Re: [R] Mismatch in logical result?
On Fri, 2008-11-07 at 15:53 +0530, Shubha Vishwanath Karanth wrote: > > Hi R, > > > > I have certain checkings, which gives FALSE, but actually it is true. Why > does this happen? Note that the equations that I am checking below are not > even the case of recurring decimals... > > > > > 1.4^2 == 1.96 > > [1] FALSE > > > > > 1.2^3==1.728 > > [1] FALSE Shubha the correct answer for us is TRUE for the computer is FALSE 1.4^2-1.96 [1] -2.220446e-16 1.2^3-1.728 [1] -2.220446e-16 but if you use "all.equal" all.equal(1.4^2,1.96) [1] TRUE all.equal(1.2^3,1.728) [1] TRUE More details in R FAQ 7.31 -- []s Tura __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] New to List - just starting to learn R - question about reading in data
Are you looking for read.fwf? I've not seen that form of data file since the days of punched cards. (That _was_ the era when SPSS was written._) (Real names and real signature blocks are preferred here.) On Fri, 7 Nov 2008, someone wrote: Hi, everyone, I'm new to R and the List, and excited to be here. I am coming from a background working in SPSS 16, primarily, which has a lot of easy options for reading in data. I am working in R in an Ubuntu Linux environment, and I'm learning with R Commander. Now, here's my question. In my line of work, I'm used to getting data in some kind of flat ASCII format with a separate file indicating which columns correspond to which variables. In the Venables and Ripley book on S/R that I'm reading, there doesn't seem to be a way to read this kind of data format into R, and I haven't found anything online. If anyone has the tine, I'd sure appreciate a little guidance. Thanks, DD __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Umlaut read from csv-file
We have no idea what you understood (you didn't tell us), but the help
says
encoding: character vector. The encoding(s) to be assumed when 'file'
is a character string: see 'file'. A possible value is
'"unknown"': see the ‘Details’.
...
This paragraph applies if 'file' is a filename (rather than a
connection). If 'encoding = "unknown"', an attempt is made to
guess the encoding. The result of 'localeToCharset()' is used as
a guide. If 'encoding' has two or more elements, they are tried
in turn until the file/URL can be read without error in the trial
encoding.
So source(encoding="latin1") says the file is encoded in Latin-1 and
should be re-encoded if necessary (e.g. in UTF-8 locale).
Setting the Encoding of parsed character strings is not mentioned.
You could have written out a data frame with write.csv() and re-read it
with read.csv(encoding = "latin1"): that was the workaround you were given
earlier (not to use source).
On Sat, 8 Nov 2008, Heinz Tuechler wrote:
At 16:52 07.11.2008, Prof Brian Ripley wrote:
On Fri, 7 Nov 2008, Peter Dalgaard wrote:
Heinz Tuechler wrote:
Dear Prof.Ripley!
Thank you very much for your attention. In the given example Encoding(),
or the encoding parameter of read.csv solve the problem. I hope your
patch will solve also the problem, when I read a spss file by
spss.get(), since this function has no encoding parameter and my real
problem originated there.
read.spss() (package foreign) does have a reencode argument, though; and
this is called by spss.get(), so it looks like an easy hack to add it
there.
Yes, older software like spss.get needs to get updated for the
internationalization age. Modifying it to have a ... argument passed to
read.spss would be a good idea (and future-proofing).
In cases like this it is likely that the SPSS file does contain its
encoding (although sometimes it does not and occasionally it is wrong), so
it is helpful to make use of the info if it is there. However, the default
is read.spss(reencode=NA) because of the problems of assuming that the info
is correct when it is not are worse.
The cause, why I tried the example below was to solve the encoding by dumping
and then re-sourcing a data.frame with the encoding parameter set to latin1.
As you can see, source(x, encoding='latin1') does not have the effect I
expected. Unfortunately I do not have any idea, what I understood wrong
regarding the meaning of encoding='latin1'.
Heinz Tüchler
us <- c("a", "b", "c", "ä", "ö", "ü")
Encoding(us)
[1] "unknown" "unknown" "unknown" "latin1" "latin1" "latin1"
dump('us', 'us_dump.txt')
rm(us)
source('us_dump.txt', encoding='latin1')
us
[1] "a" "b" "c" "ä" "ö" "ü"
Encoding(us)
[1] "unknown" "unknown" "unknown" "unknown" "unknown" "unknown"
unlink('us_dump.txt')
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595__
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[R] plot.hclust with lots of objects
Dear all, The default plotting method for hclust trees looks just fine for few objects like in the example dataset. But when it comes to many features (eg some 1000 and more - I'm trying to visualize clustered microarray data) it renders a tree, that one cannot inspect, because of overlapping text and lines. My question is, is there a way or a plotting parameter for plotting a tree which is wide enough to have all leaves separated and readable labels even for that many objects? This would produce a very big image, so I think scrolling is essential. I believe the answer is simple, but I'm unable to figure it out. Thanks in advance -- BEST Paul Murima __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [Stat related] Understanding Portmanteau test
Sorry to be off-topic. Can somebody please explain me what is Portmanteau test? Why it's name is like that? When I would say, a particular test is portmanteau test? I did some googling but got no satisfactory answer at all. Please anybody help for understanding that? Regards, -- View this message in context: http://www.nabble.com/-Stat-related--Understanding-Portmanteau-test-tp20393010p20393010.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] New to List - just starting to learn R - question about reading in data
I've not had occasion to use it yet, but perhaps read.fwf. At the R command prompt, type ?read.fwf to learn about it. --Chris Ryan Original message >Date: Fri, 07 Nov 2008 23:23:18 -0500 >From: Dirty D <[EMAIL PROTECTED]> >Subject: [R] New to List - just starting to learn R - question about reading >in data >To: r-help@r-project.org > >Hi, everyone, > >I'm new to R and the List, and excited to be here. I am coming from a >background working in SPSS 16, primarily, which has a lot of easy >options for reading in data. I am working in R in an Ubuntu Linux >environment, and I'm learning with R Commander. > >Now, here's my question. In my line of work, I'm used to getting data in >some kind of flat ASCII format with a separate file indicating which >columns correspond to which variables. In the Venables and Ripley book >on S/R that I'm reading, there doesn't seem to be a way to read this >kind of data format into R, and I haven't found anything online. > >If anyone has the tine, I'd sure appreciate a little guidance. > >Thanks, > >DD > >__ >R-help@r-project.org mailing list >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] New to List - just starting to learn R - question about reading in data
Welcome!
Dirty D wrote:
Hi, everyone,
I'm new to R and the List, and excited to be here. I am coming from a
background working in SPSS 16, primarily, which has a lot of easy
options for reading in data. I am working in R in an Ubuntu Linux
environment, and I'm learning with R Commander.
I work with R in Ubuntu also, it is a good decision I think. I use ESS
(a package for Emacs), however.
Now, here's my question. In my line of work, I'm used to getting data in
some kind of flat ASCII format with a separate file indicating which
columns correspond to which variables. In the Venables and Ripley book
on S/R that I'm reading, there doesn't seem to be a way to read this
kind of data format into R, and I haven't found anything online.
Certainly, look up read.table (?read.table in R) and in V&R, bottom of
page 21 specifically. There is also an R Data Import/Export manual on
the project's site. But, read.table is almost surely what you're
looking for.
test <- read.table("/path/to/file", header = TRUE)
This is a common example if your text file has a first row of variable
names. See pages 21-22 for more examples.
HTH!
If anyone has the tine, I'd sure appreciate a little guidance.
Thanks,
DD
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Re: [R] determining plot location in lattice
As far as R is concerned, the plot in tkrplot is being written to a file and therefore is not interactive. So functions like grid.locator and trellis.focus are not going to work. You need to use the Tk commands to determine where the mouse is and where a click occurred, then this needs to be converted to the coordinates of the trellis plot. There are some examples of the first part in the TeachingDemos package, eg TkBrush, TkIdentify, TkApprox, and a couple of others (also the play.sudoku function in the Sudoku package). These all work with base graphics rather than trellis/grid so will not help with the 2nd part. Possibly something in the grid package can tell you the coordinates of the various viewports (but you need to save this information before the end of the plotting command because the information will probably be lost when the plot finishes). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] 801.408.8111 > -Original Message- > From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] > project.org] On Behalf Of Sundar Dorai-Raj > Sent: Friday, November 07, 2008 12:43 PM > To: r-help@r-project.org > Subject: [R] determining plot location in lattice > > Hi, > > I'm dealing with a lattice plot inserted into a tk widget and would > like > to know when a user has clicked on the plot area of a plot (i.e. inside > the axes). For example, > > library(tkrplot) > library(lattice) > tt <- tktoplevel() > makePlot <- function() print(xyplot(1 ~ 1)) > printCoords <- function(x, y) print(c(x, y)) > img <- tkrplot(tt, makePlot) > tkbind(img, "<1>", printCoords) > tkpack(img) > > I would like to know when a user clicks inside the axes, but don't know > how to determine where the plot region is. Essentially, this comes down > to answering the following question: Assuming the entire plot is on a > unit square, what are the coordinates of the plotting area? > > This question is may seem unrelated to tkrplot, but outside of this > context I may not get answers that work. I have been playing with both > lattice::trellis.focus and grid::grid.locator to no avail. > > Thanks, > > --sundar > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting- > guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] New to List - just starting to learn R - question about reading in data
Hi, everyone, I'm new to R and the List, and excited to be here. I am coming from a background working in SPSS 16, primarily, which has a lot of easy options for reading in data. I am working in R in an Ubuntu Linux environment, and I'm learning with R Commander. Now, here's my question. In my line of work, I'm used to getting data in some kind of flat ASCII format with a separate file indicating which columns correspond to which variables. In the Venables and Ripley book on S/R that I'm reading, there doesn't seem to be a way to read this kind of data format into R, and I haven't found anything online. If anyone has the tine, I'd sure appreciate a little guidance. Thanks, DD __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] missing value where TRUE/FALSE needed
They may be the same length; that's not what the error message is
complaining about: it says there is a missing value (i.e., an NA) where
a TRUE/FALSE value is needed, therefore the 'if' doesn't know what to
do, since it is not TRUE or FALSE. So, try
summary(vector1)
summary(vector2)
sum(is.na(vector1))
sum(is.na(vector2))
to see if/where/why there are NAs in these vectors.
David Croll wrote:
Hello dear R people,
for my MSc thesis I need to program some functions, and some of them
simply do not work. In the following example, I made sure both vectors
have the same length (10), but R gives me the following error:
Error in if (vector1[i] == vector2[j]) { :
missing value where TRUE/FALSE needed
I googled for possible solutions, but I did not find a good explanation
for this...
The code:
test <- function() {
vector1 <- sample(1:100,10)
vector2 <- sample(1:100,10)
for (i in vector1) {
for (j in vector2) {
if (vector1[i] == vector2[j]) {
show(list(i,j))
}
}
}
}
Regards, David
__
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Re: [R] Re move specific rows from a matrix
use the trick from this thread: https://stat.ethz.ch/pipermail/r-help/2008-November/178985.html your problem shall be fixed On Fri, Nov 7, 2008 at 8:21 PM, mentor_ <[EMAIL PROTECTED]> wrote: > > Hi, > > imagine the following two matrix: > Matrix A: > a b c > 1 2 3 > 4 5 6 > 7 8 9 > > Matrix B: > a > 4 > 7 > > I would like to remove those rows from matrix A which are present in both > matrices. > So after removing the corresponding rows the matrix A should look like this: > Matrix A: > a b c > 1 2 3 > > Thanks in advance! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Sweave:How to load available colors from pkg 'color'--(Resolved)
Hello: I found a solution to my problem here, I just downloaded mystyle.sty to
the same directory where my latex file resides. Then I included
\usepackage{mystyle} in my preamble and that took care of coloring any section
of the document.
http://www.sci.usq.edu.au/staff/robertsa/LaTeX/ltxfloats.html
Felipe
Is there a shorter way to color the abstract,sections subsections and page
headings of a document. I am using the code below in my preamble to accomplish
some of it:
[EMAIL PROTECTED],dvips]{color}
\else\usepackage[usenames,dvipsnames]{color}
% and fix pdf colour problems
\IfFileExists{pdfcolmk.sty}{\usepackage{pdfcolmk}}{}
\fi
%\renewcommand{\MakeUppercase}[1]{\color{OliveGreen}\textsf{#1}}
%\renewcommand{\abstractname}{\color{blue}Abstract}
Also,'\usepackage[usenames]{color}' is suppossed to load all the available
colors in the package 'color' but for some reason it doesn't load them when
running it trhough Sweave. It loads all the colors if I run it outside R though.
Here's a dummy document:
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish & Wildlife Service
California, USA
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Re: [R] missing value where TRUE/FALSE needed
It would be useful to learn how to use debug/browser. Here is the
state of the machine when the error occurs:
> test()
Error in if (vector1[i] == vector2[j]) { :
missing value where TRUE/FALSE needed
Enter a frame number, or 0 to exit
1: test()
Selection: 1
Called from: eval(expr, envir, enclos)
Browse[1]> ls()
[1] "i" "j" "vector1" "vector2"
Browse[1]> vector1
[1] 27 37 57 89 20 86 97 62 58 6
Browse[1]> vector2
[1] 21 18 68 38 74 48 98 93 35 71
Browse[1]> vector1[27]
[1] NA
Browse[1]>
Notice that vector1 is only 10 long and then you try to address
vector1[27] you get an NA
On Fri, Nov 7, 2008 at 7:22 PM, David Croll <[EMAIL PROTECTED]> wrote:
> Hello dear R people,
>
>
> for my MSc thesis I need to program some functions, and some of them simply
> do not work. In the following example, I made sure both vectors have the
> same length (10), but R gives me the following error:
>
> Error in if (vector1[i] == vector2[j]) { :
> missing value where TRUE/FALSE needed
>
> I googled for possible solutions, but I did not find a good explanation for
> this...
>
>
> The code:
>
> test <- function() {
> vector1 <- sample(1:100,10)
> vector2 <- sample(1:100,10)
> for (i in vector1) {
> for (j in vector2) {
> if (vector1[i] == vector2[j]) {
> show(list(i,j))
> }
> }
> }
> }
>
> Regards, David
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] grouping data together
If you provide the input and the expected output, it would help a lot. You could use 'split' to partition the data monthly <- split(yourDF, yourDF$month) but I am still not sure exactly what you want to do with it, or the format that you are expecting. On Fri, Nov 7, 2008 at 6:03 PM, Swanton0822 <[EMAIL PROTECTED]> wrote: > > Hi. > i have a data, and there is 3 columns, Month, Year and Total. and there is > over 1000 rows for them because there is 87 years data for every month, so > there is month from Jan-Dec, and year from 1900-1987, > so i was wondering if i would want to make 12 groups (Jan,Feb...,Dec), > and put each year's total into the corresponding month (so every month group > will have 88 values of total) > how can i do that? > many thanks. > -- > View this message in context: > http://www.nabble.com/grouping-data-together-tp20389726p20389726.html > Sent from the R help mailing list archive at Nabble.com. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Umlaut read from csv-file
At 16:52 07.11.2008, Prof Brian Ripley wrote:
On Fri, 7 Nov 2008, Peter Dalgaard wrote:
Heinz Tuechler wrote:
Dear Prof.Ripley!
Thank you very much for your attention. In the given example Encoding(),
or the encoding parameter of read.csv solve the problem. I hope your
patch will solve also the problem, when I read a spss file by
spss.get(), since this function has no encoding parameter and my real
problem originated there.
read.spss() (package foreign) does have a reencode argument, though; and
this is called by spss.get(), so it looks like an easy hack to add it
there.
Yes, older software like spss.get needs to get
updated for the internationalization
age. Modifying it to have a ... argument passed
to read.spss would be a good idea (and future-proofing).
In cases like this it is likely that the SPSS
file does contain its encoding (although
sometimes it does not and occasionally it is
wrong), so it is helpful to make use of the info
if it is there. However, the default is
read.spss(reencode=NA) because of the problems
of assuming that the info is correct when it is not are worse.
The cause, why I tried the example below was to
solve the encoding by dumping and then
re-sourcing a data.frame with the encoding
parameter set to latin1. As you can see,
source(x, encoding='latin1') does not have the
effect I expected. Unfortunately I do not have
any idea, what I understood wrong regarding the meaning of encoding='latin1'.
Heinz Tüchler
us <- c("a", "b", "c", "ä", "ö", "ü")
Encoding(us)
[1] "unknown" "unknown" "unknown" "latin1" "latin1" "latin1"
dump('us', 'us_dump.txt')
rm(us)
source('us_dump.txt', encoding='latin1')
us
[1] "a" "b" "c" "ä" "ö" "ü"
Encoding(us)
[1] "unknown" "unknown" "unknown" "unknown" "unknown" "unknown"
unlink('us_dump.txt')
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] grouping data together
hi friend,
this is from your previous posts on Kruskal-Wallis test:)
i came up with this one:
A5 <- read.table('kew.dat' ,header=TRUE)
plot(factor(A5$Month, levels=month.abb), A5$Rain)
is that what you want?
