . US News . Association of Hand-to-Fin Carp Fisherman . Iron Chef (not to be confused with the Iron Dukes!) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] How to calculate random matrices from the multivariate normal distribution
RSiteSearch(random multivariate normal) produced 82 hits, the fourth of which was for multivariate normal distribution, function pmvnorm in library mvtnorm, which also includes a function rmvorm, that should do what you want. Buena suerte. spencer graves Juan Carlos Quiroz Espinosa wrote: Hi R users, I am trying to calculate MonteCarlo from the multivariate normal distribution. I am utilizing the parameters vector (how mean) and covariance matrix (or 1/hessian) how input. Can anyone provide guidance on how I could do this? Thank you. Juan Carlos Quiroz Instituto de Fomento Pesquero Blanco 839 Valparaiso, CHILE. Casilla 8V Office: 56+032-322497 Email: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] drop elements of vector by class
a - data.frame(a = 1:2, b = c(a, b), c = I(c(a, b))) a - a[ , !sapply(a, class) %in% factor] -Original Message- From: E. Michael Foster [mailto:[EMAIL PROTECTED] Sent: Friday, June 17, 2005 10:58 AM To: r-help@stat.math.ethz.ch Subject: [R] drop elements of vector by class i'm trying to build a little summary table for the contents of a data frame. t-sapply(macro, data.class) c-sapply(macro, length) m-sapply(macro, mean, na.rm=T, digits=2) cbind(type=t, n=c , mean=m) I want to drop the variables that are factors so I can include -max- and -min- in my table. -macro- contacts the data--how do I drop the variables according to their data.class thanks, michael foster E. Michael Foster (W) 814-865-1923 (Fax) 814-863- After 7/1 Work: Professor of Maternal and Child Health School of Public Health University of North Carolina, Chapel Hill Rosenau Hall, CB# 7445 Chapel Hill, NC 27599-7445 Home: 309 Old Larkspur Way Chapel Hill, NC 27516 UNC School of Public Health--voted #2 SPH by . US News . Association of Hand-to-Fin Carp Fisherman . Iron Chef (not to be confused with the Iron Dukes!) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Mixed model question
Have you also tried lmer in library(lme4)? This is newer and better in many ways. Unfortunately, I see only one example in the Help file, but you might be able to figure out how to use lmer from the help file and from the 125 hits on RSiteSearch(lmer). The definitive work on this subject, from my perspective, is Pinheiro and Bates (2000) Mixed-Effects Models in S and S-PLUS (Springer). This book is primarily how to use lme. I learned a lot from it (and from Doug Bates' other book with Don Watts on nonlinear regression). I couldn't find Pinheiro and Bates in the catalog for the library at Universidad de Sevilla (though Bates and Watts was listed). Feel free to submit other questions after you've considered this, but PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html first. In particular, you might have gotten a more helpful reply to this question if you had included one of your attempts with lme, asking what you did wrong or what you don't understand about the results. Buena suerte, spencer graves Alfonso M Sanchez-Lafuente wrote: Hi, I am new to this list as a poster, but a reader for some time. I've using R for several weeks now, and I have a lot of questions about certain procedures. Here I go: I want to test if there are differences in the time spent by pollinators visiting flowers of a given plant species, according to a number of experimental manipulations made on those flowers. All experimental manipulations (factor with 5 levels) are replicated within plants (i.e. plant is my sampling unit). Further, I have two populations (two level factor), and a number of pollinator groups (again, two levels, the same in both populations). The response variables in the numbe of seconds invested in each probe. Further, I have plant floral display as a covariate, as it may influence visitation rates. I think I have to analyse this desing considering population, pollinator group and their interaction as fixed effects, and treatment nested within plant, and its interaction with population and pollinator group, as random factors. In SAS terminology, the model looks like this: proc mixed data=flwfunc.visitflower covtest method=reml; class site pollclass treatm plantid; model time = site|pollclass flwinflor / chisq; random treatm site*treatm pollclass*treatm / subject=plantid; lsmeans site pollclass site*pollclass; run; I've been successfully trying lm, but I think is not suitable for random effects. Thus, I've tried lme, but no success when defining the random part or trying to interpret the results... Any help will be welcome! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Fit values for NA's in linear regression
