Re: [R] plot legend: combining filled boxes and lines
Check out: http://tolstoy.newcastle.edu.au/R/e2/help/07/05/16777.html On 9/10/07, Lauri Nikkinen [EMAIL PROTECTED] wrote: Hello, I have difficulties combining boxes and lines in plot legend. I searched previous R-posts and found this (with no solution): http://tolstoy.newcastle.edu.au/R/help/06/07/30248.html. Is there a way to avoid boxes behind the line legends? x1 - rnorm(100) x2 - rnorm(100, 2) hist(x1, main = , col = orange,ylab = density, xlab = x, freq = F, density = 55, xlim = c(-2, 5), ylim = c(0, 0.5)) par(new = T) hist(x2, main = , col = green, ylab = , xlab = ,axes = F, xlim = c(-2, 5), ylim = c(0, 0.5), density = 45, freq = F) abline(v = mean(x1), col = orange, lty = 2, lwd = 2.5) abline(v = mean(x2), col = green, lty = 2, lwd = 2.5) legend(3, 0.45, legend = c(x1, x2, mean(x1), mean(x2)), col = c(orange, green), fill=c(orange,green, 0, 0), lty = c(0, 0, 2, 2), merge = T) Thanks Lauri __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice panel.lmline problem
Why not use the more simple
xyplot(total.fat~x|variable,groups=Group,
data=tmp1,type=c(p,r))
???
See ?panel.xyplot and especially the type argument of that panel function.
Best regards
Frede Aakmann Tøgersen
Scientist
UNIVERSITY OF AARHUS
Faculty of Agricultural Sciences
Dept. of Genetics and Biotechnology
Blichers Allé 20, P.O. BOX 50
DK-8830 Tjele
Phone: +45 8999 1900
Direct: +45 8999 1878
E-mail: [EMAIL PROTECTED]
Web: http://www.agrsci.org
This email may contain information that is confidential.
Any use or publication of this email without written permission from Faculty of
Agricultural Sciences is not allowed.
If you are not the intended recipient, please notify Faculty of Agricultural
Sciences immediately and delete this email.
-Oprindelig meddelelse-
Fra: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] På vegne af Ross Darnell
Sendt: 10. september 2007 07:55
Til: r-help@stat.math.ethz.ch
Emne: [R] lattice panel.lmline problem
I am wanting to generate panels showing scatterplots with the
linear fitted line for two groups within each panel superimposed.
I have two conditioning factors, variable and Group and I
want separate panels for each level of variable
with different symbols and lmlines for each level of
Group. However all observations for a group are missing for
some variables so I would still like the points and lmline
for the observed group plotted for that variable(panel).
My attempt is
print(xyplot(total.fat~x|variable,data=tmp1,subscripts=TRUE,
scales=list(x=list(relation=free)),xlab=,groups=Group,
panel=function(x,y,subscripts,...){
panel.superpose(x,y,subscripts,...)
if(length(x[subscripts])!=0)
{panel.superpose(x,y,panel.groups=panel.lmline,subscripts,...)}}))
Which gives an error
Error in lm.fit(x, y, offset = offset, singular.ok =
singular.ok, ...) :
0 (non-NA) cases
Which occurs when the first panel with all values for one
group are missing.
The same error is returned if I replace the last line without the if
statement which obviously means it's ignored.
Of course I may be taking the wrong tack completely to get
the result I need. Any advice would be appreciated
Ross Darnell
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] plot legend: combining filled boxes and lines
Thanks Gabor, I got it! For example: x1 - rnorm(100) x2 - rnorm(100, 2) hist(x1, main = , col = orange,ylab = density, xlab = x, freq = F, density = 55, xlim = c(-2, 5), ylim = c(0, 0.5)) par(new = T) hist(x2, main = , col = green, ylab = , xlab = ,axes = F, xlim = c(-2, 5), ylim = c(0, 0.5), density = 45, freq = F) abline(v = mean(x1), col = orange, lty = 2, lwd = 2.5) abline(v = mean(x2), col = green, lty = 2, lwd = 2.5) legend(3, 0.45, legend = c(x1, x2, mean(x1), mean(x2)), col = rep(c(orange, green), 2), pch=c(15,15, NA, NA), pt.cex=2, lty = c(0, 0, 2, 2)) -Lauri 2007/9/10, Gabor Grothendieck [EMAIL PROTECTED]: Check out: http://tolstoy.newcastle.edu.au/R/e2/help/07/05/16777.html On 9/10/07, Lauri Nikkinen [EMAIL PROTECTED] wrote: Hello, I have difficulties combining boxes and lines in plot legend. I searched previous R-posts and found this (with no solution): http://tolstoy.newcastle.edu.au/R/help/06/07/30248.html. Is there a way to avoid boxes behind the line legends? x1 - rnorm(100) x2 - rnorm(100, 2) hist(x1, main = , col = orange,ylab = density, xlab = x, freq = F, density = 55, xlim = c(-2, 5), ylim = c(0, 0.5)) par(new = T) hist(x2, main = , col = green, ylab = , xlab = ,axes = F, xlim = c(-2, 5), ylim = c(0, 0.5), density = 45, freq = F) abline(v = mean(x1), col = orange, lty = 2, lwd = 2.5) abline(v = mean(x2), col = green, lty = 2, lwd = 2.5) legend(3, 0.45, legend = c(x1, x2, mean(x1), mean(x2)), col = c(orange, green), fill=c(orange,green, 0, 0), lty = c(0, 0, 2, 2), merge = T) Thanks Lauri __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problems with strsplit
unlist(strsplit(bA531F16-rep,\\-))[1] [1] bA531F16 --- Carlos Morales [EMAIL PROTECTED] wrote: Hello, I would like to know what can I do if I use strplit with a string and I want to use the middle left,I mean I have this: strsplit(bA531F16-rep,\\-) [[1]] [1] bA531F16 rep I would like to work just with bA531F16 in another variable, what could I do?, Thank you - Sé un Mejor Amante del Cine ¿Quieres saber cómo? ¡Deja que otras personas te ayuden!. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] command to plot variannce of data (like error bar plot)
Yogesh Tiwari wrote: Hi, How to plot a variance over a data point, something like error bar. Hi Yogesh, The generic method is to use the arrows function, and there are quite a few variations on this theme (in alpha order): brkdn.plot(plotrix) dispbars(plotrix) errbar(Hmisc and sfsmisc) plotCI(gplots and plotrix) plotMeans (Rcmdr) and there are surely more error bar functions of which I am unaware. Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice panel.lmline problem
Thanks Frede
I didn't know about the r type.
Ross Darnell
-Original Message-
From: Frede Aakmann Tøgersen [mailto:[EMAIL PROTECTED]
Sent: Mon 10-Sep-07 4:45 PM
To: Ross Darnell; r-help@stat.math.ethz.ch
Subject: SV: [R] lattice panel.lmline problem
Why not use the more simple
xyplot(total.fat~x|variable,groups=Group,
data=tmp1,type=c(p,r))
???
See ?panel.xyplot and especially the type argument of that panel function.
Best regards
Frede Aakmann Tøgersen
Scientist
UNIVERSITY OF AARHUS
Faculty of Agricultural Sciences
Dept. of Genetics and Biotechnology
Blichers Allé 20, P.O. BOX 50
DK-8830 Tjele
Phone: +45 8999 1900
Direct: +45 8999 1878
E-mail: [EMAIL PROTECTED]
Web: http://www.agrsci.org
This email may contain information that is confidential.
Any use or publication of this email without written permission from Faculty of
Agricultural Sciences is not allowed.
If you are not the intended recipient, please notify Faculty of Agricultural
Sciences immediately and delete this email.
-Oprindelig meddelelse-
Fra: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] På vegne af Ross Darnell
Sendt: 10. september 2007 07:55
Til: r-help@stat.math.ethz.ch
Emne: [R] lattice panel.lmline problem
I am wanting to generate panels showing scatterplots with the
linear fitted line for two groups within each panel superimposed.
I have two conditioning factors, variable and Group and I
want separate panels for each level of variable
with different symbols and lmlines for each level of
Group. However all observations for a group are missing for
some variables so I would still like the points and lmline
for the observed group plotted for that variable(panel).
My attempt is
print(xyplot(total.fat~x|variable,data=tmp1,subscripts=TRUE,
scales=list(x=list(relation=free)),xlab=,groups=Group,
panel=function(x,y,subscripts,...){
panel.superpose(x,y,subscripts,...)
if(length(x[subscripts])!=0)
{panel.superpose(x,y,panel.groups=panel.lmline,subscripts,...)}}))
Which gives an error
Error in lm.fit(x, y, offset = offset, singular.ok =
singular.ok, ...) :
0 (non-NA) cases
Which occurs when the first panel with all values for one
group are missing.
The same error is returned if I replace the last line without the if
statement which obviously means it's ignored.
Of course I may be taking the wrong tack completely to get
the result I need. Any advice would be appreciated
Ross Darnell
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] PCA IN R
prcomp() in stats handles matrices with n p well, IMO. -- Bjørn-Helge Mevik __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] artificial data matrix with 100000 rows
PS == Paul Smith [EMAIL PROTECTED] on Sun, 9 Sep 2007 12:17:32 +0100 writes: PS On 9/9/07, kevinchang [EMAIL PROTECTED] wrote: I tried to made the matrix with this size by either matrix() or array(). However, there seems to be default limit of number for rows made. I got sort of error message from R .To be specific, m--matrix(ncol=3,nrow=10) error message:[ reached getOption(max.print) -- omitted 7 rows ]] or a-array(dim=c(1,3,10)) error message:reached getOption(max.print) -- omitted 6667 row(s) and 6 matrix slice(s) ] PS That is not an error message, I guess. Definitely not, thank you, Paul! Also, they were not produced by what Kevin showed (namely assignments) but rather when he *prints* the contents of his huge matrix / array. PS When the matrices are huge, R is unable to print them PS totally on the screen, but all data are present. Not at all unable !! R protects you from accidentally overflowing your console with huge amount of non-sensical output. As the warning above mentions, you should look at ? getOption ? options and particularly the 'max.print' option Is '' reached getOption(max.print) '' too difficult to read? You *can* increase the 'max.print' option as much as you like, and that's why I said 'not at all unable' above. Regards, Martin PS For instance, m[(nrow(m)-10):nrow(m),] PS [,1] [,2] [,3] PS [1,] NA NA NA PS [2,] NA NA NA PS [3,] NA NA NA PS [4,] NA NA NA PS [5,] NA NA NA PS [6,] NA NA NA PS [7,] NA NA NA PS [8,] NA NA NA PS [9,] NA NA NA PS [10,] NA NA NA PS [11,] NA NA NA or rather just tail(m) or tail(m, 11) or head(m) or str(m) etc etc PS See PS ?getOption yes indeed. Martin PS Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in installing and loading the BradleyTerry add on package in R
However you may also need to install the package brlr, since the BradleyTerry package depends on this. For Windows users, it's usually easiest to install packages using the Packages menu in the RGui - any dependencies are then automatically installed. Heather -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jim Lemon Sent: 10 September 2007 12:07 To: [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Help in installing and loading the BradleyTerry add on package in R Kalyan Roy (DEL/MSG) wrote: How do I install and load the BradleyTerry add on package in R 2.5.1 in MSWindowsXP environment? Hi Kalyan, If R CMD INSTALL doesn't work, you can use WinZip or Zip Reader to unzip the package to: C:\Program Files\R-2.5.1\library or whatever your path to the library directory is, and then hand edit the packages.html file in: C:\Program Files\R-2.5.1\doc\html to include the new package in your HTML listing. This will allow you to access the help files and use the package. Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] artificial data matrix with 100000 rows
On 9/10/07, Martin Maechler [EMAIL PROTECTED] wrote: PS On 9/9/07, kevinchang [EMAIL PROTECTED] wrote: I tried to made the matrix with this size by either matrix() or array(). However, there seems to be default limit of number for rows made. I got sort of error message from R .To be specific, m--matrix(ncol=3,nrow=10) error message:[ reached getOption(max.print) -- omitted 7 rows ]] or a-array(dim=c(1,3,10)) error message:reached getOption(max.print) -- omitted 6667 row(s) and 6 matrix slice(s) ] PS That is not an error message, I guess. Definitely not, thank you, Paul! Also, they were not produced by what Kevin showed (namely assignments) but rather when he *prints* the contents of his huge matrix / array. PS When the matrices are huge, R is unable to print them PS totally on the screen, but all data are present. Not at all unable !! R protects you from accidentally overflowing your console with huge amount of non-sensical output. As the warning above mentions, you should look at ? getOption ? options and particularly the 'max.print' option Is '' reached getOption(max.print) '' too difficult to read? You *can* increase the 'max.print' option as much as you like, and that's why I said 'not at all unable' above. Regards, Martin PS For instance, m[(nrow(m)-10):nrow(m),] PS [,1] [,2] [,3] PS [1,] NA NA NA PS [2,] NA NA NA PS [3,] NA NA NA PS [4,] NA NA NA PS [5,] NA NA NA PS [6,] NA NA NA PS [7,] NA NA NA PS [8,] NA NA NA PS [9,] NA NA NA PS [10,] NA NA NA PS [11,] NA NA NA or rather just tail(m) or tail(m, 11) or head(m) or str(m) etc etc PS See PS ?getOption yes indeed. Thanks, Martin, for your detailed comments. I have learned something from them. Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in installing and loading the BradleyTerry add on package in R
Kalyan Roy (DEL/MSG) wrote: How do I install and load the BradleyTerry add on package in R 2.5.1 in MSWindowsXP environment? Hi Kalyan, If R CMD INSTALL doesn't work, you can use WinZip or Zip Reader to unzip the package to: C:\Program Files\R-2.5.1\library or whatever your path to the library directory is, and then hand edit the packages.html file in: C:\Program Files\R-2.5.1\doc\html to include the new package in your HTML listing. This will allow you to access the help files and use the package. Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot legend: combining filled boxes and lines
This is not quite what you want but you can try this: legend(3, 0.45, legend = c(x1, x2, mean(x1), mean(x2)), col = c(orange, green),pch = c(15,15,-1,-1), lty=c(-1,-1,2,2)) Although pch=22 should draw a filled square with a border - but it draws only the border instead Monica __ Message: 24Date: Mon, 10 Sep 2007 09:05:27 +0300From: Lauri Nikkinen [EMAIL PROTECTED]Subject: [R] plot legend: combining filled boxes and linesTo: [EMAIL PROTECTED]:[EMAIL PROTECTED]Content-Type: text/plain; charset=ISO-8859-1 Hello, I have difficulties combining boxes and lines in plot legend. Isearched previous R-posts and found this (with no solution):http://tolstoy.newcastle.edu.au/R/help/06/07/30248.html. Is there away to avoid boxes behind the line legends? x1 - rnorm(100)x2 - rnorm(100, 2)hist(x1, main = , col = orange,ylab = density, xlab = x, freq= F, density = 55, xlim = c(-2, 5), ylim = c(0, 0.5))par(new = T)hist(x2, main = , col = green, ylab = , xlab = ,axes = F, xlim= c(-2, 5), ylim = c(0, 0.5), density = 45, freq = F) abline(v = mean(x1), col = orange, lty = 2, lwd = 2.5)abline(v = mean(x2), col = green, lty = 2, lwd = 2.5)legend(3, 0.45, legend = c(x1, x2, mean(x1), mean(x2)), col =c(orange, green), fill=c(orange,green, 0, 0), lty = c(0, 0,2, 2), merge = T) ThanksLauri _ Gear up for Halo® 3 with free downloads and an exclusive offer. Its our way of saying thanks for using Windows Live. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] overlay lattice histograms with goodness-of-fit pdfs
The following is one of the examples in the help page for histogram:
histogram( ~ height | voice.part, data = singer,
xlab = Height (inches), type = density,
panel = function(x, ...) {
panel.histogram(x, ...)
panel.mathdensity(dmath = dnorm, col = black,
args = list(mean=mean(x),sd=sd(x)))
} )
This should give you some thing to start from.
Also using the subset argument of the lattice functions will make make your
code more readable. Instead of your code
histogram(~ data | month * station,
data = sta.stack[sta.stack[,type]==precip
(sta.stack[,month]==Dec | sta.stack[,month]==Jan |
sta.stack[,month]==Feb),],
xlab = Precipitation (mm))
you can use (not tested because you didn't supply a reproducable example)
histogram(~ data | month * station, data = sta.stack
subset = type==precip month %in% c(Dec, Jan, Feb),
xlab = Precipitation (mm))
Med venlig hilsen
Frede Aakmann Tøgersen
-Oprindelig meddelelse-
Fra: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] På vegne af Brad
Christoffersen
Sendt: 10. september 2007 12:08
Til: R-help@stat.math.ethz.ch
Emne: [R] overlay lattice histograms with goodness-of-fit pdfs
Hello,
I am new to R exploratory data analysis and plotting. Is
anyone aware of a way to overlay a set of conditional
histograms with conditional PDFs? Below, I generate a
lattice plot of precipitation histograms based on different
months and stations, given a subset of the dataset:
histogram(~ data | month * station,
data = sta.stack[sta.stack[,type]==precip
(sta.stack[,month]==Dec | sta.stack[,month]==Jan |
sta.stack[,month]==Feb),],
xlab = Precipitation (mm))
I previously used a combination of the low-level 'lines()'
and 'dgamma()'
functions to overlay a gamma PDF onto a single histogram.
Now what I would like to do is to do the same thing, but with
a function that allows me to specify a formula similar to
that in the histogram function above
[SomeKindOfPDF] ~ [x-range] | month * station
which will plot the PDF with the appropriate factors (month
and station).
All I'm looking for is for someone to get me going in the
right direction with a useful package or function to use.
Any help is much appreciated!
Brad Christoffersen
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Re: [R] mode or parameters of readBin
On 9/10/2007 10:26 AM, Sigbert Klinke wrote: Hi, sapply(formals(readBin), mode) con what n sizesignedendian namename numeric logical logicalcall returns for the mode of size logical. But in the documentation is said that size should be integer. Does anyone know why the mode is logical? The default value is NA, and the mode of NA is logical. (It will automatically be promoted into another type if included in a vector, e.g. c(1:2, NA) promotes it to integer, c(3.14, NA) promotes it to numeric.) More generally, I'd say reading the man page or the source is a better way to find out about the requirements for the arguments of a function. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problems with strsplit
You could directly use strsplit(bA531F16-rep,\\-)[[1]][1] and strsplit(bA531F16-rep,\\-)[[1]][2] Regards Carlos Morales-2 wrote: Hello, I would like to know what can I do if I use strplit with a string and I want to use the middle left,I mean I have this: strsplit(bA531F16-rep,\\-) [[1]] [1] bA531F16 rep I would like to work just with bA531F16 in another variable, what could I do?, Thank you - Sé un Mejor Amante del Cine ¿Quieres saber cómo? ¡Deja que otras personas te ayuden!. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/Problems-with-strsplit-tf4412525.html#a12594380 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot legend consolidation
I have recently been introduced to the ggplot package by Hadley Wickham and must say I am quite impressed so far at how easy it is to make attractive plots, but one thing I am struggling over is how to consolidate legends. It's not currently possible to consolidate them (although in the distant future that would be something nice to have), but you can turn them off: hide_colour - scale_colour_continuous() hide_colour$legend - FALSE p - ggplot(data = mtcars) + geom_point(mapping = aes(x = hp, y = mpg, colour = cyl)) + hide_colour You'll also need to twiddle your viewports a little so that you still have space for the viewport, since space will not be allocated automatically anymore. The next thing is to extract the grob for the legend itself - this is a little tricker, because there's currently no way to get at the scales after they have been trained with the data. Load get_legends from http://pastie.textmate.org/95755, and then you can do: grid.newpage(); grid.draw(get_legends(p)) If you're not familiar enough with grid to stitch all of these pieces together, please let me know, but this should be enough to get you started. Hadley __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] RWinEdt installation problems with Vista
Hi, I was trying to install the packageRWinEdt in my computer with Vista OS. O Use Tinn-R it works with Vista although has some minor issues (which are probably system specific). https://sourceforge.net/projects/tinn-r Stefan -=-=- ... Time is an illusion, lunchtime doubly so. (Ford Prefect) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Are the error messages of ConstrOptim() consisten with each other?