On Sat, Nov 8, 2008 at 7:03 AM, Swanton0822 <[EMAIL PROTECTED]> wrote:
>
> Hi.
> i have a data, and there is 3 columns, Month, Year and Total. and there is
> over 1000 rows for them because there is 87 years data for every month, so
> there is month from Jan-Dec, and year from 1900-1987,
> so i was wondering if i would want to make 12 groups (Jan,Feb...,Dec),
> and put each year's total into the corresponding month (so every month group
> will have 88 values of total)
> how can i do that?
> many thanks.
__
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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] missing value where TRUE/FALSE needed
is this what you want?
> vector1
[1] 65 1 34 100 42 20 79 43 89 10
> vector2
[1] 34 65 47 91 48 32 23 74 92 86
>
> for (i in 1:10) {
+ for (j in 1:10) {
+ if (vector1[i] == vector2[j])
+ show(c(i,j))
+ }
+ }
[1] 1 2
[1] 3 1
On Sat, Nov 8, 2008 at 8:22 AM, David Croll <[EMAIL PROTECTED]> wrote:
> Hello dear R people,
>
>
> for my MSc thesis I need to program some functions, and some of them simply
> do not work. In the following example, I made sure both vectors have the
> same length (10), but R gives me the following error:
>
> Error in if (vector1[i] == vector2[j]) { :
> missing value where TRUE/FALSE needed
>
> I googled for possible solutions, but I did not find a good explanation for
> this...
>
>
> The code:
>
> test <- function() {
> vector1 <- sample(1:100,10)
> vector2 <- sample(1:100,10)
> for (i in vector1) {
> for (j in vector2) {
> if (vector1[i] == vector2[j]) {
> show(list(i,j))
> }
> }
> }
> }
>
> Regards, David
>
> __
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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Re: [R] esoteric inconsistency -- intended or not?
2008/11/8 Bert Gunter <[EMAIL PROTECTED]>: > Is the following intended or not? > >> func<- function(y) match.call() > >> z <- func(y =2) > >> z > func(y = 2) > >> z[["a"]] <- 5 > >> z > func(y = 2, 5) ## Note that the second argument **is not** named > > ## BUT... > >> z <- func(y =2) > >> z$a <- 5 > >> z > func(y = 2, a = 5) ## The second argument **is** named > > ### End of example code ### > > The reason I ask is that the man page for [[ specifically says: > ** > Both [[ and $ select a single element of the list. The main difference is > that $ does not allow computed indices, whereas [[ does. x$name is > equivalent to x[["name", exact = FALSE]]. Also, the partial matching > behavior of [[ can be controlled using the exact argument. > > [ and [[ are sometimes applied to other recursive objects such as calls and > expressions. Pairlists are coerced to lists for extraction by [, but all > three operators can be used for replacement. > > > I (mis?)read this as saying the behavior in the code snippets above should > produce identical results. > > I note that the above inconsistency can be trivially avoided by first > coercing the call object to a list, modifying it either way, and then > coercing it back to a call object. I too have been caught by this apparent inconsistency a few times. By the way, another way to get around it is z <- quote(func(y = 2)) z["a"] <- list(5) Another inconsistency between the indexing of lists and calls is recursive indexing, e.g. ## recursive indexing of lists works > foo <- list(a = list(1, 2, 3), b = list(4, 5, 6)) > foo[[c(2,2)]] [1] 5 ## recursive indexing of calls fails > foocall <- quote(func(a = list(1, 2, 3), b = list(4, 5, 6))) > foocall[[c(3, 2)]] Error in foocall[[c(3, 2)]] : attempt to select more than one element OK, this is pretty obscure, and it is easy enough to write a function to do it (as I have done), but it would be nice to have in the long run, so that indexing is more standardised. -- Felix Andrews / 安福立 http://www.neurofractal.org/felix/ 3358 543D AAC6 22C2 D336 80D9 360B 72DD 3E4C F5D8 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] missing value where TRUE/FALSE needed
Hello dear R people,
for my MSc thesis I need to program some functions, and some of them
simply do not work. In the following example, I made sure both vectors
have the same length (10), but R gives me the following error:
Error in if (vector1[i] == vector2[j]) { :
missing value where TRUE/FALSE needed
I googled for possible solutions, but I did not find a good explanation
for this...
The code:
test <- function() {
vector1 <- sample(1:100,10)
vector2 <- sample(1:100,10)
for (i in vector1) {
for (j in vector2) {
if (vector1[i] == vector2[j]) {
show(list(i,j))
}
}
}
}
Regards, David
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Re: [R] Re gular Expression help
Here are a few more solutions. x is the input vector
of character strings.
The first is a slightly shorter version of one of Wacek's.
The next three all create an anonymous grouping variable
(using sub, substr/gsub and strapply respectively)
whose components are "p" and "q" and then tapply
is used to separate out the corresponding components
of x according to the grouping:
sapply(c(p = "^[^pq]*p", q = "^[^pq]*q"), grep, x = x, value = TRUE)
tapply(x, sub("^[^pq]*(.).*", "\\1", x), c)
tapply(x, substr(gsub("[^pq]", "", x), 1, 1), c)
library(gsubfn)
tapply(x, strapply(x, "^[^pq]*(.)", simplify = c), c)
On Fri, Nov 7, 2008 at 1:09 PM, Rajasekaramya <[EMAIL PROTECTED]> wrote:
>
> hi there
>
> I have a vector with a set of data.I just wanna seperate them based on the
> first p and q values metioned within the data.
>
> [1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
> [2] chr1q22-q24
> [3] chr1q22-q24
> [4] chr1pter-q24
> [5] chr1pter-q24
> [6] chr1pter-q24
>
> i used a regular expression [+q*] to match up the values but it matches q
> found anywhere i know i have written like that but i jus want it to match
> the first p or q values.
>
> my result should be for q and
> [2] chr1q22-q24
> [3] chr1q22-q24
>
> for p
> [1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
> [4] chr1pter-q24
> [5] chr1pter-q24
> [6] chr1pter-q24
>
>
>
> --
> View this message in context:
> http://www.nabble.com/Regular-Expression-help-tp20385971p20385971.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
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> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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Re: [R] Re gular Expression help
Wacek Kusnierczyk wrote:
>
>>
>>
>>> Rajasekaramya wrote:
>>>
>>>
>>>
hi there
I have a vector with a set of data.I just wanna seperate them based on the
first p and q values metioned within the data.
[1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
[2] chr1q22-q24
[3] chr1q22-q24
[4] chr1pter-q24
[5] chr1pter-q24
[6] chr1pter-q24
i used a regular expression [+q*] to match up the values but it matches q
found anywhere i know i have written like that but i jus want it to match
the first p or q values.
my result should be for q and
[2] chr1q22-q24
[3] chr1q22-q24
for p
[1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
[4] chr1pter-q24
[5] chr1pter-q24
[6] chr1pter-q24
>> the following will do the whole job (assuming x is your vector):
>>
>>
>>
> result = local({
>p = grep("^[^pq]*p", d)
>list(p=d[p], q=d[-p])
> })
>
oops, replace 'd' with 'x'
vQ
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Re: [R] Re gular Expression help
Wacek Kusnierczyk wrote:
> Peter Dalgaard wrote:
>
>> Rajasekaramya wrote:
>>
>>
>>> hi there
>>>
>>> I have a vector with a set of data.I just wanna seperate them based on the
>>> first p and q values metioned within the data.
>>>
>>> [1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
>>> [2] chr1q22-q24
>>> [3] chr1q22-q24
>>> [4] chr1pter-q24
>>> [5] chr1pter-q24
>>> [6] chr1pter-q24
>>>
>>> i used a regular expression [+q*] to match up the values but it matches q
>>> found anywhere i know i have written like that but i jus want it to match
>>> the first p or q values.
>>>
>>> my result should be for q and
>>> [2] chr1q22-q24
>>> [3] chr1q22-q24
>>>
>>> for p
>>> [1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
>>> [4] chr1pter-q24
>>> [5] chr1pter-q24
>>> [6] chr1pter-q24
>>>
>>>
>>>
>> Something like
>>
>> sub("[^pq]*([pq]).*","\\1",x)
>>
>> should get you the first p or q
>>
>>
>>
>
> and the following will do the whole job (assuming x is your vector):
>
> result = lapply(
>list(p='p', q='q'),
>function(letter)
> grep(paste("^[^pq]*[", "]", sep=letter), x, value=TRUE))
>
>
and this one might be slightly faster, depending on your data:
result = local({
p = grep("^[^pq]*p", d)
list(p=d[p], q=d[-p])
})
vQ
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Re: [R] Faster way to combine data sets with different date ranges
See ?merge. Also ?merge.zoo in the zoo package and sqldf
in the sqldf package (http://sqldf.googlecode.com).
Please read the last line to every message to r-help.
On Fri, Nov 7, 2008 at 6:19 PM, t c <[EMAIL PROTECTED]> wrote:
> I am trying to combine two data sets, one with daily values and one with
> weekly values. SurveyData conatins environmental data collected on a daily
> basis. sat.data contains satellite sea surface temperature that is an
> average of satellite measurements over a six day period. I would like to
> combine the two files so that my output file has the daily dates from
> SurveyData and the weekly average from sat.data that corresponds to that day.
> I have written a loop that does exactly what I want, but it takes a very
> long time. Is there a faster way to do this? What I have so far is below.
> Thanks,
> Tim
>
> #Compare dates
> for(i in 1:length(SurveyData$Date))
> {
> print(i)
> for(j in 1:length(sat.data$Date.Start))
> {
> if(
> sat.data$Date.Start[j]<=SurveyData$Date[i] &
> sat.data$Date.End[j]>=SurveyData$Date[i])
> survey.data[i]<- sat.data$data[j]
> }
> }
>
> #Create data frame with final data
> final.data<-data.frame(SurveyData$Record,SurveyData$Date,survey.data)
>
> #Write data to file
> write.csv(final.data,file="Combined.csv")
>
>
>
>
>[[alternative HTML version deleted]]
>
>
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> and provide commented, minimal, self-contained, reproducible code.
>
>
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[R] Faster way to combine data sets with different date ranges
I am trying to combine two data sets, one with daily values and one with weekly
values. SurveyData conatins environmental data collected on a daily basis.
sat.data contains satellite sea surface temperature that is an average of
satellite measurements over a six day period. I would like to combine the two
files so that my output file has the daily dates from SurveyData and the weekly
average from sat.data that corresponds to that day. I have written a loop that
does exactly what I want, but it takes a very long time. Is there a faster way
to do this? What I have so far is below.
Thanks,
Tim
#Compare dates
for(i in 1:length(SurveyData$Date))
{
print(i)
for(j in 1:length(sat.data$Date.Start))
{
if(
sat.data$Date.Start[j]<=SurveyData$Date[i] &
sat.data$Date.End[j]>=SurveyData$Date[i])
survey.data[i]<- sat.data$data[j]
}
}
#Create data frame with final data
final.data<-data.frame(SurveyData$Record,SurveyData$Date,survey.data)
#Write data to file
write.csv(final.data,file="Combined.csv")
[[alternative HTML version deleted]]
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[R] grouping data together
Hi. i have a data, and there is 3 columns, Month, Year and Total. and there is over 1000 rows for them because there is 87 years data for every month, so there is month from Jan-Dec, and year from 1900-1987, so i was wondering if i would want to make 12 groups (Jan,Feb...,Dec), and put each year's total into the corresponding month (so every month group will have 88 values of total) how can i do that? many thanks. -- View this message in context: http://www.nabble.com/grouping-data-together-tp20389726p20389726.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Re gular Expression help
Peter Dalgaard wrote:
> Rajasekaramya wrote:
>
>> hi there
>>
>> I have a vector with a set of data.I just wanna seperate them based on the
>> first p and q values metioned within the data.
>>
>> [1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
>> [2] chr1q22-q24
>> [3] chr1q22-q24
>> [4] chr1pter-q24
>> [5] chr1pter-q24
>> [6] chr1pter-q24
>>
>> i used a regular expression [+q*] to match up the values but it matches q
>> found anywhere i know i have written like that but i jus want it to match
>> the first p or q values.
>>
>> my result should be for q and
>> [2] chr1q22-q24
>> [3] chr1q22-q24
>>
>> for p
>> [1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
>> [4] chr1pter-q24
>> [5] chr1pter-q24
>> [6] chr1pter-q24
>>
>>
>
> Something like
>
> sub("[^pq]*([pq]).*","\\1",x)
>
> should get you the first p or q
>
>
and the following will do the whole job (assuming x is your vector):
result = lapply(
list(p='p', q='q'),
function(letter)
grep(paste("^[^pq]*[", "]", sep=letter), x, value=TRUE))
result$p
# those with p first
result$q
# those with q first
vQ
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[R] UTF8
Hi, I am trying to get R 2.8.0 for Mac OS from CRAN, but I thing I am doing something wrong because when R starts I have annoying error messages: During startup - Warning messages: 1: Setting LC_CTYPE failed, using "C" 2: Setting LC_COLLATE failed, using "C" 3: Setting LC_TIME failed, using "C" 4: Setting LC_MESSAGES failed, using "C" WARNING: You're using a non-UTF8 locale, therefore only ASCII characters will work. Please read R for Mac OS X FAQ (see Help) section 9 and adjust your system preferences accordingly. I have tried all suggestions in such section 9 but still get the same error message. If you can give me some another suggestion It would be very nice. Sincerely, Denise Duarte UFMG- MG- Brazil __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] two kind of Hosmer and Lemeshow’s test
leo_wa wrote:
I know that there are two method to apply the Hosmer and Lemeshow’s. One of
them is calculated based on the fixed and pre-determined cut-off points of
the estimated probability of success. One of them is calculated based on
the percentiles of estimated probabilities.
Both of these methods have been made obsolete by methods that do not
require any arbitrary groupings of predicted probabilities. See the
residuals.lrm function in the Design package and the reference in its
help file, to the Hosmer paper.
Frank
In the previous post,i find that the Hosmer and Lemeshow’s test how to use
in R.
hosmerlem <-
function (y, yhat, g = 10)
{
cutyhat <- cut(x, breaks = quantile(yhat, probs = seq(0,
1, 1/g)), include.lowest = T)
obs <- xtabs(cbind(1 - y, y) ~ cutyhat)
expect <- xtabs(cbind(1 - yhat, yhat) ~ cutyhat)
chisq <- sum((obs - expect)^2/expect)
P <- 1 - pchisq(chisq, g - 2)
c("X^2" = chisq, Df = g - 2, "P(>Chi)" = P)
}
I want to know how can i use the another method which is not use the
probability of success. i want to know how can i revise above program to
achieve an objective.