Have you considered: set.seed(1) tstDF - data.frame(x=1:4, y=c(NA, 2:4+rnorm(3))) fit - lm(y~x, tstDF) predict(fit) 234 1.678441 2.573854 3.469266 predict(fit, tstDF) 1 2 3 4 0.7830284 1.6784410 2.5738536 3.4692662 spencer graves Luca Wacker wrote: Hi, To obtain estimates for some missing values in my data I fitted a linear regression and then used the command fitted(model) to get the fitted values from the model, but R doesn't return any values for the NA's. I can calculate the fitted values from the estimates obtained from the summary of model, but that's not very handy. Is there a way to include the missing values in the analysis and get fitted values for the NA's? I tried to use with the function na.action, but that didn't work. thanks Luca __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] reading csv-data
I've struggled with this myself in the past. I've recently started using the following: File - pair.txt # File name with path if different from getwd() readLines(File, n=9) The function readLines reads the first n lines as n individual character strings. From this, you can identify extra headers, the separate characters, etc. Then I can do something like the following: plot(count.fields(File, sep=\t)) The function count.fields also has arguments to specify a number of lines to skip before it starts to process the file, which can be helpful with multiple headers. After count.fields produces a constant result consistent with what I want, then I'm ready to use read.table or one of its variants like read.csv2. hope this helps. spencer graves Don MacQueen wrote: In my experience, this has always been due to the presence of non-numeric values in the input. In the example you show, it is not obvious that there is any. I would start by first inspecting the input file very carefully, using a text editor outside of R. Since your example appears to have only one column of data, you could try reading it with the scan() function. This might produce additional information that would help you identify any non-numeric data. Using count.fields() on the data file might reveal something. If Mean1 is an element of data, then simply typing Mean1 at the prompt should produce a not found message. Yet Mean1 was found. Have you omitted something in your email, or is there another object named Mean1? -Don At 4:50 PM +0300 6/17/05, Johanna Sundvik wrote: Hi! I have had this problem for a long time. I have tried to study the manuals and search the mailing lists, but I can not solve this. I think there has to be one simple solution to this, but I just can not find it. I have saved the data in excel (csv-format). Then I read the data in R e.g. data - read.csv2(example.csv, header=TRUE) I look the data and it looks ok. E.g data Mean1 1 4.4332 2 8.5113 3 35.1624 4 9.1693 52.974 6 65.1578 7 43.2241 8 3.1278 9 5.3364 10 3.9767 However, this Mean1 is categorical when it should be real numbers. Mean1 [1] 4.4332 8.5113 35.1624 9.1693 2.974 65.1578 43.2241 3.1278 5.3364 Levels: 2.974 3.1278 35.1624 4.4332 43.2241 5.3364 65.1578 8.5113 9.1693 Why R does not understand that this should be real numbers? What am I doing wrong here? Thanks for your help. Regards, Johanna Sundvik __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Data comparison
Question : Is it possible to create a function, using a for ifelse function, inside sapply, to compare the values in one data frame to a set of upper and lower limits in another data frame, same number of columns., Take the values which meet the requirements TRUE and create a new data frame or table containing the filtered data? Can you give me a shove in the correct direction? Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Fit values for NA's in linear regression
It depends on the na.action used in lm(), which defaults to na.omit. Here's an example: x - c(1:5, NA, 7:10) y - rnorm(x) fitted(lm(y ~ x)) 1 2 3 4 5 7 8 0.68680104 0.47913875 0.27147646 0.06381417 -0.14384812 -0.55917271 -0.76683500 9 10 -0.97449729 -1.18215958 fitted(lm(y ~ x, na.action=na.exclude)) 1 2 3 4 5 6 7 0.68680104 0.47913875 0.27147646 0.06381417 -0.14384812 NA -0.55917271 8 9 10 -0.76683500 -0.97449729 -1.18215958 Andy From: Luca Wacker Hi, To obtain estimates for some missing values in my data I fitted a linear regression and then used the command fitted(model) to get the fitted values from the model, but R doesn't return any values for the NA's. I can calculate the fitted values from the estimates obtained from the summary of model, but that's not very handy. Is there a way to include the missing values in the analysis and get fitted values for the NA's? I tried to use with the function na.action, but that didn't work. thanks Luca __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] vectorization
Hi there,
I have a data frame (mydata) with 1 numeric variable (income) and 1 factor
(education). I want a new column in this data with the median income for each
education level. A obviously inneficient way to do this is
for ( k in 1: nrow(mydata) ){
l - mydata$education[k]
mydata$md[k] - median(mydata$income[mydata$education==l],na.rm=T)
}
Since mydata has nearly 30.000 rows, this will be done not untill the end of
this month. I thus need some help for vectorizing this, please.