Yuchen Luo wrote: Dear Friends. I found something very puzzling with constOptim(). When I change the parameters for ConstrOptim, the error messages do not seem to be consistent with each other: constrOptim(c(0.5,0.3,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in constrOptim(c(0.5, 0.3, 0.5), f = fit.error, gr = fit.error.grr, : initial value not feasible Not feasible means it doesn't satisfy the constraints. constrOptim(c(0.5,0.9,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in constrOptim(c(0.5, 0.9, 0.5), f = fit.error, gr = fit.error.grr, : initial value not feasible constrOptim(c(0.3,0.5,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in f(theta, ...) : argument lambda1 is missing, with no default This time your starting values satisfied the constraints, so your objective function was called, but you didn't pass it a value for lambda1. I only changed the parameters, how come the lambda1 that is not missing in the first 2 cases suddently become missing? For your convenience, I put the complete code below: Best Wishes Yuchen Luo rm(list = ls()) mat=5 rint=c(4.33,4.22,4.27,4.43,4.43,4.44,4.45,4.65,4.77,4.77) tot=rep(13319.17,10) sh=rep(1553656,10) sigmae=c(0.172239074,0.188209271,0.193703774,0.172659891,0.164427247,0.24602361,0.173555309,0.186701165,0.193150456, 0.1857315601) ss=c(56.49,56.39,56.55,57.49,57.37,55.02,56.02,54.35,54.09, 54.67) orange=rep(21.25,10) apple2=expression(rint*(1.0-rec)*(1.0-(pnorm(-lambda/2.0+log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/lambda)-((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))*pnorm(-lambda/2.0-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/lambda))+(exp(rint*(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)*ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0+0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))-sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0*(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/s! h*! 1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(-sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0+0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0*(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))-(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0+0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt((lambda*l! am! bda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh* 1000.0))-sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0*(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt((lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(-sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0+0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt((lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0*(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt((lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))/((pnorm(-lambda/2.0+log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/lambda)-((ss+(tot/sh*1! 00!
Re: [R] Survey package
On Thu, 6 Sep 2007, eugen pircalabelu wrote: Good afternoon! I'm trying to use the Survey package for a stratified sample which has 4 criteria on which the stratification is based. I would like to get the corrected weights and for every element i get a weight of 1 E.g: tipping design - svydesign (id=~1, strata= ~regiune + size_loc + age_rec_hhh + size_hh, data= tabel) and then weights(design) gives me: 1,1,1,1,1,1,1,1,1,1,1,... for each element There are two problems. The first is that you have the wrong syntax for strata. If you have one stage of sampling with multiple stratifying factors you need to create a single factor representing the strata. One way is with interaction() design - svydesign (id=~1, strata= ~interaction(regiune, size_loc, age_rec_hhh, size_hh), data= tabel) Second, you have not specified either weights or population sizes, so R has no way to work out the sampling weights. That's why you get weights of 1. You should also get a warning. -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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] the survey package
On Thu, 6 Sep 2007, Tobias Verbeke wrote: eugen pircalabelu wrote: I'm trying to use the survey package to get a better point of view for my data, but i need some piece of advice: i have some data from a survey which has been stratified using 2 criteria: region(7 values), size of locality(5 values) Using the survey pakage how can i define in a correct way this design (taking into account all 4 strata not just one as in the Survey example) snip According to ?svydesign, strata is a formula. The following should work (untested): design - svydesign(ids=~0, strata=~regiune + size_loc, data=tabel) This would be a two-stage sample, you actually need ~interaction(regiune, size_loc). [this reply is just to make sure it ends up linked in the archives]. -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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] writing complex outputs to table
Steve, This example works: x-TukeyHSD(fm1, tension, ordered = TRUE) as.table(x$tension) Steve Powers wrote: So I've come across a few cases where complex outputs from functions will not write to tables. The most recent case involves the TukeyHSD function in the stats package. If I save the TukeyHSD call and print it, that obviously goes fine, but when I try writing to a table, I get an error message that says cannot coerce class \multicomp\ into dataframe. What does this mean, and how do I work around it? Is there a more fail-safe general method of writing complex outputs to a table? The story here is I want to be able to call certain values from the TukeyHSD results to produce custom summaries. Using R version 2.4 on Windows XP.---steve __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/writing-complex-outputs-to-table-tf4407732.html#a12596104 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Are the error messages of ConstrOptim() consisten with each other?
The error message about the feasible region comes from constrOptim(), before your function is called. The error message about missing lambda1 comes from calling your function. -thomas On Sun, 9 Sep 2007, Yuchen Luo wrote: Dear Friends. I found something very puzzling with constOptim(). When I change the parameters for ConstrOptim, the error messages do not seem to be consistent with each other: constrOptim(c(0.5,0.3,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in constrOptim(c(0.5, 0.3, 0.5), f = fit.error, gr = fit.error.grr, : initial value not feasible constrOptim(c(0.5,0.9,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in constrOptim(c(0.5, 0.9, 0.5), f = fit.error, gr = fit.error.grr, : initial value not feasible constrOptim(c(0.3,0.5,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in f(theta, ...) : argument lambda1 is missing, with no default I only changed the parameters, how come the lambda1 that is not missing in the first 2 cases suddently become missing? For your convenience, I put the complete code below: Best Wishes Yuchen Luo rm(list = ls()) mat=5 rint=c(4.33,4.22,4.27,4.43,4.43,4.44,4.45,4.65,4.77,4.77) tot=rep(13319.17,10) sh=rep(1553656,10) sigmae=c(0.172239074,0.188209271,0.193703774,0.172659891,0.164427247,0.24602361,0.173555309,0.186701165,0.193150456, 0.1857315601) ss=c(56.49,56.39,56.55,57.49,57.37,55.02,56.02,54.35,54.09, 54.67) orange=rep(21.25,10) apple2=expression(rint*(1.0-rec)*(1.0-(pnorm(-lambda/2.0+log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/lambda)-((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))*pnorm(-lambda/2.0-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/lambda))+(exp(rint*(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)*ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0+0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))-sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0*(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/s! h*! 1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(-sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0+0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0*(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))-(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0+0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt((lambda*la! m! bda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh* 1000.0))-sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0*(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt((lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(-sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0+0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt((lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0*(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))*sqrt((lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))/((pnorm(-lambda/2.0+log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda)))/lambda)-((ss+(tot/sh*1! 00!
Re: [R] Survey package
On Sun, 9 Sep 2007, eugen pircalabelu wrote: A short example: stratum id weight nh Nh y sex 1 1 3 5 15 23 1 1 2 3 5 15 25 1 1 3 3 5 15 27 2 1 4 3 5 15 21 2 1 5 3 5 15 22 1 2 6 4 3 12 33 1 2 7 4 3 12 27 1 2 8 4 3 12 29 2 where nh is size of sample stratum and Nh the corresponding population value, and y is metric variable. Now if i let design - svydesign( id=~1, data=age, strata=~stratum, fpc=~Nh) then weights(design) gives me 3,3,3,3,3,4,4,4. If i then let x- postStratify( design, strata=~sex, data.frame(sex=c(1,2), freq=c(10,15))) the weights become 123456 78 2.17 2.17 5.35 5.352.171.731.73 4.28 If i define design - svydesign( id=~1, data=age ) x- postStratify( design, strata=~sex, data.frame(sex=c(1,2), freq=c(10,15))) weights become 2 2 5 5 2 2 2 5 The question: does poststratify recognize that i have already stratified in the first design by stratum and then it post stratifies by sex? and why is that? (because i don't have the full joint distribution, the sex*stratum crossing, in order to apply correctly the post stratify function) I see that Mr Lumley uses the postStratify function when the design does not include strata (eg from ?poststratify: This gives you a design stratified by stratum and post-stratified by sex, which is not the same as stratifying by stratum*sex or post-stratifying by stratum*sex. In this case you should probably rake() on stratum and sex rather than just post-stratifying. Post-stratifying on sex is equivalent to one iteration of the iterative proportional fitting algorithm used in raking. -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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] write.csv / string extraction and field limits
This example works fine: test-matrix(c(1,2,'VOICIUNPETITTES',3),ncol=2,nrow=2) write.csv(test,file='C:/xavier/test.csv') Could you provide the same small example when it doesn't work? kwaj wrote: Hello, I have a peculiar problem which I am hoping I can get help on. I am using the write.csv command to write a matrix structure to a file, which I later read in excel. The command works quite well for most strings and numerical values in the matrix structure. However, I have found that when a field in the matrix contains a string of long length, when the matrix is finally written the file - the field shows up as NA. I am assuming write.csv has a limit on the field size? Maybe 16 characters? Assuming the above is correct - I tried to extract a portion of the string using the 'substring' command and enter the extracted portion into the field before using the write.csv command. However I find, that when a string is extracted, the output from write.csv generates a NA in the file output. My questions are: 1) Does write.csv have a limit on the size of strings in the matrix fields? Is there anyway to place large strings in the field? 2) Is there anyway to make the substring command or an alternative but similar command, compatible with write.csv? I have tried 'as.character(substring(phrase, min, max)' and that does not seem to work cheers -- View this message in context: http://www.nabble.com/write.csv---string-extraction-and-field-limits-tf4395535.html#a12596551 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mode or parameters of readBin
On Mon, 10 Sep 2007, Sigbert Klinke wrote: Hi, sapply(formals(readBin), mode) con what n sizesignedendian namename numeric logical logicalcall returns for the mode of size logical. But in the documentation is said that size should be integer. Does anyone know why the mode is logical? Because NA is a logical constant. -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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot legend consolidation
Hi Hadley,
I just tried out your suggestion, but it does not look like the
get_legends function is working correctly. Instead of returning a grob
back to me it returns NULL.
Here is my modified code and the results of running it.
Any help would be appreciated. I believe that once I can get the legend
in grob form then I can figure out how to deconstruct it myself.
Thanks,
Kaom
p.legend - get_legends(p)
grid.draw(p.legend)
Error in grid.draw(p.legend) : no applicable method for grid.draw
p.legend
NULL
BEGIN CODE
## Obtained from http://pastie.textmate.org/95755
get_legends - function(plot) {
if (length(plot$layers) == 0) stop(No layers to plot, call.=FALSE)
# Apply function to layer and matching data
dlapply - function(f) mapply(f, data, layers, SIMPLIFY=FALSE)
plot - plot_clone(plot)
layers - plot$layers
scales - plot$scales
facet - plot$facet
cs - plot$coordinates
# Evaluate aesthetics
data - lapply(layers, function(x) x$make_aesthetics(plot))
# Facet
data - mapply(function(d, p) facet$stamp_data(d), data, layers,
SIMPLIFY=FALSE)
# Transform scales where possible. Also need to train so statisics
# (e.g. stat_smooth) have access to info
data - dlapply(function(d, p) p$scales_transform(d, scales))
dlapply(function(d, p) p$scales_train(d, scales))
# Apply statistics
data - dlapply(function(d, p) p$calc_statistics(d, scales))
data - dlapply(function(d, p) p$map_statistics(d, plot))
# Adjust position before scaling
data - dlapply(function(d, p) p$adjust_position(d, scales, before))
# Transform, train and map scales
# data - dlapply(function(d, p) p$scales_transform(d, scales))
dlapply(function(d, p) p$scales_train(d, scales, adjust=TRUE))
data - dlapply(function(d, p) p$scales_map(d, scales))
# Adjust position after scaling
data - dlapply(function(d, p) p$adjust_position(d, scales, after))
scales - scales$minus(plot$scales$get_scales(c(x, y, z)))
legends(scales, FALSE)
}
library(ggplot2)
data(mtcars)
grid.newpage()
hide_colour - scale_colour_continuous()
hide_colour$legend - FALSE
pushViewport(viewport(layout = grid.layout(2, 2)))
p - ggplot(data = mtcars) +
geom_point(mapping = aes(x = hp, y = mpg, colour = cyl)) +
hide_colour
pushViewport(viewport(layout.pos.col = 1,
layout.pos.row = 1))
print(p, vp = current.viewport())
upViewport()
p - ggplot(data = mtcars) +
geom_point(mapping = aes(x = drat, y = disp, colour = cyl)) +
hide_colour
pushViewport(viewport(layout.pos.col = 2,
layout.pos.row = 1))
print(p, vp = current.viewport())
upViewport()
p - ggplot(data = mtcars) +
geom_point(mapping = aes(x = qsec, y = mpg, colour = cyl)) +
hide_colour
pushViewport(viewport(layout.pos.col = 1,
layout.pos.row = 2))
print(p, vp = current.viewport())
upViewport()
pushViewport(viewport(layout.pos.col = 2,
layout.pos.row = 2))
grid.rect()
p.legend - get_legends(p)
grid.draw(p.legend)
--END CODE
-Original Message-
From: hadley wickham [mailto:[EMAIL PROTECTED]
Sent: Monday, September 10, 2007 7:58 AM
To: Te, Kaom
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] ggplot legend consolidation
I have recently been introduced to the ggplot package by Hadley
Wickham and must say I am quite impressed so far at how easy it is to
make attractive plots, but one thing I am struggling over is how to
consolidate legends.
It's not currently possible to consolidate them (although in the distant
future that would be something nice to have), but you can turn them off:
hide_colour - scale_colour_continuous() hide_colour$legend - FALSE
p - ggplot(data = mtcars) +
geom_point(mapping = aes(x = hp, y = mpg, colour = cyl)) +
hide_colour
You'll also need to twiddle your viewports a little so that you still
have space for the viewport, since space will not be allocated
automatically anymore.
The next thing is to extract the grob for the legend itself - this is a
little tricker, because there's currently no way to get at the scales
after they have been trained with the data. Load get_legends from
http://pastie.textmate.org/95755, and then you can do:
grid.newpage(); grid.draw(get_legends(p))
If you're not familiar enough with grid to stitch all of these pieces
together, please let me know, but this should be enough to get you
started.
Hadley
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] off-topic: better OS for statistical computing
You want whatever all the people you are working with are using to make it as easy as possible to work together with them. On 9/10/07, Wensui Liu [EMAIL PROTECTED] wrote: Good morning, everyone, I am sorry for this off-topic post but think I can get great answer from this list. My question is what is the best OS on PC (laptop) for statistical computing and why. I really appreciate your insight. Have a nice day. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] finding the minimum positive value of some data
Try this: min(diff(sort(x))[diff(sort(x))0]) -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 10/09/2007, dxc13 [EMAIL PROTECTED] wrote: useRs, I am looking to find the minimum positive value of some data I have. Currently, I am able to find the minimum of data after I apply some other functions to it: x [1] 1 0 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10 sort(x) [1] 0 1 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10 diff(sort(x)) [1] 1 0 1 1 0 1 1 0 0 1 1 1 0 1 0 1 0 min(diff(sort(x))) [1] 0 The minimum is given as zero, which is clearly true, but I am interested in only the positive minimum, which is 1. Can I find this by using only 1 line of code, like I have above? Thanks! dxc13 -- View this message in context: http://www.nabble.com/finding-the-minimum-positive-value-of-some-data-tf4417250.html#a12599319 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] finding the minimum positive value of some data
On Mon, 2007-09-10 at 11:20 -0700, dxc13 wrote: useRs, I am looking to find the minimum positive value of some data I have. Currently, I am able to find the minimum of data after I apply some other functions to it: x [1] 1 0 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10 sort(x) [1] 0 1 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10 diff(sort(x)) [1] 1 0 1 1 0 1 1 0 0 1 1 1 0 1 0 1 0 min(diff(sort(x))) [1] 0 The minimum is given as zero, which is clearly true, but I am interested in only the positive minimum, which is 1. Can I find this by using only 1 line of code, like I have above? Thanks! dxc13 It's not clear to me which vector you wish to get the minimum for, but the basic premise would be along the lines of: x [1] 1 0 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10 min(x[which(x 0)]) [1] 1 or min(which(diff(sort(x)) 0)) [1] 1 See ?which HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot legend consolidation
Sorry, I should have mentioned that get_legend won't work on the plots
that you are actually plotting - you have turned their legends off!
You'll need a plot which isn't plotted, but is used to produce the
legends.
Hadley
On 9/10/07, Te, Kaom [EMAIL PROTECTED] wrote:
Hi Hadley,
I just tried out your suggestion, but it does not look like the
get_legends function is working correctly. Instead of returning a grob
back to me it returns NULL.
Here is my modified code and the results of running it.
Any help would be appreciated. I believe that once I can get the legend
in grob form then I can figure out how to deconstruct it myself.