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
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[R] unable to add a new workbook with rcom
__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] easy way to cap a data frame to a max value
A few more. These each have the advantage of not destroying the original data frame: # based on Erik's DF2 <- replace(DF, DF > 10, 10) # based on my previous one DF2 <- replace(DF, TRUE, pmin(10, unlist(DF))) DF2 <- (DF + 10)/2 - abs(DF - 10)/2 On Fri, Nov 7, 2008 at 4:38 PM, Gabor Grothendieck <[EMAIL PROTECTED]> wrote: > That should be pmin: > > On Fri, Nov 7, 2008 at 4:25 PM, Gabor Grothendieck > <[EMAIL PROTECTED]> wrote: >> Assuming the data frame is all numeric: >> >> DF[] <- pmax(10, unlist(DF)) >> >> >> On Fri, Nov 7, 2008 at 4:16 PM, Grey Moran <[EMAIL PROTECTED]> wrote: >>> Hello, >>> >>> I have some rather large matrices. Is there a way (without having to >>> loop) to cap all the values of a data frame to a given ceiling? >>> E.g. >>> junk <- cbind(c(1,2,3,4,5),c(2,4,6,8,10)) junk >>> [,1] [,2] >>> [1,]12 >>> [2,]24 >>> [3,]36 >>> [4,]48 >>> [5,]5 10 >>> " replace anything over the value of 5 with 5..." >>> >>> Thank you all, >>> >>> Grey >>> >>> __ >>> R-help@r-project.org mailing list >>> https://stat.ethz.ch/mailman/listinfo/r-help >>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >>> and provide commented, minimal, self-contained, reproducible code. >>> >> > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] easy way to cap a data frame to a max value
On Fri, Nov 7, 2008 at 4:46 PM, Grey Moran <[EMAIL PROTECTED]> wrote: > Thanks to all who replied - lots of good ideas. > The one I prefered at the end was: > junk[junk > 5] <- 5 > > Grey > > On Fri, Nov 7, 2008 at 4:38 PM, Gabor Grothendieck > <[EMAIL PROTECTED]> wrote: >> That should be pmin: >> >> On Fri, Nov 7, 2008 at 4:25 PM, Gabor Grothendieck >> <[EMAIL PROTECTED]> wrote: >>> Assuming the data frame is all numeric: >>> >>> DF[] <- pmax(10, unlist(DF)) >>> >>> >>> On Fri, Nov 7, 2008 at 4:16 PM, Grey Moran <[EMAIL PROTECTED]> wrote: Hello, I have some rather large matrices. Is there a way (without having to loop) to cap all the values of a data frame to a given ceiling? E.g. junk <- cbind(c(1,2,3,4,5),c(2,4,6,8,10)) > junk [,1] [,2] [1,]12 [2,]24 [3,]36 [4,]48 [5,]5 10 >" replace anything over the value of 5 with 5..." Thank you all, Grey __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. >>> >> > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] easy way to cap a data frame to a max value
Thanks to all who replied - lots of good ideas. The one I prefered at the end was: junk[junk > 5] <- 5 Grey On Fri, Nov 7, 2008 at 4:38 PM, Gabor Grothendieck <[EMAIL PROTECTED]> wrote: > That should be pmin: > > On Fri, Nov 7, 2008 at 4:25 PM, Gabor Grothendieck > <[EMAIL PROTECTED]> wrote: >> Assuming the data frame is all numeric: >> >> DF[] <- pmax(10, unlist(DF)) >> >> >> On Fri, Nov 7, 2008 at 4:16 PM, Grey Moran <[EMAIL PROTECTED]> wrote: >>> Hello, >>> >>> I have some rather large matrices. Is there a way (without having to >>> loop) to cap all the values of a data frame to a given ceiling? >>> E.g. >>> junk <- cbind(c(1,2,3,4,5),c(2,4,6,8,10)) junk >>> [,1] [,2] >>> [1,]12 >>> [2,]24 >>> [3,]36 >>> [4,]48 >>> [5,]5 10 >>> " replace anything over the value of 5 with 5..." >>> >>> Thank you all, >>> >>> Grey >>> >>> __ >>> R-help@r-project.org mailing list >>> https://stat.ethz.ch/mailman/listinfo/r-help >>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >>> and provide commented, minimal, self-contained, reproducible code. >>> >> > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] easy way to cap a data frame to a max value
That should be pmin: On Fri, Nov 7, 2008 at 4:25 PM, Gabor Grothendieck <[EMAIL PROTECTED]> wrote: > Assuming the data frame is all numeric: > > DF[] <- pmax(10, unlist(DF)) > > > On Fri, Nov 7, 2008 at 4:16 PM, Grey Moran <[EMAIL PROTECTED]> wrote: >> Hello, >> >> I have some rather large matrices. Is there a way (without having to >> loop) to cap all the values of a data frame to a given ceiling? >> E.g. >> junk <- cbind(c(1,2,3,4,5),c(2,4,6,8,10)) >>> junk >> [,1] [,2] >> [1,]12 >> [2,]24 >> [3,]36 >> [4,]48 >> [5,]5 10 >> >>>" replace anything over the value of 5 with 5..." >> >> Thank you all, >> >> Grey >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] nls: Fitting two models at once?
Dear Heather!
Thank you very, very much for your solution which also reproduces exactly
the results I get from the SigmaPlot analysises. :)
Martin
Heather Turner schrieb:
> Dear Martin,
>
> You can use the same idea of concatenating the data in R. The following
> reproduces your example:
>
> spec <- c(asfe, dias)
> n1 <- length(asfe)
> n2 <- length(dias)
> CYT <- c(28 * CYTF, 24.6 * 2 * CYTB6)
> B559HP <- c(B559HP, numeric(n2))
> B559LP <- c(numeric(n1), B559LP)
> C550 <- c(numeric(n1), C550)
>
> spec.fit <-
> nls(
> spec ~ cyt.v*CYT + hp.v*B559HP + lp.v*B559LP + c550.v*C550,
> start = list(cyt.v = 0.5, hp.v = 0.03, lp.v = 1,
>c550.v = 1) # arbitrary
> );
>
> or more efficiently, using lm
>
> spec.fit <- lm(spec ~ 0 + CYT + B559HP + B559LP + C550)
>
> # draw stuff
> plot(
> 1, 2,
> type="n",
> xlim = c(540, 575),
> ylim=c(-0.002, 0.008),
> );
>
> # first spectrum and fit
> lines(wl, asfe, type="b", pch=19); # solid circles
> lines(wl, fitted(spec.fit)[1:n1], col = "red");
>
> # second spectrum and fit
> lines(wl, dias, type="b");
> lines(wl, fitted(spec.fit)[(n1 + 1):(n1 + n2)], col = "blue");
>
> Best wishes,
>
> Heather
>
> Dr H Turner
> Senior Research Fellow
> Dept. of Statistics
> The University of Warwick
> Coventry
> CV4 7AL
>
> Tel: 024 76575870
> Fax: 024 76524532
> Url: www.warwick.ac.uk/go/heatherturner
>
>
> Martin Ballaschk wrote:
>> Hello,
>>
>> I'm still a newbie user and struggling to automate some analyses from
>> SigmaPlot using R. R is a great help for me so far!
>>
>> But the following problem makes me go nuts.
>>
>> I have two spectra, both have to be fitted to reference data. Problem: the
>> both spectra are connected in some way: the stoichiometry of coefficients
>> "cytf.v"/"cytb.v" is 1/2.
>> {{In the SigmaPlot workflow one has to copy the two spectra into one column
>> beneath each other and the two spectra are somehow treated as one curve -
>> like in http://home.arcor.de/ballaschk/cytbf-help/sigmaplot%20formula.png}}
>>
>> Can anybody help? :(
>>
>> I tried to condense everything to the "minimum" R script below to give an
>> impression of what I'm talking about.
>>
>> Martin
>>
>>
>>
>>
>> # "Minimal" R script reading remote data for convenience
>>
>> ### READ IN DATA
>> # first spectrum
>> asfe <- read.table("http://home.arcor.de/ballaschk/cytbf-help/asfe.csv";)[,
>> 1];
>> # second spectrum
>> dias <- read.table("http://home.arcor.de/ballaschk/cytbf-help/dias.csv";)[,
>> 1];
>>
>> # reference data for fit, wavelength = wl
>> ref <- read.table("http://home.arcor.de/ballaschk/cytbf-help/reference.csv";,
>> sep="\t", dec=".", header=T);
>> attach(ref);
>>
>> ### FITTING, problem: 2*cytf.v == cytb.v
>>
>> # fit first spectrum to two reference spectra
>> asfe.fit <-
>> nls(
>> asfe ~ (cytf.v * 28) * CYTF + hp.v * B559HP,
>> start = list(cytf.v = 0.5, hp.v = 0.03) # arbitrary
>> );
>>
>> # fit second spectrum to three reference spectra
>> dias.fit <-
>> nls(
>> dias ~ (cytb.v * 24.6 * 2) * CYTB6 + lp.v * B559LP + c550.v * C550,
>> start = list(cytb.v = 1, lp.v = 1, c550.v = 1) # arbitrary
>> );
>>
>>
>> # draw stuff
>> plot(
>> 1, 2,
>> type="n",
>> xlim = c(540, 575),
>> ylim=c(-0.002, 0.008),
>> );
>>
>> # first spectrum and fit
>> lines(wl, asfe, type="b", pch=19); # solid circles
>> lines(wl, fitted(asfe.fit), col = "red");
>>
>> # second spectrum and fit
>> lines(wl, dias, type="b");
>> lines(wl, fitted(dias.fit), col = "blue");
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
__
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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] easy way to cap a data frame to a max value
Are they objects of class "matrix" or "data.frame"? You seem to make reference to both, but know there is a difference. Try: junk[junk > 5] <- 5 Grey Moran wrote: Hello, I have some rather large matrices. Is there a way (without having to loop) to cap all the values of a data frame to a given ceiling? E.g. junk <- cbind(c(1,2,3,4,5),c(2,4,6,8,10)) junk [,1] [,2] [1,]12 [2,]24 [3,]36 [4,]48 [5,]5 10 " replace anything over the value of 5 with 5..." Thank you all, Grey __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] easy way to cap a data frame to a max value
Assuming the data frame is all numeric: DF[] <- pmax(10, unlist(DF)) On Fri, Nov 7, 2008 at 4:16 PM, Grey Moran <[EMAIL PROTECTED]> wrote: > Hello, > > I have some rather large matrices. Is there a way (without having to > loop) to cap all the values of a data frame to a given ceiling? > E.g. > junk <- cbind(c(1,2,3,4,5),c(2,4,6,8,10)) >> junk > [,1] [,2] > [1,]12 > [2,]24 > [3,]36 > [4,]48 > [5,]5 10 > >>" replace anything over the value of 5 with 5..." > > Thank you all, > > Grey > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] easy way to cap a data frame to a max value
Hello, I have some rather large matrices. Is there a way (without having to loop) to cap all the values of a data frame to a given ceiling? E.g. junk <- cbind(c(1,2,3,4,5),c(2,4,6,8,10)) > junk [,1] [,2] [1,]12 [2,]24 [3,]36 [4,]48 [5,]5 10 >" replace anything over the value of 5 with 5..." Thank you all, Grey __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] negative binomial predicted probabilities
I estimated a negative binomial model using zelig. z.out<- zelig(NEWBHC~ PW80 + CHNGBLK + XBLK,data=data, model="negbin") How do I calculate predicted probabilities for this model? Is it the same process as a poisson regression? Thanks in advance Joe [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Agent-based social simulation and R
On Friday 07 November 2008 (19:45:04), you wrote: > Martin Elff wrote: > > Hi Tom, > > > > my package 'memisc' contains a sort of an infrastructure for doing > > simulations. As a fun exercise I also used it to create a 'toy' agent > > based simulation of Schelling's neighbourhood model. Although it is not a > > serious application, at least it shows that agent based simulation is > > possible in R. > > > > Just run 'demo(schelling)' after loading 'memisc'. > > Interesting. I'll have to look trough the code. How far away do the > agents see? Anything beyond their immidiate neighbors? In the demo, dissatisfied agents jump at random to free places. This puts the neighborhood model to the 'hardest' test. Yet the demo code also contains some other possibilities. Best, Martin -- - Dr. Martin Elff Department of Social Sciences University of Mannheim A5, Room 328 68131 Mannheim Germany Phone: ++49-621-181-2093 Fax: ++49-621-181-2099 E-Mail: [EMAIL PROTECTED] Homepage: http://webrum.uni-mannheim.de/sowi/elff http://www.sowi.uni-mannheim.de/lspwivs/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] For Loop - loading 10 sets of data and calculating
# I would put this in a list in the following manner
Bin <- lapply(1:10, function(.file){
#---
#Loads bin data frame from csv files with acres and TAZ data
fileName <-
paste("I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin",
.file, "_lookup.csv", sep="")
Bin_main <- read.csv(file=fileName,head=FALSE);
#Separates Acres data from main data and converts acres to square feet
Bin_Acres=Bin_main[[1]]*43560
#Separates TAZ data from main data
Bin_TAZ=Bin_main[[2]]
#Separates TAZ data from main data and converts acres to square feet
Bin_TAZvacant=Bin_main[[3]]*43560
#Sums each parcel acreage data of the bin
BinAcres_sum=sum(Bin_Acres)
#Creates data frame of cumlative percentages of each parcel of bin
Bin_cumper=cumsum(Bin_Acres/BinAcres_sum)
#Calculates the probability of choosing particular parcel from bin
Bin_parprob=abs(1-Bin_cumper)
#Combines parcel acreage data and cumlative percentage data
cbind(Bin_Acres,Bin_parprob,Bin_TAZ,Bin_TAZvacant)
})
On Fri, Nov 7, 2008 at 1:48 PM, PDXRugger <[EMAIL PROTECTED]> wrote:
>
> I am trying to simplify my code by adding a for loop that will load and
> compute a sequence of code 10 time. They way i run it now is that the same
> 8 lines of code are basically reproduced 10 times. I would like to replace
> the numeric value in the code (e.g. Bin1, Bin2Bin10) each time the loop
> goes around. Below i tried doing this with a simple for loop and adding the
> string character before each numeric value. I need to first load the data
> then calculate, im sure this is possible as i have seen it done but cant
> seem to reproduce it in my own code. Hope my question is clear and i hope
> someone can offer some guidance.
>
> Cheers,
> JR
>
> for (i in 1:10) {
>
> #---
>
> #Loads bin data frame from csv files with acres and TAZ data
> Bin$i_main <-
> read.csv(file="I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin$i_lookup.csv",head=FALSE);
>
> #Separates Acres data from main data and converts acres to square feet
> Bin$i_Acres=Bin$i_main[[1]]*43560
>
> #Separates TAZ data from main data
> Bin$i_TAZ=Bin$i_main[[2]]
>
> #Separates TAZ data from main data and converts acres to square feet
> Bin$i_TAZvacant=Bin$i_main[[3]]*43560
>
> #Sums each parcel acreage data of the bin
> Bin$iAcres_sum=sum(Bin$i_Acres)
>
> #Creates data frame of cumlative percentages of each parcel of bin
> Bin$i_cumper=cumsum(Bin$i_Acres/Bin$iAcres_sum)
>
> #Calculates the probability of choosing particular parcel from bin
> Bin$i_parprob=abs(1-Bin$i_cumper)
>
> #Combines parcel acreage data and cumlative percentage data
> Bin$iMain.data = cbind(Bin%i_Acres,Bin$i_parprob,Bin$i_TAZ,Bin$i_TAZvacant)
> }
>
> --
> View this message in context:
> http://www.nabble.com/For-Loop---loading-10-sets-of-data-and-calculating-tp20386673p20386673.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
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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] Rmpi task-pull
Hi,
there is a new mailing list for R and HPC: [EMAIL PROTECTED]
This is probably a better list for your question. Do not forget, first
of all you have to register: https://stat.ethz.ch/mailman/listinfo/r-sig-hpc
I tried your code and it is working!
Please send us your sessionInfo() output. Probably you use some old
package versions?
Is this your first Rmpi code? Is other code working?
You can try something like this to get output from all nodes.
library("Rmpi")
mpi.spawn.Rslaves(nslaves=3)
mpi.remote.exec(paste("I am
node",mpi.comm.rank(),"of",mpi.comm.size(),"on",Sys.info()["nodename"]))
mpi.remote.exec(sessionInfo())
mpi.close.Rslaves()
mpi.quit()
For some more debugging you can start your cluster with log output:
mpi.spawn.Rslaves(nslaves=3, needlog=TRUE)
Then there should be logfiles for every node in your working directory.
Best
Markus
Daniel Ferrara wrote:
> Hi, I'm testing the efficiency of the Rmpi package regarding parallelization
> using a cluster.
> I've found and tried the task pull programming method, but even if it is
> described as the best method, it seems to cause deadlock, anyone could help
> me in using this method?
> here is the code I've found and tried:
>
>
> # Initialize MPI
> library("Rmpi")
>
> # Notice we just say "give us all the slaves you've got."
> mpi.spawn.Rslaves()
>
> if (mpi.comm.size() < 2) {
> print("More slave processes are required.")
> mpi.quit()
> }
>
> .Last <- function(){
> if (is.loaded("mpi_initialize")){
> if (mpi.comm.size(1) > 0){
> print("Please use mpi.close.Rslaves() to close slaves.")
> mpi.close.Rslaves()
> }
> print("Please use mpi.quit() to quit R")
> .Call("mpi_finalize")
> }
> }
>
> # Function the slaves will call to perform a validation on the
> # fold equal to their slave number.
> # Assumes: thedata,fold,foldNumber,p
> foldslave <- function() {
> # Note the use of the tag for sent messages:
> # 1=ready_for_task, 2=done_task, 3=exiting
> # Note the use of the tag for received messages:
> # 1=task, 2=done_tasks
> junk <- 0
>
> done <- 0
> while (done != 1) {
> # Signal being ready to receive a new task
> mpi.send.Robj(junk,0,1)
>
> # Receive a task
> task <- mpi.recv.Robj(mpi.any.source(),mpi.any.tag())
> task_info <- mpi.get.sourcetag()
> tag <- task_info[2]
>
> if (tag == 1) {
> foldNumber <- task$foldNumber
>
> rss <- double(p)
> for (i in 1:p) {
> # produce a linear model on the first i variables on
> # training data
> templm <- lm(y~.,data=thedata[fold!=foldNumber,1:(i+1)])
>
> # produce predicted data from test data
> yhat <-
> predict(templm,newdata=thedata[fold==foldNumber,1:(i+1)])
>
> # get rss of yhat-y
> localrssresult <- sum((yhat-thedata[fold==foldNumber,1])^2)
> rss[i] <- localrssresult
> }
>
> # Send a results message back to the master
> results <- list(result=rss,foldNumber=foldNumber)
> mpi.send.Robj(results,0,2)
> }
> else if (tag == 2) {
> done <- 1
> }
> # We'll just ignore any unknown messages
> }
>
> mpi.send.Robj(junk,0,3)
> }
>
> # We're in the parent.
