Re: [R] customizing RGui/Rconsole
Philip A. Viton wrote: Is it possible to change the behavior of RGui/RConsole under MS Windows so that, if you submit a bunch of commands from a script window, when they've finished, focus is given to the Console, and not (as now) to the script? That's normally the job of the sending side, because RGui does not know, when it's finished. You might have a look at the source code of Tinn-R which has the required feature as an option. Dieter -- View this message in context: http://www.nabble.com/customizing-RGui-Rconsole-tp25106128p2579.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] Generate random sample interval
Hi, R users, I have a problem about how to generate random sample interval from a duration. For example, during a time duration 0-70s, I want to generate a sample which last 10s. the sample could be 0-10 or 30-40s or 25-35s etc How could I do it in R`?? Thanks a lot. Tammy _ Share your memories online with anyone you want. http://www.microsoft.com/middleeast/windows/windowslive/products/photos-share.aspx?tab=1 [[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] Odp: When factor is better than other types, such as vector and frame?
Hi r-help-boun...@r-project.org napsal dne 23.08.2009 05:00:11: Hi, It is easy to understand the types vector and frame. But I am wondering why the type factor is designed in R. What is the advantage of factor compare with other data types in R? Can somebody give an example in which case the type factor is much better than other data types? Although your expressions do not correspond much with naming conventions in R, usage of factor is sometimes preferable to character values. consider e.g. set.seed(111) df-data.frame(1:5, fac=sample(letters[1:2], 5, replace=T)) plot(df[,1], pch=as.numeric(df[,2])) df[,2]-as.character(df[,2]) plot(df[,1], pch=as.numeric(df[,2])) Warning message: In plot.xy(xy, type, ...) : NAs introduced by coercion Another advantage is simple and straightforward manipulation with levels. levels(df[,2])-c(yes, no) df X1.5 fac 11 no 22 no 33 yes 44 no 55 yes together with easy ordering option of levels and subsequent plotting order in boxplots and similar. factor(df$fac, levels=levels(df$fac)) [1] no no yes no yes Levels: yes no factor(df$fac, levels=levels(df$fac)[2:1]) [1] no no yes no yes Levels: no yes You need to get used to some features which are sometimes surprising but has a reason like levels persisting in subset. str(df[df$fac==no,]) 'data.frame': 3 obs. of 2 variables: $ X1.5: int 1 2 4 $ fac : Factor w/ 2 levels yes,no: 2 2 2 Regards Petr Regards, Peng __ 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] Convert list to data frame while controlling column types
Hi
r-help-boun...@r-project.org napsal dne 23.08.2009 17:29:48:
On 8/23/2009 9:58 AM, David Winsemius wrote:
I still have problems with this statement. As I understand R, this
should be impossible. I have looked at both you postings and neither of
them clarify the issues. How can you have blanks or spaces in an R
numeric vector?
Just because I search numeric columns doesn't mean that my regex matches
them! I posted some info on my data frame in an earlier email:
str(final_dataf)
'data.frame': 1127 obs. of 43 variables:
$ block : Factor w/ 1 level 2: 1 1 1 1 1 1 1 1 1 1 ...
$ treatment : Factor w/ 4 levels I,M,N,T: 1 1 1 1 1 1 ...
$ transect : Factor w/ 1 level 4: 1 1 1 1 1 1 1 1 1 1 ...
$ tag: chr NA 121AL 122AL 123AL ...
...
$ h1 : num NA NA NA NA NA NA NA NA NA NA ...
...
You can see that I do indeed have some numeric columns. And while I
Well, AFAICS you have a data frame with 3 columns which are factors and 1
which is character. I do not see any numeric column. If you want to change
block and transect to numeric you can use
df$block - as.numeric(as.character(df$block))
search them for spaces, I only do so because my dataset isn't so large
as to require me to exclude them from the search. If my dataset grows
too big at some point, I will exclude numeric columns, and other columns
which cannot hold blanks or spaces.
To clarify further with an example:
df = data.frame(a=c(1,2,3,4,5),b=c(a,,c,d, ))
df = as.data.frame(lapply(df, function(x){ is.na(x) -
+ grep('^\\s*$',x); return(x) }), stringsAsFactors = FALSE)
df
ab
1 1a
2 2 NA
3 3c
4 4d
5 5 NA
which can be done also by
df[,2] - levels(df[,2])[1:2]-NA
but maybe with less generalization
str(df)
'data.frame': 5 obs. of 2 variables:
$ a: num 1 2 3 4 5
$ b: Factor w/ 5 levels , ,a,c,..: 3 NA 4 5 NA
And one final clarification: I left out as.data.frame in my previous
solution. So it now becomes:
final_dataf = as.data.frame(lapply(final_dataf, function(x){ is.na(x)
+ - grep('^\\s*$',x); return(x) }), stringsAsFactors = FALSE)
Again not too much of clarification, in your first data frame second
column is a factor with some levels you want to convert to NA, which can
be done by different approaches.
Your final_dataf is same as df.
Columns which shall be numeric and are read as factor/character by
read.table likely contain some values which strictly can not be considered
numeric. You can see them quite often in Excel like programs and some
examples are
1..2, o.5, 12.o5 and or spaces, - e.t.c.
and you usually need handle them by hand.
Regards
Petr
Hope that clarifies things, and thanks for your help.
Thanks,
Allie
On 8/23/2009 9:58 AM, David Winsemius wrote:
On Aug 23, 2009, at 2:47 AM, Alexander Shenkin wrote:
On 8/21/2009 3:04 PM, David Winsemius wrote:
On Aug 21, 2009, at 3:41 PM, Alexander Shenkin wrote:
Thanks everyone for their replies, both on- and off-list. I should
clarify, since I left out some important information. My original
dataframe has some numeric columns, which get changed to character
by
gsub when I replace spaces with NAs.