Thanks,
Kaom
p.legend - get_legends(p)
grid.draw(p.legend)
Error in grid.draw(p.legend) : no applicable method for grid.draw
p.legend
NULL
BEGIN CODE
## Obtained from http://pastie.textmate.org/95755
get_legends - function(plot) {
if (length(plot$layers) == 0) stop(No layers to plot, call.=FALSE)
# Apply function to layer and matching data
dlapply - function(f) mapply(f, data, layers, SIMPLIFY=FALSE)
plot - plot_clone(plot)
layers - plot$layers
scales - plot$scales
facet - plot$facet
cs - plot$coordinates
# Evaluate aesthetics
data - lapply(layers, function(x) x$make_aesthetics(plot))
# Facet
data - mapply(function(d, p) facet$stamp_data(d), data, layers,
SIMPLIFY=FALSE)
# Transform scales where possible. Also need to train so statisics
# (e.g. stat_smooth) have access to info
data - dlapply(function(d, p) p$scales_transform(d, scales))
dlapply(function(d, p) p$scales_train(d, scales))
# Apply statistics
data - dlapply(function(d, p) p$calc_statistics(d, scales))
data - dlapply(function(d, p) p$map_statistics(d, plot))
# Adjust position before scaling
data - dlapply(function(d, p) p$adjust_position(d, scales, before))
# Transform, train and map scales
# data - dlapply(function(d, p) p$scales_transform(d, scales))
dlapply(function(d, p) p$scales_train(d, scales, adjust=TRUE))
data - dlapply(function(d, p) p$scales_map(d, scales))
# Adjust position after scaling
data - dlapply(function(d, p) p$adjust_position(d, scales, after))
scales - scales$minus(plot$scales$get_scales(c(x, y, z)))
legends(scales, FALSE)
}
library(ggplot2)
data(mtcars)
grid.newpage()
hide_colour - scale_colour_continuous()
hide_colour$legend - FALSE
pushViewport(viewport(layout = grid.layout(2, 2)))
p - ggplot(data = mtcars) +
geom_point(mapping = aes(x = hp, y = mpg, colour = cyl)) +
hide_colour
pushViewport(viewport(layout.pos.col = 1,
layout.pos.row = 1))
print(p, vp = current.viewport())
upViewport()
p - ggplot(data = mtcars) +
geom_point(mapping = aes(x = drat, y = disp, colour = cyl)) +
hide_colour
pushViewport(viewport(layout.pos.col = 2,
layout.pos.row = 1))
print(p, vp = current.viewport())
upViewport()
p - ggplot(data = mtcars) +
geom_point(mapping = aes(x = qsec, y = mpg, colour = cyl)) +
hide_colour
pushViewport(viewport(layout.pos.col = 1,
layout.pos.row = 2))
print(p, vp = current.viewport())
upViewport()
pushViewport(viewport(layout.pos.col = 2,
layout.pos.row = 2))
grid.rect()
p.legend - get_legends(p)
grid.draw(p.legend)
--END CODE
-Original Message-
From: hadley wickham [mailto:[EMAIL PROTECTED]
Sent: Monday, September 10, 2007 7:58 AM
To: Te, Kaom
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] ggplot legend consolidation
I have recently been introduced to the ggplot package by Hadley
Wickham and must say I am quite impressed so far at how easy it is to
make attractive plots, but one thing I am struggling over is how to
consolidate legends.
It's not currently possible to consolidate them (although in the distant
future that would be something nice to have), but you can turn them off:
hide_colour - scale_colour_continuous() hide_colour$legend - FALSE
p - ggplot(data = mtcars) +
geom_point(mapping = aes(x = hp, y = mpg, colour = cyl)) +
hide_colour
You'll also need to twiddle your viewports a little so that you still
have space for the viewport, since space will not be allocated
automatically anymore.
The next thing is to extract the grob for the legend itself - this is a
little tricker, because there's currently no way to get at the scales
after they have been trained with the data. Load get_legends from
http://pastie.textmate.org/95755, and then you can do:
grid.newpage(); grid.draw(get_legends(p))
If you're not familiar enough with grid to stitch all of these pieces
together, please let me know, but this should be enough to get you
started.
Hadley
--
http://had.co.nz/
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Re: [R] S-Plus resample package and associated functions
http://www.burns-stat.com/pages/Tutor/bootstrap_resampling.html includes a synopsis of R packages that do bootstrapping. It is brief and incomplete, but hopefully useful. Patrick Burns [EMAIL PROTECTED] +44 (0)20 8525 0696 http://www.burns-stat.com (home of S Poetry and A Guide for the Unwilling S User) Robert A. LaBudde wrote: Are there any packages in R that reproduce the package resample of S-Plus? The sample() function in R doesn't provide equivalent flexibility of bootstrap() and bootstrap2(). Robert A. LaBudde, PhD, PAS, Dpl. ACAFS e-mail: [EMAIL PROTECTED] Least Cost Formulations, Ltd.URL: http://lcfltd.com/ 824 Timberlake Drive Tel: 757-467-0954 Virginia Beach, VA 23464-3239Fax: 757-467-2947 Vere scire est per causas scire __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Too many warnings when updating R
A Lenzo wrote: Hello friends, I loaded R 2.4.1 onto a Fedora Core 6 Linux box (taking all defaults). Then I ran these commands from within R: options(CRAN=http://cran.stat.ucla.edu;) install.packages(CRAN.packages()[,1]) As a new user of R, I was shocked when I finished loading R and discovered the following message: There were 50 or more warnings (use warnings() to see the first 50) Let me get this straight: You install last year's R on last year's Fedora, then install over 1000 unspecified packages and you are shocked that you get warnings? In addition to this, I saw errors such as this one: ERROR: lazy loading failed for package 'PerformanceAnalytics' What is this lazy loading? More importantly, do I have to worry about all these warnings? I am intimidated by the idea that I have to go back and fix each and every one in order to have a clean R update. Shouldn't the update with CRAN just work? Or is there something really important that I am missing? Well, you need to know what you're doing. At the very least, notice what the warnings say and decide whether they point to real trouble or are just what they say they are: warnings. If you are worried about investigating all the packages, maybe install what you really need first. And no, you can't expect a repository like CRAN to keep track of all versions of R on all versions of all OS's. In each individual case, a human maintainer is responsible for fixing problems and he/she may or may not be around to fix issues. -- O__ Peter Dalgaard Øster 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice panel.lmline problem
On 9/10/07, Ross Darnell [EMAIL PROTECTED] wrote:
Thanks Frede
I didn't know about the r type.
For the record, this is probably what you wanted:
xyplot(...
panel = panel.superpose,
panel.groups = function(x, y, ...) {
panel.xyplot(x, y, ...)
if (length(x) 0) panel.lmline(x, y, ...)
})
x[subscripts] inside the panel function doesn't give anything
meaningful ('x' is already subscripted).
-Deepayan
Ross Darnell
-Original Message-
From: Frede Aakmann Tøgersen [mailto:[EMAIL PROTECTED]
Sent: Mon 10-Sep-07 4:45 PM
To: Ross Darnell; r-help@stat.math.ethz.ch
Subject: SV: [R] lattice panel.lmline problem
Why not use the more simple
xyplot(total.fat~x|variable,groups=Group,
data=tmp1,type=c(p,r))
???
See ?panel.xyplot and especially the type argument of that panel function.
[...]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] off-topic: better OS for statistical computing
On 11/09/2007, at 4:22 AM, Wensui Liu wrote:
Good morning, everyone,
I am sorry for this off-topic post but think I can get great answer
from this list.
My question is what is the best OS on PC (laptop) for statistical
computing and why.
I really appreciate your insight.
Have a nice day.
Linux. It's best for ***everything***.
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confidenti...{{dropped}}
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] off-topic: better OS for statistical computing
Linux! Mac OS is ok to me2. On 9/10/07, Wensui Liu [EMAIL PROTECTED] wrote: Good morning, everyone, I am sorry for this off-topic post but think I can get great answer from this list. My question is what is the best OS on PC (laptop) for statistical computing and why. I really appreciate your insight. Have a nice day. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. Did you always know? No, I did not. But I believed... ---Matrix III __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Are the error messages of ConstrOptim() consisten with each other?
Dear Professor Murdoch. Thank you for your help! 1. I believe c(0.5,0.3,0.5) satisfies the constrain because I did the following experiment ui=-1*ui ci=-1*ci constrOptim(c(0.5,0.3,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) The same error message pops up. Any theta ( in this case, c(0.5,0.3,0.5)) cannot violate both ui%*%theta=ci and -ui%*%theta=-ci. 2. There is lambda1 available. The 0.3 in c(0.5,0.3,0.5) is lambda1. If you plug c(0.5,0.3,0.5) into fit.error and fit.error.grr by fit.error(0.5,0.3,0.5) fit.error.grr(0.5,0.3,0.5) It works. Best Wishes Yuchen Luo On 9/10/07, Duncan Murdoch [EMAIL PROTECTED] wrote: Yuchen Luo wrote: Dear Friends. I found something very puzzling with constOptim(). When I change the parameters for ConstrOptim, the error messages do not seem to be consistent with each other: constrOptim(c(0.5,0.3,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in constrOptim(c(0.5, 0.3, 0.5), f = fit.error, gr = fit.error.grr , : initial value not feasible Not feasible means it doesn't satisfy the constraints. constrOptim(c(0.5,0.9,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in constrOptim(c(0.5, 0.9, 0.5), f = fit.error, gr = fit.error.grr , : initial value not feasible constrOptim(c(0.3,0.5,0.5), f=fit.error, gr=fit.error.grr, ui=ui,ci=ci) Error in f(theta, ...) : argument lambda1 is missing, with no default This time your starting values satisfied the constraints, so your objective function was called, but you didn't pass it a value for lambda1. I only changed the parameters, how come the lambda1 that is not missing in the first 2 cases suddently become missing? For your convenience, I put the complete code below: Best Wishes Yuchen Luo rm(list = ls()) mat=5 rint=c(4.33,4.22,4.27,4.43,4.43,4.44,4.45,4.65,4.77,4.77) tot=rep(13319.17,10) sh=rep(1553656,10) sigmae=c(0.172239074,0.188209271,0.193703774,0.172659891,0.164427247, 0.24602361,0.173555309,0.186701165,0.193150456, 0.1857315601) ss=c(56.49,56.39,56.55,57.49,57.37,55.02,56.02,54.35,54.09, 54.67) orange=rep(21.25,10) apple2=expression(rint*(1.0-rec)*(1.0- (pnorm(-lambda/2.0+log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0 )/lbar*exp(lambda*lambda)))/lambda)-((ss+(tot/sh*1000.0)*lbar)/(tot/sh* 1000.0)/lbar*exp(lambda*lambda))*pnorm(-lambda/2.0-log(((ss+(tot/sh*1000.0 )*lbar)/(tot/sh*1000.0 )/lbar*exp(lambda*lambda)))/lambda))+(exp(rint*(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh* 1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)*ss+(tot/sh*1000.0 )*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(sqrt(0.25+2.0*rint/ (sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 +0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0 )/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))-sqrt(0.25+2.0*rint/ (sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 *(sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*! 1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+((ss+(tot/sh*1000.0 )*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(-sqrt(0.25+2.0*rint/ (sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 +0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0 )/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+sqrt(0.25+2.0*rint/ (sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 *(sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))*sqrt(mat+(lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))-(((ss+(tot/sh*1000.0 )*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(sqrt(0.25+2.0*rint/ (sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 +0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0 )/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))*sqrt((lambda*lam! bda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh* 1000.0))-sqrt(0.25+2.0*rint/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0*(sigmae*ss/(ss+lbar*(tot/sh* 1000.0)))*sqrt((lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+((ss+(tot/sh*1000.0 )*lbar)/(tot/sh*1000.0)/lbar*exp(lambda*lambda))^(-sqrt(0.25+2.0*rint/ (sigmae*ss/(ss+lbar*(tot/sh*1000.0)))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 +0.5)*pnorm(-log(((ss+(tot/sh*1000.0)*lbar)/(tot/sh*1000.0 )/lbar*exp(lambda*lambda)))/((sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))*sqrt((lambda*lambda/(sigmae*ss/(ss+lbar*(tot/sh*1000.0 )))/(sigmae*ss/(ss+lbar*(tot/sh*1000.0))+sqrt(0.25+2.0*rint/
Re: [R] MLE Function
Terence Broderick wrote:
I am just trying to teach myself how to use the mle function in R because it
is much better than what is provided in MATLAB. I am following tutorial
material from the internet, however, it gives the following errors, does
anybody know what is happening to cause such errors, or does anybody know any
better tutorial material on this particular subject.
x.gam-rgamma(200,rate=0.5,shape=3.5)
x-x.gam
library(stats4)
ll-function(lambda,alfa){n-200;x-x.gam
-n*alfa*log(lambda)+n*log(gamma(alfa))-9alfa-1)*sum(log(x))+lambda*sum(x)}
Error: syntax error, unexpected SYMBOL, expecting '\n' or ';' or '}' in
ll-function(lambda,alfa){n-200;x-x.gam
-n*alfa*log(lambda)+n*log(gamma(alfa))-9alfa
ll-function(lambda,alfa){n-200;x-x.gam
-n*alfa*log(lambda)+n*log(gamma(alfa))-(alfa-1)*sum(log(x))+lambda*sum(x)}
est-mle(minuslog=ll,start=list(lambda=2,alfa=1))
Error in optim(start, f, method = method, hessian = TRUE, ...) :
objective function in optim evaluates to length 200 not 1
Er, not what I get. Did your version have that linefeed after x - x.gam
? If not, then you'll get your negative log-likelihood added to x.gam
and the resulting likelihood becomes a vector of length 200 instead of
a scalar.
In general, the first piece of advice for mle() is to check that the
likelihood function really is what it should be. Otherwise there is no
telling what the result might mean...
Secondly, watch out for parameter constraints. With your function, it
very easily happens that alfa tries to go negative in which case the
gamma function in the likelihood will do crazy things.
A common trick in such cases is to reparametrize by log-parameters, i.e.
ll - function(lambda,alfa){n-200; x-x.gam
-n*alfa*log(lambda)+n*lgamma(alfa)-(alfa-1)*sum(log(x))+lambda*sum(x)}
ll2 - function(llam, lalf) ll(exp(llam),exp(lalf))
est - mle(minuslog=ll2,start=list(llam=log(2),lalf=log(1)))
par(mfrow=c(2,1))
plot(profile(est))
Notice, incidentally, the use of lgamma rather than log(gamma(.)), which
is prone to overflow.
In fact, you could also write this likelihood directly as
-sum(dgamma(x, rate=lambda, shape=alfa, log=T))
audaces fortuna iuvat
-
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--
O__ Peter Dalgaard Øster 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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] off-topic: better OS for statistical computing
My sense is that R users are even split between UNIX and Windows
users so either will do in terms of the larger community.
Some R packages may not be avaliable on every platform or will
be available on one platform before another or there will be
certain platform-specific issues. So in the end its easiest to
have the same thing everyone else that you work with does.
Also if you run into
problems then you can ask others whereas if you are the lone
person with something different you have no one to turn to.
Also associated software may be, for example, Microsoft Office in
a Microsoft environment and LaTeX in a UNIX environment. And
networking will be simplified in a consistent environment too.
Certainly there is Open Office, Samba and putty but the easiest
is just not to have to worry about getting everything to work
together by just having the same thing in the first place.
Neither Linux nor Windows is superior to the other. People
making such representations generally know one much better
than the other and its more a reflection of their own experience
than anything else. I personally have used both UNIX and
Windows since their inception and find that I tend to have a
slight preference for whatever I used last. Technical merits of
one vs. the other are basically irrelevant for most purposes.
On 9/10/07, Patrick Connolly [EMAIL PROTECTED] wrote:
On Mon, 10-Sep-2007 at 12:26PM -0400, Gabor Grothendieck wrote:
| You want whatever all the people you are working with are using
| to make it as easy as possible to work together with them.
Assuming you're using R, there is negligible difficulty using a
different OS from what your colleagues use (apart from the
inconsistencies you get between different versions of Windows, but
even that has little effect on R). The standard .RData binary files
work with Windows and Linux (and probably OS X).
The only issue I come across is that Linux can't create WMF files as
readily as Windows can, and that is more than made up for by the
greater flexibility that Linux offers. It's easier in Linux to
produce Excel files from dataframes and matrices using a perl script
posted to this list by Marc Schwartz. Thanks again Marc.
Best
Patrick
|
| On 9/10/07, Wensui Liu [EMAIL PROTECTED] wrote:
| Good morning, everyone,
| I am sorry for this off-topic post but think I can get great answer
| from this list.
| My question is what is the best OS on PC (laptop) for statistical
| computing and why.
| I really appreciate your insight.
| Have a nice day.
|
| __
| R-help@stat.math.ethz.ch mailing list
| https://stat.ethz.ch/mailman/listinfo/r-help
| PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
| and provide commented, minimal, self-contained, reproducible code.
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_Middle minds discuss events
(:_~*~_:)Small minds discuss people
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Re: [R] Generating Replicate Datasets (using loops or other means)
Hi Jonathan,
What exactly do you mean by replication?
Do you want to keep a1,b1,c1,... unchanged but have 30
different sets of random numbers?
Regards,
Moshe.
--- VTLT1999 [EMAIL PROTECTED] wrote:
Hello All,
I have searched many help forums, message boards,
etc. and I just can't
apply the comments to what I need my program to do.
I am running R 2.5.1 on
an XP system, and my desire is to produce replicate
datasets for a
simulation study I am running. Essentially, I have
sets of parameters (a's,
b's, and c's) that define a function which produces
a decimal value. This
value is compared to a random uniform value, and is
coded a 1 if the
function is greater than the uniform value, 0 if it
is = to the uniform
value. My code thus far works great, but I just
need it to run several
times.Here we go:
library(mvtnorm)
library(sm)
library(ltm)
library(irtoys)
k- 5000
set.seed(271828)
t -
rmvnorm(n=k,mean=c(-1,0,1),sigma=matrix(c(1,.8,.5,.8,1,.8,.5,.8,1),3,3))
#Using mv here because of the likely association
of ability (theta = t)
across time.
t1-as.matrix(t[,1])
t2-as.matrix(t[,2])
t3-as.matrix(t[,3])
set.seed(271828)
# Population item parameters (n=54) from which we
will select relevant
items
# These are the parameters that are used in the
function
a - c(1.18120, 0.92613, 0.96886, 0.80503, 1.12384,
0.84073, 0.85544, 0.86801, 1.01054, 0.82278,
1.10353, 0.78865, 0.98421, 1.76071, 0.89603,
0.84671, 0.89737, 0.74775, 0.32190, 0.69730,
0.72059, 1.16762, 1.29257, 1.32902, 0.59540,
0.51022, 0.59259, 0.93951, 0.68568, 0.55649,
0.88084, 0.52940, 0.45735, 0.57560, 1.11779,
0.96984, 1.19692, 0.99102, 1.25847, 1.62555,
0.63049, 1.07807, 1.04897, 1.23138, 1.14014,
1.25230, 1.14844, 0.59287, 0.83143, 0.81723,
0.52141, 0.61980, 0.49945, 1.02749)
b - c(-2.51737, -1.95897, -1.72667, -0.82988,
-0.36093,
0.72554, 0.91442, 0.78061, 0.06088,
0.75733,
-0.76371, 0.24552, -0.42050, 0.88232,
-0.81761,
0.06466, -0.43866, -0.46042, 0.21636,
-0.73147,
-1.44086, -1.03718, 0.07275, -0.17197,
1.53796,
-0.45631, -1.69826, -0.66506, 0.98921,
0.30714,
-0.62245, 0.97253, 1.95894, 0.21277,
1.96346,
1.18825, 1.59917, -0.28401, -1.23530,
-0.09671,
-0.31581, -0.66149, -0.81284, -0.35399,
-0.07623,
1.06442, -0.68559, 1.07591, 0.97458,
0.06436,
1.25622, 1.73954, 1.75052, 2.34088)
c - c(0.0, 0.0, 0.0, 0.0, 0.19648,
0.31302, 0.26454, 0.19714, 0.06813, 0.21344,
0.0, 0.03371, 0.0, 0.16581, 0.11054,
0.08756, 0.07115, 0.26892, 0.0, 0.06883,
0.0, 0.14815, 0.32389, 0.19616, 0.17597,
0.0, 0.0, 0.04337, 0.19949, 0.20377,
0.0, 0.06243, 0.13639, 0.0, 0.18166,
0.15996, 0.20184, 0.08331, 0.24453, 0.26114,
0.16434, 0.20750, 0.32658, 0.31870, 0.45227,
0.35039, 0.31178, 0.17999, 0.22774, 0.21675,
0.10153, 0.17764, 0.15205, 0.19858)
# Item parameters for generating 3PL data for all
five testing occasions:
# This selects the relevant parameters for a
particular data generation run
# Only parameters for the first testing occasion
are shown to save space
a1 - as.matrix(a[c(1:5,15:20,22:24,38:44)])
b1 - as.matrix(b[c(1:5,15:20,22:24,38:44)])
c1 - as.matrix(c[c(1:5,15:20,22:24,38:44)])
# Here is where I would like to begin my
replications, but don't know how
to make R do it.