Thanks,
Dimitri
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Re: [R] Fit values for NA's in linear regression
I posted a bad example. The question would only make sense if NAs are all in the response. Here's try #2: I believe you can not get fitted values for cases where y is missing directly from the model object. You can, however, use predict(lm.object, newdata=mydata[!complete.cases(mydata)]) to get the predictions for those with missing y. Andy From: Luca Wacker Hi, To obtain estimates for some missing values in my data I fitted a linear regression and then used the command fitted(model) to get the fitted values from the model, but R doesn't return any values for the NA's. I can calculate the fitted values from the estimates obtained from the summary of model, but that's not very handy. Is there a way to include the missing values in the analysis and get fitted values for the NA's? I tried to use with the function na.action, but that didn't work. thanks Luca __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] logistic regression - using polys and products of features
On Fri, 17 Jun 2005 07:39:30 +1000 Stephen Choularton [EMAIL PROTECTED] wrote: Hi I can get all my features by doing this: logistic.model = glm(similarity ~ ., family=binomial, data = cData[3001:3800,]) I can get the product of all my features by this: logistic.model = glm(similarity ~ . ^ 2, family=binomial, data = cData[3001:3800,]) I don't seem to be able to get polys by doing this: logistic.model = glm(similarity ~ poly(.,2), family=binomial, data = cData[3001:3800,]) Error in poly(., 2) : Object . not found How can I get polys? What do the warnings mean when I do this: logistic.model = glm(similarity ~ . + . ^ 2, family=binomial, data = cData[3001:3800,]) Warning messages: 1: Algorithm did not converge in: glm.fit(x = X, y = Y, weights = weights, start = start, etastart = etastart, 2: fitted probabilities numerically 0 or 1 occurred in: glm.fit(x = X, y = Y, weights = weights, start = start, etastart = etastart, .^2 means all the 2-order and 1-order terms,so .+.^2 is meaningless. How can I do this? logistic.model = glm(similarity ~ . + . ^ 2 + poly(.,2), family=binomial, data = cData[3001:3800,]) Thanks Stephen -- Internal Virus Database is out-of-date. Checked by AVG Anti-Virus. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Department of Sociology Fudan University,Shanghai Blog:http://sociology.yculblog.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] trim a string
How to trim the leading and trailing white space off of a string? If the variable is E I need to convert it to E. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Mixed model question
and the newest R-new in the www.r-project.org has an article about how to use the lmer function. On Fri, 17 Jun 2005 08:43:41 -0700 Spencer Graves [EMAIL PROTECTED] wrote: Have you also tried lmer in library(lme4)? This is newer and better in many ways. Unfortunately, I see only one example in the Help file, but you might be able to figure out how to use lmer from the help file and from the 125 hits on RSiteSearch(lmer). The definitive work on this subject, from my perspective, is Pinheiro and Bates (2000) Mixed-Effects Models in S and S-PLUS (Springer). This book is primarily how to use lme. I learned a lot from it (and from Doug Bates' other book with Don Watts on nonlinear regression). I couldn't find Pinheiro and Bates in the catalog for the library at Universidad de Sevilla (though Bates and Watts was listed). Feel free to submit other questions after you've considered this, but PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html first. In particular, you might have gotten a more helpful reply to this question if you had included one of your attempts with lme, asking what you did wrong or what you don't understand about the results. Buena suerte, spencer graves Alfonso M Sanchez-Lafuente wrote: Hi, I am new to this list as a poster, but a reader for some time. I've using R for several weeks now, and I have a lot of questions about certain procedures. Here I go: I want to test if there are differences in the time spent by pollinators visiting flowers of a given plant species, according to a number of experimental manipulations made on those flowers. All experimental manipulations (factor with 5 levels) are replicated within plants (i.e. plant is my sampling unit). Further, I have two populations (two level factor), and a number of pollinator groups (again, two levels, the same in both populations). The response variables in the numbe of seconds invested in each probe. Further, I have plant floral display as a covariate, as it may influence visitation rates. I think I have to analyse this desing considering population, pollinator group and their interaction as fixed effects, and treatment nested within plant, and its interaction with population and pollinator group, as random factors. In SAS terminology, the model looks like this: proc mixed data=flwfunc.visitflower covtest method=reml; class site pollclass treatm plantid; model time = site|pollclass flwinflor / chisq; random treatm site*treatm pollclass*treatm / subject=plantid; lsmeans site pollclass site*pollclass; run; I've been successfully trying lm, but I think is not suitable for random effects. Thus, I've tried lme, but no success when defining the random part or trying to interpret the results... Any help will be welcome! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- Department of Sociology Fudan University,Shanghai Blog:http://sociology.yculblog.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] How to get the values of a vector having the indices?
Hi I want to get the values of a vector which I have its indices. How it is possible? For example after clustering , I can access to the indices of the first cluster using: first- which(clusters$clustering==1) first give me the indices, but how can I access to the values? Thanks a lot and have a fun. Amir __ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] vectorization
Here I go again with ave():
mydata$md - ave(mydata$income, mydata$education, FUN=median, na.rm=TRUE)
IMHO it's one of the most under-rated helper functions in R.
Andy
From: Dimitri Joe
Hi there,
I have a data frame (mydata) with 1 numeric variable (income)
and 1 factor (education). I want a new column in this data
with the median income for each education level. A obviously
inneficient way to do this is
for ( k in 1: nrow(mydata) ){
l - mydata$education[k]
mydata$md[k] - median(mydata$income[mydata$education==l],na.rm=T)
}
Since mydata has nearly 30.000 rows, this will be done not
untill the end of this month. I thus need some help for
vectorizing this, please.