> # first make some data
> n <- 1000# number of obs
> p <- 30# number of variables
>
> # Create data as a set of n samples of p independent variables,
> # make a "random" beta with higher weights in the front.
> # Generate y's as y = beta*x + random
> x <- matrix(rnorm(n*p),n,p)
> beta <- c(rnorm(p/2,0,5),rnorm(p/2,0,.25))
> y <- x %*% beta + rnorm(n,0,20)
> thedata <- data.frame(y=y,x=x)
>
> fold <- rep(1:10,length=n)
> fold <- sample(fold)
>
> summary(lm(y~x))
>
> # Now, send the data to the slaves
> mpi.bcast.Robj2slave(thedata)
> mpi.bcast.Robj2slave(fold)
> mpi.bcast.Robj2slave(p)
>
> # Send the function to the slaves
> mpi.bcast.Robj2slave(foldslave)
>
> # Call the function in all the slaves to get them ready to
> # undertake tasks
> mpi.bcast.cmd(foldslave())
>
>
> # Create task list
> tasks <- vector('list')
> for (i in 1:10) {
> tasks[[i]] <- list(foldNumber=i)
> }
>
> # Create data structure to store the results
> rssresult = matrix(0,p,10)
>
> junk <- 0
> closed_slaves <- 0
> n_slaves <- mpi.comm.size()-1
>
> while (closed_slaves < n_slaves) {
> # Receive a message from a slave
> message <- mpi.recv.Robj(mpi.any.source(),mpi.any.tag())
> message_info <- mpi.get.sourcetag()
> slave_id <- message_info[1]
> tag <- message_info[2]
>
> if (tag == 1) {
> # slave is ready for a task. Give it the next task, or tell it tasks
>
> # are done if there are none.
> if (length(tasks) > 0) {
> # Send a task, and then remo
Re: [R] Help With Graphs
?text On Fri, Nov 7, 2008 at 12:42 PM, Alex99 <[EMAIL PROTECTED]> wrote: > > Hi guys, > > I am truing to draw a circle and choose 10 points on the circle and put a > number from 1 to 10 above each point. I don't have a problem with drawing a > circle and choosing 10 points. my problem is with numbering . is it even > possible to put a number above each chosen point? if so any idea how to do > it? > > ps: the circle has red dots and the selected points are in black. > > here is my code: > > D=cos(2*pi*(0:9)/10) > C=sin(2*pi*(0:9)/10) > > x1=(sin((0:100)*2*pi/100)) > y1=(cos((0:100)*2*pi/100)) > plot(x1,y1,col="red") > > points(C,D) > > Thanks, > -- > View this message in context: > http://www.nabble.com/Help-With-Graphs-tp20385465p20385465.html > Sent from the R help mailing list archive at Nabble.com. > -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] rmpi/snow/sge/openmpi help
Hi, there is a new mailing list for R and HPC: [EMAIL PROTECTED] This is probably a better list for your question. Do not forgett, first of all you have to register: https://stat.ethz.ch/mailman/listinfo/r-sig-hpc Did you tried to run openmpi and R without SGE? Like this: "orterun -n 1 R --no-save" I think openmpi / orterun do not get the correct hostfile. Therefore it starts with a default hostfile and this is only for the local machine! So all nodes are at the same machine. It normally works like this that SGE creates a hostfile and puts the path of the hostfile to a special (environment) variable. Now you have to make sure that the correct hostfile will be used. So SGE should call something like this: "ortrun -n 1 --hostfile $PATH_TO_HOSTFILE R --so-save" Best Markus jk jk wrote: > Forgive the spam. Let me try that e-mail again. > > > I'm trying to get snow working with openmpi and sge. Everything > appears to work, except the program only runs on 1 node. If i tell > snow to run on 5 nodes, it spawns 5 processes on a single node, 6 > nodes = 6 procs on 1 node and so on. I'm at a loss for ideas. Any > help would be appreciated. > > Setup: Rocks Cluster, OpenMPI, R 2.8.0, Snow, Rmpi. > > > Please ask for specifics. Thanks in advance. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- Dipl.-Tech. Math. Markus Schmidberger Ludwig-Maximilians-Universität München IBE - Institut für medizinische Informationsverarbeitung, Biometrie und Epidemiologie Marchioninistr. 15, D-81377 Muenchen URL: http://www.ibe.med.uni-muenchen.de Mail: Markus.Schmidberger [at] ibe.med.uni-muenchen.de Tel: +49 (089) 7095 - 4599 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help With Graphs
Hi guys, I am truing to draw a circle and choose 10 points on the circle and put a number from 1 to 10 above each point. I don't have a problem with drawing a circle and choosing 10 points. my problem is with numbering . is it even possible to put a number above each chosen point? if so any idea how to do it? ps: the circle has red dots and the selected points are in black. here is my code: D=cos(2*pi*(0:9)/10) C=sin(2*pi*(0:9)/10) x1=(sin((0:100)*2*pi/100)) y1=(cos((0:100)*2*pi/100)) plot(x1,y1,col="red") points(C,D) Thanks, -- View this message in context: http://www.nabble.com/Help-With-Graphs-tp20385465p20385465.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help with prediction of GLM
So, I have training data, and testing data
however, when I try to predict values for the testing data, it gives me
values for the training data. what gives?
the following is my code:
train = read.table(train.txt, header = TRUE, sep = " ")
test = read.table(test.txt, header = TRUE, sep = " ")
model = glm(formula = train[,10]~ train[2] + train[3] + ... + train[9],
family = binomial("logit"))
pred = predict(model, newdata = test.txt, type = "response")
and then I get an warning:
"Warning messages:
1: 'newdata' had 45014 rows but variable(s) found have 45001 rows
2: In predict.lm(object, newdata, se.fit, scale = 1, type = ifelse(type ==
:
prediction from a rank-deficient fit may be misleading
"
and it seems it's not reading the newdata at all, as the pred is exactly the
same whether I put
pred = predict(model, newdata = test.txt, type = "response")
or
pred = predict(model, type = "response")
thanks in advance for any help
--
View this message in context:
http://www.nabble.com/help-with-prediction-of-GLM-tp20383039p20383039.html
Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] help with syntax of random formula
Dear R-users Thanks to Jose Pinheiro, Douglas Bates and coworkes for providing R with the nlme package. Could someone help me, please, to specify a correct random formula for a mixed model, that specifies no random effect on a higher level? I have the following dataset of timeseries of respiration measurements (column resp) of biomass including a parameter initial biomass x0. Respiration measurements have been recorded in several experiments, each consisting of several replicates. str(rd) 'data.frame': 3229 obs. of 5 variables: $ suite : Factor w/ 3 levels "Face","Fal","Pushchino": 1 1 1 1 1 1 1 1 1 1 ... $ experiment: Factor w/ 44 levels "1","2","3","4",..: 1 1 1 1 1 1 1 1 1 1 ... $ replicate : Factor w/ 3 levels "1","2","3": 1 1 1 1 1 1 1 1 1 1 ... $ time : num 1.84 2.84 3.84 4.84 5.84 ... $ resp : num 4.09 3.94 3.70 3.88 4.03 ... rd <- groupedData( resp ~ time | experiment/replicate, data=rd) Time is the first level at the same level as the dependent variable resp, replicate is the second level, experiment the third. (In future suite will be used as another level. For simplicity here, I first try with data of only one suite) I want to fit a mixed model to the data which includes random effects on the replicate level but no random effects on the experiment level. The random effects should have a distribution around 0 within each experiment. I succeded to include random effects on both levels. random=list( experiment=x0~1, replicate=x0~1 ) I succeded also to fit a model to only replicate level: # replicate 1 in experiment i has nothing to do with replicate 1 in other experiment rd$exprep <- paste( rd$experiment, rd$replicate, sep="_") random=list( exprep=x0~1 ) However, this gives a single distribution across all replicates of all experiments. Instead, I want to have a distribution of random effects across only replicates of an experiment with mean zero for each experiment I tried amongst others the following syntax without success: random=list( experiment=~1, exprep=x0~1 ) random=list( experiment=1~1, exprep=x0~1 ) random=list( experiment=NULL, exprep=x0~1 ) I am using nlme version 3.1-86 on R 2.6.1 on Windows XP. Is it possible to fit such a model with nlme? What is the correct syntax? Do I need the exprep- variable or is it clear with the grouping that replicate=1 in experiment=1 is different from replicate=1 in other experiments!=1? Best regards Thomas Wutzler __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] For Loop - loading 10 sets of data and calculating
I am trying to simplify my code by adding a for loop that will load and
compute a sequence of code 10 time. They way i run it now is that the same
8 lines of code are basically reproduced 10 times. I would like to replace
the numeric value in the code (e.g. Bin1, Bin2Bin10) each time the loop
goes around. Below i tried doing this with a simple for loop and adding the
string character before each numeric value. I need to first load the data
then calculate, im sure this is possible as i have seen it done but cant
seem to reproduce it in my own code. Hope my question is clear and i hope
someone can offer some guidance.
Cheers,
JR
for (i in 1:10) {
#---
#Loads bin data frame from csv files with acres and TAZ data
Bin$i_main <-
read.csv(file="I:/Research/Samba/urb_transport_modeling/LUSDR/Workspace/BizLandPrice/data/Bin_lookup_values/Bin$i_lookup.csv",head=FALSE);
#Separates Acres data from main data and converts acres to square feet
Bin$i_Acres=Bin$i_main[[1]]*43560
#Separates TAZ data from main data
Bin$i_TAZ=Bin$i_main[[2]]
#Separates TAZ data from main data and converts acres to square feet
Bin$i_TAZvacant=Bin$i_main[[3]]*43560
#Sums each parcel acreage data of the bin
Bin$iAcres_sum=sum(Bin$i_Acres)
#Creates data frame of cumlative percentages of each parcel of bin
Bin$i_cumper=cumsum(Bin$i_Acres/Bin$iAcres_sum)
#Calculates the probability of choosing particular parcel from bin
Bin$i_parprob=abs(1-Bin$i_cumper)
#Combines parcel acreage data and cumlative percentage data
Bin$iMain.data = cbind(Bin%i_Acres,Bin$i_parprob,Bin$i_TAZ,Bin$i_TAZvacant)
}
--
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[R] determining plot location in lattice
Hi, I'm dealing with a lattice plot inserted into a tk widget and would like to know when a user has clicked on the plot area of a plot (i.e. inside the axes). For example, library(tkrplot) library(lattice) tt <- tktoplevel() makePlot <- function() print(xyplot(1 ~ 1)) printCoords <- function(x, y) print(c(x, y)) img <- tkrplot(tt, makePlot) tkbind(img, "<1>", printCoords) tkpack(img) I would like to know when a user clicks inside the axes, but don't know how to determine where the plot region is. Essentially, this comes down to answering the following question: Assuming the entire plot is on a unit square, what are the coordinates of the plotting area? This question is may seem unrelated to tkrplot, but outside of this context I may not get answers that work. I have been playing with both lattice::trellis.focus and grid::grid.locator to no avail. Thanks, --sundar __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] rmpi/snow/sge/openmpi help
Forgive the spam. Let me try that e-mail again. I'm trying to get snow working with openmpi and sge. Everything appears to work, except the program only runs on 1 node. If i tell snow to run on 5 nodes, it spawns 5 processes on a single node, 6 nodes = 6 procs on 1 node and so on. I'm at a loss for ideas. Any help would be appreciated. Setup: Rocks Cluster, OpenMPI, R 2.8.0, Snow, Rmpi. Please ask for specifics. Thanks in advance. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Election Maps
On Fri, Nov 7, 2008 at 4:02 PM, hadley wickham <[EMAIL PROTECTED]> wrote:
> The source code (in C) for this type of cartogram ("Diffusion-based
> method for producing density equalizing maps") is available from here:
> http://www-personal.umich.edu/~mejn/cart/download/
>
>From the documentation [1]:
"If you wish to make more sophisticated use of this software, such as
incorporating it into one of your own programs, then you should read
this page [2], which describes the workings in detail."
[1] http://www-personal.umich.edu/~mejn/cart/doc/
[2] http://www-personal.umich.edu/~mejn/cart/doc/description.html
Liviu
--
Do you know how to read?
http://www.alienetworks.com/srtest.cfm
Do you know how to write?
http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail
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[R] esoteric inconsistency -- intended or not?
Is the following intended or not? > func<- function(y) match.call() > z <- func(y =2) > z func(y = 2) > z[["a"]] <- 5 > z func(y = 2, 5) ## Note that the second argument **is not** named ## BUT... > z <- func(y =2) > z$a <- 5 > z func(y = 2, a = 5) ## The second argument **is** named ### End of example code ### The reason I ask is that the man page for [[ specifically says: ** Both [[ and $ select a single element of the list. The main difference is that $ does not allow computed indices, whereas [[ does. x$name is equivalent to x[["name", exact = FALSE]]. Also, the partial matching behavior of [[ can be controlled using the exact argument. [ and [[ are sometimes applied to other recursive objects such as calls and expressions. Pairlists are coerced to lists for extraction by [, but all three operators can be used for replacement. I (mis?)read this as saying the behavior in the code snippets above should produce identical results. I note that the above inconsistency can be trivially avoided by first coercing the call object to a list, modifying it either way, and then coercing it back to a call object. I doubt if it makes a difference, but: > version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status Patched major 2 minor 8.0 year 2008 month 10 day23 svn rev46779 language R version.string R version 2.8.0 Patched (2008-10-23 r46779) Cheers, Bert Gunter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] rmpi/snow/sge/openmpi help
I'm trying to snow working with openmpi and sge. Everything appears to work, except, the program only runs on 2 node. If i tell snow to run on 9 nodes, it spawns 9 processes on a single node. I'm at a loss of ideas. Any help would be appreciated. Setup: Rocks Cluster, OpenMPI, R 2.8.0, Snow, Rmpi. Please ask for specifics. Thanks in advance. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Agent-based social simulation and R
Martin Elff wrote: Hi Tom, my package 'memisc' contains a sort of an infrastructure for doing simulations. As a fun exercise I also used it to create a 'toy' agent based simulation of Schelling's neighbourhood model. Although it is not a serious application, at least it shows that agent based simulation is possible in R. Just run 'demo(schelling)' after loading 'memisc'. Interesting. I'll have to look trough the code. How far away do the agents see? Anything beyond their immidiate neighbors? Tom __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Sweave: How to load available colors from pkg 'color'
Hi: Is there a shorter way to color the abstract,sections subsections and page
headings of a document. I am using the code below in my preamble to accomplish
some of it:
[EMAIL PROTECTED],dvips]{color}
\else\usepackage[usenames,dvipsnames]{color}
% and fix pdf colour problems
\IfFileExists{pdfcolmk.sty}{\usepackage{pdfcolmk}}{}
\fi
%\renewcommand{\MakeUppercase}[1]{\color{OliveGreen}\textsf{#1}}
%\renewcommand{\abstractname}{\color{blue}Abstract}
Also,'\usepackage[usenames]{color}' is suppossed to load all the available
colors in the package 'color' but for some reason it doesn't load them when
running it trhough Sweave. It loads all the colors if I run it outside R though.
Here's a dummy document:
\documentclass[12pt]{article}
\usepackage{floatflt,graphicx,times,babel}
\usepackage[usenames]{color}
\usepackage{verbatim}
\usepackage[ps2pdf,
bookmarks=true,
bookmarksnumbered=false,
bookmarksopen=false,
colorlinks=true,
linkcolor=blue]{hyperref}
\pdfbookmark[1]{Contents}{table}
\title{BROOD-YEAR 2008 PINK SALMON RUN}
\author{Felipe D. Carrillo}
\pagestyle{headings}
\begin{document}
\setkeys{Gin}{width=0.9\textwidth}
\maketitle
\begin{abstract}
This is my abstract
\end{abstract}
\tableofcontents
\section{list of figures}
\section{introduction}
here's my introduction
\section{study area}
This is my study area
\section{methods}
subsection{sampling gear}
I used this sampling gear
\subsection{data collection}
\end{document}
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish & Wildlife Service
California, USA
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Re: [R] Re gular Expression help
Rajasekaramya wrote:
> hi there
>
> I have a vector with a set of data.I just wanna seperate them based on the
> first p and q values metioned within the data.
>
> [1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
> [2] chr1q22-q24
> [3] chr1q22-q24
> [4] chr1pter-q24
> [5] chr1pter-q24
> [6] chr1pter-q24
>
> i used a regular expression [+q*] to match up the values but it matches q
> found anywhere i know i have written like that but i jus want it to match
> the first p or q values.