If you used is.na() - that would not happen to a true _numeric_
vector
(but, of course, a numeric vector in a data.frame could not have
spaces,
so you are probably not using precise terminology).
I do have true numeric columns, but I loop through my entire
dataframe
looking for blanks and spaces for convenience.
I still have problems with this statement. As I understand R, this
should be impossible. I have looked at both you postings and neither
of
them clarify the issues. How can you have blanks or spaces in an R
numeric vector?
You would be well
advised to include the actual code rather than applying loose
terminology subject you your and our misinterpretation.
I did include code in my previous email. Perhaps you were looking
for
different parts.
?is.na
I am guessing that you were using read.table() on the original data,
in
which case you should look at the colClasses parameter.
yep - I use read.csv, and I do use colClasses. But as I mentioned
earlier, gsub converts those columns to characters. Thanks for the
tip
about is.na() -. I'm now using the following, thus side-stepping
the
whole controlling as.data.frame's column conversion issue:
final_dataf = lapply(final_dataf, function(x){ is.na(x) -
+ grep('^\\s*$',x); return(x) })
Good that you have a solution.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] Generate random sample interval
I need a simple algorithm to generate random ranges. From: metal_lical...@live.com To: r-help@r-project.org Date: Mon, 24 Aug 2009 09:55:35 +0300 Subject: [R] Generate random sample interval Hi, R users, I have a problem about how to generate random sample interval from a duration. For example, during a time duration 0-70s, I want to generate a sample which last 10s. the sample could be 0-10 or 30-40s or 25-35s etc How could I do it in R`?? Thanks a lot. Tammy _ Share your memories online with anyone you want. http://www.microsoft.com/middleeast/windows/windowslive/products/photos-share.aspx?tab=1 [[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. _ Show them the way! Add maps and directions to your party invites. http://www.microsoft.com/windows/windowslive/products/events.aspx [[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] Generate random sample interval
On Aug 24, 2009, at 2:55 AM, Tammy Ma wrote: Hi, R users, I have a problem about how to generate random sample interval from a duration. For example, during a time duration 0-70s, I want to generate a sample which last 10s. the sample could be 0-10 or 30-40s or 25-35s etc How could I do it in R`?? ?runif Wouldn't you just generate a runif variable for the starting time that was in the range of 0-60 and then add 10 to each such start time? If the problem is any more difficult, then you will need to be more precise. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to add 95% confidential interval as vertical lines to x axe in density plot
Dear R-help listers, I want to add 95% confidential interval as vertical lines to x axe in density plot. I have found the library(hdrcde) can do this work, but I do not know how to handle functions of this library when I used ggplot2 to draw the graph. Thank you in advance. The data and codes followed: # dummy data factor-rep(c(Alice,Jone,Mike),each=100) factor-factor(factor) traits1-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits2-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits3-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits4-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits5-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits6-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits7-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits8-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits9-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits10-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits11-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits12-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits13-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits14-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits15-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits16-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits17-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits18-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits19-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits20-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits21-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) myda-data.frame(factor,traits1,traits2,traits3,traits4,traits5,traits6,traits7,traits8,traits9,traits10,traits11,traits12,traits13,traits14,traits15,traits16,traits17,traits18, traits19,traits20,traits21) library(ggplot2) d = melt(myda, id = factor) str(d) pdf(test33.pdf) p = ggplot(data=d, mapping=aes(x=value, y=..density..)) + facet_wrap(~variable)+ stat_density(aes(fill=factor), alpha=0.5, col=NA, position = 'identity') + stat_density(aes(colour = factor), geom=path, position = 'identity') print(p) dev.off() Mao J-F __ 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 in building new function
Hi: I've installed the precompiled binary for Windows. I need to use an existing function, but I want to introduce some slight changes to it. 1. Is there a way for me to find the source files through windows explorer? I know I can see it using edit(object name) but I want to know if I can see it via explorer in some location under the R directory. 2. I don´t want to modify the code of that function until I'm sure that my changes are not causing any harm. So, I got the R-2.9.1.tar.gz file and open the arima.R file. I change the name of the function to myarima, the name of the file to myarima.R, save it and the loaded it using the source command. So far, so good. When I try to execute my function I get the error message Error in Delta %+% c(1, -1) : object 'R_TSconv' not found. So far, the only change I made is add a Hello world in the first line, so my change is not the source of the problem. Looks like my function (although it is exactly the same) is not being able to see this object. Can someone help me out? Am I missing something? Thanks a lot in advance. Regards, Ana [[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] Help in building new function
1. Just enter arima at the R console to see its source code (without comments). The source tar.gz for R is found by googling for R, clicking on CRAN in left column and choosing mirror. Or to view it online or get it via svn: https://svn.r-project.org/R/ 2. You want myarima's free variables to be found in the stats package so set the environment of myarima like this: environment(myarima) - asNamespace(stats) On Mon, Aug 24, 2009 at 2:18 AM, Ana Paula Moraanamor...@gmail.com wrote: Hi: I've installed the precompiled binary for Windows. I need to use an existing function, but I want to introduce some slight changes to it. 1. Is there a way for me to find the source files through windows explorer? I know I can see it using edit(object name) but I want to know if I can see it via explorer in some location under the R directory. 2. I don´t want to modify the code of that function until I'm sure that my changes are not causing any harm. So, I got the R-2.9.1.tar.gz file and open the arima.R file. I change the name of the function to myarima, the name of the file to myarima.R, save it and the loaded it using the source command. So far, so good. When I try to execute my function I get the error message Error in Delta %+% c(1, -1) : object 'R_TSconv' not found. So far, the only change I made is add a Hello world in the first line, so my change is not the source of the problem. Looks like my function (although it is exactly the same) is not being able to see this object. Can someone help me out? Am I missing something? Thanks a lot in advance. Regards, Ana [[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.