Thanks,
Dimitri
[[alternative HTML version deleted]]
___
Instale o discador agora! http://br.acesso.yahoo.com/
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] trim a string
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/14493.html Andy From: Omar Lakkis How to trim the leading and trailing white space off of a string? If the variable is E I need to convert it to E. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] vectorization
You can use tapply() to compute the medians, as in
meds - tapply(mydata$inc,INDEX=mydata$ed,FUN=median)
then create a new column with the medians as
medianEd - meds[mydata$ed]
Reid Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dimitri Joe
Sent: Friday, June 17, 2005 1:01 PM
To: R-Help
Subject: [R] vectorization
Hi there,
I have a data frame (mydata) with 1 numeric variable (income) and 1 factor
(education). I want a new column in this data with the median income for
each education level. A obviously inneficient way to do this is
for ( k in 1: nrow(mydata) ){
l - mydata$education[k]
mydata$md[k] - median(mydata$income[mydata$education==l],na.rm=T)
}
Since mydata has nearly 30.000 rows, this will be done not untill the end of
this month. I thus need some help for vectorizing this, please.
Thanks,
Dimitri
[[alternative HTML version deleted]]
___
Instale o discador agora! http://br.acesso.yahoo.com/
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] vectorization
These two lines worked for me:
rst - tapply(mydata$income, mydata$education, median)
mydata$md - rst[mydata$education]
Here's my cheesy example:
mydata - data.frame(income= round(rnorm(3, 55000, 1)),
+ education = letters[rbinom(3, 4, 1/2)+1])
rst - tapply(mydata$income, mydata$education, median)
mydata$md - rst[mydata$education]
head(mydata)
income education md
1 66223 e 55094.5
2 56830 c 54966.0
3 58035 b 54937.5
4 74045 a 55213.5
5 61327 b 54937.5
6 64150 b 54937.5
Is this what you wanted?
Kevin
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dimitri Joe
Sent: Friday, June 17, 2005 10:01 AM
To: R-Help
Subject: [R] vectorization
Hi there,
I have a data frame (mydata) with 1 numeric variable (income) and 1
factor (education). I want a new column in this data with the median
income for each education level. A obviously inneficient way to do this
is
for ( k in 1: nrow(mydata) ){
l - mydata$education[k]
mydata$md[k] - median(mydata$income[mydata$education==l],na.rm=T)
}
Since mydata has nearly 30.000 rows, this will be done not untill the
end of this month. I thus need some help for vectorizing this, please.
Thanks,
Dimitri
[[alternative HTML version deleted]]
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Re: [R] vectorization
try this:
x.1 - data.frame(income=runif(100)*1,
educ=sample(c('hs','col','none'),100,T))
x.1
income educ
1 5930.30882 col
2 5528.83222 hs
3 5967.04041 hs
4 3926.30682 hs
5 2603.75924 none
...
x.2 - tapply(x.1$income, x.1$educ, mean)
x.2
col hs none
5575.310 4994.921 5481.962
x.1$median - x.2[x.1$educ]
x.1
income educ median
1 5930.30882 col 5575.310
2 5528.83222 hs 4994.921
3 5967.04041 hs 4994.921
4 3926.30682 hs 4994.921
5 2603.75924 none 5481.962
6 7398.83325 col 5575.310
7265.06895 hs 4994.921
.
Jim
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James HoltmanWhat is the problem you are trying to solve?
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Dimitri Joe
[EMAIL PROTECTED]To: R-Help
r-help@stat.math.ethz.ch
.br cc:
Sent by: Subject: [R] vectorization
[EMAIL PROTECTED]
ath.ethz.ch
06/17/2005 14:00
Hi there,
I have a data frame (mydata) with 1 numeric variable (income) and 1 factor
(education). I want a new column in this data with the median income for
each education level. A obviously inneficient way to do this is
for ( k in 1: nrow(mydata) ){
l - mydata$education[k]
mydata$md[k] - median(mydata$income[mydata$education==l],na.rm=T)
}
Since mydata has nearly 30.000 rows, this will be done not untill the end
of this month. I thus need some help for vectorizing this, please.
Thanks,
Dimitri
[[alternative HTML version deleted]]
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Re: [R] trim a string
RSiteSearch(trim) will give you a lot of answers. You cal also use the
higher level function trim{R.oo} i.e.:
library(R.oo)
x=e
trim(x)
[1] e
From: Omar Lakkis [EMAIL PROTECTED]
Reply-To: Omar Lakkis [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject: [R] trim a string
Date: Fri, 17 Jun 2005 13:34:36 -0400
How to trim the leading and trailing white space off of a string?
If the variable is E I need to convert it to E.