>
> my result should be for q and
> [2] chr1q22-q24
> [3] chr1q22-q24
>
> for p
> [1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3
> [4] chr1pter-q24
> [5] chr1pter-q24
> [6] chr1pter-q24
>
Something like
sub("[^pq]*([pq]).*","\\1",x)
should get you the first p or q
--
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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[R] Rmpi task-pull
Hi, I'm testing the efficiency of the Rmpi package regarding parallelization
using a cluster.
I've found and tried the task pull programming method, but even if it is
described as the best method, it seems to cause deadlock, anyone could help
me in using this method?
here is the code I've found and tried:
# Initialize MPI
library("Rmpi")
# Notice we just say "give us all the slaves you've got."
mpi.spawn.Rslaves()
if (mpi.comm.size() < 2) {
print("More slave processes are required.")
mpi.quit()
}
.Last <- function(){
if (is.loaded("mpi_initialize")){
if (mpi.comm.size(1) > 0){
print("Please use mpi.close.Rslaves() to close slaves.")
mpi.close.Rslaves()
}
print("Please use mpi.quit() to quit R")
.Call("mpi_finalize")
}
}
# Function the slaves will call to perform a validation on the
# fold equal to their slave number.
# Assumes: thedata,fold,foldNumber,p
foldslave <- function() {
# Note the use of the tag for sent messages:
# 1=ready_for_task, 2=done_task, 3=exiting
# Note the use of the tag for received messages:
# 1=task, 2=done_tasks
junk <- 0
done <- 0
while (done != 1) {
# Signal being ready to receive a new task
mpi.send.Robj(junk,0,1)
# Receive a task
task <- mpi.recv.Robj(mpi.any.source(),mpi.any.tag())
task_info <- mpi.get.sourcetag()
tag <- task_info[2]
if (tag == 1) {
foldNumber <- task$foldNumber
rss <- double(p)
for (i in 1:p) {
# produce a linear model on the first i variables on
# training data
templm <- lm(y~.,data=thedata[fold!=foldNumber,1:(i+1)])
# produce predicted data from test data
yhat <-
predict(templm,newdata=thedata[fold==foldNumber,1:(i+1)])
# get rss of yhat-y
localrssresult <- sum((yhat-thedata[fold==foldNumber,1])^2)
rss[i] <- localrssresult
}
# Send a results message back to the master
results <- list(result=rss,foldNumber=foldNumber)
mpi.send.Robj(results,0,2)
}
else if (tag == 2) {
done <- 1
}
# We'll just ignore any unknown messages
}
mpi.send.Robj(junk,0,3)
}
# We're in the parent.
# first make some data
n <- 1000# number of obs
p <- 30# number of variables
# Create data as a set of n samples of p independent variables,
# make a "random" beta with higher weights in the front.
# Generate y's as y = beta*x + random
x <- matrix(rnorm(n*p),n,p)
beta <- c(rnorm(p/2,0,5),rnorm(p/2,0,.25))
y <- x %*% beta + rnorm(n,0,20)
thedata <- data.frame(y=y,x=x)
fold <- rep(1:10,length=n)
fold <- sample(fold)
summary(lm(y~x))
# Now, send the data to the slaves
mpi.bcast.Robj2slave(thedata)
mpi.bcast.Robj2slave(fold)
mpi.bcast.Robj2slave(p)
# Send the function to the slaves
mpi.bcast.Robj2slave(foldslave)
# Call the function in all the slaves to get them ready to
# undertake tasks
mpi.bcast.cmd(foldslave())
# Create task list
tasks <- vector('list')
for (i in 1:10) {
tasks[[i]] <- list(foldNumber=i)
}
# Create data structure to store the results
rssresult = matrix(0,p,10)
junk <- 0
closed_slaves <- 0
n_slaves <- mpi.comm.size()-1
while (closed_slaves < n_slaves) {
# Receive a message from a slave
message <- mpi.recv.Robj(mpi.any.source(),mpi.any.tag())
message_info <- mpi.get.sourcetag()
slave_id <- message_info[1]
tag <- message_info[2]
if (tag == 1) {
# slave is ready for a task. Give it the next task, or tell it tasks
# are done if there are none.
if (length(tasks) > 0) {
# Send a task, and then remove it from the task list
mpi.send.Robj(tasks[[1]], slave_id, 1);
tasks[[1]] <- NULL
}
else {
mpi.send.Robj(junk, slave_id, 2)
}
}
else if (tag == 2) {
# The message contains results. Do something with the results.
# Store them in the data structure
foldNumber <- message$foldNumber
rssresult[,foldNumber] <- message$result
}
else if (tag == 3) {
# A slave has closed down.
closed_slaves <- closed_slaves + 1
}
}
# plot the results
plot(apply(rssresult,1,mean))
mpi.close.Rslaves()
mpi.quit(save="no")
Thanks for your help!!!
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[R] Re gular Expression help
hi there I have a vector with a set of data.I just wanna seperate them based on the first p and q values metioned within the data. [1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3 [2] chr1q22-q24 [3] chr1q22-q24 [4] chr1pter-q24 [5] chr1pter-q24 [6] chr1pter-q24 i used a regular expression [+q*] to match up the values but it matches q found anywhere i know i have written like that but i jus want it to match the first p or q values. my result should be for q and [2] chr1q22-q24 [3] chr1q22-q24 for p [1] chr10p15.3 /// chr3q29 /// chr4q35 /// chr9q34.3 [4] chr1pter-q24 [5] chr1pter-q24 [6] chr1pter-q24 -- View this message in context: http://www.nabble.com/Regular-Expression-help-tp20385971p20385971.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vectorizing sample()
Hi,
I'm not quite sure I understood everything but is this something close?
d <- read.table(textConnection("Dad_ID SpouseYN NKids NSick
1 10 1
2 02 2
3 10 2
4 13 3"), header=TRUE)
mapply(sample, d$NKids+d$SpouseYN+1, d$NSick)
# d$NKids+d$SpouseYN+1 is supposed to be family size
Best regards,
Kenn
On Fri, Nov 7, 2008 at 7:00 PM, Stephen Collins <[EMAIL PROTECTED]>wrote:
> I am simulating sickness among a group of families. Part of the task is
> to randomly draw who in the family will be sick, randomly drawing from
> family ID's where Dad =1, Mom = 2, Kid1 = 3, Kid2 = 4., etc. My census of
> Dads is of the form shown below.
>
> Dad_ID Spouse (Y=1;N=0)#Kids #People_Becoming_Sick
> 1 1 0 1
> 2 0 2 2
> 3 1 0 2
> 4 1 3 3
> ...
>
> The end output needed is if 3 people in a family are to be sick, was it
> the dad and two kids, with random family ID's = {1,3,4}, or the mom, dad,
> and one kid, with random family ID's = {2,1,4}, etc.. The complication
> is that length of the family ID's to choose from and the associated
> sampling probabilities -- changes with each family. I could loop through
> the Dads, from i in 1:nrow(census), but is there a way I could vectorize
> sample() to get at the same objective?
>
> My attempts to use the apply-based functions have dead ended. Other ideas
> to vectorize this problem are warmly welcomed.
>
>
>
> Regards,
>
> Stephen Collins, MPP | Analyst
> Health & Benefits | Aon Consulting
> 200 East Randolph, Suite 900, Chicago, IL
> Tel: 312-381-2578 | Fax: 312-381-0136
> Email: [EMAIL PROTECTED]
>
> Aon Consulting selected by the readers of Business Insurance as the âBest
> Employee Benefit Consulting Firmâ in 2006, 2007, and 2008
>
> NOTE: The information contained in this transmission, including any
> attachment(s) is only for the use of the intended individual(s) or entity,
> and may contain information that is privileged and confidential. If the
> reader of this message is not an intended recipient, you are hereby
> notified that any dissemination, distribution, disclosure, or copying of
> this information is unauthorized and strictly prohibited. If you have
> received this communication in error, please contact the sender
> immediately by reply email and destroy all copies of the original message.
>
>
>[[alternative HTML version deleted]]
>
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
>
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Re: [R] help with maps : center on the dateline
On 11/7/2008 12:08 PM, [EMAIL PROTECTED] wrote: Hi, I'm trying to plot a map of the pacific ocean, centered on the dateline, using the maps package. library(maps) # Basic library to draw maps library(mapdata)# Library with specialized maps library(mapproj) map(database = "world", fill = TRUE, col = 1, plot = TRUE,add=F, xlim = c(120,300), ylim = c(-20,40)) I tried several different combinations of coordinates, but whatever I try, it doesn't plot anything past the dateline. Is there a way to plot a map like the one attached, with this package ? Use the world2 (or world2Hires, from mapdata) database. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Vectorizing sample()
On 11/7/2008 12:00 PM, Stephen Collins wrote:
I am simulating sickness among a group of families. Part of the task is
to randomly draw who in the family will be sick, randomly drawing from
family ID's where Dad =1, Mom = 2, Kid1 = 3, Kid2 = 4., etc. My census of
Dads is of the form shown below.
Dad_ID Spouse (Y=1;N=0)#Kids #People_Becoming_Sick
1 1 0 1
2 0 2 2
3 1 0 2
4 1 3 3
...
The end output needed is if 3 people in a family are to be sick, was it
the dad and two kids, with random family ID's = {1,3,4}, or the mom, dad,
and one kid, with random family ID's = {2,1,4}, etc.. The complication
is that length of the family ID's to choose from and the associated
sampling probabilities -- changes with each family. I could loop through
the Dads, from i in 1:nrow(census), but is there a way I could vectorize
sample() to get at the same objective?
My attempts to use the apply-based functions have dead ended. Other ideas
to vectorize this problem are warmly welcomed.
You might want to transform runif instead of using sample(). For
example, if you want to generate M integers from 1:n_i, where n_i varies
from sample to sample, you could use
gen <- ceiling(runif(M, 0, n))
(where n is a vector of length M giving the upper limits).
Duncan Murdoch
Regards,
Stephen Collins, MPP | Analyst
Health & Benefits | Aon Consulting
200 East Randolph, Suite 900, Chicago, IL
Tel: 312-381-2578 | Fax: 312-381-0136
Email: [EMAIL PROTECTED]
Aon Consulting selected by the readers of Business Insurance as the “Best
Employee Benefit Consulting Firm� in 2006, 2007, and 2008
NOTE: The information contained in this transmission, including any
attachment(s) is only for the use of the intended individual(s) or entity,
and may contain information that is privileged and confidential. If the
reader of this message is not an intended recipient, you are hereby
notified that any dissemination, distribution, disclosure, or copying of
this information is unauthorized and strictly prohibited. If you have
received this communication in error, please contact the sender
immediately by reply email and destroy all copies of the original message.
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[R] help with maps : center on the dateline
Hi, I'm trying to plot a map of the pacific ocean, centered on the dateline, using the maps package. library(maps) # Basic library to draw maps library(mapdata)# Library with specialized maps library(mapproj) map(database = "world", fill = TRUE, col = 1, plot = TRUE,add=F, xlim = c(120,300), ylim = c(-20,40)) I tried several different combinations of coordinates, but whatever I try, it doesn't plot anything past the dateline. Is there a way to plot a map like the one attached, with this package ? thanks ! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] grouped and stacked plot
Dear R users, Beside lattice and vcd packages, what is my option if I want to create a grouped and stacked categorical plots? With lattice, what I usually do is breaking my data from a wide form to a long form (using subset and rbind), such as: TypeID - Var 1 - Var 2 - Var 3 - Val to TypeID+Var1 - Val TypeID+Var2 - Val ... therefore, if I use auto.key = TRUE, I will have permutation as Type* Number of Vars. I think I have to use similar process with vcd as well, using doubledecker or mosaic plot. Or, is there a different way/setting to create grouped and stacked using lattice or vcd that I am not accustomed (which is very like, since I am such as newbie with R, much less with R-graphics). Any idea is appreciated, Thanks, Ferry [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] AIC value in lmer
Dear R Users, May be this message should be directy send to Douglas Bates ... I just want to know if I can use the AIC value given in the output of an lmer model to classify my logistic models. I heard that the AIC value given in GLIMMIX output (SAS) is false because it come from a calculation based on pseudo-likelyhood. Is it the same for lmer ??? thanks, Arnaud Arnaud MOSNIER Biologiste Ph.D. Département de Biologie Université du Québec à Rimouski 300 Allée des Ursulines Rimouski, Québec, G5L 3A1 Tél. : 723-1986 poste 1653 E-mail : [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Vectorizing sample()
I am simulating sickness among a group of families. Part of the task is
to randomly draw who in the family will be sick, randomly drawing from
family ID's where Dad =1, Mom = 2, Kid1 = 3, Kid2 = 4., etc. My census of
Dads is of the form shown below.
Dad_ID Spouse (Y=1;N=0)#Kids #People_Becoming_Sick
1 1 0 1
2 0 2 2
3 1 0 2
4 1 3 3
...
The end output needed is if 3 people in a family are to be sick, was it
the dad and two kids, with random family ID's = {1,3,4}, or the mom, dad,
and one kid, with random family ID's = {2,1,4}, etc.. The complication
is that length of the family ID's to choose from and the associated
sampling probabilities -- changes with each family. I could loop through
the Dads, from i in 1:nrow(census), but is there a way I could vectorize
sample() to get at the same objective?
My attempts to use the apply-based functions have dead ended. Other ideas
to vectorize this problem are warmly welcomed.
Regards,
Stephen Collins, MPP | Analyst
Health & Benefits | Aon Consulting
200 East Randolph, Suite 900, Chicago, IL
Tel: 312-381-2578 | Fax: 312-381-0136
Email: [EMAIL PROTECTED]
Aon Consulting selected by the readers of Business Insurance as the âBest
Employee Benefit Consulting Firmâ in 2006, 2007, and 2008
NOTE: The information contained in this transmission, including any
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Re: [R] Bootstrapping gnls models
Dear all,
Here comes a reproducible example, with the original data added.
The error message when running boot() is:
Error in gnls(response.variable ~ a * LD/(b + LD), params <- list(a + :
Step halving factor reduced below minimum in NLS step
boot() in this case only seems to work for very small values of R (say, within [1..5]), which is of
course not desirable. Maybe this is due to problems with model convergence.
I would very much appreciate any help.
###
LD=c(4.087462841,2.321928095,4.087462841,1,1,4.087462841,2.321928095,1,1,2.321928095,4.087462841,2.321928095,5.930737338,2.321928095,5.930737338,1,1,2.321928095,2.321928095,4.087462841,1,1,2.321928095,4.087462841,4.087462841,1,2.321928095,1,4.087462841,2.321928095,1,2.321928095,5.930737338,4.087462841,1,4.087462841,2.321928095,4.087462841,5.930737338,4.087462841,1,2.321928095,2.321928095,1,2.321928095,1,1,4.087462841,4.087462841,2.321928095)
L=c(1,1,0,1,0,0,1,0,0,0,1,0,1,1,1,0,0,1,0,0,0,1,1,1,1,0,1,0,0,0,1,0,1,1,0,1,1,1,1,0,0,1,1,1,1,0,0,1,1,0)
response.variable=c(0.335737179487179,0.391025641025641,0.391826923076923,0.487980769230769,0.294070512820513,0.507211538461538,0.3958333,0.0825320512820513,0.443108974358974,0.290064102564103,0.59775641025641,0.514423076923077,0.65625,0.193910256410256,0.4479167,0,0.2604167,0.407852564102564,0.44150641025641,0.596153846153846,0.0600961538461538,0.21875,0.256410256410256,0.3645833,0.435096153846154,0.0793269230769231,0.249198717948718,0.0304487179487179,0.230769230769231,0.485576923076923,0.684294871794872,0.0737179487179487,0.490384615384615,0.599358974358974,0.215544871794872,0.219551282051282,0.602564102564103,0.907852564102564,0.391025641025641,0.43349358974359,0.0384615384615385,0.337339743589744,0.502403846153846,0.405448717948718,1,0.362980769230769,0.116185897435897,0.459134615384615,0.661057692307692,0.0769230769230769)
a=data.frame(LD,L,response.variable)
model=gnls(model = response.variable ~ a * LD/(b + LD), data = a,
params = list(a + b ~ L), start = c(1, 1, 1, 1))
df<-cbind(a,fit<-as.numeric(predict(model,list(LD=1,L=0.5
rs<-scale(resid(model),scale=F)
model.bootfunc<-function(rs,i){
df$response.variable<-df$fit+rs[i]
as.numeric(predict(gnls(response.variable ~ a * LD/(b + LD),
params <- list(a + b ~ L), start = coef(model), data=df)))
}
(model.boot<-boot(rs,model.bootfunc,R=100))
booted<-boot.ci(model.boot,index=1,type=c("norm"))
booted$t0
booted$normal
###
Best wishes,
Christoph.