[R] R 2.9.2 is released
I've rolled up R-2.9.2.tar.gz a short while ago. This is a maintenance release and fixes a number of mostly minor issues. See the full list of changes below. You can get it from http://cran.r-project.org/src/base/R-2/R-2.9.2.tar.gz or wait for it to be mirrored at a CRAN site nearer to you. Binaries for various platforms will appear in due course (Duncan Murdoch is out of town, so Windows binaries may take a few days longer than usual). For the R Core Team Peter Dalgaard These are the md5sums for the freshly created files, in case you wish to check that they are uncorrupted: 70447ae7f2c35233d3065b004aa4f331 INSTALL 433182754c05c2cf7a04ad0da474a1d0 README 4f004de59e24a52d0f500063b4603bcb OONEWS ff4bd9073ef440b1eb43b1428ce96872 ONEWS 17cbe5399dc80f39bd8c8b3c244d6a50 NEWS 7abcbbc7480df75a11a00bb09783db90 THANKS 070cca21d9f8a6af15f992edb47a24d5 AUTHORS a6f89e2100d9b6cdffcea4f398e37343 COPYING.LIB eb723b61539feef013de476e68b5c50a COPYING 020479f381d5f9038dcb18708997f5da RESOURCES 4767b922bb20620946ee9a4275a6ee6c FAQ 112e2a1306cf71320e45d14e87e5b913 R-2.9.2.tar.gz 112e2a1306cf71320e45d14e87e5b913 R-latest.tar.gz Here is the relevant part of the NEWS file: CHANGES IN R VERSION 2.9.2 NEW FEATURES o install.packages(NULL) now lists packages only once even if they occur in more than one repository (as the latest compatible version of those available will always be downloaded). o approxfun() and approx() now accept a 'rule' of length two, for easy specification of different interpolation rules on left and right. They no longer segfault for invalid zero-length specification of 'yleft, 'yright', or 'f'. o seq_along(x) is now equivalent to seq_len(length(x)) even where length() has an S3/S4 method; previously it (intentionally) always used the default method for length(). o PCRE has been updated to version 7.9 (for bug fixes). o agrep() uses 64-bit ints where available on 32-bit platforms and so may do a better job with complex matches. (E.g. PR#13789, which failed only on 32-bit systems.) DEPRECATED DEFUNCT o R CMD Rd2txt is deprecated, and will be removed in 2.10.0. (It is just a wrapper for R CMD Rdconv -t txt.) o tools::Rd_parse() is deprecated and will be removed in 2.10.0 (which will use only Rd version 2). BUG FIXES o parse_Rd() still did not handle source reference encodings properly. o The C utility function PrintValue no longer attempts to print attributes for CHARSXPs as those attributes are used internally for the CHARSXP cache. This fixes a segfault when calling it on a CHARSXP from C code. o PDF graphics output was producing two instances of anything drawn with the symbol font face. (Report from Baptiste Auguie.) o length(x) - newval and grep() could cause memory corruption. (PR#13837) o If model.matrix() was given too large a model, it could crash R. (PR#13838, fix found by Olaf Mersmann.) o gzcon() (used by load()) would re-open an open connection, leaking a file descriptor each time. (PR#13841) o The checks for inconsistent inheritance reported by setClass() now detect inconsistent superclasses and give better warning messages. o print.anova() failed to recognize the column labelled P(|Chi|) from a Poisson/binomial GLM anova as a p-value column in order to format it appropriately (and as a consequence it gave no significance stars). o A missing PROTECT caused rare segfaults during calls to load(). (PR#13880, fix found by Bill Dunlap.) o gsub() in a non-UTF-8 locale with a marked UTF-8 input could in rare circumstances overrun a buffer and so segfault. o R CMD Rdconv --version was not working correctly. o Missing PROTECTs in nlm() caused random errors. (PR#13381 by Adam D.I. Kramer, analysis and suggested fix by Bill Dunlap.) o Some extreme cases of pbeta(log.p = TRUE) are more accurate (finite values -700 rather than -Inf). (PR#13786) pbeta() now reports on more cases where the asymptotic expansions lose accuracy (the underlying TOMS708 C code was ignoring some of these, including the PR#13786 example). o new.env(hash = TRUE, size = NA) now works the way it has been documented to for a long time. o tcltk::tk_choose.files(multi = TRUE) produces better-formatted output with filenames containing spaces. (PR#13875) o R CMD check --use-valgrind did not run valgrind on the package tests. o The tclvalue() and the print() and as.xxx methods for class tclObj crashed R with an invalid object -- seen with an object saved from an earlier session. o R CMD BATCH garbled options -d debugger
[R] Using R CMD Batch similarly to gnuplot pause mouse
I have a gnuplot script: plot 'gp.dat' using 1:2 with line pause mouse and would like to have R CMD BATCH script that works similarily... x-read.table(gp.dat) options(device=windows) plot(x[,1],x[,2]) what I am looking for now is an R functions which behaves like pause mouse in gnuplot, i.e. pause - command displays any text associated with the command and then waits a specified amount of time or until the carriage return is pressed. If the current terminal supports mousing, then pause mouse will terminate on either a mouse click or on ctrl-C. Thanks in advance -- Witold Eryk Wolski Heidmark str 5 D-28329 Bremen tel.: 04215261837 www.brementango.de [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to add 95% confidential interval as vertical lines to x axein density plot