# The code below produces a matrix of 0's and 1's
(which will be used by
another program)
# I would like to nest this in a do loop such
that, say, 30 replicate
datasets are produced using the
#same parameters.
N - nrow(t1) # number of examinees
n - nrow(a1) # number of items
d - 1.7
theta - t1
response - matrix (0,N,n)
uni - matrix (runif(N*n),nrow = N)
for (i in 1:N)
{
for (j in 1:n)
{
if (
c1[j]+(1-c1[j])/(1+exp(-d*a1[j]*(theta[i]-b1[j])))
uni[i,j] )
response[i,j] = 1
else
response[i,j] = 0
}
}
write.table(response, file=C:/responses.dat, sep=
,row.names=FALSE,
col.names=FALSE)
I tried earlier nesting this in another for loop,
but that indexes elements
of matrices and vectors, and doesn't seem to apply
to a global loop
methodology. I am attempting to use replicate as we
speak, but
documentation is sparse (help(replicate) is nested
in lapply information).
Any guidance is greatly appreciated.
Thanks in advance,
Jonathan Beard
--
View this message in context:
http://www.nabble.com/Generating-Replicate-Datasets-%28using-loops-or-other-means%29-tf4418768.html#a12603580
Sent from the R help mailing list archive at
Nabble.com.
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Re: [R] finding the minimum positive value of some data
Here are some solutions each of which
1. has only one line,
2. x only occurs once so you can just plug in a complex expression
3. no temporary variables are left
min(sapply(x, function(z) if (z 0) z else Inf))
(function(z) min(ifelse(z 0, z, Inf))) (x)
with(list(z = x), min(z[z 0]))
local({ z - x; min(z[z 0]) })
On 9/10/07, dxc13 [EMAIL PROTECTED] wrote:
useRs,
I am looking to find the minimum positive value of some data I have.
Currently, I am able to find the minimum of data after I apply some other
functions to it:
x
[1] 1 0 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10
sort(x)
[1] 0 1 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10
diff(sort(x))
[1] 1 0 1 1 0 1 1 0 0 1 1 1 0 1 0 1 0
min(diff(sort(x)))
[1] 0
The minimum is given as zero, which is clearly true, but I am interested in
only the positive minimum, which is 1. Can I find this by using only 1 line
of code, like I have above? Thanks!
dxc13
--
View this message in context:
http://www.nabble.com/finding-the-minimum-positive-value-of-some-data-tf4417250.html#a12599319
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] what am I missing
Its a FAQ:
http://hermes.sdu.dk/Rdoc/faq.html#Why%20does%20outer()%20behave%20strangely%20with%20my%20function%3f
On 9/10/07, Jan de Leeuw [EMAIL PROTECTED] wrote:
x-seq(-1,1,length=10)
y-seq(-1,1,length=10)
a-matrix(c(1,2,2,1),2,2)
b-matrix(c(2,1,1,2),2,2)
fv-function(x,y) {
m-x*a+y*b
t-m[1,1]+m[2,2]; d-m[1,1]*m[2,2]-m[1,2]^2
return((t-sqrt(t^2-4*d))/2)
}
gv-function(x,y) {
t-x*(a[1,1]+a[2,2])+y*(b[1,1]+b[2,2])
d-(x*a[1,1]+y*b[1,1])*(x*a[2,2]+y*b[2,2])-(x*a[1,2]+y*b[1,2])^2
return((t-sqrt(t^2-4*d))/2)
}
now outer(x,y,gv) works as expected, outer(x,y,fv) bombs. But
z-matrix(0,10,10); for (i in 1:10) for (j in 1:10) z[i,j]-fv(x[i],y
[j])
works fine. Must be something in outer().
==
Jan de Leeuw, 11667 Steinhoff Rd, Frazier Park, CA 93225, 661-245-1725
.mac: jdeleeuw ++ aim: deleeuwjan ++ skype: j_deleeuw
homepages: http://www.cuddyvalley.org and http://gifi.stat.ucla.edu
==
A bath when you're born,
a bath when you die,
how stupid. (Issa 1763-1827)
[[alternative HTML version deleted]]
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Re: [R] install packages automatically
Try ?update.packages Ross Darnell -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Wensui Liu Sent: Tuesday, 11 September 2007 11:45 AM To: r-help@stat.math.ethz.ch Subject: [R] install packages automatically Dear Listers, I am a little tired of installing all packages I want every time when I instill a new version of R. Say, if I have a list of packages I need to use, is it possible to tell R to install them all for me automatically rather than I install them one by one? 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 and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] overlay lattice histograms with goodness-of-fit pdfs
Mange tak!
FYI, this is the way it is able to run (I was going to attach
station.precip.R, but I read that attaching files is not recommended
- let me
know if you would like it)
x - dget(file=C://Documents and Settings/Bradley/My
Documents/Arizona/CourseResources/ATMO529/station.precip.R)
histogram(~ data | month * station, data = sta.stack,
subset = type==precip month %in% c(Dec,Jan,Feb),
xlab = Precipitation (mm),
type = density,
panel = function(x, ...) {
panel.histogram(x, ...)
panel.mathdensity(dmath = dnorm, col = black,
args = list(mean =
mean(sta.stack$data), sd = sd(sta.stack$data)))
panel.mathdensity(dmath = dgamma, col = black,
args = list(shape =
(mean(sta.stack$data))^2 / (stdev(sta.stack$data))^2,
scale =
(stdev(sta.stack$data))^2 / mean(sta.stack$data)))
})
Now, what would be great is to be able to reference the different calls to
panel.mathdensity() so that the corresponding probability * histogram area ( =
counts) in each bin can be used to compute a simple chi-square
goodness-of-fit.
I tried calling panel.mathdensity() outside of histogram(), but I don't think
this is right - it returns NULL. I also looked at chisq.test, but this
doesn't
support trellis formulas. Any thoughts or leads?
Thanks,
Brad Christoffersen
Quoting Frede Aakmann Tøgersen [EMAIL PROTECTED]:
The following is one of the examples in the help page for histogram:
histogram( ~ height | voice.part, data = singer,
xlab = Height (inches), type = density,
panel = function(x, ...) {
panel.histogram(x, ...)
panel.mathdensity(dmath = dnorm, col = black,
args = list(mean=mean(x),sd=sd(x)))
} )
This should give you some thing to start from.
Also using the subset argument of the lattice functions will make
make your code more readable. Instead of your code
histogram(~ data | month * station,
data = sta.stack[sta.stack[,type]==precip
(sta.stack[,month]==Dec | sta.stack[,month]==Jan |
sta.stack[,month]==Feb),],
xlab = Precipitation (mm))
you can use (not tested because you didn't supply a reproducable example)
histogram(~ data | month * station, data = sta.stack
subset = type==precip month %in% c(Dec, Jan, Feb),
xlab = Precipitation (mm))
Med venlig hilsen
Frede Aakmann Tøgersen
-Oprindelig meddelelse-
Fra: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] På vegne af Brad
Christoffersen
Sendt: 10. september 2007 12:08
Til: R-help@stat.math.ethz.ch
Emne: [R] overlay lattice histograms with goodness-of-fit pdfs
Hello,
I am new to R exploratory data analysis and plotting. Is
anyone aware of a way to overlay a set of conditional
histograms with conditional PDFs? Below, I generate a
lattice plot of precipitation histograms based on different
months and stations, given a subset of the dataset:
histogram(~ data | month * station,
data = sta.stack[sta.stack[,type]==precip
(sta.stack[,month]==Dec | sta.stack[,month]==Jan |
sta.stack[,month]==Feb),],
xlab = Precipitation (mm))
I previously used a combination of the low-level 'lines()'
and 'dgamma()'
functions to overlay a gamma PDF onto a single histogram.
Now what I would like to do is to do the same thing, but with
a function that allows me to specify a formula similar to
that in the histogram function above
[SomeKindOfPDF] ~ [x-range] | month * station
which will plot the PDF with the appropriate factors (month
and station).
All I'm looking for is for someone to get me going in the
right direction with a useful package or function to use.
Any help is much appreciated!
Brad Christoffersen
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Re: [R] finding the minimum positive value of some data
Either min(diff(sort(x))[diff(sort(x))0]) or min(diff(sort(unique(x --- dxc13 [EMAIL PROTECTED] wrote: useRs, I am looking to find the minimum positive value of some data I have. Currently, I am able to find the minimum of data after I apply some other functions to it: x [1] 1 0 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10 sort(x) [1] 0 1 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10 diff(sort(x)) [1] 1 0 1 1 0 1 1 0 0 1 1 1 0 1 0 1 0 min(diff(sort(x))) [1] 0 The minimum is given as zero, which is clearly true, but I am interested in only the positive minimum, which is 1. Can I find this by using only 1 line of code, like I have above? Thanks! dxc13 -- View this message in context: http://www.nabble.com/finding-the-minimum-positive-value-of-some-data-tf4417250.html#a12599319 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using bootstrap for tree selection step in rpart
Fiona Callaghan wrote: I was wondering if someone could help me with an rpart problem. I can see that cross-validation is the default for tree selection in rpart -- has a bootstrap method been implemented anywhere? I think this is a different thing to 'bagging' or 'boosting' -- I still want 'one' tree at the end, I just would like it chosen using a bootstrap method. Any ideas??? Hi Fiona, I'm not sure if I understand you correctly. To get one single rpart tree trained on one bootstrap sample, try bagging() from the 'ipred' package and set nbagg=1. Bye, Chris __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] artificial data matrix with 100000 rows
On 9/9/07, kevinchang [EMAIL PROTECTED] wrote: I tried to made the matrix with this size by either matrix() or array(). However, there seems to be default limit of number for rows made. I got sort of error message from R .To be specific, m--matrix(ncol=3,nrow=10) error message:[ reached getOption(max.print) -- omitted 7 rows ]] or a-array(dim=c(1,3,10)) error message:reached getOption(max.print) -- omitted 6667 row(s) and 6 matrix slice(s) ] That is not an error message, I guess. When the matrices are huge, R is unable to print them totally on the screen, but all data are present. For instance, m[(nrow(m)-10):nrow(m),] [,1] [,2] [,3] [1,] NA NA NA [2,] NA NA NA [3,] NA NA NA [4,] NA NA NA [5,] NA NA NA [6,] NA NA NA [7,] NA NA NA [8,] NA NA NA [9,] NA NA NA [10,] NA NA NA [11,] NA NA NA See ?getOption Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem in installing packages on linux machine...
Your problem is not related to missing file permissions but to a problem with the linker (ld). As you can see from the error message, ld cannot find a compatible libR.so so compilation fails. You did not specify which architecture you're on but in case it is a 64-bit platform, the following thread might be helpful: http://tolstoy.newcastle.edu.au/R/e2/help/07/07/22071.html BTW, the correct command for running R as root is su -c R or sudo R. Furthermore, lib is optional and defaults to .libPaths()[1], see ?install.packages tkobayas at indiana.edu writes: Hi, Still got the same error message. I did su R when I got the error message for the first time. I have never seen this error message. I will be googling for solutions as well... Thank you. Quoting Henrique Dallazuanna wwwhsd at gmail.com: Hi, try install packages whit 'sudo'. $sudo R -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 08/09/2007, tkobayas at indiana.edu tkobayas at indiana.edu wrote: Hi, I am trying to install RSQLite package on my Fedora workstation. I tried to install other packages as well, but each time I got the same error messages saying compilation error and non zero exit status. Do I have to specify lib=? I never specified the library path before when I was using Fedora Core 6. Warning in install.packages(RSQLite) : argument 'lib' is missing: using '/usr/lib/R/library' --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done trying URL 'http://cran.cnr.Berkeley.edu/src/contrib/RSQLite_0.5-6.tar.gz' Content type 'application/x-gzip' length 710241 bytes opened URL == downloaded 693Kb gcc -std=gnu99 -shared -L/usr/local/lib -o RSQLite.so RS-DBI.o RS-SQLite.o sqlite-all.o -L/usr/lib/R/lib -lR /usr/bin/ld: skipping incompatible /usr/lib/R/lib/libR.so when searching for -lR /usr/bin/ld: cannot find -lR collect2: ld returned 1 exit status make: *** [RSQLite.so] Error 1 chmod: cannot access `/usr/lib/R/library/RSQLite/libs/*': No such file or directory ERROR: compilation failed for package 'RSQLite' ** Removing '/usr/lib/R/library/RSQLite' The downloaded packages are in /tmp/RtmpHQ5Y7C/downloaded_packages Warning message: installation of package 'RSQLite' had non-zero exit status in: install.packages(RSQLite) I appreciate your help. Thank you very much Taka __ R-help at stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Takatsugu Kobayashi PhD Student Indiana University, Dept. Geography __ R-help at stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] What does it mean by initial value not available?
On 09/09/2007 7:01 AM, Yuchen Luo wrote:
Dear friends.
I use ConstrOptim( ) and got error message initial value not available.
My understanding of initial value not available is that one of the
following 3 cases happens:
1.The objective function is not well defined at the point of the initial
value.
2. The differentiation of the objective function is not well defined at the
point of the initial value.
3. The initial value violate the constrain of ui %*% theta - ci = 0
But my situation does not belong to any of the above cases.
I have attached my code bellow and could you please help me take a look?
I haven't tried your code, but there's one obvious error:
...
fit.error=function(rec,lambda,lbar)
{sum((eval(apple)*1000-orange)^2/(orange^2))
}
The function optimizes over the elements of the first parameter. You've
got two other parameters there, and I think you're trying to optimize
over them as well. Put them all into one vector.
The documentation for constrOptim doesn't make this as clear as it
should; I'll clarify (by copying the docs from ?optim).
Duncan Murdoch
fit.error.grr=function(rec,lambda, lbar)
{drec=sum(eval(D(apple,'rec'))*(eval(apple)*1000-orange)/(orange^2))
dlambda=sum(eval(D(apple,'lambda'))*(eval(apple)*1000-orange)/(orange^2))
dlbar=sum(eval(D(apple,'lbar'))*(eval(apple)*1000-orange)/(orange^2))
c(drec,dlambda,dlbar)
}
rr=c(4.33,4.22,4.27,4.43,4.43,4.44,4.45,4.65,4.77,4.77)
tot=rep(13319.17,10)
sh=rep(1553656,10)
sigmae=c(0.172239074,0.188209271,0.193703774,0.172659891,0.164427247,
0.24602361,0.173555309,0.186701165,0.193150456)
ss=c(56.49,56.39,56.55,57.49,57.37,55.02,56.02,54.35,54.09)
orange=rep(21.25,10)
constrOptim(c(0.5,0.3,0.5), fit.error, fit.error.grr, ui=ui,ci=ci)
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Re: [R] Sweave figure aspect ratio
On 09/09/2007 7:24 AM, Werner Wernersen wrote:
Hello,
using Sweave, is there any option to preserve the
original aspect ratio of plots generated from R code?
Consider this Sweave chunk:
test,echo=F,fig=T,width=2,height=2=
x - 1:10
y - sin(x)
par(mar=c(4,4,0,4))
plot(x,y,
xlab=x label,
ylab=y label
)
@
In Latex, I want to produce a plot of width 2 but
don't care about the height. If I put the width and
height like shown on the example, the box isn't square
any more. But if I put only width or height in the
Sweave option, then the values then the appearance is
totally off.
Is there any option for Sweave or other possibility to
preserve the ratio of the sides of plots?
The width and height args in the header of the chunk are passed to the
graphics device (postscript or pdf), but they don't appear in the LaTeX
output. To modify the appearance in LaTeX, you need to set a graphicx
package parameter, e.g. something like
\setkeys{Gin}{width=2in}
test,echo=F,fig=T=
x - 1:10
y - sin(x)
par(mar=c(4,4,0,4))
plot(x,y,
xlab=x label,
ylab=y label
)
@
If you only specify width, then the aspect ratio will be maintained.
The \setkeys{} changes are persistent; as far as I know there's no way
to say go back to the default afterwards. So what I'd suggest is to
define a macro (e.g. \Gwidth) to give the default width, and then call
\setkeys{Gin}{width=\Gwidth}
at the start of your document and after any change like the one above.
Duncan Murdoch
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] stacking data frames with different variables
Have a look at rbind.fill in the reshape package. Hadley On 9/9/07, Muenchen, Robert A (Bob) [EMAIL PROTECTED] wrote: Hi All, If I need to stack two data frames, I can use rbind, but it requires that all variables exist in both sets. I can make that happen, but other stat packages would figure out where the differences were, add the missing variables to each, set their values to missing and stack them. Is there a more automatic way to do that in R? Below is an example program. Thanks, Bob # Top data frame has two variables. x - c(1,2) y - c(1,2) top - data.frame(x,y) top # Bottom data frame has only one of them. x - c(3,4) bottom - data.frame(x) bottom # So rbind won't work. rbind(top, bottom) # After figuring out where the mismatches are I can # make the two DFs the same manually. bottom - data.frame( bottom, y=NA) bottom # Now I get the desired result. both - rbind(top,bottom) both = Bob Muenchen (pronounced Min'-chen), Manager Statistical Consulting Center U of TN Office of Information Technology 200 Stokely Management Center, Knoxville, TN 37996-0520 Voice: (865) 974-5230 FAX: (865) 974-4810 Email: [EMAIL PROTECTED] Web: http://oit.utk.edu/scc, News: http://listserv.utk.edu/archives/statnews.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 and provide commented, minimal, self-contained, reproducible code. -- http://had.co.nz/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] stacking data frames with different variables
Perfect. Thanks Hadley! -Original Message- From: hadley wickham [mailto:[EMAIL PROTECTED] Sent: Sunday, September 09, 2007 10:11 AM To: Muenchen, Robert A (Bob) Cc: r-help@stat.math.ethz.ch Subject: Re: [R] stacking data frames with different variables Have a look at rbind.fill in the reshape package. Hadley On 9/9/07, Muenchen, Robert A (Bob) [EMAIL PROTECTED] wrote: Hi All, If I need to stack two data frames, I can use rbind, but it requires that all variables exist in both sets. I can make that happen, but other stat packages would figure out where the differences were, add the missing variables to each, set their values to missing and stack them. Is there a more automatic way to do that in R? Below is an example program. Thanks, Bob # Top data frame has two variables. x - c(1,2) y - c(1,2) top - data.frame(x,y) top # Bottom data frame has only one of them. x - c(3,4) bottom - data.frame(x) bottom # So rbind won't work. rbind(top, bottom) # After figuring out where the mismatches are I can # make the two DFs the same manually. bottom - data.frame( bottom, y=NA) bottom # Now I get the desired result. both - rbind(top,bottom) both = Bob Muenchen (pronounced Min'-chen), Manager Statistical Consulting Center U of TN Office of Information Technology 200 Stokely Management Center, Knoxville, TN 37996-0520 Voice: (865) 974-5230 FAX: (865) 974-4810 Email: [EMAIL PROTECTED] Web: http://oit.utk.edu/scc, News: http://listserv.utk.edu/archives/statnews.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 and provide commented, minimal, self-contained, reproducible code. -- http://had.co.nz/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] confusion matrix - better code?