Prof Brian Ripley schrieb:
On Fri, 7 Nov 2008, Christoph Scherber wrote:
Dear all,
I am trying to bootstrap predictions from gnls models using the
following code:
# a is the dataframe with which I am working; it contains the variables
# response.variable,LD,L,G,P and F
And without it your code is not reproducible.
###
model=gnls(response.variable ~ a * LD/(b + LD),
params = list(a + b ~ L), start = c(1,1,1,1), data=a)
df=cbind(a,fit=predict(model,list(LD=1,L=0.5,G=0.5,P=0.46,F=2.2)))
model.bootfunc=function(rs,i){
df$response.variable=df$fit+rs[i]
as.numeric(predict(gnls(response.variable ~ a * LD/(b + LD),
params = list(a + b ~ L), start = coef(model), data=df)))
}
rs=scale(resid(model),scale=F)
(model.boot=boot(rs,model.bootfunc,R=1))
booted=boot.ci(model.boot,index=1,type=c("norm","basic","perc","bca"))
Do please try to make your code readable, using spaces and <- for
assignments. I would have spotted the problem much sooner with legible,
reproducible code.
###
The problem is that this code yields "NA" for the s.e. of the
bootstrap statistics:
Bootstrap Statistics :
original biasstd. error
t1* 0.1651658 -0.020663364 NA
t2* 0.1669592 -0.021759335 NA
t3* 0.1676765 -0.001858686 NA
t4* 0.1726982 -0.025321349 NA
t5* 0.1658092 0.024721214 NA
And hence the boot.ci function and others don?t work.
Does anyone have an idea on that?
Yes: how can you estimate standard errors from a single sample (you set
R=1)?
Many thanks and best wishes
Christoph
--
Dr. rer.nat. Christoph Scherber
University of Goettingen
DNPW, Agroecology
Waldweg 26
D-37073 Goettingen
Germany
phone +49 (0)551 39 8807
fax +49 (0)551 39 8806
Homepage http://www.gwdg.de/~cscherb1
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Re: [R] Bootstrapping gnls models
On Fri, 7 Nov 2008, Christoph Scherber wrote:
Dear all,
I am trying to bootstrap predictions from gnls models using the following
code:
# a is the dataframe with which I am working; it contains the variables
# response.variable,LD,L,G,P and F
And without it your code is not reproducible.
###
model=gnls(response.variable ~ a * LD/(b + LD),
params = list(a + b ~ L), start = c(1,1,1,1), data=a)
df=cbind(a,fit=predict(model,list(LD=1,L=0.5,G=0.5,P=0.46,F=2.2)))
model.bootfunc=function(rs,i){
df$response.variable=df$fit+rs[i]
as.numeric(predict(gnls(response.variable ~ a * LD/(b + LD),
params = list(a + b ~ L), start = coef(model), data=df)))
}
rs=scale(resid(model),scale=F)
(model.boot=boot(rs,model.bootfunc,R=1))
booted=boot.ci(model.boot,index=1,type=c("norm","basic","perc","bca"))
Do please try to make your code readable, using spaces and <- for
assignments. I would have spotted the problem much sooner with legible,
reproducible code.
###
The problem is that this code yields "NA" for the s.e. of the bootstrap
statistics:
Bootstrap Statistics :
original biasstd. error
t1* 0.1651658 -0.020663364 NA
t2* 0.1669592 -0.021759335 NA
t3* 0.1676765 -0.001858686 NA
t4* 0.1726982 -0.025321349 NA
t5* 0.1658092 0.024721214 NA
And hence the boot.ci function and others don?t work.
Does anyone have an idea on that?
Yes: how can you estimate standard errors from a single sample (you set
R=1)?
Many thanks and best wishes
Christoph
--
Dr. rer.nat. Christoph Scherber
University of Goettingen
DNPW, Agroecology
Waldweg 26
D-37073 Goettingen
Germany
phone +49 (0)551 39 8807
fax +49 (0)551 39 8806
Homepage http://www.gwdg.de/~cscherb1
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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.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] Unexpected behavior of clocktime related to daylight savings time
On Fri, 7 Nov 2008, Dennis Fisher wrote:
Colleagues,
I submitted this several days ago and no one responded, so I am trying
again, trying a different subject line:
Well, you posted something that indicated you had not studied the relevant
help pages, without the information requested in the R posting guide, and
with an HTML posting.
I just encountered some unexpected behavior of difftime in
relationship to the change from daylight savings to standard time.
My understanding is that DST and ST take effect at 2AM. However, the
results below suggests that R (version 2.8.0 in OS X) implements the
change at 2:16AM:
The transition time depends on the country (and in some cases, the year).
In the EU it is at 2am (and always has been, not that the EU is very old).
Expected:
difftime("2008-11-02 02:01:00", "2008-11-02 00:59:00")
Time difference of 2.03 hours
difftime("2008-11-02 01:16:00", "2008-11-02 01:15:00")
Time difference of 1 mins
difftime("2008-11-02 01:18:00", "2008-11-02 01:17:00")
Time difference of 1 mins
Not expected:
difftime("2008-11-02 01:17:00", "2008-11-02 01:16:00")
Time difference of 1.016667 hours
Can anyone explain this?
Yes, and you should have been able to do so from the information in the
help pages. As ?strptime says
Remember that in most timezones some times do not occur and some
occur twice because of transitions to/from summer time. What
happens in those cases is OS-specific.
See also the note on ?Sys.timezone. No OS I tried did this, not even Mac
OS X set to EST5EDT. But in EST5EDT 2008-11-02 01:17:00 occurred twice:
as.POSIXct("2008-11-02 01:17:00")
[1] "2008-11-02 01:17:00 EDT"
as.POSIXct("2008-11-02 01:17:00") + 3600
[1] "2008-11-02 01:17:00 EST"
so 1.016667 hours would be one of two correct answers in that timezone.
However, you didn't tell us what timezone you are in and hence we cannot
know when DST transitions occur in that timezone.
Dennis
Dennis Fisher MD
P < (The "P Less Than" Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-415-564-2220
www.PLessThan.com
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] clogit and small sample sizes: what to do?
-BEGIN PGP SIGNED MESSAGE- Hash: SHA1 Conditional logistic regression and small sample sizes: what to do? Dear R-Gurus, [I run R 2.8.0 on WinXP. I have no formal training in statistics.] Please feel free to skip the 'Blah blah part' for questions below. - ---Blah blah part Why I use clogit (package 'survival')?: I am looking for risk factors for cases during a recent small "flare-up" of disease in a hospital. To this end I matched 11 cases (no more of them, sorry!) with 33 controls based on sex. The next step was to perform conditional logistic regression to evaluate risk 'exposures' for cases. Now clogit() (Package 'survival') produces some conflicting results: Briefly, 10 of 11 cases have a certain exposure, compared to 9 of 33 controls. The odds ratio is significant (p-value = 0.000311) while the output of clogit() is inconclusive (.95 confidence intervals for risk factor range from 0 to infinity, conflicting 'likelihood ratio test' and 'Wald test') plus produces following warning message, "Warning message: In fitter(X, Y, strats, offset, init, control, weights = weights,: Ran out of iterations and did not converge." Tampering with the data (i.e. reducing the number of cases with the exposure to 7 of 11) renders the odds ratio unsignificant (p-value = 0.06681), yet makes clogit() assign a significant p-value to the exposure in question (0.043). Furthermore, the warning message is gone after the tampering. Interestingly reducing (instead of increasing) cases with the risk factor yields a significant result with clogit(). I believe now, that this behaviour might be due to small sample size. Now I am looking for an R package to maybe circumvent this limitation of clogit(). (Alternatively, I could just chuck the project and order a Margerita. Maybe the risk factor is plainly not significant after all. But I digress...) Six years ago, the following message was posted to this list: -old post 2002 Dear R folks, We completed a matched case-control study that was analyzed using conditional logistic regression. Because of the small sample size we need to calculate exact confidence intervals. The quick solution is to purchase LogExact by Cytel. However, we'd like to do this in R. Anyone have experience with this? Many thanks, Tomas Tomas Aragon, MD, DrPH -old post 2002 end At that time, the answers were not too encouraging. Anyway, I might have the very same query 2008. Ok, I tried to do some homework and started to look for packages maybe implementing different methods. I did run across 'elrm' (Zamar, D., McNeney, B. & Graham, J., 2007), but I am unsure whether this package is appropriate for my original 1:3 matched design. - ---Blah blah part end - ---Questions, finally!--- So my questions to you are: 1) Is the problem I describe above likely related to small sample sizes? 2) Do you have experience with 'elrm' or packages finding a solution for the above problem? Is 1:n matching as in clogit() possible? - ---Questions, finally! end--- Thank you for your time and effort, Johannes -BEGIN PGP SIGNATURE- Version: GnuPG v1.4.9 (MingW32) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org iEYEARECAAYFAkkUZLsACgkQeixPRMMDSjthxwCgpe0ACd6qrMVCz60yzVfzML2K EpMAnRnrfbWDrZ29TLFKMCe+bH/YrNZI =hb1P -END PGP SIGNATURE- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Umlaut read from csv-file
On Fri, 7 Nov 2008, Peter Dalgaard wrote: Heinz Tuechler wrote: Dear Prof.Ripley! Thank you very much for your attention. In the given example Encoding(), or the encoding parameter of read.csv solve the problem. I hope your patch will solve also the problem, when I read a spss file by spss.get(), since this function has no encoding parameter and my real problem originated there. read.spss() (package foreign) does have a reencode argument, though; and this is called by spss.get(), so it looks like an easy hack to add it there. Yes, older software like spss.get needs to get updated for the internationalization age. Modifying it to have a ... argument passed to read.spss would be a good idea (and future-proofing). In cases like this it is likely that the SPSS file does contain its encoding (although sometimes it does not and occasionally it is wrong), so it is helpful to make use of the info if it is there. However, the default is read.spss(reencode=NA) because of the problems of assuming that the info is correct when it is not are worse. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] [newbie] scatterplot with marginal histograms (done) and axes labels
Hello,
sometimes it helps if I formulate my problem. I tried the following after I
posted my previous post.
scatterhist=function(x,y, xlab="", ylab=""){
#oldpar = par(no.readonly = TRUE) # save default, for resetting...
zones=matrix(c(2,0,1,3),ncol=2,byrow=TRUE)
layout(zones,widths=c(4/5,1/5),heights=c(1/5,4/5))
xhist <- hist(x, plot=FALSE)
yhist <- hist(y, plot=FALSE)
top <- max(c(xhist$counts, yhist$counts))
par(mar=c(3,3,1,1))
plot(x,y)
par(mar=c(0,3,1,1))
barplot(xhist$counts, axes=FALSE, ylim=c(0, top), space=0)
par(mar=c(3,0,1,1))
barplot(yhist$counts, axes=FALSE, xlim=c(0, top), space=0, horiz=TRUE)
par(oma=c(3,3,0,0))
mtext(xlab, side=1, line=1, outer=TRUE, adj=0)
mtext(ylab, side=2, line=1, outer=TRUE, adj=0)
#par(oldpar) # reset to default
}
This works quite nice in the R Gui (the labels appear, but it is not easy
for me to center them). However, when I use the function in a Sweave
context, i.e.
\begin{Scode}{fig=true, echo=false, results=hide}
print(scatterhist(x,y, "x", "y"))
\end{Scode}
the labels still do not appear...
Kind regards,
Karsten.
KarstenW wrote:
>
> Hello,
>
> I am stuck when I want to add axes labels to my scatterplot with
> histograms. I guess it must be something with par(mar=) or so, but could
> someone give me a hint?
>
> Here is what I got so far:
>
> # adapted from
> http://www.stat.ucl.ac.be/ISdidactique/comment/fichiers/r/scatterhist.rs
> and from ?layout
> #
> # see also:
> # - http://biom1.univ-lyon1.fr/ADE-4/ade4-html/s.hist.html (a function
> called s.hist in package ade4)
> # - http://www.statmethods.net/graphs/scatterplot.html (shows how to
> do some nice scatterplots,
> # e.g. with the function scatterplot in package car)
>
> scatterhist=function(x,y, xlab="", ylab=""){
> #oldpar = par(no.readonly = TRUE) # save default, for resetting...
> zones=matrix(c(2,0,1,3),ncol=2,byrow=TRUE)
> layout(zones,widths=c(4/5,1/5),heights=c(1/5,4/5))
> xhist <- hist(x, plot=FALSE)
> yhist <- hist(y, plot=FALSE)
> top <- max(c(xhist$counts, yhist$counts))
> par(mar=c(3,3,1,1))
> plot(x,y, xlab=xlab, ylab=ylab)
> par(mar=c(0,3,1,1))
> barplot(xhist$counts, axes=FALSE, ylim=c(0, top), space=0)
> par(mar=c(3,0,1,1))
> barplot(yhist$counts, axes=FALSE, xlim=c(0, top), space=0, horiz=TRUE)
> #par(oldpar) # reset to default
> }
>
> Now if I do
>> x=rnorm(50)
>> y=rnorm(50)
>> scatterhist(x,y,xlab="x", ylab="y")
>
> there are no axes labels.
> I also wonder why R complains if I uncomment the first and last line in
> the function ("invalid value
> for parameter fig specified").
>
> Any hint appreciated,
> Karsten.
>
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[R] [newbie] scatterplot with marginal histograms (done) and axes labels
Hello,
I am stuck when I want to add axes labels to my scatterplot with histograms.
I guess it must be something with par(mar=) or so, but could someone give
me a hint?
Here is what I got so far:
# adapted from
http://www.stat.ucl.ac.be/ISdidactique/comment/fichiers/r/scatterhist.rs and
from ?layout
#
# see also:
# - http://biom1.univ-lyon1.fr/ADE-4/ade4-html/s.hist.html (a function
called s.hist in package ade4)
# - http://www.statmethods.net/graphs/scatterplot.html (shows how to do
some nice scatterplots,
# e.g. with the function scatterplot in package car)
scatterhist=function(x,y, xlab="", ylab=""){
#oldpar = par(no.readonly = TRUE) # save default, for resetting...
zones=matrix(c(2,0,1,3),ncol=2,byrow=TRUE)
layout(zones,widths=c(4/5,1/5),heights=c(1/5,4/5))
xhist <- hist(x, plot=FALSE)
yhist <- hist(y, plot=FALSE)
top <- max(c(xhist$counts, yhist$counts))
par(mar=c(3,3,1,1))
plot(x,y, xlab=xlab, ylab=ylab)
par(mar=c(0,3,1,1))
barplot(xhist$counts, axes=FALSE, ylim=c(0, top), space=0)
par(mar=c(3,0,1,1))
barplot(yhist$counts, axes=FALSE, xlim=c(0, top), space=0, horiz=TRUE)
#par(oldpar) # reset to default
}
Now if I do
> x=rnorm(50)
> y=rnorm(50)
> scatterhist(x,y,xlab="x", ylab="y")
there are no axes labels.
I also wonder why R complains if I uncomment the first and last line in the
function ("invalid value
for parameter fig specified").
Any hint appreciated,
Karsten.
--
View this message in context:
http://www.nabble.com/-newbie--scatterplot-with-marginal-histograms-%28done%29-and-axes-labels-tp20381450p20381450.html
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and provide commented, minimal, self-contained, reproducible code.
[R] x axe values
Dear list, I have to draw a simple plot. On y axe some numerical values that correspond to various categories on axe x. The table I am reading looks like: cat Obj1 Obj2 Obj3 max 23 27 34 ave 21 25 32 min 19 23 30 In order to avoid that the first column is reordered alphabetically I used: (found here http://tolstoy.newcastle.edu.au/R/help/06/09/33808.html) a <- as.character(table$cat) table$cat = factor(a, levels=a) However if I use plot() I get thick lines instead of points because of the reason explained in the link above. If I use plot.default() I get number on the x axe instead of the name of the categories. 1) first question: any solution? An alternative could be to use plot.default() and then to change the value on the axe. 2) second question: how to chnage the value of the axes? not the label, but the values themself? Thank you very much in advance Laura [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Election Maps
On Fri, Nov 7, 2008 at 7:53 AM, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 11/7/2008 8:31 AM, roger koenker wrote:
>>
>> Those of you with an interest in the US election and/or
>> statistical graphics may find the maps at:
>>
>>http://www-personal.umich.edu/~mejn/election/2008/
>>
>> interesting.
>
>
> Nice stuff. Do you know if anyone has ported the cartogram code to R? I see
> a question on the list a couple of years ago
>
> https://stat.ethz.ch/pipermail/r-help/2006-May/106501.html
>
> but I don't see a positive answer...