Show us how you extract the confidence interval from the functions in the hdrcde library and then we might be able to help you. HTH, Thierry ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest Cel biometrie, methodologie en kwaliteitszorg / Section biometrics, methodology and quality assurance Gaverstraat 4 9500 Geraardsbergen Belgium tel. + 32 54/436 185 thierry.onkel...@inbo.be www.inbo.be To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey -Oorspronkelijk bericht- Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens Mao Jianfeng Verzonden: maandag 24 augustus 2009 9:40 Aan: r-help@r-project.org Onderwerp: [R] how to add 95% confidential interval as vertical lines to x axein density plot Dear R-help listers, I want to add 95% confidential interval as vertical lines to x axe in density plot. I have found the library(hdrcde) can do this work, but I do not know how to handle functions of this library when I used ggplot2 to draw the graph. Thank you in advance. The data and codes followed: # dummy data factor-rep(c(Alice,Jone,Mike),each=100) factor-factor(factor) traits1-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits2-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits3-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits4-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits5-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits6-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits7-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits8-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits9-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits10-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits11-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits12-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits13-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits14-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits15-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits16-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits17-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits18-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits19-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits20-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) traits21-c(rnorm(100, mean=1, sd=1), rnorm(100, mean=3, sd=3), rnorm(100, mean=6, sd=6)) myda-data.frame(factor,traits1,traits2,traits3,traits4,traits5,traits6, traits7,traits8,traits9,traits10,traits11,traits12,traits13,traits14,tra its15,traits16,traits17,traits18, traits19,traits20,traits21) library(ggplot2) d = melt(myda, id = factor) str(d) pdf(test33.pdf) p = ggplot(data=d, mapping=aes(x=value, y=..density..)) + facet_wrap(~variable)+ stat_density(aes(fill=factor), alpha=0.5, col=NA, position = 'identity') + stat_density(aes(colour = factor), geom=path, position = 'identity') print(p) dev.off() Mao J-F __ 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. Druk dit bericht a.u.b. niet onnodig af. Please do not print this message unnecessarily. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. __ 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
[R] Lattice xyplot: modify line width of plot lines
# Hi all, # I want to increase the line width of the plotted lines # in a xy-lattice plot. My own attempts were all in vain. # Without the group option the line width is modified - # with the option it is funnily enough not. # Please have a look at my syntax. # # Many thanks in advance # Udo library(lattice) data - data.frame(cbind(1:2,c(1,1,2,2), c(0.5,0.9,1.0,1.8))) names(data) - c(BMI,time,Choline) data$BMI - factor(data$BMI) levels(data$BMI) - c(=17.5,17.5) data$time - factor(data$time) levels(data$time) - c(Admission,Discharge) #Show names of settings names(trellis.par.get()) #Try to set the line width of the two plotted colored lines line-trellis.par.get(plot.line) line line$lwd=3 trellis.par.set(plot.line, line) line #Without group option: Line width is changed xyplot(Choline ~ time, data=data, type=l, scales=list(relation=free), auto.key=list(title=BMI Group, border=FALSE), xlab=c(Point in Time), ylab=c(Concentration of Choline)) #With group option: Line width is not changed xyplot(Choline ~ time, group=BMI, data=data, type=l, scales=list(relation=free), auto.key=list(title=BMI Group, border=FALSE), xlab=c(Point in Time), ylab=c(Concentration of Choline)) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Lattice xyplot: modify line width of plot lines
On 8/24/2009 4:47 AM, ukoe...@med.uni-marburg.de wrote: # Hi all, # I want to increase the line width of the plotted lines # in a xy-lattice plot. My own attempts were all in vain. # Without the group option the line width is modified - # with the option it is funnily enough not. # Please have a look at my syntax. # # Many thanks in advance # Udo You need to change the superpose.line setting: xyplot(Choline ~ time, groups=BMI, data=data, type=l, scales=list(relation=free), auto.key=list(title=BMI Group, border=FALSE, lines=TRUE, points=FALSE), xlab=c(Point in Time), ylab=c(Concentration of Choline), par.settings = list(superpose.line = list(lwd=3))) library(lattice) data - data.frame(cbind(1:2,c(1,1,2,2), c(0.5,0.9,1.0,1.8))) names(data) - c(BMI,time,Choline) data$BMI - factor(data$BMI) levels(data$BMI) - c(=17.5,17.5) data$time - factor(data$time) levels(data$time) - c(Admission,Discharge) #Show names of settings names(trellis.par.get()) #Try to set the line width of the two plotted colored lines line-trellis.par.get(plot.line) line line$lwd=3 trellis.par.set(plot.line, line) line #Without group option: Line width is changed xyplot(Choline ~ time, data=data, type=l, scales=list(relation=free), auto.key=list(title=BMI Group, border=FALSE), xlab=c(Point in Time), ylab=c(Concentration of Choline)) #With group option: Line width is not changed xyplot(Choline ~ time, group=BMI, data=data, type=l, scales=list(relation=free), auto.key=list(title=BMI Group, border=FALSE), xlab=c(Point in Time), ylab=c(Concentration of Choline)) __ 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. -- Chuck Cleland, Ph.D. NDRI, Inc. (www.ndri.org) 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ 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] Lattice xyplot: modify line width of plot lines