Wolfgang,
This looks great and certainly puts to shame my code .
Thanks,
Monica
Date: Fri, 7 Sep 2007 21:04:32 +0100 From: [EMAIL PROTECTED] To: [EMAIL
PROTECTED] CC: r-help@stat.math.ethz.ch Subject: Re: [R] confusion matrix -
better code? Dear Monica, try this: cm = table(tr, pr) cm pr tr 1
2 3 1 2 1 0 2 2 1 0 3 0 0 3 4 0 1 0 rowSums(cm) colSums(cm) Best
wishes Wolfgang Huber Monica Pisica ha scritto: Hi,I�ve
written some code to obtain a confusion matrix when the true classification
and the predicted classification are known. Suppose true classification is
called �tr� and predicted classification is �pr�. I have 4 classes in
tr, but only 3 classes out of 4 are predicted in �pr�. Following is my
code, but looks quite �clunky� to me. I wonder if you have any
suggestions to improve it.Thanks,Monica
-tr - c(1,2,2,3,3,3,2,4,1,1)
pr-c(1,2,1,3,3,3,1,2,1,2) dat - data.frame(tr, pr) class -
c(1:length(tr)) m - max(c(length(unique(tr)), length(unique(pr
for(i in 1:length(class)) { class[i] - sub('
','',paste(dat[i,1],dat[i,2])) } dat - data.frame(dat, class) mat -
matrix(0, nrow=m, ncol=m) for (i in 1:m){ for (j in 1:m){ mat[i,j]
- sub(' ','',paste(i,j)) }} cat - matrix(0, nrow=(m+1), ncol=(m+1))
for (i in 1:m){ for(j in 1:m){ cat[i,j]-
nrow(dat[dat$class==mat[i,j],]) }} for (i in 1:m){
cat[(m+1),i]-sum(cat[1:m,i]) cat[i,(m+1)]- sum(cat[i,1:m])
cat[(m+1),(m+1)] - sum(cat[1:m,(m+1)]) } cat [,1] [,2] [,3] [,4]
[,5] [1,] 2 1 0 0 3 [2,] 2 1 0 0 3 [3,] 0 0 3 0 3 [4,] 0 1 0 0 1
[5,] 4 3 3 0 10The 5th row / col represents the sum on each row /
col respectively.
_
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Re: [R] loop execution problem
On 09/09/2007 1:36 PM, kevinchang wrote: This is bizzare. The for loop I set seems to go infinitly because the hourglass cursor never disappers and I never get the result. But when I click on X of the window try to exit R, the execution finishes right after I abort exiting when RGui asks me to save the workplace or not. Please help me out. In Windows you can also abort execution by hitting Esc, or the menu entry Misc|Stop current computation. Duncan Murdoch __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] fitdistr()
I assume that you want to do the fitdistr on one of the columns of the dataframe that you have read in. What does 'str(ONES3)' show? If the data is in the first column, try: fitdistr(ONES3[[1]],chi-squared) On 9/9/07, Terence Broderick [EMAIL PROTECTED] wrote: I am trying to fit the chi-squared distribution to a set of data using the fitdistr function found in the MASS4 library, the data set is called ONES3, I have loaded it using the command ONES3-read.table(ONES3.pdf,header=TRUE,na=NA) I print out the dataset ONES3 to the screen to make sure it has loaded Then I try to fit this data using the command fitdistr fitdistr(ONES3,chi-squared) and it returns the comment Error in fitdistr(ONES3, chi-squared) : 'x' must be a non-empty numeric vector Can anybody help with this, I imagine it is a common mistake for beginners like myself audaces fortuna iuvat - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] What does it mean by initial value not available?
Dear Professor Mordoch.
Thank you very much for your help! Your time is highly appreciated!
I do intend to optimize over 3 parameters and the way I did it is
constrOptim(c(0.5,0.3,0.5), fit.error, fit.error.grr, ui=ui,ci=ci).
Also, There is a missing value for both sigmae and ss. (The last 3rd
and 4th line) Please correct them as:
sigmae=c(0.172239074,0.188209271,0.193703774,0.172659891,0.164427247,0.24602361,0.173555309,0.186701165,0.193150456,
0.1857315601)
ss=c(56.49,56.39,56.55,57.49 ,57.37,55.02,56.02,54.35,54.09,54.67)
Best Wishes
Yuchen Luo
On 9/9/07, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 09/09/2007 7:01 AM, Yuchen Luo wrote:
Dear friends.
I use ConstrOptim( ) and got error message initial value not
available.
My understanding of initial value not available is that one of the
following 3 cases happens:
1.The objective function is not well defined at the point of the initial
value.
2. The differentiation of the objective function is not well defined at
the
point of the initial value.
3. The initial value violate the constrain of ui %*% theta - ci = 0
But my situation does not belong to any of the above cases.
I have attached my code bellow and could you please help me take a look?
I haven't tried your code, but there's one obvious error:
...
fit.error=function(rec,lambda,lbar)
{sum((eval(apple)*1000-orange)^2/(orange^2))
}
The function optimizes over the elements of the first parameter. You've
got two other parameters there, and I think you're trying to optimize
over them as well. Put them all into one vector.
The documentation for constrOptim doesn't make this as clear as it
should; I'll clarify (by copying the docs from ?optim).
Duncan Murdoch
fit.error.grr=function(rec,lambda, lbar)
{drec=sum(eval(D(apple,'rec'))*(eval(apple)*1000-orange)/(orange^2))
dlambda=sum(eval(D(apple,'lambda'))*(eval(apple)*1000-orange)/(orange^2))
dlbar=sum(eval(D(apple,'lbar'))*(eval(apple)*1000-orange)/(orange^2))
c(drec,dlambda,dlbar)
}
rr=c(4.33,4.22,4.27,4.43,4.43,4.44,4.45,4.65,4.77,4.77)
tot=rep(13319.17,10)
sh=rep(1553656,10)
sigmae=c(0.172239074,0.188209271,0.193703774,0.172659891,0.164427247,
0.24602361,0.173555309,0.186701165,0.193150456)
ss=c(56.49,56.39,56.55,57.49,57.37,55.02,56.02,54.35,54.09)
orange=rep(21.25,10)
constrOptim(c(0.5,0.3,0.5), fit.error, fit.error.grr, ui=ui,ci=ci)
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Re: [R] Sweave figure aspect ratio
Many thanks for the quick reply, Duncan.
Now I see that something is going wrong in R already.
I overlooked this because the R window was so big. I
have to persuade plotViewport() not to create a square
area but one which takes the different margins into
account.
Best regards,
Werner
--- Duncan Murdoch [EMAIL PROTECTED] schrieb:
On 09/09/2007 7:24 AM, Werner Wernersen wrote:
Hello,
using Sweave, is there any option to preserve the
original aspect ratio of plots generated from R
code?
Consider this Sweave chunk:
test,echo=F,fig=T,width=2,height=2=
x - 1:10
y - sin(x)
par(mar=c(4,4,0,4))
plot(x,y,
xlab=x label,
ylab=y label
)
@
In Latex, I want to produce a plot of width 2 but
don't care about the height. If I put the width
and
height like shown on the example, the box isn't
square
any more. But if I put only width or height in the
Sweave option, then the values then the appearance
is
totally off.
Is there any option for Sweave or other
possibility to
preserve the ratio of the sides of plots?
The width and height args in the header of the chunk
are passed to the
graphics device (postscript or pdf), but they don't
appear in the LaTeX
output. To modify the appearance in LaTeX, you need
to set a graphicx
package parameter, e.g. something like
\setkeys{Gin}{width=2in}
test,echo=F,fig=T=
x - 1:10
y - sin(x)
par(mar=c(4,4,0,4))
plot(x,y,
xlab=x label,
ylab=y label
)
@
If you only specify width, then the aspect ratio
will be maintained.
The \setkeys{} changes are persistent; as far as I
know there's no way
to say go back to the default afterwards. So what
I'd suggest is to
define a macro (e.g. \Gwidth) to give the default
width, and then call
\setkeys{Gin}{width=\Gwidth}
at the start of your document and after any change
like the one above.
Duncan Murdoch
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] NA in AR model residuals
Thanks all! On 09/09/2007, Nick Chorley [EMAIL PROTECTED] wrote: Hi, This may be a stupid question, but why when creating an AR model, why does R set the first n components of the ar object's resid vector to NA (where n is the order of the model)? Regards, NC [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] SQL like function?
Hi Gabor, Wow, this is awesome although I eventually should learn MySQL for integrating it on web-based DB management using PHP or Perl, this is a very helpful tool for me to start with! Thank you very much Gabor Grothendieck wrote: Others have already pointed out %in% but regarding your comment about SQL, you can use SQL to manipulate R data frames using the sqldf package which provides an interface to lower level RSQLite (and RMySQL in the future) routines. The following examples use SQLite underneath: DF - data.frame(observation = c(1,2,3,4,5)) ID - data.frame(ID = c(1, 3, 4)) library(sqldf) sqldf(select observation, observation in (select * from ID) `ID?` from DF) # or sqldf(select observation, observation in (1, 3, 4) `ID?` from DF) See home page at: http://sqldf.googlecode.com On 9/7/07, Takatsugu Kobayashi [EMAIL PROTECTED] wrote: Hi RUsers, I am wonder if I can search observations whose IDs matches any of the values in another vector, such as in MySQL. While I am learing MySQL for future database management, I appreciate if anyone could give me a hint. Suppose I have one 5*1 vector containing observation IDs and frequencies, and one 3*1 vector containing observation IDs. observation-c(1,2,3,4,5) ID-c(1,3,4) Then, I would like to program a code that returns a results showing matched observations like result: TRUE FALSE TRUE TRUE FALSE I am reading S programming, but I cannot find a way to do this. Thank you very much. Taka __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] SQL like function?
Great. Regarding the web, note that there are actually quite a few R web projects as well: http://www.lmbe.seu.edu.cn/CRAN/doc/FAQ/R-FAQ.html#R-Web-Interfaces I have used rpad (www.rpad.org) which has an integrated web server right in the R package making setup a non-issue. On 9/8/07, Takatsugu Kobayashi [EMAIL PROTECTED] wrote: Hi Gabor, Wow, this is awesome although I eventually should learn MySQL for integrating it on web-based DB management using PHP or Perl, this is a very helpful tool for me to start with! Thank you very much Gabor Grothendieck wrote: Others have already pointed out %in% but regarding your comment about SQL, you can use SQL to manipulate R data frames using the sqldf package which provides an interface to lower level RSQLite (and RMySQL in the future) routines. The following examples use SQLite underneath: DF - data.frame(observation = c(1,2,3,4,5)) ID - data.frame(ID = c(1, 3, 4)) library(sqldf) sqldf(select observation, observation in (select * from ID) `ID?` from DF) # or sqldf(select observation, observation in (1, 3, 4) `ID?` from DF) See home page at: http://sqldf.googlecode.com On 9/7/07, Takatsugu Kobayashi [EMAIL PROTECTED] wrote: Hi RUsers, I am wonder if I can search observations whose IDs matches any of the values in another vector, such as in MySQL. While I am learing MySQL for future database management, I appreciate if anyone could give me a hint. Suppose I have one 5*1 vector containing observation IDs and frequencies, and one 3*1 vector containing observation IDs. observation-c(1,2,3,4,5) ID-c(1,3,4) Then, I would like to program a code that returns a results showing matched observations like result: TRUE FALSE TRUE TRUE FALSE I am reading S programming, but I cannot find a way to do this. Thank you very much. Taka __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lisp-like primitives in R
François Pinard wrote: [Roland Rau] [François Pinard] I wonder what happened, for R to hide the underlying Scheme so fully, at least at the level of the surface language (despite there are hints). To further foster portability, we chose to write R in ANSI C Yes, of course. Scheme is also (often) implemented in C. I meant that R might have implemented a Scheme engine (or part of a Scheme engine, extended with appropriate data types) with a surface language (nearly the S language) which is purposely not Scheme, but could have been. If the gap is not extreme, one could dare dreaming that the Scheme engine in R be completed, and Scheme offered as an alternate extension language. If you allow me to continue dreaming awake -- they told me they will let me free as long as I do not get dangerous! :-) -- part of the interest lies in the fact there are excellent Scheme compilers. If we could only find or devise some kind of marriage between a mature Scheme and R, so to speed up the non-vectorisable parts of R scripts... Well, depending on what you want, this is either trivial or impossible... The internal storage of R is still pretty much equivalent to scheme. E.g. try this: r2scheme - function(e) if (!is.recursive(e)) deparse(e) else c((, unlist(lapply(as.list(e), r2scheme)), )) paste(r2scheme(quote(for(i in 1:4)print(i))), collapse= ) [1] ( for i ( : 1 4 ) ( print i ) ) and a parser that parses a similar language to R internal format is not a very hard exercise (some care needed in places). However, replacing the front-end is not going to make anything faster, and the evaluation engine in R does a couple of tricks which are not done in Scheme, notably lazy evaluation, and other forms of non-local evaluation, which drives optimizers crazy. Look up the writings of Luke Tierney on the matter to learn more. If we are lucky and one of the original authors reads this thread they might explain the situation further and better [...]. In r-devel, maybe! We would be lucky if the authors really had time to read r-help. :-) -- O__ Peter Dalgaard Øster 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plotting lines to sets of points
# Create a matrix of ball locations # You'd do this using the calls within your points function balls - matrix(c(0,50,25,-150,-100,-50), ncol=2, byrow=F) # Draw a line from the origin to each ball location apply(balls, 1, function(x) lines(c(125, x[1]), c(-210, x[2]), col='red')) A more complete example might loop over all the unique elements of framename$hit_traj, and then run this procedure for each value with a different colour, plotting both ball points and trajectories. lawnboy34 wrote: I am using R to plot baseball spray charts from play-by-play data. I have used the following command to plot the diamond: plot (0:250, -250:0, type=n, bg=white) lines(c(125,150,125,100,125),c(-210,-180,-150,-180,-210), col=c(black)) I have also plotted different hit locations using commands such as the following: points(subset(framename$hit_x, framename$hit_traj==line_drive), subset(-framename$hit_y, framename$hit_traj==line_drive), pch=20, col=c(red)) My question: Is there any easy way to plot a line from the origin (home plate) to each point on the graph? Preferably the line would share the same color as the dot that denotes where the ball landed. I have tried searching Google and these forums, and most graphing questions have to do with scatterplots or other varieties of graphs I am not using. Thanks very much in advance. -Jason -- View this message in context: http://www.nabble.com/Plotting-lines-to-sets-of-points-tf4404235.html#a12564959 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] enable object name to be called as object (a dataset)
a - 1:3
b - 11:13
c - 21:23
names - c('a','b','c')
do.call(data.frame, list(sapply(names, function(x) get(x
runner wrote:
What I am trying to do is as follows:
- I have listed names of all wanted objects (datasets A,B,C... ) in
current workspace as a vector:
obj - c('A','B','C')
- then i need to use these objects, say to extract all the 1st columns and
bind to an existing dataset ('data'):
for ( i in 1:3){
newdata - obj[i]
data - cbind(data,newdata [[1]] )
}
Obviously, it doesn't work since obj[i] is just a string of dataset name.
Here is my question: how to call it as a original dataset? Thanks.
--
View this message in context:
http://www.nabble.com/enable-object-name-to-be-called-as-object-%28a-dataset%29-tf4403933.html#a12564175
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] FW: variable format
Frank E Harrell Jr wrote:
Martin Becker wrote:
Dear Cory,
I am not familiar with SAS, but is this what you are looking for?
divisionTable - matrix(c(1, New England,
2, Middle Atlantic,
3, East North Central,
4, West North Central,
5, South Atlantic,
6, East South Central,
7, West South Central,
8, Mountain,
9, Pacific),
ncol=2, byrow=T)
How about just divisionTable - c('New England', 'Middle Atlantic',
...) then factor(old, 1:9, divisionTable) ?
Frank
Of course, this solution is more elegant, but my intention was
1. to provide a solution which makes use of the exisiting object
divisionTable
2. to reproduce the output from the working example (-conversion to
character)
Maybe I should have emphasized that I was quoting the existing
definition of divisionTable from the original email (for the sake of
providing self-contained code) and not introducing a unnecessarily
complicated new definition of divisionTable.
Regards,
Martin
a - NULL
a$divisionOld - c(0,1,2,3,4,5)
a$divisionNew -
as.character(factor(a$divisionOld,levels=divisionTable[,1],labels=divisionTable[,2]))
a$divisionNew
[1] NA New EnglandMiddle Atlantic [4]
East North Central West North Central South Atlantic
Kind regards,
Martin
Cory Nissen schrieb:
Anybody?
From: Cory Nissen
Sent: Tue 9/4/2007 9:30 AM
To: r-help@stat.math.ethz.ch
Subject: variable format
Okay, I want to do something similar to SAS proc format.
I usually do this...
a - NULL
a$divisionOld - c(1,2,3,4,5)
divisionTable - matrix(c(1, New England,
2, Middle Atlantic,
3, East North Central,
4, West North Central,
5, South Atlantic),
ncol=2, byrow=T)
a$divisionNew[match(a$divisionOld, divisionTable[,1])] -
divisionTable[,2]
But how do I handle the case where...
a$divisionOld - c(0,1,2,3,4,5) #no format available for 0, this
throws an error.
OR
divisionTable - matrix(c(1, New England,
2, Middle Atlantic,
3, East North Central,
4, West North Central,
5, South Atlantic,
6, East South Central,
7, West South Central,
8, Mountain,
9, Pacific),
ncol=2, byrow=T) There are extra formats
available... this throws a warning.