The source code (in C) for this type of cartogram ("Diffusion-based
method for producing density equalizing maps") is available from here:
http://www-personal.umich.edu/~mejn/cart/download/
Hadley
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Re: [R] Umlaut read from csv-file
At 13:34 07.11.2008, Peter Dalgaard wrote:
Heinz Tuechler wrote:
> Dear Prof.Ripley!
>
> Thank you very much for your attention. In the given example Encoding(),
> or the encoding parameter of read.csv solve the problem. I hope your
> patch will solve also the problem, when I read a spss file by
> spss.get(), since this function has no encoding parameter and my real
> problem originated there.
read.spss() (package foreign) does have a reencode argument, though; and
this is called by spss.get(), so it looks like an easy hack to add it
there.
Thank you, that means, I have to change spss.get
to make it accept the reencode argument and pass
it to read.spss. At the moment I prefer to step
back to R 2.7.2 and to wait for a more general
solution, because to me, there seem to be still strange effects of encoding.
In the following example the encoding gets lost
by dumping and rereading, even if I use the
encoding parameter of source(). But may be, I
don't understand what this parameter should do.
Heinz Tüchler
us <- c("a", "b", "c", "ä", "ö", "ü")
Encoding(us)
[1] "unknown" "unknown" "unknown" "latin1" "latin1" "latin1"
dump('us', 'us_dump.txt')
rm(us)
source('us_dump.txt', encoding='latin1')
us
[1] "a" "b" "c" "ä" "ö" "ü"
Encoding(us)
[1] "unknown" "unknown" "unknown" "unknown" "unknown" "unknown"
unlink('us_dump.txt')
--
O__ Peter Dalgaard Ãster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907
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Re: [R] Applying a function to a list of arguments ...
Try this:
func <- function(f, ...) f(...)
# e.g.
func(sin, 0) # same as sin(0)
func(max, 1, 2) # same as max(1, 2)
On Fri, Nov 7, 2008 at 5:21 AM, <[EMAIL PROTECTED]> wrote:
> How can I apply function f, that I get as an argument as in
>
> func <- function(f, ...) {
> .
> .
> .
> }
>
> to a list of arguments list(a, b, c) (eg the ... argument of func above)
> in order to obtain
>
> f(a, b, c)
>
> Thanks a lot,
>
> Roberto
>
>[[alternative HTML version deleted]]
>
> __
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>
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Re: [R] Election Maps
On 11/7/2008 8:31 AM, roger koenker wrote: Those of you with an interest in the US election and/or statistical graphics may find the maps at: http://www-personal.umich.edu/~mejn/election/2008/ interesting. Nice stuff. Do you know if anyone has ported the cartogram code to R? I see a question on the list a couple of years ago https://stat.ethz.ch/pipermail/r-help/2006-May/106501.html but I don't see a positive answer... Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Unexpected behavior of clocktime related to daylight savings time
Colleagues,
I submitted this several days ago and no one responded, so I am trying
again, trying a different subject line:
I just encountered some unexpected behavior of difftime in
relationship to the change from daylight savings to standard time.
My understanding is that DST and ST take effect at 2AM. However, the
results below suggests that R (version 2.8.0 in OS X) implements the
change at 2:16AM:
Expected:
> > difftime("2008-11-02 02:01:00", "2008-11-02 00:59:00")
> Time difference of 2.03 hours
> > difftime("2008-11-02 01:16:00", "2008-11-02 01:15:00")
> Time difference of 1 mins
> > difftime("2008-11-02 01:18:00", "2008-11-02 01:17:00")
> Time difference of 1 mins
Not expected:
> > difftime("2008-11-02 01:17:00", "2008-11-02 01:16:00")
> Time difference of 1.016667 hours
Can anyone explain this?
Dennis
Dennis Fisher MD
P < (The "P Less Than" Company)
Phone: 1-866-PLessThan (1-866-753-7784)
Fax: 1-415-564-2220
www.PLessThan.com
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Re: [R] Applying a function to a list of arguments ...
On 11/7/2008 5:40 AM, baptiste auguie wrote:
perhaps something like,
func <- function(f, ...) {
do.call(f, ...)
}
func(rnorm, list(n=3, mean=2, sd=3))
Alternatively, if the caller doesn't want to put the args in a list,
your func can do it:
func2 <- function(f, ...) {
do.call(f, list(...))
}
func2(rnorm, n=3, mean=2, sd=3)
baptiste
On 7 Nov 2008, at 10:21, [EMAIL PROTECTED] wrote:
How can I apply function f, that I get as an argument as in
func <- function(f, ...) {
.
.
.
}
to a list of arguments list(a, b, c) (eg the ... argument of func
above)
in order to obtain
f(a, b, c)
Thanks a lot,
Roberto
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_
Baptiste Auguié
School of Physics
University of Exeter
Stocker Road,
Exeter, Devon,
EX4 4QL, UK
Phone: +44 1392 264187
http://newton.ex.ac.uk/research/emag
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Re: [R] method to return rgb values from pixels of an image
On 11/7/2008 3:24 AM, Hans-Joachim Klemmt wrote: hello, i am looking for a method to return rgb-values of predifined pixels of jpg images. can anybody help me? See the rimage package, with function read.jpeg. It will read the whole file into a large array, with separate red, green, blue entries. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Re move specific rows from a matrix
Hi, imagine the following two matrix: Matrix A: a b c 1 2 3 4 5 6 7 8 9 Matrix B: a 4 7 I would like to remove those rows from matrix A which are present in both matrices. So after removing the corresponding rows the matrix A should look like this: Matrix A: a b c 1 2 3 Thanks in advance! -- View this message in context: http://www.nabble.com/Remove-specific-rows-from-a-matrix-tp20379566p20379566.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] two kind of Hosmer and Lemeshow’s test
I know that there are two method to apply the Hosmer and Lemeshow’s. One of
them is calculated based on the fixed and pre-determined cut-off points of
the estimated probability of success. One of them is calculated based on
the percentiles of estimated probabilities.
In the previous post,i find that the Hosmer and Lemeshow’s test how to use
in R.
hosmerlem <-
function (y, yhat, g = 10)
{
cutyhat <- cut(x, breaks = quantile(yhat, probs = seq(0,
1, 1/g)), include.lowest = T)
obs <- xtabs(cbind(1 - y, y) ~ cutyhat)
expect <- xtabs(cbind(1 - yhat, yhat) ~ cutyhat)
chisq <- sum((obs - expect)^2/expect)
P <- 1 - pchisq(chisq, g - 2)
c("X^2" = chisq, Df = g - 2, "P(>Chi)" = P)
}
I want to know how can i use the another method which is not use the
probability of success. i want to know how can i revise above program to
achieve an objective.
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[R] Election Maps
Those of you with an interest in the US election and/or statistical graphics may find the maps at: http://www-personal.umich.edu/~mejn/election/2008/ interesting. url:www.econ.uiuc.edu/~rogerRoger Koenker email[EMAIL PROTECTED]Department of Economics vox: 217-333-4558University of Illinois fax: 217-244-6678Champaign, IL 61820 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Bootstrapping gnls models
Dear all,
I am trying to bootstrap predictions from gnls models using the following code:
# a is the dataframe with which I am working; it contains the variables
# response.variable,LD,L,G,P and F
###
model=gnls(response.variable ~ a * LD/(b + LD),
params = list(a + b ~ L), start = c(1,1,1,1), data=a)
df=cbind(a,fit=predict(model,list(LD=1,L=0.5,G=0.5,P=0.46,F=2.2)))
model.bootfunc=function(rs,i){
df$response.variable=df$fit+rs[i]
as.numeric(predict(gnls(response.variable ~ a * LD/(b + LD),
params = list(a + b ~ L), start = coef(model), data=df)))
}
rs=scale(resid(model),scale=F)
(model.boot=boot(rs,model.bootfunc,R=1))
booted=boot.ci(model.boot,index=1,type=c("norm","basic","perc","bca"))
###
The problem is that this code yields "NA" for the s.e. of the bootstrap
statistics:
Bootstrap Statistics :
original biasstd. error
t1* 0.1651658 -0.020663364 NA
t2* 0.1669592 -0.021759335 NA
t3* 0.1676765 -0.001858686 NA
t4* 0.1726982 -0.025321349 NA
t5* 0.1658092 0.024721214 NA
And hence the boot.ci function and others don?t work.
Does anyone have an idea on that?
Many thanks and best wishes
Christoph
--
Dr. rer.nat. Christoph Scherber
University of Goettingen
DNPW, Agroecology
Waldweg 26
D-37073 Goettingen
Germany
phone +49 (0)551 39 8807
fax +49 (0)551 39 8806
Homepage http://www.gwdg.de/~cscherb1
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Re: [R] Agent-based social simulation and R
Simone Gabbriellini wrote: Tom, I don't know if there are better ways, but this is the way I do: I use Python for building the AB model, and RPy as an interface to R for statistical analysis. One of the best package for SNA in R is igraph, which has a nice Python version. But if you prefere statnet (which is great too), you can simply handle it via RPy. I've learned this strategy from Pietro Terna - http://web.econ.unito.it/terna/ Thank you for useful information and suggestions. I will certainly look into what you mention. I am at the moment looking at the igraph package, which seems to have what I need, and includes visualization as well. As for Python, that is nice, but if I can do most of what I want in R, I would prefer that. Tom ++ | Tom Backer Johnsen, Psychometrics Unit, Faculty of Psychology | | University of Bergen, Christies gt. 12, N-5015 Bergen, NORWAY | | Tel : +47-5558-9185Fax : +47-5558-9879 | | Email : [EMAIL PROTECTED]URL : http://www.galton.uib.no/ | ++ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] need help in plotting barchart
"Kurapati, Ravichandra \(Ravichandra\)" <[EMAIL PROTECTED]> napsal dne 07.11.2008 11:06:28: > Hi >I want NA counts also. Well barley[5:15, 1] <- NA barchart(yield ~ variety | site, data = barley, groups = year, layout = c(1,6), ylab = "Barley Yield (bushels/acre)", scales = list(x = list(abbreviate = TRUE, minlength = 5))) works and bars with NA values are missing but all labels are preserved. So obviously everything works. Without reproducible example what you did and what and how something is not as you expect it is rather difficult to provide suitable advice. Regards Petr > > K.Ravi > > -Original Message- > From: Petr PIKAL [mailto:[EMAIL PROTECTED] > Sent: Friday, November 07, 2008 3:31 PM > To: Kurapati, Ravichandra (Ravichandra) > Cc: r-help@r-project.org > Subject: Odp: [R] need help in plotting barchart > > Hi > > I did not see any response and will not give you any answer too, just a > hint. Your code can not be reproduced and it is also difficult to say > what > you really want. > > Eg. this is rather confusing. > > > On a generated plot I observed that what ever records have counts as > > "NA", those are not suppressed. but I don't want them to be > suppressed. > > Do you want NA to be suppressed or not? > > I would > > ensure my data frame has variables with correct class and then I would > select those with non zero and non NA counts by e.g. > > Df.OK <- Df[which(Df$counts),] > > if you want to drop unused levels of factor you can use e.g. drop=T in > subseting this factor > > Regards > Petr > > > [EMAIL PROTECTED] napsal dne 06.11.2008 12:30:13: > > > Df contains > > > > Session_Setup DCT RevDataVols_bincounts > > comp > > > > 1Session_Setup RLL 1NA > > Session_Setup+RLL+1 > > > > 2Session_Setup RLL 2NA > > Session_Setup+RLL+2 > > > > 3Session_Setup RLL 3NA > > Session_Setup+RLL+3 > > > > 4Session_Setup RLL 4NA > > Session_Setup+RLL+4 > > > > 5Session_Setup RLL 5NA > > Session_Setup+RLL+5 > > > > 6Session_Setup RLL 6NA > > Session_Setup+RLL+6 > > > > 7Session_Setup RLL 7NA > > Session_Setup+RLL+7 > > > > 8Session_Setup RLL 8NA > > Session_Setup+RLL+8 > > > > 9Session_Setup RLL 9NA > > Session_Setup+RLL+9 > > > > 10 Session_Setup RLL 10NA > > Session_Setup+RLL+10 > > > > 11 Session_Setup RLL 11NA > > Session_Setup+RLL+11 > > > > 12 Session_Setup RLL 12NA > > Session_Setup+RLL+12 > > > > 13 Session_Setup RLL 13NA > > Session_Setup+RLL+13 > > > > 14 Session_Setup RLL 14NA > > Session_Setup+RLL+14 > > > > 15 Session_Setup RLL 15NA > > Session_Setup+RLL+15 > > > > 16 Session_Setup RLL 16NA > > Session_Setup+RLL+16 > > > > 17 Session_Setup RLL 17NA > > Session_Setup+RLL+17 > > > > 18 Session_Setup RLL 18NA > > Session_Setup+RLL+18 > > > > 19 Session_Setup RLL 19NA > > Session_Setup+RLL+19 > > > > 20 Session_Setup RLL 20NA > > Session_Setup+RLL+20 > > > > 21 Session_Setup RLL 21NA > > Session_Setup+RLL+21 > > > > 22 Session_Setup RLL 22NA > > Session_Setup+RLL+22 > > > > 23 Session_Setup RLL 23NA > > Session_Setup+RLL+23 > > > > 24 Session_Setup RLL 24NA > > Session_Setup+RLL+24 > > > > 25 Session_Setup RLL 25NA > > Session_Setup+RLL+25 > > > > 26 Session_Setup RLL 26NA > > Session_Setup+RLL+26 > > > > 27 Session_Setup RLL 27NA > > Session_Setup+RLL+27 > > > > 28 Session_Setup RLL 28NA > > Session_Setup+RLL+28 > > > > 29 Session_Setup RLL 29NA > > Session_Setup+RLL+29 > > > > 30 Session_Setup RLL 30NA > > Session_Setup+RLL+30 > > > > 31 User_Initiated RLL 1 8.487840 > > User_Initiated+RLL+1 > > > > 32 User_Initiated RLL 2 6.066089 > > User_Initiated+RLL+2 > > > > 33 User_Initiated RLL 3NA > > User_Initiated+RLL+3 > > > > 34 User_Initiated RLL 4NA > > User_Initiated+RLL+4 > > > > 35 User_Initiated RLL 5 5.906891 > > User_Initiated+RLL+5 > > > > 36 User_Initiated RLL 6NA > > User_Initiated+RLL+6 > > > > 37 User_Initiated RLL 7NA > > User_Initiated+RLL+7 > > > > 38 User_Initiated RLL 8NA > > User_Initiated+RLL+8 > >
Re: [R] Umlaut read from csv-file
Heinz Tuechler wrote: > Dear Prof.Ripley! > > Thank you very much for your attention. In the given example Encoding(), > or the encoding parameter of read.csv solve the problem. I hope your > patch will solve also the problem, when I read a spss file by > spss.get(), since this function has no encoding parameter and my real > problem originated there. read.spss() (package foreign) does have a reencode argument, though; and this is called by spss.get(), so it looks like an easy hack to add it there. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Agent-based social simulation and R
Tom, I don't know if there are better ways, but this is the way I do: I use Python for building the AB model, and RPy as an interface to R for statistical analysis. One of the best package for SNA in R is igraph, which has a nice Python version. But if you prefere statnet (which is great too), you can simply handle it via RPy. I've learned this strategy from Pietro Terna - http://web.econ.unito.it/terna/ hope it helps, Simone Gabbriellini Il giorno 07/nov/08, alle ore 13:06, Tom Backer Johnsen ha scritto: Do anyone know anything about the use of R for agent-based social simulation? It should be possible, and would be convenient for the simple reason that there are several nice packages containing useful stuff for SNA (Social Network Analysis). Information about packages, web sites, experienced persons in the field, etc. would be very welcome. Tom ++ | Tom Backer Johnsen, Psychometrics Unit, Faculty of Psychology | | University of Bergen, Christies gt. 12, N-5015 Bergen, NORWAY | | Tel : +47-5558-9185Fax : +47-5558-9879 | | Email : [EMAIL PROTECTED]URL : http://www.galton.uib.no/ | ++ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. dott. Simone Gabbriellini - PhD Student Department of Social Sciences University of Pisa [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [solved] sapply and median, possible or not ?