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Re: [R] trim a string
Better code for this purpose is in example(grep). `white space' and `a blank' are not necessarily the same thing. On Fri, 17 Jun 2005, Liaw, Andy wrote: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/14493.html Andy From: Omar Lakkis How to trim the leading and trailing white space off of a string? If the variable is E I need to convert it to E. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] vectorization
Hi,
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dimitri Joe
Sent: Friday, June 17, 2005 7:01 PM
To: R-Help
Subject: [R] vectorization
Hi there,
I have a data frame (mydata) with 1 numeric variable (income)
and 1 factor (education). I want a new column in this data
with the median income for each education level. A obviously
inneficient way to do this is
I guess the attached code (incl. simulating your data structure) is not
the most efficient way to do this, but at least (I hope so!) it does
what you wanted it to do:
### Beginning of Example Code
income - runif(100)
education - as.factor(sample(c(high, middle, low),
size=length(income), replace=TRUE))
mydata - data.frame(inc=income, edu=education)
mymedians - tapply(X=mydata$inc, INDEX=mydata$edu, FUN=median)
mydata$medians - ifelse(mydata$edu==high, mymedians[high], 0)
mydata$medians - ifelse(mydata$edu==middle, mymedians[middle],
mydata$medians)
mydata$medians - ifelse(mydata$edu==low, mymedians[low],
mydata$medians)
head(mydata)
mymedians
### End of Example Code
Maybe one can increase the speed, but I think it is sufficient for your
case of 30,000 cases as you can see from the timing on my desktop
computer here (WinXP Pro SP2, P4, 3GHz, 512MB RAM):
time.check - function(){
+ income - runif(3)
+ education - as.factor(sample(c(high, middle, low),
size=length(income), replace=TRUE))
+ mydata - data.frame(inc=income, edu=education)
+
+ mymedians - tapply(X=mydata$inc, INDEX=mydata$edu, FUN=median)
+
+ mydata$medians - ifelse(mydata$edu==high, mymedians[high], 0)
+ mydata$medians - ifelse(mydata$edu==middle, mymedians[middle],
mydata$medians)
+ mydata$medians - ifelse(mydata$edu==low, mymedians[low],
mydata$medians)
+ return(NULL)
+ }
system.time(time.check())
[1] 0.36 0.02 0.38 NA NA
version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status beta
major2
minor1.0
year 2005
month04
day 04
language R
Best,
Roland
+
This mail has been sent through the MPI for Demographic Rese...{{dropped}}
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Re: [R] trim a string
How to trim the leading and trailing white space off of a string?
If the variable is E I need to convert it to E.
gsub('^[[:space:]]+', '',E )
gsub('[[:space:]]+$', '',E )
as in R-2.1.0/library/base/html/grep.html
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Re: [R] Data comparison
On 6/17/2005 12:33 PM, [EMAIL PROTECTED] wrote: Question : Is it possible to create a function, using a for ifelse function, inside sapply, to compare the values in one data frame to a set of upper and lower limits in another data frame, same number of columns., Take the values which meet the requirements TRUE and create a new data frame or table containing the filtered data? Can you give me a shove in the correct direction? Thanks I think you don't need a function within sapply. You just want something like this: df1 - data.frame(x = runif(20)) y - runif(20) df2 - data.frame(lower = y, upper = y + runif(20)) df3 - cbind(df1, df2) df3[(df2$lower df1$x) (df1$x df2$upper), ] x lower upper 5 0.6688050 0.4357477 0.6786472 6 0.5608836 0.4649370 0.8596602 8 0.5109508 0.2654933 1.0573998 9 0.5966776 0.3035084 0.9834681 19 0.3787230 0.1894318 0.6783048 20 0.2826356 0.2321261 1.0913582 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] another aov results interpretation question
I commend you to (a) the recent article by Doug Bates on Fitting nonlinear mixed models in R pp. 27-30 in the latest issue of R News available from www.r-project.org - Newsletter and (b) Doug's book with Pinheiro (2000) Mixed-Effects Models in S and S-PLUS (Springer). I suggest you try the same analysis using in lmer, library(lme4), and lme, library(nlme), with method = ML, as explained in Pinheiro and Bates. If you have trouble with this, please post another question on this, preferably using either a standard data set distributed with R or one of the standard packages or a very simple made-up data set with very few observations that you can distribute with your question in a short sequence of R commands illustrating something you tried that either didn't work or that gave results you don't understand. I can't do much more with the example you've provided below, because I don't know how to access the your data. (And PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html if you haven't already.) hope this helps. spencer graves RenE J.V. Bertin wrote: Hello, I'm trying to understand how to interpret the differences in results between two versions of a 2-factor ANOVA with (slightly?) different models, of an observable y, a within-subject factor 'indep' and a grouping factor 'cond' (and a subject 'factor' Snr): summary( aov( y~cond + indep + Error(Snr/indep) ) ) # example results: Error: Snr Df Sum Sq Mean Sq F value Pr(F) cond 1 103.1 103.1 1.425 0.248 indep 5 159.832.0 0.442 0.813 Residuals 18 1301.672.3 Error: Snr:indep Df Sum Sq Mean Sq F value Pr(F) indep 5 20.814.16 3.167 0.0104 * Residuals 111 145.891.31 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Error: Within Df Sum Sq Mean Sq F value Pr(F) Residuals 137 22.178 0.162 summary( aov( y~cond * indep + Error(Snr/indep) ) ) # example results: Error: Snr Df Sum Sq Mean Sq F value Pr(F) cond1 174.6 174.6 1.689 0.213 indep 5 201.940.4 0.391 0.848 cond:indep 5 124.024.8 0.240 0.939 Residuals 15 1550.8 103.4 Error: Snr:indep Df Sum Sq Mean Sq F value Pr(F) indep5 73.16 14.63 8.601 5e-07 *** cond:indep 5 21.324.26 2.507 0.0336 * Residuals 125 212.641.70 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Error: Within Df Sum Sq Mean Sq F value Pr(F) Residuals 464 507.5 1.1 I would like to understand a bit better what the cond:indep line under the second Error:Snr:indep can mean. If I understood correctly, this represents some higher-order interaction, but not a real indep/cond interaction. What I also do not grasp is why the indep effect's F and significance is so different between the two models. Finally, what does it mean when significant effects are listed under the Error:Within line? Is there a good resource available (web, or if not printed) which discusses this kind of question in a way accessible to non statisticians? The last time I checked, manuals like R for Psychologists do not really enter into this level of detail... Thanks very much in advance, R. Bertin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] an operator for contains
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not contained) }
is therre a simple way to test if a vector/list contains a particular value?