Thanks
Cory
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Re: [R] Matlab's lsqnonlin
KateM == Katharine Mullen [EMAIL PROTECTED] on Fri, 7 Sep 2007 20:07:41 +0200 (CEST) writes: KateM The thread you linked to regarding Levenberg-Marquardt's supposed lack of KateM availability is from 2001; it has been possible to get KateM to the MINPACK implementation of Levenberg-Marquardt within R via the KateM package minpack.lm KateM (http://cran.r-project.org/src/contrib/Descriptions/minpack.lm.html) since KateM 2005. Thanks a lot, Kate. I'm wondering about experiences: Do you know of cases where minpack.lm's nls.lm() solved a (real) problem that nls() would have a problem with ? Beware however -- one of the main things I learned about this field from Doug Bates, co-author of Bates_and_Watts and prinicipal author of S's and R's nls() : It's a *feature* that nls() does not converge sometimes when other methods do falsely claim convergence! Martin Maechler, ETH Zurich KateM KateM Katharine Mullen KateM mail: Department of Physics and Astronomy, Faculty of Sciences KateM Vrije Universiteit Amsterdam, de Boelelaan 1081 KateM 1081 HV Amsterdam, The Netherlands KateM room: T.1.06 KateM tel: +31 205987870 KateM fax: +31 205987992 KateM e-mail: [EMAIL PROTECTED] KateM homepage: http://www.nat.vu.nl/~kate/ KateM On Fri, 7 Sep 2007, Jose Luis Aznarte M. wrote: Hi! I'm translating some code from Matlab to R and I found a problem. I need to translate Matlab's function 'lsqnonlin' (http://www-ccs.ucsd.edu/matlab/toolbox/optim/lsqnonlin.html) into R, and at the beginning I thought it would be the same as R's 'optim'. But then I looked at the definition of 'lsqnonlin' and I don't quite see how to make 'optim' to do the same thing. Does anyone have an idea? This is apart from the fact that I would like to use the Levenberg Marquardt algorithm which is not implemented in R (some discussion about this: http://tolstoy.newcastle.edu.au/R/help/00b/2492.html). Thank you! All the best, -- -- Jose Luis Aznarte M. http://decsai.ugr.es/~jlaznarte Department of Computer Science and Artificial Intelligence Universidad de Granada Tel. +34 - 958 - 24 04 67 GRANADA (Spain) Fax: +34 - 958 - 24 00 79 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. KateM __ KateM R-help@stat.math.ethz.ch mailing list KateM https://stat.ethz.ch/mailman/listinfo/r-help KateM PLEASE do read the posting guide http://www.R-project.org/posting-guide.html KateM and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] argument 'lib' is missing: using '/usr/lib/R/library'
Paul, what is the question? If the question is why you get this warning message, the reason is that the 'lib' argument is missing and install.packages is using '/usr/lib/R/library'. If you want to get rid of the warning supply the 'lib' argument. Gabor On Sat, Sep 08, 2007 at 11:26:44AM +0100, Paul Smith wrote: Dear All, When installing packages, I get the following warning: install.packages(sqldf) Warning in install.packages(sqldf) : argument 'lib' is missing: using '/usr/lib/R/library' Any ideas? The details of my R installation are: version _ platform i386-redhat-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 2 minor 5.1 year 2007 month 06 day27 svn rev42083 language R version.string R version 2.5.1 (2007-06-27) R is here installed on Fedora 7. Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Csardi Gabor [EMAIL PROTECTED]MTA RMKI, ELTE TTK __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R survey package again
On 7/9/07 11:42 PM, eugen pircalabelu wrote: I have a sample from a survey where household were interviewed. The sample has 4 criteria on which the stratification was based: REGION, SIZE OF HOUSEHOLD, SIZE OF LOCALITY, AGE OF HEAD OF HOUSEHOLD. Since i don't have the whole information in each cell of the cross region*sizehh*sizeloc*age i can't use the postStratify function from Survey package. Is that correct? (I think so but i need a competent answer) The only additional info that i have is the size of a cell from a 2*2 crossing (eg: I know the population size for all the strata defined by region*sizehh, region*sizeloc, sizeloc*age) so i have the behaviour of the population but in a 2 by 2 cross for each of these criteria. You're right, poststratification can't work from two-way marginal distributions, but raking or calibration can. However it seems odd that you only have this much information, since the full joint distribution would have been needed for stratification. Usually these details would be documented as part of the sample design. Can you get this information from those responsible for the sample design? It would also be good to check your understanding of the design. A sampling frame listing details of household size and age of household head would have been needed to do the four-way stratification you mention, but in my experience such frames aren't very common. James -- James Reilly Department of Statistics, University of Auckland Private Bag 92019, Auckland, New Zealand __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matlab's lsqnonlin
I'm wondering about experiences: Do you know of cases where minpack.lm's nls.lm() solved a (real) problem that nls() would have a problem with ? In short, no. However, I looked at this question in the limited context of fitting the parameters of a linear superposition of 2 exponentials with Gaussian noise. A simulation study showed nearly identical performance for the range of parameter values/noise levels that are of practical interest to us. Are there problems for which steepest descent gets you in the neighborhood of a solution whereas GN does not? If such problems exist then there would be reason to apply LM instead of GN, but I don't know of any. Beware however -- one of the main things I learned about this field from Doug Bates, co-author of Bates_and_Watts and prinicipal author of S's and R's nls() : It's a *feature* that nls() does not converge sometimes when other methods do falsely claim convergence! Martin Maechler, ETH Zurich KateM KateM Katharine Mullen KateM mail: Department of Physics and Astronomy, Faculty of Sciences KateM Vrije Universiteit Amsterdam, de Boelelaan 1081 KateM 1081 HV Amsterdam, The Netherlands KateM room: T.1.06 KateM tel: +31 205987870 KateM fax: +31 205987992 KateM e-mail: [EMAIL PROTECTED] KateM homepage: http://www.nat.vu.nl/~kate/ KateM On Fri, 7 Sep 2007, Jose Luis Aznarte M. wrote: Hi! I'm translating some code from Matlab to R and I found a problem. I need to translate Matlab's function 'lsqnonlin' (http://www-ccs.ucsd.edu/matlab/toolbox/optim/lsqnonlin.html) into R, and at the beginning I thought it would be the same as R's 'optim'. But then I looked at the definition of 'lsqnonlin' and I don't quite see how to make 'optim' to do the same thing. Does anyone have an idea? This is apart from the fact that I would like to use the Levenberg Marquardt algorithm which is not implemented in R (some discussion about this: http://tolstoy.newcastle.edu.au/R/help/00b/2492.html). Thank you! All the best, -- -- Jose Luis Aznarte M. http://decsai.ugr.es/~jlaznarte Department of Computer Science and Artificial Intelligence Universidad de Granada Tel. +34 - 958 - 24 04 67 GRANADA (Spain) Fax: +34 - 958 - 24 00 79 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. KateM __ KateM R-help@stat.math.ethz.ch mailing list KateM https://stat.ethz.ch/mailman/listinfo/r-help KateM PLEASE do read the posting guide http://www.R-project.org/posting-guide.html KateM and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with color coded bar graph
Luis Naver wrote: I have a list of observations that are -1, 1 or 0. I would like to represent them in a horizontal bar color coded based on value like a stacked bar graph. I can achieve this in the form of a png with the following code: A = floor(runif(10)*3) - 1 png(width=100, height=10) par(mar=c(0,0,0,0)) image(matrix(A), col=grey(c(0.1, 0.5, 0.9))) dev.off() However I would like to do this with one of the standard plotting tools (i.e. barplot) to take advantage of labels and multiple series. Any help would be appreciated. Hi Luis, I understood your request as wanting a single horizontal bar with 10 segments, each colored according to the value of A. If this is correct, you might want: library(plotrix) plot(1,xlim=c(-1,1),ylim=c(-1,1),xlab=,ylab=,type=n,axes=FALSE) gradient.rect(-1,-0.1,1,0.1,col=grey(c(0.1,0.5,0.9))[A+2]) Jim __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] argument 'lib' is missing: using '/usr/lib/R/library'
On 9/8/07, Gabor Csardi [EMAIL PROTECTED] wrote: Paul, what is the question? If the question is why you get this warning message, the reason is that the 'lib' argument is missing and install.packages is using '/usr/lib/R/library'. If you want to get rid of the warning supply the 'lib' argument. Thanks, Gabor. Yes, I want to get rid of the warning message. Since the warning did not appear before, even without supplying the 'lib' argument, I thought that something was wrong here. Paul On Sat, Sep 08, 2007 at 11:26:44AM +0100, Paul Smith wrote: Dear All, When installing packages, I get the following warning: install.packages(sqldf) Warning in install.packages(sqldf) : argument 'lib' is missing: using '/usr/lib/R/library' Any ideas? The details of my R installation are: version _ platform i386-redhat-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 2 minor 5.1 year 2007 month 06 day27 svn rev42083 language R version.string R version 2.5.1 (2007-06-27) R is here installed on Fedora 7. Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Csardi Gabor [EMAIL PROTECTED]MTA RMKI, ELTE TTK __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] genoud problem
Hi Shubha, genoud does not return the initial fit value. But you could easily obtain it by passing your starting values to your function directly. Alternatively, one can have genoud print out the entire initial population (or the entire population as is evolves), and one can then decide to report whatever summary of this one would like. Note that the best fit in generation zero is printed by default. See the project.path and print.level options for details. Cheers, Jas. === Jasjeet S. Sekhon Associate Professor Travers Department of Political Science Survey Research Center UC Berkeley http://sekhon.berkeley.edu/ V: 510-642-9974 F: 617-507-5524 === Shubha Vishwanath Karanth writes: Hi R users, genoud function of rgenoud package will optimize my function. If opt = genoud(fn,2,max=TRUE,starting.value=c(1,10),) opt$value will give the optimized value of the function, fn. My problem is from the same opt, can I get the value of the function at the initial parameter values? I need the initial value of the function for reporting purposes. BR, Shubha [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running a PERL script from R
Dear Ken. You could also try and use RSPerl (http://www.omegahat.org/RSPerl/). It allows one to use R commands in Perl and vice-versa. regards, Paul Pierce, Ken schreef: Is there a way to run a simple perl script from R? Kenneth B. Pierce Jr. Research Ecologist Landscape Ecology, Modeling, Mapping and Analysis Team PNW Research Station - USDA-FS 3200 SW Jefferson Way, Corvallis, OR 97331 [EMAIL PROTECTED] 541 750-7393 http://www.fsl.orst.edu/lemma/gnnfire http://www.fsl.orst.edu/R_users/index.php [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Drs. Paul Hiemstra Department of Physical Geography Faculty of Geosciences University of Utrecht Heidelberglaan 2 P.O. Box 80.115 3508 TC Utrecht Phone: +31302535773 Fax:+31302531145 http://intamap.geo.uu.nl/~paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lisp-like primitives in R
On 9/8/07, Peter Dalgaard [EMAIL PROTECTED] wrote: François Pinard wrote: [Roland Rau] [François Pinard] I wonder what happened, for R to hide the underlying Scheme so fully, at least at the level of the surface language (despite there are hints). To further foster portability, we chose to write R in ANSI C Yes, of course. Scheme is also (often) implemented in C. I meant that R might have implemented a Scheme engine (or part of a Scheme engine, extended with appropriate data types) with a surface language (nearly the S language) which is purposely not Scheme, but could have been. If the gap is not extreme, one could dare dreaming that the Scheme engine in R be completed, and Scheme offered as an alternate extension language. If you allow me to continue dreaming awake -- they told me they will let me free as long as I do not get dangerous! :-) -- part of the interest lies in the fact there are excellent Scheme compilers. If we could only find or devise some kind of marriage between a mature Scheme and R, so to speed up the non-vectorisable parts of R scripts... Well, depending on what you want, this is either trivial or impossible... The internal storage of R is still pretty much equivalent to scheme. E.g. try this: r2scheme - function(e) if (!is.recursive(e)) deparse(e) else c((, unlist(lapply(as.list(e), r2scheme)), )) paste(r2scheme(quote(for(i in 1:4)print(i))), collapse= ) [1] ( for i ( : 1 4 ) ( print i ) ) Also see showTree in codetools: library(codetools) showTree(quote(for(i in 1:4)print(i))) (for i (: 1 4) (print i)) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running a PERL script from R
On 7 September 2007 at 14:04, Pierce, Ken wrote: | I've tried various configurations of .script, system and shell to no | avail. It seems to pause and run something but then no output is | created. Make sure you read the help page for system, and understand the options. Esp on Windows, you will need them to display/capture what is happening. Dirk | -Original Message- | From: Dirk Eddelbuettel [mailto:[EMAIL PROTECTED] | Sent: Friday, September 07, 2007 1:11 PM | To: Pierce, Ken | Cc: r-help | Subject: Re: [R] Running a PERL script from R | | On Fri, Sep 07, 2007 at 12:15:51PM -0700, Pierce, Ken wrote: | Is there a way to run a simple perl script from R? | | ?system | | Hth, Dirk | | -- | Three out of two people have difficulties with fractions. -- Three out of two people have difficulties with fractions. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] confusion matrix - better code?
Michael,
Thank you very much. My code is certainly put to shame by yours. I promise to
read about factor to see how you use it and why ;-))
I really appreciate your help.
Monica
Subject: RE: [R] confusion matrix - better code? Date: Fri, 7 Sep 2007
15:36:00 -0500 From: [EMAIL PROTECTED] To: [EMAIL PROTECTED] How about
this?
tab-table(factor(tr,levels=unique(tr)),factor(pr,levels=unique(tr)))
tab-rbind(tab,colSums(tab)) tab-cbind(tab,rowSums(tab)) tab 1 2 3 4
1 2 1 0 0 3 2 2 1 0 0 3 3 0 0 3 0 3 4 0 1 0 0 1 4 3 3 0 10 Of
course you can add some dimnames for the 5th row and 5th column if you
want. dimnames(tab)[[1]][5]-Total dimnames(tab)[[2]][5]-Total
tab 1 2 3 4 Total 1 2 1 0 0 3 2 2 1 0 0 3 3 0 0 3 0 3 4 0 1 0 0 1 Total
4 3 3 0 10 Michael Conklin Chief Methodologist - Advanced Analytics
MarketTools, Inc. 6465 Wayzata Blvd. Suite 170 Minneapolis, MN
55426 Tel: 952.417.4719 | Mobile:612.201.8978 [EMAIL PROTECTED]
MarketTools(r) http://www.markettools.comThis e-mail and any
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If you are not the intended recipient, be aware that any review, copying, or
distribution of this e-mail or any attachment is strictly prohibited. If you
have received this e-mail in error, please return it to the sender
immediately, and permanently delete the original and any copies from your
system. Thank you for your cooperation.-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Monica
Pisica Sent: Friday, September 07, 2007 2:53 PM To:
r-help@stat.math.ethz.ch Subject: [R] confusion matrix - better code?
Importance: High Hi, I've written some code to obtain a confusion
matrix when the true classification and the predicted classification are
known. Suppose true classification is called tr and predicted
classification is pr. I have 4 classes in tr, but only 3 classes out of 4
are predicted in pr. Following is my code, but looks quite clunky to me.
I wonder if you have any suggestions to improve it. Thanks, Monica
- tr - c(1,2,2,3,3,3,2,4,1,1)
pr-c(1,2,1,3,3,3,1,2,1,2) dat - data.frame(tr, pr) class -
c(1:length(tr)) m - max(c(length(unique(tr)), length(unique(pr for(i
in 1:length(class)) { class[i] - sub(' ','',paste(dat[i,1],dat[i,2])) }
dat - data.frame(dat, class) mat - matrix(0, nrow=m, ncol=m) for (i in
1:m){ for (j in 1:m){ mat[i,j] - sub(' ','',paste(i,j)) }} cat -
matrix(0, nrow=(m+1), ncol=(m+1)) for (i in 1:m){ for(j in 1:m){
cat[i,j]- nrow(dat[dat$class==mat[i,j],]) }} for (i in 1:m){
cat[(m+1),i]-sum(cat[1:m,i]) cat[i,(m+1)]- sum(cat[i,1:m])
cat[(m+1),(m+1)] - sum(cat[1:m,(m+1)]) } cat [,1] [,2] [,3] [,4] [,5]
[1,] 2 1 0 0 3 [2,] 2 1 0 0 3 [3,] 0 0 3 0 3 [4,] 0 1 0 0 1 [5,] 4 3 3 0
10 The 5th row / col represents the sum on each row / col respectively.
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Re: [R] Lisp-like primitives in R
[Peter Dalgaard] [François Pinard] I meant that R might have implemented a Scheme engine [...] with a surface language [...] which is purposely not Scheme, but could have been. [...] one could dare dreaming that the Scheme engine in R be completed, and Scheme offered as an alternate extension language. [...] there are excellent Scheme compilers. [...] Well, depending on what you want, this is either trivial or impossible... I'm more leaning on the impossible side :-). The internal storage of R is still pretty much equivalent to scheme. R needs a few supplementary data types, and it motivated the R authors into re-implementing their own Scheme engine instead of relying on an existing implementation of a Scheme system. r2scheme - function(e) [...] Nice exercise! :-) a parser that parses a similar language to R internal format is not a very hard exercise (some care needed in places). However, replacing the front-end is not going to make anything faster, Of course. The idea is nothing more than to please people starving to use Scheme instead of S as a surface language, here and there in scripts. I merely thought that if the gap is small enough (so to not require an extraordinary effort), it would be worth the leap. One immediate difficulty to foresee is the name clashes between R and RnRS. There might also be missing things in R (like continuations, say). To make anything faster, and this is a totally different idea, one might consider replacing the back-end, not the front-end. Writing good optimizing Scheme compilers is quite an undertaking, and if one only considers type inference (as a subproblem), this still is an active research area. The Scheme engine in R was written as to quickly get a working S (non-obstant lexical scoping and some library issues). My ramble was about switching this quick base of R to some solid Scheme implementation, than to re-address separately compiling issues for R. and the evaluation engine in R does a couple of tricks which are not done in Scheme, notably lazy evaluation, Promises? Aren't they already part of Scheme? The main difference I saw is their systematic use in R argument passing. All aspects of mere argument passing would require a lot of thought. As you wrote, variable scope is another difficulty. Offering a compatible C API, and library interface in general, might be a frightening but necessary challenge. It's all more of a dream than a thought, actually... :-) Look up the writings of Luke Tierney on the matter to learn more. Thanks for this interesting reference. -- François Pinard http://pinard.progiciels-bpi.ca __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R first.id last.id function error
Hi Jim,
Thanks for going to the trouble of writing the function. I'll copy and try it.
Gerard
At 06:30 PM 9/7/2007, jim holtman wrote:
This function should do it for you:
file1 - read.table(textConnection( id rx week dv1
+ 1 1 11 1
+ 2 1 12 1
+ 3 1 13 2
+ 4 2 11 3
+ 5 2 12 4
+ 6 2 13 1
+ 7 3 11 2
+ 8 3 12 3
+ 9 3 13 4
+ 10 4 11 2
+ 11 4 12 6
+ 12 4 13 5
+ 13 5 21 7
+ 14 5 22 8
+ 15 5 23 5
+ 16 6 21 2
+ 17 6 22 4
+ 18 6 23 6
+ 19 7 21 7
+ 20 7 22 8
+ 21 8 21 9
+ 22 9 21 4
+ 23 9 22 5), header=TRUE)
mark.function -
+ function(df){
+ df - df[order(df$id, df$week),]
+ # create 'diff' of 'id' to determine where the breaks are
+ breaks - diff(df$id)
+ # the first entry will be TRUE, and then every occurance of
non-zero in breaks
+ df$first.id - c(TRUE, breaks != 0)
+ # the last entry is TRUE and every non-zero breaks
+ df$last.id - c(breaks != 0, TRUE)
+ df
+ }
mark.function(file1)
id rx week dv1 first.id last.id
1 1 11 1 TRUE FALSE
2 1 12 1FALSE FALSE
3 1 13 2FALSETRUE
4 2 11 3 TRUE FALSE
5 2 12 4FALSE FALSE
6 2 13 1FALSETRUE
7 3 11 2 TRUE FALSE
8 3 12 3FALSE FALSE
9 3 13 4FALSETRUE
10 4 11 2 TRUE FALSE
11 4 12 6FALSE FALSE
12 4 13 5FALSETRUE
13 5 21 7 TRUE FALSE
14 5 22 8FALSE FALSE
15 5 23 5FALSETRUE
16 6 21 2 TRUE FALSE
17 6 22 4FALSE FALSE
18 6 23 6FALSETRUE
19 7 21 7 TRUE FALSE
20 7 22 8FALSETRUE
21 8 21 9 TRUETRUE
22 9 21 4 TRUE FALSE
23 9 22 5FALSETRUE
On 9/7/07, Gerard Smits [EMAIL PROTECTED] wrote:
Hi R users,
I have a test dataframe (file1, shown below) for which I am trying
to create a flag for the first and last ID record (equivalent to SAS
first.id and last.id variables.