Thanks a lot Keith : your function does the job (but I don't understand the trick) ! Have a nice week-end (Thanks to you, I can have a good one :-)), Ptit Bleu. - median.data.frame <- function(x, ...) sapply(x, median, ...) I haven't tried it, but it might work hth Keith J -- View this message in context: http://www.nabble.com/sapply-and-median%2C-possible-or-not---tp20378222p20379667.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Agent-based social simulation and R
Do anyone know anything about the use of R for agent-based social simulation? It should be possible, and would be convenient for the simple reason that there are several nice packages containing useful stuff for SNA (Social Network Analysis). Information about packages, web sites, experienced persons in the field, etc. would be very welcome. Tom ++ | Tom Backer Johnsen, Psychometrics Unit, Faculty of Psychology | | University of Bergen, Christies gt. 12, N-5015 Bergen, NORWAY | | Tel : +47-5558-9185Fax : +47-5558-9879 | | Email : [EMAIL PROTECTED]URL : http://www.galton.uib.no/ | ++ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] is there any way to run R method as a background process from R interface
Hello, Two approaches depending when you want to trigger this background calculation: 1) It is enough to trigger the background computation after each top-level instruction entered at the command line: use ?addTaskCallback 2) You want to trigger the background calculation at a given time, or redo it every xxx ms. This is a little bit more complicate. I have success using the tcltk package and the 'after' Tcl function. Best, Philippe Grosjean ..<°}))>< ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons-Hainaut University, Belgium ( ( ( ( ( .. Kurapati, Ravichandra (Ravichandra) wrote: Hi , can some body tell to me "how to run a R method /function as a background process from R interface" Thanks K.Ravichandra [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sapply and median, possible or not ?
I haven't looked at the detail, but I guess the answer is that mean works on a data frame while median doesn't. ?mean For a data frame, a named vector with the appropriate method being applied column by column. - I guess to use median you'll need nested '[l/s]apply's, the outer working through the list of dataframes and the inner working through the columns of each dataframe. Or perhaps, by analogy with mean.data.frame you could just define - median.data.frame <- function(x, ...) sapply(x, median, ...) I haven't tried it, but it might work hth Keith J - "Ptit_Bleu" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] > > Unfortunately, I have the same error message. > lapply(rowsplit, function(x)mean(x[,sapply(x, is.numeric)])) works but not > with median. > Strange, isn't it? > > Any other idea? > > Thanks in advance, > Ptit Bleu. > > > Henrique Dallazuanna wrote: >> >> Try this: >> >> lapply(l, function(x)median(x[,sapply(x, is.numeric)])) >> >> >> -- >> Henrique Dallazuanna >> Curitiba-Paraná-Brasil >> 25° 25' 40" S 49° 16' 22" O >> >> [[alternative HTML version deleted]] >> >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> >> > > -- > View this message in context: > http://www.nabble.com/sapply-and-median%2C-possible-or-not---tp20378222p20378663.html > Sent from the R help mailing list archive at Nabble.com. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ordinal logistic model with pre-defined coefficients
On Fri, 7 Nov 2008, [EMAIL PROTECTED] wrote: Hi, I'm trying to fit a proportional ordinal logistic model using function polr() (package MASS). Is there a way to fix certain betas in the regression (e.g. function arima() allows this by defining fixed ) Maybe there is another function than polr() which allows that? See ?offset, the standard tool in regression-like fitting in R (and described in the book MASS supports). Thanks Kazys -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sapply and median, possible or not ?
You can provide a example of your data? On Fri, Nov 7, 2008 at 9:14 AM, Ptit_Bleu <[EMAIL PROTECTED]> wrote: > > Unfortunately, I have the same error message. > lapply(rowsplit, function(x)mean(x[,sapply(x, is.numeric)])) works but not > with median. > Strange, isn't it? > > Any other idea? > > Thanks in advance, > Ptit Bleu. > > > Henrique Dallazuanna wrote: > > > > Try this: > > > > lapply(l, function(x)median(x[,sapply(x, is.numeric)])) > > > > > > -- > > Henrique Dallazuanna > > Curitiba-Paraná-Brasil > > 25° 25' 40" S 49° 16' 22" O > > > > [[alternative HTML version deleted]] > > > > > > __ > > R-help@r-project.org mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > > > > > -- > View this message in context: > http://www.nabble.com/sapply-and-median%2C-possible-or-not---tp20378222p20378663.html > Sent from the R help mailing list archive at Nabble.com. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40" S 49° 16' 22" O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sapply and median, possible or not ?
Unfortunately, I have the same error message. lapply(rowsplit, function(x)mean(x[,sapply(x, is.numeric)])) works but not with median. Strange, isn't it? Any other idea? Thanks in advance, Ptit Bleu. Henrique Dallazuanna wrote: > > Try this: > > lapply(l, function(x)median(x[,sapply(x, is.numeric)])) > > > -- > Henrique Dallazuanna > Curitiba-Paraná-Brasil > 25° 25' 40" S 49° 16' 22" O > > [[alternative HTML version deleted]] > > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > > -- View this message in context: http://www.nabble.com/sapply-and-median%2C-possible-or-not---tp20378222p20378663.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to manipulate the time data without the date?
On Fri, Nov 7, 2008 at 3:32 AM, Barry Rowlingson
<[EMAIL PROTECTED]> wrote:
> 2008/11/7 tedzzx <[EMAIL PROTECTED]>:
>>
>> The problem is that: There is some rounding problems, for example
>> Library(chron)
>> any(times("4:00:01")==times("4:00:00")+times("00:00:01")))
>> False
>>
>> But,it should be true
>
> FAQ 7.31 in disguise!
>
> chron stores date-times as fractions, so you're comparing two
> floating point numbers here. FAQ 7.31 applies.
>
> Use 'all.equal', which has a method for chron that uses a 1 second
> tolerance to match times:
>
That solution works but just to be precise, note that a "times"
class object need not be a "dates" or "chron" class object and
while there is a chron:::all.equal.dates there is no chron:::all.equal.times.
The reason all.equal works here is that all.equal.default works here and
it works here by ultimately calling all.equal.numeric to do the real work.
> y <- times("4:00:00")+times("00:00:01")
> class(y) # no "dates" or "chron" in class vector
[1] "times"
Also all.equal.dates uses a tenth of a second as the tolerance:
> chron:::all.equal.dates
function (..., tolerance = 1/(10 * 24 * 60 * 60))
NextMethod("all.equal", ..., tolerance = tolerance)
>> all.equal(times("4:00:01"),times("4:00:00")+times("00:00:01"))
> [1] TRUE
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] sapply and median, possible or not ?
Try this: lapply(l, function(x)median(x[,sapply(x, is.numeric)])) On Fri, Nov 7, 2008 at 8:42 AM, Ptit_Bleu <[EMAIL PROTECTED]> wrote: > > Hello, > > I have a list of data.frame > rowsplit : List of 15 > $ (0,0.025] :'data.frame': 169 obs. of 7 variables: > $ (0.025,0.05]:'data.frame': 174 obs. of 7 variables: > $ (0.05,0.075]:'data.frame': 92 obs. of 7 variables: > $ (0.075,0.1] :'data.frame': 76 obs. of 7 variables: > $ (0.1,0.125] :'data.frame': 37 obs. of 7 variables: > $ (0.125,0.15]:'data.frame': 32 obs. of 7 variables: > $ (0.15,0.175]:'data.frame': 45 obs. of 7 variables: > $ (0.175,0.2] :'data.frame': 56 obs. of 7 variables: > $ (0.2,0.225] :'data.frame': 36 obs. of 7 variables: > $ (0.225,0.25]:'data.frame': 47 obs. of 7 variables: > $ (0.25,0.275]:'data.frame': 34 obs. of 7 variables: > $ (0.275,0.3] :'data.frame': 43 obs. of 7 variables: > $ (0.3,0.325] :'data.frame': 29 obs. of 7 variables: > $ (0.325,0.35]:'data.frame': 29 obs. of 7 variables: > $ (0.35,0.375]:'data.frame': 17 obs. of 7 variables: > > And I would like to get a data.frame gathering the median value of each > variable for each intervall. > as.data.frame(t(sapply(rowsplit, mean)) works well but I would prefer to > use > median but with sapply(rowsplit, median), I get the following message > "Error > in median.default(X[[1L]], ...) : need numeric data". > > Any idea ? > > Thanks in advance for your help, > Have a nice week-end, > Ptit Bleu. > > PS :I'm under XP and use R2.7.2. > -- > View this message in context: > http://www.nabble.com/sapply-and-median%2C-possible-or-not---tp20378222p20378222.html > Sent from the R help mailing list archive at Nabble.com. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40" S 49° 16' 22" O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] nls: Fitting two models at once?
Dear Martin,
You can use the same idea of concatenating the data in R. The following
reproduces your example:
spec <- c(asfe, dias)
n1 <- length(asfe)
n2 <- length(dias)
CYT <- c(28 * CYTF, 24.6 * 2 * CYTB6)
B559HP <- c(B559HP, numeric(n2))
B559LP <- c(numeric(n1), B559LP)
C550 <- c(numeric(n1), C550)
spec.fit <-
nls(
spec ~ cyt.v*CYT + hp.v*B559HP + lp.v*B559LP + c550.v*C550,
start = list(cyt.v = 0.5, hp.v = 0.03, lp.v = 1,
c550.v = 1) # arbitrary
);
or more efficiently, using lm
spec.fit <- lm(spec ~ 0 + CYT + B559HP + B559LP + C550)
# draw stuff
plot(
1, 2,
type="n",
xlim = c(540, 575),
ylim=c(-0.002, 0.008),
);
# first spectrum and fit
lines(wl, asfe, type="b", pch=19); # solid circles
lines(wl, fitted(spec.fit)[1:n1], col = "red");
# second spectrum and fit
lines(wl, dias, type="b");
lines(wl, fitted(spec.fit)[(n1 + 1):(n1 + n2)], col = "blue");
Best wishes,
Heather
Dr H Turner
Senior Research Fellow
Dept. of Statistics
The University of Warwick
Coventry
CV4 7AL
Tel: 024 76575870
Fax: 024 76524532
Url: www.warwick.ac.uk/go/heatherturner
Martin Ballaschk wrote:
> Hello,
>
> I'm still a newbie user and struggling to automate some analyses from
> SigmaPlot using R. R is a great help for me so far!
>
> But the following problem makes me go nuts.
>
> I have two spectra, both have to be fitted to reference data. Problem: the
> both spectra are connected in some way: the stoichiometry of coefficients
> "cytf.v"/"cytb.v" is 1/2.
> {{In the SigmaPlot workflow one has to copy the two spectra into one column
> beneath each other and the two spectra are somehow treated as one curve -
> like in http://home.arcor.de/ballaschk/cytbf-help/sigmaplot%20formula.png}}
>
> Can anybody help? :(
>
> I tried to condense everything to the "minimum" R script below to give an
> impression of what I'm talking about.
>
> Martin
>
>
>
>
> # "Minimal" R script reading remote data for convenience
>
> ### READ IN DATA
> # first spectrum
> asfe <- read.table("http://home.arcor.de/ballaschk/cytbf-help/asfe.csv";)[, 1];
> # second spectrum
> dias <- read.table("http://home.arcor.de/ballaschk/cytbf-help/dias.csv";)[, 1];
>
> # reference data for fit, wavelength = wl
> ref <- read.table("http://home.arcor.de/ballaschk/cytbf-help/reference.csv";,
> sep="\t", dec=".", header=T);
> attach(ref);
>
> ### FITTING, problem: 2*cytf.v == cytb.v
>
> # fit first spectrum to two reference spectra
> asfe.fit <-
> nls(
> asfe ~ (cytf.v * 28) * CYTF + hp.v * B559HP,
> start = list(cytf.v = 0.5, hp.v = 0.03) # arbitrary
> );
>
> # fit second spectrum to three reference spectra
> dias.fit <-
> nls(
> dias ~ (cytb.v * 24.6 * 2) * CYTB6 + lp.v * B559LP + c550.v * C550,
> start = list(cytb.v = 1, lp.v = 1, c550.v = 1) # arbitrary
> );
>
>
> # draw stuff
> plot(
> 1, 2,
> type="n",
> xlim = c(540, 575),
> ylim=c(-0.002, 0.008),
> );
>
> # first spectrum and fit
> lines(wl, asfe, type="b", pch=19); # solid circles
> lines(wl, fitted(asfe.fit), col = "red");
>
> # second spectrum and fit
> lines(wl, dias, type="b");
> lines(wl, fitted(dias.fit), col = "blue");
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
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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] Updating Packages with Personal Library
Test. On Fri, Nov 7, 2008 at 2:40 AM, Alan Lue <[EMAIL PROTECTED]> wrote: > I'm having trouble running `updates.packages()' and installing into a > personal library. > > Setup: > 1. .Renviron file contains: R_LIBS_USER="~/lib/R/%p-library/%v" > 2. Hence "personal library" is: ~/lib/R/i486-pc-linux-gnu-library/2.6/ > 3. Library path upon starting R: > > > .libPaths() > [1] "/home/johnny/lib/R/i486-pc-linux-gnu-library/2.6" > [2] "/usr/local/lib/R/site-library" > [3] "/usr/lib/R/site-library" > [4] "/usr/lib/R/library" > > That's the setup. The problem is that `update.packages()' does not > consider packages in my personal library when determining what to update. > To elaborate: > > Here's some output when I update packages: > > > update.packages(instlib=.libPaths()[1]) > KernSmooth : > Version 2.22-21 installed in /usr/lib/R/library > Version 2.22-22 available at http://cran.cnr.Berkeley.edu > Update (y/N/c)? > ... > > I answer "yes" to everything, and the update succeeds; updated packages are > installed into my personal library, > `~/lib/R/i486-pc-linux-gnu-library/2.6/'. But if I run the package update > command again, immediately after having just run it, I get the exact same > output: > > > update.packages(instlib=.libPaths()[1]) > KernSmooth : > Version 2.22-21 installed in /usr/lib/R/library > Version 2.22-22 available at http://cran.cnr.Berkeley.edu > Update (y/N/c)? > > R looked in `/usr/lib/R/library', saw that there was a package that needed > updating, and neglected the fact that the updated package was already in my > personal library. > > How can I get R to check packages in my personal library while determining > what to update? > > Alan > [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] sapply and median, possible or not ?
Hello, I have a list of data.frame rowsplit : List of 15 $ (0,0.025] :'data.frame': 169 obs. of 7 variables: $ (0.025,0.05]:'data.frame': 174 obs. of 7 variables: $ (0.05,0.075]:'data.frame': 92 obs. of 7 variables: $ (0.075,0.1] :'data.frame': 76 obs. of 7 variables: $ (0.1,0.125] :'data.frame': 37 obs. of 7 variables: $ (0.125,0.15]:'data.frame': 32 obs. of 7 variables: $ (0.15,0.175]:'data.frame': 45 obs. of 7 variables: $ (0.175,0.2] :'data.frame': 56 obs. of 7 variables: $ (0.2,0.225] :'data.frame': 36 obs. of 7 variables: $ (0.225,0.25]:'data.frame': 47 obs. of 7 variables: $ (0.25,0.275]:'data.frame': 34 obs. of 7 variables: $ (0.275,0.3] :'data.frame': 43 obs. of 7 variables: $ (0.3,0.325] :'data.frame': 29 obs. of 7 variables: $ (0.325,0.35]:'data.frame': 29 obs. of 7 variables: $ (0.35,0.375]:'data.frame': 17 obs. of 7 variables: And I would like to get a data.frame gathering the median value of each variable for each intervall. as.data.frame(t(sapply(rowsplit, mean)) works well but I would prefer to use median but with sapply(rowsplit, median), I get the following message "Error in median.default(X[[1L]], ...) : need numeric data". Any idea ? Thanks in advance for your help, Have a nice week-end, Ptit Bleu. PS :I'm under XP and use R2.7.2. -- View this message in context: http://www.nabble.com/sapply-and-median%2C-possible-or-not---tp20378222p20378222.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Updating Packages with Personal Library
I'm having trouble running `updates.packages()' and installing into a personal library. Setup: 1. .Renviron file contains: R_LIBS_USER="~/lib/R/%p-library/%v" 2. Hence "personal library" is: ~/lib/R/i486-pc-linux-gnu-library/2.6/ 3. Library path upon starting R: > .libPaths() [1] "/home/johnny/lib/R/i486-pc-linux-gnu-library/2.6" [2] "/usr/local/lib/R/site-library" [3] "/usr/lib/R/site-library" [4] "/usr/lib/R/library" That's the setup. The problem is that `update.packages()' does not consider packages in my personal library when determining what to update. To elaborate: Here's some output when I update packages: > update.packages(instlib=.libPaths()[1]) KernSmooth : Version 2.22-21 installed in /usr/lib/R/library Version 2.22-22 available at http://cran.cnr.Berkeley.edu Update (y/N/c)? ... I answer "yes" to everything, and the update succeeds; updated packages are installed into my personal library, `~/lib/R/i486-pc-linux-gnu-library/2.6/'. But if I run the package update command again, immediately after having just run it, I get the exact same output: > update.packages(instlib=.libPaths()[1]) KernSmooth : Version 2.22-21 installed in /usr/lib/R/library Version 2.22-22 available at http://cran.cnr.Berkeley.edu Update (y/N/c)? R looked in `/usr/lib/R/library', saw that there was a package that needed updating, and neglected the fact that the updated package was already in my personal library. How can I get R to check packages in my personal library while determining what to update? Alan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