Now it works. Many thanks, Chuck! Quoting Chuck Cleland cclel...@optonline.net: On 8/24/2009 4:47 AM, ukoe...@med.uni-marburg.de wrote: # Hi all, # I want to increase the line width of the plotted lines # in a xy-lattice plot. My own attempts were all in vain. # Without the group option the line width is modified - # with the option it is funnily enough not. # Please have a look at my syntax. # # Many thanks in advance # Udo You need to change the superpose.line setting: xyplot(Choline ~ time, groups=BMI, data=data, type=l, scales=list(relation=free), auto.key=list(title=BMI Group, border=FALSE, lines=TRUE, points=FALSE), xlab=c(Point in Time), ylab=c(Concentration of Choline), par.settings = list(superpose.line = list(lwd=3))) library(lattice) data - data.frame(cbind(1:2,c(1,1,2,2), c(0.5,0.9,1.0,1.8))) names(data) - c(BMI,time,Choline) data$BMI - factor(data$BMI) levels(data$BMI) - c(=17.5,17.5) data$time - factor(data$time) levels(data$time) - c(Admission,Discharge) #Show names of settings names(trellis.par.get()) #Try to set the line width of the two plotted colored lines line-trellis.par.get(plot.line) line line$lwd=3 trellis.par.set(plot.line, line) line #Without group option: Line width is changed xyplot(Choline ~ time, data=data, type=l, scales=list(relation=free), auto.key=list(title=BMI Group, border=FALSE), xlab=c(Point in Time), ylab=c(Concentration of Choline)) #With group option: Line width is not changed xyplot(Choline ~ time, group=BMI, data=data, type=l, scales=list(relation=free), auto.key=list(title=BMI Group, border=FALSE), xlab=c(Point in Time), ylab=c(Concentration of Choline)) __ 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. -- Chuck Cleland, Ph.D. NDRI, Inc. (www.ndri.org) 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ 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] compare observed and fitted GAM values
You haven't given quite enough information to be sure, but I would guess that this is not really a problem, but rather the interesting proporty of GLMs fitted with a canonical link described in e.g. section 2.1.8 of Wood (2006) Generalized additive models: and introduction with R, or at the beginning of the GLM chapter in Venables and Ripley MASS. On Friday 21 August 2009 22:03, Lucía Rueda wrote: Hi, I am comparing the observed and fitted values of my GAM model, which includes the explanatory variables: longitude, depth, ssh, year and month. When I compare observed and fitted values for longitude, depth and ssh it works. But when I try to do it for month and year (which are as factors in the GAM model) it doesn't work. My observed and fitted values are exactly the same.. How is that possible?� Thanks Obs_factor1-aggregate(x=albdata$turtles,by=list(albdata$Year),FUN=mean) names(Obs_factor1)=c(Bin,Observed) Obs_factor1 ��� Bin��� Observed 1� 1997 0.017094017 2� 1998 0.010652463 3� 1999 0.02300 4� 2000 0.017167382 5� 2001 0.030465950 6� 2002 0.007446809 7� 2003 0.010568032 8� 2004 0.011450382 9� 2005 0.016270338 10 2006 0.017006803 11 2007 0.030969031 12 2008 0.066455696 Fit_factor1-aggregate(x=predict(gam.def.lon,type=response),by=list(alb data$Year),FUN=mean) names(Fit_factor1)=c(Bin,Fitted) Fit_factor1 ��� Bin� Fitted 1� 1997 0.017094017 2� 1998 0.010652463 3� 1999 0.02300 4� 2000 0.017167382 5� 2001 0.030465950 6� 2002 0.007446809 7� 2003 0.010568032 8� 2004 0.011450382 9� 2005 0.016270338 10 2006 0.017006803 11 2007 0.030969031 12 2008 0.066455696 [[alternative HTML version deleted]] -- Simon Wood, Mathematical Sciences, University of Bath, Bath, BA2 7AY UK +44 1225 386603 www.maths.bath.ac.uk/~sw283 __ 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] natural sorting a data frame /vector by row
How to NATURAL sort a vector or data frame* by row* , in ascending order ? V1 V2V3 V4 i1 5.00e-01 1.036197e-17 4.825338e+16 0. i104.001692e-18 1.365740e-17 2.930053e-01 0.76973827 i12 -1.052843e-17 1.324484e-17 -7.949081e-01 0.42735000 i132.571236e-17 1.357336e-17 1.894325e+00 0.05922715 i2-5.630739e-18 1.638267e-17 -3.437010e-01 0.73133282 i3 4.291387e-18 1.207522e-17 3.553879e-01 0.72257050 i4 1.472662e-17 1.423051e-17 1.034863e+00 0.30163897 i5 5.00e-01 1.003323e-17 4.983441e+16 0. i6 5.147966e-18 1.569095e-17 3.280850e-01 0.74309614 i7 1.096044e-17 1.555829e-17 7.044760e-01 0.48173041 i8-1.166290e-18 1.287370e-17 -9.059482e-02 0.92788026 i9 1.627371e-17 1.540567e-17 1.056345e+00 0.29173427 recmeanC2 9.275880e-17 6.322780e-17 1.467057e+00 0.14349903 NA NANA NA recmeanC3 1.283534e-17 2.080644e-17 6.168929e-01 0.53781390 recmeanC4-3.079466e-17 2.565499e-17 -1.200338e+00 0.23103743 I want a sequence of rows as :-- *recmeanC2 ,recmeanC3,recmeanC4* and the *NA row in the third position from the top*(presently it's third from down) -- Thanks Moumita [[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] image() generates many border lines in pdf, not on screen (quartz) - R 2.9.1 GUI 1.28 Tiger build 32-bit (5444) - OS X 10.5.8
On 23 Aug 2009, at 20:26, Uwe Ligges wrote: Since it looks like nobody answered so far: Your code is not reproducible, we do not have rfc, y, zVals nor NoCols. It's much easier to reproduce: just type in the first example from the image help page x - y - seq(-4*pi, 4*pi, len=27) r - sqrt(outer(x^2, y^2, +)) image(z = z - cos(r^2)*exp(-r/6), col=gray((0:32)/32)) then save from the quartz() display (I used the menu) and view with Adobe Reader 9 (I seem to have 9.0.0). Instead of the fine white lines you always get with Preview.app and other inaccurate PDF renderers, there are now huge gaps between the pixels (around 1/10th of pixel width). This is very probably a bug in the Quartz device (or Quartz itself), as the lines go away if you save the plot with dev.copy2pdf(), which I normally use. @OP: Do you have any particular reason for using quartz.save() or the menu item instead of dev.copy2pdf()? You could also try to place a screenshot somewhere on a webpage including the info about the settings of the corresponding viewer. I've tried switching off _all_ of the numerous anti-aliasing options of Adobe Reader 9; absolutely no difference. Hope this helps, Stefan __ 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] calculating probability