Dump of file1:
file1
id rx week dv1
1 1 11 1
2 1 12 1
3 1 13 2
4 2 11 3
5 2 12 4
6 2 13 1
7 3 11 2
8 3 12 3
9 3 13 4
10 4 11 2
11 4 12 6
12 4 13 5
13 5 21 7
14 5 22 8
15 5 23 5
16 6 21 2
17 6 22 4
18 6 23 6
19 7 21 7
20 7 22 8
21 8 21 9
22 9 21 4
23 9 22 5
I have written code that correctly assigns the first.id and
last.id variabes:
require(Hmisc) #for Lags
#ascending order to define first dot
file1- file1[order(file1$id, file1$week),]
file1$first.id - (Lag(file1$id) != file1$id)
file1$first.id[1]-TRUE #force NA to TRUE
#descending order to define last dot
file1- file1[order(-file1$id,-file1$week),]
file1$last.id - (Lag(file1$id) != file1$id)
file1$last.id[1]-TRUE #force NA to TRUE
#resort to original order
file1- file1[order(file1$id,file1$week),]
I am now trying to get the above code to work as a function, and am
clearly doing something wrong:
first.last - function (df, idvar, sortvars1, sortvars2)
+ {
+ #sort in ascending order to define first dot
+ df- df[order(sortvars1),]
+ df$first.idvar - (Lag(df$idvar) != df$idvar)
+ #force first record NA to TRUE
+ df$first.idvar[1]-TRUE
+
+ #sort in descending order to define last dot
+ df- df[order(-sortvars2),]
+ df$last.idvar - (Lag(df$idvar) != df$idvar)
+ #force last record NA to TRUE
+ df$last.idvar[1]-TRUE
+
+ #resort to original order
+ df- df[order(sortvars1),]
+ }
Function call:
first.last(df=file1, idvar=file1$id,
sortvars1=c(file1$id,file1$week), sortvars2=c(-file1$id,-file1$week))
R Error:
Error in as.vector(x, mode) : invalid argument 'mode'
I am not sure about the passing of the sort strings. Perhaps this is
were things are off. Any help greatly appreciated.
Thanks,
Gerard
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What is the problem you are trying to solve?
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Re: [R] Suppress 'x' when appending to a csv file
On Sat, 2007-09-08 at 11:41 -0400, John Kane wrote: Is there any convenient way to supress the x that appears in csv export files? I would like to be able to export a file and add a comment to it yet still be able to read it back into R. I don't see any way to get rid of the x that seperates the different appended parts. Thanks EXAMPLE x 1 2 3 4 5 x #Results from file SSS.r R.2.5.1 Windows XP John, Try this: write.table(head(iris), sep = ,, row.names = FALSE, col.names = FALSE) 5.1,3.5,1.4,0.2,setosa 4.9,3,1.4,0.2,setosa 4.7,3.2,1.3,0.2,setosa 4.6,3.1,1.5,0.2,setosa 5,3.6,1.4,0.2,setosa 5.4,3.9,1.7,0.4,setosa You may need to explicitly adjust other arguments that are otherwise set to defaults when using write.csv(). HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using clustering functions
uv wrote: Hi. I need to use a few different clustering functions. I managed to run the kmeans() one which is in my stats library, but I can't use any function, such as agnes(), that is in my cluster library. Any idea how to access other libraries? Thanks! Both stats and cluster are packages, not libraries. You need to call library(cluster) in order to load package cluster from your library before using agnes() and her friends. 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem in installing packages on linux machine...
Hi, I am trying to install RSQLite package on my Fedora workstation. I tried to install other packages as well, but each time I got the same error messages saying compilation error and non zero exit status. Do I have to specify lib=? I never specified the library path before when I was using Fedora Core 6. Warning in install.packages(RSQLite) : argument 'lib' is missing: using '/usr/lib/R/library' --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done trying URL 'http://cran.cnr.Berkeley.edu/src/contrib/RSQLite_0.5-6.tar.gz' Content type 'application/x-gzip' length 710241 bytes opened URL == downloaded 693Kb gcc -std=gnu99 -shared -L/usr/local/lib -o RSQLite.so RS-DBI.o RS-SQLite.o sqlite-all.o -L/usr/lib/R/lib -lR /usr/bin/ld: skipping incompatible /usr/lib/R/lib/libR.so when searching for -lR /usr/bin/ld: cannot find -lR collect2: ld returned 1 exit status make: *** [RSQLite.so] Error 1 chmod: cannot access `/usr/lib/R/library/RSQLite/libs/*': No such file or directory ERROR: compilation failed for package 'RSQLite' ** Removing '/usr/lib/R/library/RSQLite' The downloaded packages are in /tmp/RtmpHQ5Y7C/downloaded_packages Warning message: installation of package 'RSQLite' had non-zero exit status in: install.packages(RSQLite) I appreciate your help. Thank you very much Taka __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem in installing packages on linux machine...
Hi, try install packages whit 'sudo'. $sudo R -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 08/09/2007, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Hi, I am trying to install RSQLite package on my Fedora workstation. I tried to install other packages as well, but each time I got the same error messages saying compilation error and non zero exit status. Do I have to specify lib=? I never specified the library path before when I was using Fedora Core 6. Warning in install.packages(RSQLite) : argument 'lib' is missing: using '/usr/lib/R/library' --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done trying URL 'http://cran.cnr.Berkeley.edu/src/contrib/RSQLite_0.5-6.tar.gz' Content type 'application/x-gzip' length 710241 bytes opened URL == downloaded 693Kb gcc -std=gnu99 -shared -L/usr/local/lib -o RSQLite.so RS-DBI.o RS-SQLite.o sqlite-all.o -L/usr/lib/R/lib -lR /usr/bin/ld: skipping incompatible /usr/lib/R/lib/libR.so when searching for -lR /usr/bin/ld: cannot find -lR collect2: ld returned 1 exit status make: *** [RSQLite.so] Error 1 chmod: cannot access `/usr/lib/R/library/RSQLite/libs/*': No such file or directory ERROR: compilation failed for package 'RSQLite' ** Removing '/usr/lib/R/library/RSQLite' The downloaded packages are in /tmp/RtmpHQ5Y7C/downloaded_packages Warning message: installation of package 'RSQLite' had non-zero exit status in: install.packages(RSQLite) I appreciate your help. Thank you very much Taka __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem in installing packages on linux machine...
Hi, Still got the same error message. I did su R when I got the error message for the first time. I have never seen this error message. I will be googling for solutions as well... Thank you. Quoting Henrique Dallazuanna [EMAIL PROTECTED]: Hi, try install packages whit 'sudo'. $sudo R -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 08/09/2007, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Hi, I am trying to install RSQLite package on my Fedora workstation. I tried to install other packages as well, but each time I got the same error messages saying compilation error and non zero exit status. Do I have to specify lib=? I never specified the library path before when I was using Fedora Core 6. Warning in install.packages(RSQLite) : argument 'lib' is missing: using '/usr/lib/R/library' --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done trying URL 'http://cran.cnr.Berkeley.edu/src/contrib/RSQLite_0.5-6.tar.gz' Content type 'application/x-gzip' length 710241 bytes opened URL == downloaded 693Kb gcc -std=gnu99 -shared -L/usr/local/lib -o RSQLite.so RS-DBI.o RS-SQLite.o sqlite-all.o -L/usr/lib/R/lib -lR /usr/bin/ld: skipping incompatible /usr/lib/R/lib/libR.so when searching for -lR /usr/bin/ld: cannot find -lR collect2: ld returned 1 exit status make: *** [RSQLite.so] Error 1 chmod: cannot access `/usr/lib/R/library/RSQLite/libs/*': No such file or directory ERROR: compilation failed for package 'RSQLite' ** Removing '/usr/lib/R/library/RSQLite' The downloaded packages are in /tmp/RtmpHQ5Y7C/downloaded_packages Warning message: installation of package 'RSQLite' had non-zero exit status in: install.packages(RSQLite) I appreciate your help. Thank you very much Taka __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Takatsugu Kobayashi PhD Student Indiana University, Dept. Geography __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Suppress 'x' when appending to a csv file
Thanks Marc. It works. I had not thought of using col.names = FALSE as I wanted to keep the colnames. I see that I will just have to do another write.table command to do this. Humm, actually it took a a bit of juggling to do the names but it's looking fine now. --- Marc Schwartz [EMAIL PROTECTED] wrote: On Sat, 2007-09-08 at 11:41 -0400, John Kane wrote: Is there any convenient way to supress the x that appears in csv export files? I would like to be able to export a file and add a comment to it yet still be able to read it back into R. I don't see any way to get rid of the x that seperates the different appended parts. Thanks EXAMPLE x 1 2 3 4 5 x #Results from file SSS.r R.2.5.1 Windows XP John, Try this: write.table(head(iris), sep = ,, row.names = FALSE, col.names = FALSE) 5.1,3.5,1.4,0.2,setosa 4.9,3,1.4,0.2,setosa 4.7,3.2,1.3,0.2,setosa 4.6,3.1,1.5,0.2,setosa 5,3.6,1.4,0.2,setosa 5.4,3.9,1.7,0.4,setosa You may need to explicitly adjust other arguments that are otherwise set to defaults when using write.csv(). HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 'initial value not feasible' in constrOptim
Hi, On Sep 7, 2007, at 11:00 PM, Yuchen Luo wrote: constrOptim(c(0.5,0.3,0.5), fit.error, fit.error.grr, ui=-1*ui,ci=-1*ci) and I am confronted with error message initial value not feasible I plug in the initial value of (0.5,0.3,0.5) to function fit.error and fit.error.grr and have pretty reasonable result. That doesn't mean the constraint is satisfied, which depends on the values of ui and ci but they are not provided so it's hard to tell where the problem is. hth, Ingmar I inequality ui %*% theta - ci = 0 as suggested in the R manual and it is satisfied. In case that this is a typo of the manual, I let ui=-ui and ci=-ci and try constrOptim again but the same warning message pops up. Could you please point me a way out of this? I am actually trying to translate a fortran code to R and the function I want to replace is DBCPOL, which used 'the complex method described in Nelder and Mead (1965) and Gill et al. (1981). I believe contrOptim is better than it because it is newer, is it? Best Wishes Yuchen Luo [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] 'singular gradient matrix? when using nls() and how to make the program skip nls( ) and run on
On Wed, Sep 05, 2007 at 04:43:19PM -0700, Yuchen Luo wrote: Dear friends. I use nls() and encounter the following puzzling problem: I have a function f(a,b,c,x), I have a data vector of x and a vectory y of realized value of f. Case1 I tried to estimate c with (a=0.3, b=0.5) fixed: nls(y~f(a,b,c,x), control=list(maxiter = 10, minFactor=0.5 ^2048),start=list(c=0.5)). The error message is: number of iterations exceeded maximum of 10 Case2 I then think maybe the value of a and be are not reasonable. So, I let nls() estimate (a,b,c) altogether: nls(y~f(a,b,c,x), control=list(maxiter = 10, minFactor=0.5 ^2048),start=list(a=0.3,b=0.5,c=0.5)). The error message is: singular gradient matrix at initial parameter estimates. This is what puzzles me, if the initial parameter of (a=0.3,b=0.5,c=0.5) can create 'singular gradient matrix', then why doesn't this 'singular gradient matrix' appear in Case1? I have tried to change the initial value of (a,b,c) around but the problem persists. I am wondering if there is a way out. My another question is, I need to run 220 of nls() in my program with different y and x. When one of the nls() encounter a problem, the whole program stops. In my case, the 3rd nls() runs into a problem. I would still need the program to run the remaining 217 nls( )! Is there a way to make the program skip the problematic nls() and complete the ramaining nls()'s? ?try Your help will be highly appreciated! Yuchen Luo [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] negative value for AIC and BIC
Oliver,
I am attaching an HTML document in which I have plotted -2Log(x) vs. x. If you
examine the plot you will see that -2Log(x) can be negative. Since -2Log(x) is
part of AIC and BIC, AIC and BIC can be negative.
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
Baltimore VA Medical Center GRECC,
University of Maryland School of Medicine Claude D. Pepper OAIC,
University of Maryland Clinical Nutrition Research Unit, and
Baltimore VA Center Stroke of Excellence
University of Maryland School of Medicine
Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
[EMAIL PROTECTED]
Hannu Kahra [EMAIL PROTECTED] 09/07/07 4:32 AM
Olivier,
type ?AIC and have a look at the description
Description:
Generic function calculating the Akaike information criterion for
one or several fitted model objects for which a log-likelihood
value can be obtained, according to the formula -2*log-likelihood
+ k*npar, where npar represents the number of parameters in the
fitted model, and k = 2 for the usual AIC, or k = log(n) (n the
number of observations) for the so-called BIC or SBC (Schwarz's
Bayesian criterion).
AIC = -2*log-likelihood + k*npar can be negative as SBC, too.
Hannu
On 9/7/07, Olivier MARTIN [EMAIL PROTECTED] wrote:
Hi all,
I obtained negative values for AIC and BIC criteria for a particular
model that I have
developped...
I don't remember to have negative values for these crietria for others
applications, so I am a
little suprised... Could anyone tell me if something is wrong or his
conclusion concerning my model?
Best regards,
Olivier.
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y = -2Log[x]
Out[20]=
-2 Log[x]
In[22]:=
RowBox[{Plot, [, RowBox[{y, ,, RowBox[{{, RowBox[{x, ,, 0.01, ,, 4}],
}}], ,, AxesLabel#62754; {x, -2Log[x]}}], ]}]
[Graphics:HTMLFiles/-2Log_10.gif]
Out[22]=
-Graphics-
__
Created by [1]Mathematica (September 7, 2007)
References
1. http://www.wolfram.com/
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Re: [R] Running geeglm unstructured corstr
Dear Niccolo, a) your example program works for me and on a Pentium 4, 3 Ghz, 1GB ram, The model fit uses user system elapsed 146.920.76 143.32 library(geepack) set.seed(299) header - rep.int(seq(1:615),sample(seq(1:19),size=615,replace=T)) inr - rlnorm(length(header),0.8434359,0.3268392) group_cod - sample(c(0,1),size=length(header),replace=T) inside - sample(c(0,1),size=length(header),replace=T) gee.frame - data.frame(header,inr,group_cod,inside) geeglm.model- geeglm(inside~group_cod,family=binomial, data=gee.frame ,id=header,corstr=unstructured) Version of_ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 5.1 year 2007 month 06 day27 svn rev42083 language R version.string R version 2.5.1 (2007-06-27) b) for costr= ar1 In my dataset I also have a variable weeks, that I use to specify waves in the geeglm. When I use this, the output for an autoregressive gee model gives me only the estimates, when I don't use it everything's alright. The problem's that I have unbalanced observations, so the use of waves could take this into account while estimating parameters. Is this a conflict between these options, or what? The argument waves should work in conjunction with the ar1 working correlation. Make sure that the observations are orderd wrt. cluster-id as it is said in ?geeglm for the 'id' argument c) if you build your own fixed-working correlation then there are two issues: 1) you must leave out clusters of size 1 in the construction of the zcor matrix: if the other clusters have the same size you can use geeglm with that zcor matrix Example: # using corstr=fixed the cluster with only 1 observation must be # left out from the construction of the zcor-vector. # the following example is for the case where clusters of size larger then 1 # have no missing observations and are of the same size library(geepack) data(seizure) ## Diggle, Liang, and Zeger (1994) pp166-168, compare Table 8.10 seiz.l - reshape(seizure, varying=list(c(base,y1, y2, y3, y4)), v.names=y, times=0:4, direction=long) seiz.l - seiz.l[order(seiz.l$id, seiz.l$time),] seiz.l$t - ifelse(seiz.l$time == 0, 8, 2) seiz.l$x - ifelse(seiz.l$time == 0, 0, 1) ## defining fixed correlation matrix cor.fixed - matrix(c(1, 0.5, 0.25, 0.125, 0.125, 0.5, 1, 0.25, 0.125, 0.125, 0.25, 0.25, 1, 0.5, 0.125, 0.125, 0.125, 0.5, 1, 0.125, 0.125, 0.125, 0.125, 0.125, 1), 5, 5) zcor - rep(cor.fixed[lower.tri(cor.fixed)], 59) g1 - geeglm(y ~ offset(log(t)) + x + trt + x:trt, id = id, data = seiz.l, family = poisson, corstr = fixed, zcor = zcor) #reducing clusters 1 ,3 and 58 to only one observation seiz.reduc-subset(seiz.l,!( (id==1 | id==3 | id==58) time0)) # zcor is only constructed for clusters with size larger than 1 n.larger.one-sum(table(seiz.reduc$id)1) zcor -c( rep(cor.fixed[lower.tri(cor.fixed)], n.larger.one)) g2- geeglm(y ~ offset(log(t)) + x + trt + x:trt, id = id, data = seiz.reduc, family = poisson, corstr = fixed, zcor = zcor) 2) if the clusters have different size than you must construct the zcor-matrix according to the missing structure, the usage of the wave-argument to specify the missing structure does not work. Example: # using corstr=fixed the for clusters with unbalanced number of observations # the wave argument does not work, the zcor-matrix must be constructed 'by-hand' library(geepack) data(seizure) ## Diggle, Liang, and Zeger (1994) pp166-168, compare Table 8.10 seiz.l - reshape(seizure, varying=list(c(base,y1, y2, y3, y4)), v.names=y, times=0:4, direction=long) seiz.l - seiz.l[order(seiz.l$id, seiz.l$time),] seiz.l$t - ifelse(seiz.l$time == 0, 8, 2) seiz.l$x - ifelse(seiz.l$time == 0, 0, 1) # transform time such that the initial time is 1 seiz.l$time-seiz.l$time+1 #taking only a subset of the data such the data are unbalanced wrt cluster -size set.seed(88) seiz-subset(seiz.l,!( (id==1 | id==3 | id==58) time2)) ## Construction of a fixed correlation matrix cor.fixed - matrix(c(1, 0.7, 0.5, 0.25, 0.12, 0.7, 1, 0.71, 0.125, 0.125, 0.5, 0.71, 1, 0.29, 0.123, 0.25, 0.125, 0.29, 1, 0.119, 0.120, 0.125, 0.123, 0.119, 1), 5, 5) # The zcor-vector is constructed only for times which exist in the data
Re: [R] SQL like function?