for example an operator, along the lines of: ==
more generally, is the a documentaion page that lists/describes all
such operators?
lastly, if you didn't know the answer to my question, how would you have
gone about searching for an answer? I tried RSiteSearch() using various
terms, and I opened R-2.1.0/library/base/html/00Index.html and searched
for various terms.
TIA
==
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major2
minor1.0
year 2005
month04
day 18
language R
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Re: [R] an operator for contains
Yes, %in% or is.element().
Reid Huntsinger
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Mike R
Sent: Friday, June 17, 2005 4:32 PM
To: r-help@stat.math.ethz.ch
Subject: [R] an operator for contains
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not
contained) }
is therre a simple way to test if a vector/list contains a particular value?
for example an operator, along the lines of: ==
more generally, is the a documentaion page that lists/describes all
such operators?
lastly, if you didn't know the answer to my question, how would you have
gone about searching for an answer? I tried RSiteSearch() using various
terms, and I opened R-2.1.0/library/base/html/00Index.html and searched
for various terms.
TIA
==
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major2
minor1.0
year 2005
month04
day 18
language R
__
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Re: [R] an operator for contains
Does '?%in%' or '?match' meet your needs?
spencer graves
Mike R wrote:
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not contained)
}
is therre a simple way to test if a vector/list contains a particular value?
for example an operator, along the lines of: ==
more generally, is the a documentaion page that lists/describes all
such operators?
lastly, if you didn't know the answer to my question, how would you have
gone about searching for an answer? I tried RSiteSearch() using various
terms, and I opened R-2.1.0/library/base/html/00Index.html and searched
for various terms.
TIA
==
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major2
minor1.0
year 2005
month04
day 18
language R
__
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Re: [R] an operator for contains
See match(). Also intersect().
The documentation for operators is, I think mostly at the top of
the index page for the base package, the one you searched. The
relevant one is %in%. I guess Value matching didn't ring the
right bell.
On 06/17/05 13:31, Mike R wrote:
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not contained) }
is therre a simple way to test if a vector/list contains a particular value?
for example an operator, along the lines of: ==
more generally, is the a documentaion page that lists/describes all
such operators?
lastly, if you didn't know the answer to my question, how would you have
gone about searching for an answer? I tried RSiteSearch() using various
terms, and I opened R-2.1.0/library/base/html/00Index.html and searched
for various terms.
TIA
==
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major2
minor1.0
year 2005
month04
day 18
language R
__
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--
Jonathan Baron, Professor of Psychology, University of Pennsylvania
Home page: http://www.sas.upenn.edu/~baron
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Re: [R] an operator for contains
%in%
R k - 1:9
R 3 %in% k
[1] TRUE
R 33 %in% k
[1] FALSE
Arne
On Friday 17 June 2005 22:31, Mike R wrote:
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not
contained) }
is therre a simple way to test if a vector/list contains a particular
value?
for example an operator, along the lines of: ==
more generally, is the a documentaion page that lists/describes all
such operators?
lastly, if you didn't know the answer to my question, how would you have
gone about searching for an answer? I tried RSiteSearch() using various
terms, and I opened R-2.1.0/library/base/html/00Index.html and searched
for various terms.
TIA
==
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
major2
minor1.0
year 2005
month04
day 18
language R
__
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--
Arne Henningsen
Department of Agricultural Economics
University of Kiel
Olshausenstr. 40
D-24098 Kiel (Germany)
Tel: +49-431-880 4445
Fax: +49-431-880 1397
[EMAIL PROTECTED]
http://www.uni-kiel.de/agrarpol/ahenningsen/
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Re: [R] an operator for contains
On 6/17/05, Mike R [EMAIL PROTECTED] wrote:
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not contained)
}
is therre a simple way to test if a vector/list contains a particular value?