Hi all, I've a trivial question. If (q) is a continous variable,actually a vector of 1000 values. how to calculate the probability that q is greater than a specific value, i.e. P(q45)?? Thanks Maram [[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] Import/export ENVI files
Hi! I'm a beginner with this webpage so, I don't know if I'm sending my question to the correct site. Anyway, I'm working with R and I need to import and export ENVI files, (*.HDR files). A colleague told me that there is a package to import/export envi files but I haven't found that package, so does anyone know something about this? thank you so much :) . Ciaooo _ [[elided Hotmail spam]] [[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] R with MPI
Hello, I plan to use R with my cluster with OpenMPI. I need the packaged 'snow' and 'Rmpi' for that, however, I get an error while downloading and installing them: When I do a: install.packages(Rmpi, dependencies=T) I get this error: checking for mpi.h... no Try to find libmpi.so or libmpich.a checking for main in -lmpi... no libmpi not found. exiting... However, mpi.h is present via the openmpi-devel package on my RHEL 5.3. Some of those packages need sprng 2.0 (rsprng, for instance, which is a dependency for another MPI-related package). Sprng 2.0, however, isn't in developement for years, I wonder how I am supposed to keep my software up to date... Any ideas on how to workaround that mpi.h problem? Please help, --polemon [[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] hdf5 package segfault when processing large data
Hi there, I am currently working on something that uses hdf5 library. I think hdf5 is a great data format, I've used it somewhat extensively in python via PyTables. I was looking for something similar to that in R. The closest I can get is this library: hdf5. While it does not work the same way as PyTables did, but it's good enough to let them exchange data via hdf5 file. There is just 1 problem, I keep getting Segfault error when trying to process large files (10MB), although this is by no mean large when we talk about hdf5 capabilities. I have included the example code and data below. I have tried with different OS (WinXP and Ubuntu 8.04), architecture (32 and 64bit) and R versions (2.7.1, 2.72, and 2.9.1), but all of them present the same problem. I was wondering if anyone have any clue as to what's going on here and maybe can advice me to handle it. Thank you, appreciate any help i can get. Cheers, Budi The example script library(hdf5) fileName - sample.txt myTable - read.table(fileName,header=TRUE,sep=\t,as.is=TRUE) hdf5save(test.hdf, myTable) The data example, the list continue for more than 250,000 rows: sample.txt DateTimef1 f2 f3 f4 f5 2007032807:56 463 463.07 462.9 463.01 1100 2007032807:57 463.01 463.01 463.01 463.01 200 __ 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] transforming data glm
Dear sir, I am fitting a glm with default identity link: model-glm(timetoacceptsecs~maleage*maletub*relweight*malemobtrue*femmobtrue) the model is overdisperesed and plot model shows a low level of linearity of the residuals. The overdispersion and linearity of residulas on the normal Q-Q plot is corrected well by using: model-glm(log(timetoacceptsecs)~maleage*maletub*relweight*malemobtrue*femmobtrue)) Boxcox of my model also suggests that the log transformation is what i should do. I ask how i am able to do this by changing the link function or error family of my glm and not diretly taking the log of the response variable. For instance: model-glm(log(timetoacceptsecs)~maleage*maletub*relweight*malemobtrue*femmobtrue, family=poisson)) does not improve my model in terms of overdispersion etc as much as taking the log. Thank you __ 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] Copy Paste from tktext on Mac
Hi there, a text Window is supposed to map the shortcuts for copying and pasting (Ctrl-C, Ctrl-V) automatically. I'm working under Mac OS X and my text window doesn't really map these functions automatically - it works fine under Windows. Is there an easy way to map copypaste functions to a text window under Mac OS X? This is what I'm doing with my text window: tt - tktoplevel() txt - tktext(tt, bg = white, height=30, width=100, borderwidth=2) scr - tkscrollbar(tt, orient = vertical, repeatinterval = 1, command = function(...) tkyview(txt, ...)) tkconfigure(txt, yscrollcommand = function(...) tkset(scr, ...)) tkgrid(txt, column=0, row=0, columnspan=2, sticky=nwse) Session Info: R version 2.9.1 (2009-06-26) i386-apple-darwin8.11.1 locale: de_DE.UTF-8/de_DE.UTF-8/C/C/de_DE.UTF-8/de_DE.UTF-8 attached base packages: [1] tcltk stats graphics grDevices utils datasets methods base other attached packages: [1] tkrplot_0.0-18 rpart_3.1-44 relimp_1.0-1 Thanks so much! -- Anne Skoeries __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] error in creating gantt chart.
hi every one, i have a excel sheet like this labels starts ends 1 first task 1-Jan-04 3-Mar-04 2 second task 2-Feb-04 5-May-04 3 third task 3-Mar-04 6-Jun-04 4 fourth task 4-Apr-04 8-Aug-04 5 fifth task 5-May-04 9-Sep-04 now i converted this excel sheet into csv file and i read the csv file into R with the below code. my.gantt.info-read.csv(C:/Documents and Settings/balakrishna/Desktop/one.csv). and for create gantt chart i used below code. gantt.chart(my.gantt.info). if i run this above code i am getting the error like this Error in x$starts : $ operator is invalid for atomic vectors. can anybody help in this aspect it would be very appreciable. Thanks in Advance. -- View this message in context: http://www.nabble.com/error-in-creating-gantt-chart.-tp25115102p25115102.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] Text in barplot.