Others have already pointed out %in% but regarding your comment about SQL, you can use SQL to manipulate R data frames using the sqldf package which provides an interface to lower level RSQLite (and RMySQL in the future) routines. The following examples use SQLite underneath: DF - data.frame(observation = c(1,2,3,4,5)) ID - data.frame(ID = c(1, 3, 4)) library(sqldf) sqldf(select observation, observation in (select * from ID) `ID?` from DF) # or sqldf(select observation, observation in (1, 3, 4) `ID?` from DF) See home page at: http://sqldf.googlecode.com On 9/7/07, Takatsugu Kobayashi [EMAIL PROTECTED] wrote: Hi RUsers, I am wonder if I can search observations whose IDs matches any of the values in another vector, such as in MySQL. While I am learing MySQL for future database management, I appreciate if anyone could give me a hint. Suppose I have one 5*1 vector containing observation IDs and frequencies, and one 3*1 vector containing observation IDs. observation-c(1,2,3,4,5) ID-c(1,3,4) Then, I would like to program a code that returns a results showing matched observations like result: TRUE FALSE TRUE TRUE FALSE I am reading S programming, but I cannot find a way to do this. Thank you very much. Taka __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] text in boxplots
Hi Greg, Actually, I'm using the scatterplot function in the car namespace. My understanding is that the scatterplot is turned into a boxplot when the x variable is a factor, but I can't see that in the code of the function. I've successfully modified the function to output regression stats (R^^2, etc.) with the method 'text', but I can't seem to get any text printed when the result of a call to scatterplot is a bunch of boxplots. I don't know how boxplots end up being generated (I like it though) in a call to car:scatterplot. The same data in ggobi would show the standard scatterplot with data aligned vertically in lines along the values of the x (factor) variable but car:scatterplot draws nice boxes around the data as a bonus. I guess I'll need to follow with some debugging tool. I find it odd that mtext works though. Or maybe some graphics parameter is set such that text doesn't show. Thanx Greg. -- View this message in context: http://www.nabble.com/text-in-boxplots-tf4394528.html#a12553104 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help on replacing values
Your columns are factors, not character strings. Use as.is = TRUE as an argument to read.table. Also its a bit dangerous to use T although not wrong. Its safer to use TRUE. On 9/7/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Dear List, I have a newbie question. I have read in a data.frame as follows: data = read.table(table.txt, header = T) data X1 X2 X3 X4 A AB AC AB AC B AB AC AA AB C AA AB AA AB D AA AB AB AC E AB AA AA AB F AB AA AB AC B AB AC AB AA I would like to replace AA values by BB in column X2. I have tried using replace() with no success, although I am not sure this is the right function. This is the code I have used: data$X2 - replace(data$X2, data$X2 ==AA,BB) Warning message: invalid factor level, NAs generated in: `[-.factor`(`*tmp*`, list, value = BB) What is wrong with the code? How can I get this done? how about changing AA values by BB in all 4 columns simultaneously? Actually this is a small example dataframe, the real one would have about 1000 columns. Extendind this, I found a similar thread dated July 2006 that used replace() on iris dataset, but I have tried reproducing it obtaining same warning message iris$Species - replace(iris$Species, iris$Species == setosa,NewName) Warning message: invalid factor level, NAs generated in: `[-.factor`(`*tmp*`, list, value = NewName) Thanks in advance your help, David __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] variable format
A matrix is for situations where every element is of the same class but your columns have different classes so use a data frame: DF - data.frame(a = 11:15, b = letters[1:5], stringsAsFactors = FALSE) subset(DF, a %in% 11:13) subset(DF, a %in% c(0, 11:13)) # same Suggest you review the Introduction to R manual and look at ?data.frame, ?subset and ?%in% On 9/4/07, Cory Nissen [EMAIL PROTECTED] wrote: Okay, I want to do something similar to SAS proc format. I usually do this... a - NULL a$divisionOld - c(1,2,3,4,5) divisionTable - matrix(c(1, New England, 2, Middle Atlantic, 3, East North Central, 4, West North Central, 5, South Atlantic), ncol=2, byrow=T) a$divisionNew[match(a$divisionOld, divisionTable[,1])] - divisionTable[,2] But how do I handle the case where... a$divisionOld - c(0,1,2,3,4,5) #no format available for 0, this throws an error. OR divisionTable - matrix(c(1, New England, 2, Middle Atlantic, 3, East North Central, 4, West North Central, 5, South Atlantic, 6, East South Central, 7, West South Central, 8, Mountain, 9, Pacific), ncol=2, byrow=T) There are extra formats available... this throws a warning. Thanks Cory [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Finding convex hull? [Broadcast]
From: Dong-hyun Oh
Dear UseRs,
I would like to know which function is the most efficient in finding
convex hull of points in 3(or 2)-dimensional case?
Functions for finding convex hull is the following:
convex.hull (tripack), chull (grDevices), in.chull (sgeostat),
convhulln (geometry), convexhull.xy (spatstat), calcConvexHull
(PBSmapping).
I also would like to know if there is a function that can be
used for
finding convex hull in multi-dimensional case, that is more than 3-
dimension.
If you had look a bit more carefully, you should have seen that
convhulln (geometry) will handle more than 3 dimensions.
Andy
Thank you in advance.
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Re: [R] ploting missing data
Try this: library(zoo) plot(na.approx(zoo(as.matrix(data[-1]), data[,1])), plot.type = single) See ?na.approx, ?plot.zoo, ?xyplot.zoo and vignette(zoo) On 9/7/07, Markus Schmidberger [EMAIL PROTECTED] wrote: Hello, I have this kind of dataframe and have to plot it. data - data.frame(sw= c(1,2,3,4,5,6,7,8,9,10,11,12,15), zehn = c(33.44,20.67,18.20,18.19,17.89,19.65,20.05,19.87,20.55,22.53,NA,NA,NA), zwanzig = c(61.42,NA,26.60,23.28,NA,24.90,24.47,24.53,26.41,28.26,NA,29.80,35.49), fuenfzig = c(162.51,66.08,49.55,43.40,NA,37.77,35.53,36.46,37.25,37.66,NA,42.29,47.80) ) The plot should have lines: lines(fuenfzig~sw, data=data) lines(zwanzig~sw, data=data) But now I have holes in my lines for the missing values (NA). How to plot the lines without the holes? The missing values should be interpolated or the left and right point directly connected. The function approx interpolates the whole dataset. Thats not my goal! Is there no plotting function to do this directly? Best Markus -- Dipl.-Tech. Math. Markus Schmidberger Ludwig-Maximilians-Universität München IBE - Institut für medizinische Informationsverarbeitung, Biometrie und Epidemiologie Marchioninistr. 15, D-81377 Muenchen URL: http://ibe.web.med.uni-muenchen.de Mail: Markus.Schmidberger [at] ibe.med.uni-muenchen.de __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help on replacing values
Thanks a lot Gabor, that was very helpful. All sorted now! Best David Your columns are factors, not character strings. Use as.is = TRUE as an argument to read.table. Also its a bit dangerous to use T although not wrong. Its safer to use TRUE. On 9/7/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Dear List, I have a newbie question. I have read in a data.frame as follows: data = read.table(table.txt, header = T) data X1 X2 X3 X4 A AB AC AB AC B AB AC AA AB C AA AB AA AB D AA AB AB AC E AB AA AA AB F AB AA AB AC B AB AC AB AA I would like to replace AA values by BB in column X2. I have tried using replace() with no success, although I am not sure this is the right function. This is the code I have used: data$X2 - replace(data$X2, data$X2 ==AA,BB) Warning message: invalid factor level, NAs generated in: `[-.factor`(`*tmp*`, list, value = BB) What is wrong with the code? How can I get this done? how about changing AA values by BB in all 4 columns simultaneously? Actually this is a small example dataframe, the real one would have about 1000 columns. Extendind this, I found a similar thread dated July 2006 that used replace() on iris dataset, but I have tried reproducing it obtaining same warning message iris$Species - replace(iris$Species, iris$Species == setosa,NewName) Warning message: invalid factor level, NAs generated in: `[-.factor`(`*tmp*`, list, value = NewName) Thanks in advance your help, David __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] text in boxplots
The posting guide (a link at the bottom of every e-mail from the list)
suggests including a small example of what you are trying to do.
Without an example, we have to guess at what you are trying to do.
I am still guessing since you still did not include an example. It
appears that part of the problem may be that the scatterplot function
from the car package does not leave the user coordinate system in a
state that matches intuitively with the data. Your call to text is
probably placing the text outside of the plotting region where it is not
visible (possibly even in someone elses cubicle/office). The mtext
function is not affected as much by this since it uses a different
coordinate system (based on the margins rather than the user
coordinates).
To place text on the plot, you need to find the cooresponding
coordinates. There are a few different ways to do this:
1. use par('usr') to find out what the current user coordinates are and
compute your coordinates based on that.
2. use par(usr=c(...)) to set the user coordinates to match where you
want to plot.
3. use the locator function to interactively choose the locations and
use the resulting coordinates.
4. use the cnvrt.coords function from the TeachingDemos package to
convert from figure or device coordinates to the current user coordinate
system.
5. contact the author of the function and ask him to set the user
coordinates to something more intuitive before exiting and wait for the
change.
Look at the help for par for more information, specifically look at the
'usr' entry and possibly the 'xpd' entry if you want to add text to the
marginal boxplots.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Yves Moisan
Sent: Friday, September 07, 2007 5:05 AM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] text in boxplots
Hi Greg,
Actually, I'm using the scatterplot function in the car
namespace. My understanding is that the scatterplot is
turned into a boxplot when the x variable is a factor, but I
can't see that in the code of the function.
I've successfully modified the function to output regression
stats (R^^2,
etc.) with the method 'text', but I can't seem to get any
text printed when the result of a call to scatterplot is a
bunch of boxplots.
I don't know how boxplots end up being generated (I like it
though) in a call to car:scatterplot. The same data in ggobi
would show the standard
scatterplot with data aligned vertically in lines along the
values of the x
(factor) variable but car:scatterplot draws nice boxes around
the data as a bonus. I guess I'll need to follow with some
debugging tool. I find it odd that mtext works though. Or
maybe some graphics parameter is set such that text doesn't show.
Thanx Greg.
--
View this message in context:
http://www.nabble.com/text-in-boxplots-tf4394528.html#a12553104
Sent from the R help mailing list archive at Nabble.com.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] ploting missing data
Markus Schmidberger wrote: Hello, I have this kind of dataframe and have to plot it. data - data.frame(sw= c(1,2,3,4,5,6,7,8,9,10,11,12,15), zehn = c(33.44,20.67,18.20,18.19,17.89,19.65,20.05,19.87,20.55,22.53,NA,NA,NA), zwanzig = c(61.42,NA,26.60,23.28,NA,24.90,24.47,24.53,26.41,28.26,NA,29.80,35.49), fuenfzig = c(162.51,66.08,49.55,43.40,NA,37.77,35.53,36.46,37.25,37.66,NA,42.29,47.80) ) The plot should have lines: lines(fuenfzig~sw, data=data) lines(zwanzig~sw, data=data) But now I have holes in my lines for the missing values (NA). How to plot the lines without the holes? The missing values should be interpolated or the left and right point directly connected. The function approx interpolates the whole dataset. Thats not my goal! Is there no plotting function to do this directly? Just get rid of the NAs: lines(fuenfzig~sw, data=data, subset=!is.na(fuenfzig)) -- O__ Peter Dalgaard Øster 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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] computing distance in miles or km between 2 street
Phil, If you have only a list of addresses and nothing else - i have to recognize i am lost. But if you have more info you can use a GIS software and it will calculate your distance automatically for all your addresses. To have your distance in miles or Km you need to have a projection in your data which will establish your units. The very basic principle is that for each address you have a set of coordinates, x, y and you actually use these to calculate distances. If your file is for exaple a point shape file (standard for ESRI ArcGIS products) you can very simply perform this task using either shortest driving route or as the crow flies - again depending on what other info you have in the file. A shape file is actually a group of files with with data attributes, projection info and so on - so even if you don't see xy coordinates, they actually are imbeded in the file. I suspect Mapquest or any other mapping web product has a database with all those addresses with xy coordinates. The user needs to know only the address and behind doors there is an sql process which selects particular addresses together with all the other attributes attached to it and calculate distances because it knows coordinates, a network of roads and so on. I am not sure this helps but at least maybe gives you some ideas where to look next. Monica Message: 53Date: Thu, 06 Sep 2007 14:42:32 -0400From: Philip James Smith [EMAIL PROTECTED]Subject: [R] computing distance in miles or km between 2 streetaddressesTo: [EMAIL PROTECTED]: [EMAIL PROTECTED]Content-Type: text/plain; charset=ISO-8859-1; format=flowed Hi R-ers: I need to compute the distance between 2 street addresses in either km or miles. I do not care if the distance is a shortest driving route or if it is as the crow flies. Does anybody know how to do this? Can it be done in R? I have thousands of addresses, so I think that Mapquest is out of the question! Please rely to: [EMAIL PROTECTED] Thank you!Phil Smith _ Hotmail®. NOW with 5GB storage. ration_HM_mini_5G_0907 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] text in boxplots
6. set the reset.par argument to FALSE in calling scatterplot as it states in the documentation for scatterplot. (still look at the help on 'xpd' argment to par if you want to add text to the margins). Hi Greg, That seems to have done it partly. I set it to false once and then I could see some text afterwards, even with reset.par set to TRUE. About my not following posting guidelines, I'm sorry but I thought my question didn't require me to post code. I suspected graphics parameter to be off, so the question was why a difference between mtext and text. Also, I mentioned earlier that I had succeeded in writing text to the plot for regression stats (in the reg function of scatterplot.formula) where I did use what seem to be the normal tricks [e.g. par(usr)]. I couldn't figure out why setting new graph variables in the main scatterplot.formula function in the form of usrGlobal - par(usr) then calling text(usrGlobal[2]-50,usrGlobal[4]-50,paste(Moyenne = ,r ...) did not show any text whereas it did in the reg function. But of course, pasting code could have been shorter for people to read ;-). Thanx for your help. -- View this message in context: http://www.nabble.com/text-in-boxplots-tf4394528.html#a12558160 Sent from the R help mailing list archive at Nabble.com. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] FW: variable format
Dear Cory, I am not familiar with SAS, but is this what you are looking for? divisionTable - matrix(c(1, New England, 2, Middle Atlantic, 3, East North Central, 4, West North Central, 5, South Atlantic, 6, East South Central, 7, West South Central, 8, Mountain, 9, Pacific), ncol=2, byrow=T) a - NULL a$divisionOld - c(0,1,2,3,4,5) a$divisionNew - as.character(factor(a$divisionOld,levels=divisionTable[,1],labels=divisionTable[,2])) a$divisionNew [1] NA New EnglandMiddle Atlantic [4] East North Central West North Central South Atlantic Kind regards, Martin Cory Nissen schrieb: Anybody? From: Cory Nissen Sent: Tue 9/4/2007 9:30 AM To: r-help@stat.math.ethz.ch Subject: variable format Okay, I want to do something similar to SAS proc format. I usually do this... a - NULL a$divisionOld - c(1,2,3,4,5) divisionTable - matrix(c(1, New England, 2, Middle Atlantic, 3, East North Central, 4, West North Central, 5, South Atlantic), ncol=2, byrow=T) a$divisionNew[match(a$divisionOld, divisionTable[,1])] - divisionTable[,2] But how do I handle the case where... a$divisionOld - c(0,1,2,3,4,5) #no format available for 0, this throws an error. OR divisionTable - matrix(c(1, New England, 2, Middle Atlantic, 3, East North Central, 4, West North Central, 5, South Atlantic, 6, East South Central, 7, West South Central, 8, Mountain, 9, Pacific), ncol=2, byrow=T) There are extra formats available... this throws a warning. Thanks Cory [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] text in boxplots
Looking at the help for scatterplot, it looks like there is a 6th option
(which is probably the preferred method):
6. set the reset.par argument to FALSE in calling scatterplot as it
states in the documentation for scatterplot. (still look at the help on
'xpd' argment to par if you want to add text to the margins).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Greg Snow
Sent: Friday, September 07, 2007 9:01 AM
To: Yves Moisan; r-help@stat.math.ethz.ch
Subject: Re: [R] text in boxplots
The posting guide (a link at the bottom of every e-mail from
the list) suggests including a small example of what you are
trying to do.
Without an example, we have to guess at what you are trying to do.
I am still guessing since you still did not include an
example. It appears that part of the problem may be that the
scatterplot function from the car package does not leave the
user coordinate system in a state that matches intuitively
with the data. Your call to text is probably placing the
text outside of the plotting region where it is not visible
(possibly even in someone elses cubicle/office). The mtext
function is not affected as much by this since it uses a
different coordinate system (based on the margins rather than
the user coordinates).
To place text on the plot, you need to find the cooresponding
coordinates. There are a few different ways to do this:
1. use par('usr') to find out what the current user
coordinates are and compute your coordinates based on that.
2. use par(usr=c(...)) to set the user coordinates to match
where you want to plot.
3. use the locator function to interactively choose the
locations and use the resulting coordinates.
4. use the cnvrt.coords function from the TeachingDemos
package to convert from figure or device coordinates to the
current user coordinate system.
5. contact the author of the function and ask him to set the
user coordinates to something more intuitive before exiting
and wait for the change.
Look at the help for par for more information, specifically
look at the 'usr' entry and possibly the 'xpd' entry if you
want to add text to the marginal boxplots.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Yves Moisan
Sent: Friday, September 07, 2007 5:05 AM
To: r-help@stat.math.ethz.ch
Subject: Re: [R] text in boxplots
Hi Greg,
Actually, I'm using the scatterplot function in the car
namespace. My
understanding is that the scatterplot is turned into a boxplot when
the x variable is a factor, but I can't see that in the code of the
function.
I've successfully modified the function to output regression stats
(R^^2,
etc.) with the method 'text', but I can't seem to get any
text printed
when the result of a call to scatterplot is a bunch of boxplots.
I don't know how boxplots end up being generated (I like it
though) in a call to car:scatterplot. The same data in ggobi would
show the standard
scatterplot with data aligned vertically in lines along the
values of
the x
(factor) variable but car:scatterplot draws nice boxes
around the data
as a bonus. I guess I'll need to follow with some
debugging tool. I
find it odd that mtext works though. Or maybe some
graphics parameter
is set such that text doesn't show.
Thanx Greg.
--
View this message in context:
http://www.nabble.com/text-in-boxplots-tf4394528.html#a12553104
Sent from the R help mailing list archive at Nabble.com.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Delete query in sqldf?
Yes but delete does not return anything so its not useful. In the devel version of sqldf you can pass multiple command so try this using the builtin data frame BOD noting that the record with demand = 8.3 was removed: library(sqldf) Loading required package: RSQLite Loading required package: DBI Loading required package: gsubfn Loading required package: proto # overwrite with devel version of the sqldf.R file source(http://sqldf.googlecode.com/svn/trunk/R/sqldf.R;) sqldf(c(delete from BOD where demand = 8.3, select * from BOD)) Time__1 demand 1 2 10.3 2 3 19.0 3 4 16.0 4 5 15.6 5 7 19.8 On 9/7/07, Paul Smith [EMAIL PROTECTED] wrote: Dear All, Is sqldf equipped with delete queries? I have tried delete queries but with no success. Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