Yes, several. Here's one:
k - 1:9
if(any(k == 3)) {
cat(is an element\n) } else {
cat(not an element\n) }
I don't recommend that for floating point numbers, though.
more generally, is the a documentaion page that lists/describes all
such operators?
help(==) would get you the help page for the actual binary
operators. Since any() isn't an operator, that won't help.
lastly, if you didn't know the answer to my question, how would you have
gone about searching for an answer?
There are a variety of ways to find answers in R, although if you
aren't certain what your keyword should be, it may take a few tries.
help(), help.search(), and apropos() may all be useful.
Sarah
--
Sarah Goslee
http://www.stringpage.com
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Re: [R] an operator for contains
On Fri, 17 Jun 2005, Mike R wrote:
k = c(1:9)
if( length( which(k==3) ) ){ print(contained) }else{ print(not contained)
}
is therre a simple way to test if a vector/list contains a particular value?
value %in% vector
more generally, is the a documentaion page that lists/describes all
such operators?
No. You can find binary operators fairly effectively by looking at the
html help, because they have to have non-alphabetic names (either single
characters or beginning and ending with %).
However, there are functions such as setdiff() that you might think of as
binary operators that you wouldn't find this way.
-thomas
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[R] adjusted R^2 vs. ordinary R^2
I thought the point of adjusting the R^2 for degrees of freedom is to allow comparisons about goodness of fit between similar models with different numbers of data points. Someone has suggested to me off-list that this might not be the case. Is an ADJUSTED R^2 for a four-parameter, five-point model reliably comparable to the adjusted R^2 of a four-parameter, 100-point model? If such values can't be reliably compared with one another, then what is the reasoning behind adjusting R^2 for degrees of freedom? What are the good published authorities on this topic? Sincerely, James Salsman __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] 3D Scatter Plot
Hello: I would like to be able to do a 3D scatter plot from 3 variables, 2 independent and 1 dependent. The closest R function I could find for this is cloud. However cloud uses, as input, a matrix where the value of each matrix element is the dependent variable value at that matrix coordinate. My problem is that the independent variable values are floating point and can be of any value. Consequently some of the matrix bins may not have a value assigned. An example of the sort of data I may have is as follows Independent 1 Independent 2Dependent 0.145674 0.526482534 1.676986 0.325634 0.326385237 2.384384 0.235267 0.352653288 0.356483 Is there any way to do a 3D scatter plot with this sort of data? Thanks very much in advance, Peter Lauren. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] an operator for contains
wow. thanks everyone for the multitude of suggestions ! __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] adjusted R^2 vs. ordinary R^2
James Salsman [EMAIL PROTECTED] writes: I thought the point of adjusting the R^2 for degrees of freedom is to allow comparisons about goodness of fit between similar models with different numbers of data points. Someone has suggested to me off-list that this might not be the case. Is an ADJUSTED R^2 for a four-parameter, five-point model reliably comparable to the adjusted R^2 of a four-parameter, 100-point model? If such values can't be reliably compared with one another, then what is the reasoning behind adjusting R^2 for degrees of freedom? Well, the adjusted R^2 is the percent variance explained by covariates. So it compares the conditional variance (given covariates) to the marginal variance. This is less sensitive to DF issues than the usual R^2, but it does still require that both quantities make sense. This is not a given, and in particular the R^2 (either one) is quite dubious when the covariates are chosen by design. What are the good published authorities on this topic? Dunno. Common sense should really suffice in this matter. -- O__ Peter Dalgaard ster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] axis labels vertically
2. I wasn't aware that there is an online-help. (was using the tutorial on the web, which is nat at all detailed. help(), help.search(), and apropos() may all be useful.(thanks Sarah !!) and RSiteSearch() You might find this handy too example(axis)(thanks Brian !!) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] axis labels vertically
2. I wasn't aware that there is an online-help. (was using the tutorial on the web, which is nat at all detailed. help(), help.search(), and apropos() may all be useful.(thanks Sarah !!) and RSiteSearch() You might find this handy too example(axis)(thanks Brian !!) forgot to mention . there are a number of excellent contributed tutorials here: http://cran.r-project.org/other-docs.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] one factor multiple level anova
I am having trouble figuring out this one. I want to do a one way anvoa with 13 levels. mydata is in a vector with length 65. each level has 5 repeats. but it contains NA. I made mygroup-gl(13, 5, 65, labels=(...)) anova(lm(mydata ~ mygroup)) it gives following error: Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid variable type Can you please help?? Millions of thanks. Lei Jiang Department of Chemsitry University of Washington Box 351700 Seattle, WA 98195 Phone: 206-616-6882 Fax: 206-685-8665 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] aggregate?