Dear all, i want to display a text in a barplot. Actual I simply work with the following lines: text(0.3,0.2,substring(teammembers.paint,1,140),cex=0.45,pos=4,srt=90) text(0.48,0.2,substring(teammembers.paint,141,1000),cex=0.45,pos=4,srt=90) The problem is, that teammembers.paint could be very long, so I split text with substring - but this is no good solution because the separation must be fixed. Is there any other possibility to display a long text in a nicer way, e.g in a box with the possibility of a line-break? Thank you very much in advance, Jens. -- View this message in context: http://www.nabble.com/Text-in-barplot.-tp25111937p25111937.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] Text in barplot.
?strwrap On Mon, Aug 24, 2009 at 3:53 AM, kojjens.k...@gmx.li wrote: Dear all, i want to display a text in a barplot. Actual I simply work with the following lines: text(0.3,0.2,substring(teammembers.paint,1,140),cex=0.45,pos=4,srt=90) text(0.48,0.2,substring(teammembers.paint,141,1000),cex=0.45,pos=4,srt=90) The problem is, that teammembers.paint could be very long, so I split text with substring - but this is no good solution because the separation must be fixed. Is there any other possibility to display a long text in a nicer way, e.g in a box with the possibility of a line-break? Thank you very much in advance, Jens. -- View this message in context: http://www.nabble.com/Text-in-barplot.-tp25111937p25111937.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] discriminant analysis
Beatriz Yannicelli wrote: Dear all: Is it possible to conduct a discriminant analysis in R with categorical and continuous variables as predictors? Beatriz Beatriz, Simply doing this in the R console: RSiteSearch(discriminant) yields many promising links. In particular, check documentation of package mda. HTH Rubén __ 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] image() generates many border lines in pdf, not on screen (quartz) - R 2.9.1 GUI 1.28 Tiger build 32-bit (5444) - OS X 10.5.8
Stefan Evert wrote: On 23 Aug 2009, at 20:26, Uwe Ligges wrote: Since it looks like nobody answered so far: Your code is not reproducible, we do not have rfc, y, zVals nor NoCols. It's much easier to reproduce: just type in the first example from the image help page x - y - seq(-4*pi, 4*pi, len=27) r - sqrt(outer(x^2, y^2, +)) image(z = z - cos(r^2)*exp(-r/6), col=gray((0:32)/32)) then save from the quartz() display Well, not possible for me on Windows . Anyway, please use the pdf() device directly and see if it works: pdf(path/to/file.pdf) x - y - seq(-4*pi, 4*pi, len=27) r - sqrt(outer(x^2, y^2, +)) image(z = z - cos(r^2)*exp(-r/6), col=gray((0:32)/32)) dev.off() (I used the menu) and view with Adobe Reader 9 (I seem to have 9.0.0). You should upgrade (even just for security reasons and many bugfixes), 9.1.3 is recent on Windows. Uwe Ligges Instead of the fine white lines you always get with Preview.app and other inaccurate PDF renderers, there are now huge gaps between the pixels (around 1/10th of pixel width). This is very probably a bug in the Quartz device (or Quartz itself), as the lines go away if you save the plot with dev.copy2pdf(), which I normally use. @OP: Do you have any particular reason for using quartz.save() or the menu item instead of dev.copy2pdf()? You could also try to place a screenshot somewhere on a webpage including the info about the settings of the corresponding viewer. I've tried switching off _all_ of the numerous anti-aliasing options of Adobe Reader 9; absolutely no difference. Hope this helps, Stefan __ 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] Problem with passing a string to subset
Thanks Baptiste The eval(parse()) combination is just what I need. baptiste auguie wrote: Try this, mystr -c==1 subset(foo, eval(parse(text = mystr)) ) library(fortunes) fortune(parse) # try several times # I prefer this, but there is probably a better way mycond- quote(c==1) subset(foo, eval(bquote(.(mycond))) ) HTH, baptiste 2009/8/21 Sebastien Bihorel sebastien.biho...@cognigencorp.com: Dear R-users, The following question bothered me for the whole afternoon: how can one pass a string as the conditioning argument to subset? I tried plain mystr, eval(mystr), expression(mystr), etc... I don't to be able to find the correct syntax foo - data.frame(a=1:10,b=10:1,c=rep(1:2,5)) mystr-c==1 subset(foo,c==1) a b c 1 1 10 1 3 3 8 1 5 5 6 1 7 7 4 1 9 9 2 1 subset(foo,mystr) Error in subset.data.frame(foo, mystr) : 'subset' must evaluate to logical Any help would be greatly appreciated. Sebastien __ 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] calculating probability
maram salem wrote: Hi all, I've a trivial question. If (q) is a continous variable,actually a vector of 1000 values. how to calculate the probability that q is greater than a specific value, i.e. P(q45)?? Do you want to estimate any distribution or do you just want the empirical information on the number of values greater than 45? Foir the latter: mean(q 45) Uwe Ligges Thanks Maram [[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] calculating probability