Dear all: Here is my problem: Example data: dat-data.frame(x=rep(c(a,b,c,d),2),y=c(10:17)) If I wanted to aggregate each level of column dat$x I could use: aggregate(dat$y,list(x=dat$x),sum) But I just want to aggregate two levels (?c? and ?d?) to obtain a new level ?e? I am expecting something like: x y 1 a 10 2 b 11 3 e 25 4 a 14 5 b 15 6 e 33 How can I make it? Thanks in advance and best for all A. Diaz - Email Enviado utilizando o servio MegaMail __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] hist single block plot issue
# CASE 1 # The following plots a single cell or block for all three location 0,1,2. y - rep(2,8) hist(y)# why is this a single block? hist(y,xlim=c(0,2))# same thing hist(y,breaks=2) # same # CASE 2 # adding a different value, plots as expected y - append(y,0) hist(y)# plots as expected In most cases of the data I have variances in the data so this is not an issue. In some situations case #1 appears and I would like to differentiate it among the other values. Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] one factor multiple level anova
Test that: mydata - as.numeric(mydata) anova(lm(mydata ~ mygroup)) 2005/6/18, Lei Jiang [EMAIL PROTECTED]: I am having trouble figuring out this one. I want to do a one way anvoa with 13 levels. mydata is in a vector with length 65. each level has 5 repeats. but it contains NA. I made mygroup-gl(13, 5, 65, labels=(...)) anova(lm(mydata ~ mygroup)) it gives following error: Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid variable type Can you please help?? Millions of thanks. Lei Jiang Department of Chemsitry University of Washington Box 351700 Seattle, WA 98195 Phone: 206-616-6882 Fax: 206-685-8665 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- ADDRESS: Bioinformatics Center, Shanghai Institutes for Biological Sciences, Chinese Academy of Sciences 320 Yueyang Road, Shanghai 200031, P.R.China TELEPHONE: 86-21-54920086 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] 3D Scatter Plot
On 6/17/05, Lauren, Peter [EMAIL PROTECTED] wrote: Hello: I would like to be able to do a 3D scatter plot from 3 variables, 2 independent and 1 dependent. The closest R function I could find for this is cloud. However cloud uses, as input, a matrix where the value of each matrix element is the dependent variable value at that matrix coordinate. What makes you think that? My problem is that the independent variable values are floating point and can be of any value. Consequently some of the matrix bins may not have a value assigned. An example of the sort of data I may have is as follows Independent 1 Independent 2Dependent 0.145674 0.526482534 1.676986 0.325634 0.326385237 2.384384 0.235267 0.352653288 0.356483 Is there any way to do a 3D scatter plot with this sort of data? The following should work fine: df - data.frame(x = c(0.145674, 0.325634, 0.235267), y = c(0.526482534, 0.326385237, 0.352653288), z = c(1.676986, 2.384384, 0.356483)) cloud(z ~ x * y, df) Deepayan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] hist single block plot issue
In case #1, the argument 'breaks' can break the histogram cells: hist(y, breaks=c(0, 0.5, 1.0, 1.5, 2.0)) # not the same ^_^ 2005/6/18, ap [EMAIL PROTECTED]: # CASE 1 # The following plots a single cell or block for all three location 0,1,2. y - rep(2,8) hist(y) # why is this a single block? hist(y,xlim=c(0,2)) # same thing hist(y,breaks=2) # same # CASE 2 # adding a different value, plots as expected y - append(y,0) hist(y) # plots as expected In most cases of the data I have variances in the data so this is not an issue. In some situations case #1 appears and I would like to differentiate it among the other values. Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- ADDRESS: Bioinformatics Center, Shanghai Institutes for Biological Sciences, Chinese Academy of Sciences 320 Yueyang Road, Shanghai 200031, P.R.China TELEPHONE: 86-21-54920086 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] aggregate?
On 6/17/05, alex diaz [EMAIL PROTECTED] wrote: Dear all: Here is my problem: Example data: dat-data.frame(x=rep(c(a,b,c,d),2),y=c(10:17)) If I wanted to aggregate each level of column dat$x I could use: aggregate(dat$y,list(x=dat$x),sum) But I just want to aggregate two levels (c and d) to obtain a new level e I am expecting something like: x y 1 a 10 2 b 11 3 e 25 4 a 14 5 b 15 6 e 33 In the example - dat$y[3:4] are summed and - dat$y[7:8] are summed so we assume that what is being requested is that d is to be replaced by c and runs of any level are to be summed. To do that: - create xx such that a, b, c and d in dat$x are replaced with with 1, 2, 3 and 3 in xx. - in the second statement calculate a running sum except if the last observation was the same as the current observation then the Last Observation is Carried Forward (locf) so that all entries in a run have the same number. e.g. in this case locf is c(1, 2, 3, 3, 4, 5, 6, 6) - Finally the 'by' collapses dat using locf rbinds the resulting rows together to create a data frame. xx - ifelse(dat$x == d, 3, dat$x) locf - cumsum(c(TRUE, xx[-1] != xx[-length(xx)])) f - function(x) data.frame(x=x[1,1], y=sum(x[,2])) dat2 - do.call(rbind, by(dat, locf, f)) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