On Aug 24, 2009, at 8:11 AM, maram salem wrote: Hi all, I've a trivial question. If (q) is a continous variable,actually a vector of 1000 values. how to calculate the probability that q is greater than a specific value, i.e. P(q45)?? sum(q45)/1000 # if no NA's in vector sum(q45, na.rm=TRUE)/ sum(!is.na(q)) # if NA's in vector -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] natural sorting a data frame /vector by row
Moumita Das ha scritto: How to NATURAL sort a vector or data frame* by row* , in ascending order ? V1 V2V3 V4 i1 5.00e-01 1.036197e-17 4.825338e+16 0. i104.001692e-18 1.365740e-17 2.930053e-01 0.76973827 i12 -1.052843e-17 1.324484e-17 -7.949081e-01 0.42735000 i132.571236e-17 1.357336e-17 1.894325e+00 0.05922715 i2-5.630739e-18 1.638267e-17 -3.437010e-01 0.73133282 i3 4.291387e-18 1.207522e-17 3.553879e-01 0.72257050 i4 1.472662e-17 1.423051e-17 1.034863e+00 0.30163897 i5 5.00e-01 1.003323e-17 4.983441e+16 0. i6 5.147966e-18 1.569095e-17 3.280850e-01 0.74309614 i7 1.096044e-17 1.555829e-17 7.044760e-01 0.48173041 i8-1.166290e-18 1.287370e-17 -9.059482e-02 0.92788026 i9 1.627371e-17 1.540567e-17 1.056345e+00 0.29173427 recmeanC2 9.275880e-17 6.322780e-17 1.467057e+00 0.14349903 NA NANA NA recmeanC3 1.283534e-17 2.080644e-17 6.168929e-01 0.53781390 recmeanC4-3.079466e-17 2.565499e-17 -1.200338e+00 0.23103743 I want a sequence of rows as :-- *recmeanC2 ,recmeanC3,recmeanC4* and the *NA row in the third position from the top*(presently it's third from down) I do not understand what NATURAL stand for, but I'm not mothertongue in English. Is this the order you want ? recmeanC2 9.275880e-17 6.322780e-17 1.467057e+00 0.14349903 recmeanC3 1.283534e-17 2.080644e-17 6.168929e-01 0.53781390 recmeanC4-3.079466e-17 2.565499e-17 -1.200338e+00 0.23103743 NA NANA NA i1 5.00e-01 1.036197e-17 4.825338e+16 0. i2-5.630739e-18 1.638267e-17 -3.437010e-01 0.73133282 i3 4.291387e-18 1.207522e-17 3.553879e-01 0.72257050 i4 1.472662e-17 1.423051e-17 1.034863e+00 0.30163897 i5 5.00e-01 1.003323e-17 4.983441e+16 0. i6 5.147966e-18 1.569095e-17 3.280850e-01 0.74309614 i7 1.096044e-17 1.555829e-17 7.044760e-01 0.48173041 i8-1.166290e-18 1.287370e-17 -9.059482e-02 0.92788026 i9 1.627371e-17 1.540567e-17 1.056345e+00 0.29173427 i104.001692e-18 1.365740e-17 2.930053e-01 0.76973827 ***no i11 ? *** i12 -1.052843e-17 1.324484e-17 -7.949081e-01 0.42735000 i132.571236e-17 1.357336e-17 1.894325e+00 0.05922715 If so I'm afraid there's no a simple way to do it. This a possible solution df.newdata - cbind.data.frame(df.yourdata, foo=c(13, 15, 16, 17, 1, 2:12, 13:15) df.newdataOrdered - df.newdata[sort(df.newdata$foo),] another solution could be to rename the items in column 1 -- Ottorino __ 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] Copy Paste from tktext on Mac
On Aug 24, 2009, at 8:21 AM, Anne Skoeries wrote: Hi there, a text Window is supposed to map the shortcuts for copying and pasting (Ctrl-C, Ctrl-V) automatically. I'm working under Mac OS X and my text window doesn't really map these functions automatically - it works fine under Windows. That is a bit vague. When discussing user interface behavior, it would make sense to be more explicit about keystrokes. It's not clear here that you aware that the Mac (which did start delivering this GUI pizzazz long before MS-Windows was even halfway stable) uses cmd-C and cmd-V as the to- and from-clipboard key operations, rather than crtl-C and ctrl-V? Is there an easy way to map copypaste functions to a text window under Mac OS X? ??clipboard Brings up two hits on my machine one for connections and one for TkCommands. The first has a section on functions to access the Clipboard from R. The second help page should appear if you have tcltk loaded, although I don't at the moment so I cannot test that. ?connections ?TkCommands # probably There is a specific help list for Mac questions and a lot of the really knowledgeable people read it more regularly than r-help. This is what I'm doing with my text window: tt - tktoplevel() txt - tktext(tt, bg = white, height=30, width=100, borderwidth=2) scr - tkscrollbar(tt, orient = vertical, repeatinterval = 1, command = function(...) tkyview(txt, ...)) tkconfigure(txt, yscrollcommand = function(...) tkset(scr, ...)) tkgrid(txt, column=0, row=0, columnspan=2, sticky=nwse) Session Info: R version 2.9.1 (2009-06-26) i386-apple-darwin8.11.1 locale: de_DE.UTF-8/de_DE.UTF-8/C/C/de_DE.UTF-8/de_DE.UTF-8 attached base packages: [1] tcltk stats graphics grDevices utils datasets methods base other attached packages: [1] tkrplot_0.0-18 rpart_3.1-44 relimp_1.0-1 Thanks so much! -- Anne Skoeries __ 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. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to specify between group variance in lme
Hello r-help, I am using lme with two specs for the variance func varComb(varFixed(~1/n)),varPower(~Age)) this produces worse forecasts than the lm model with simple weights=n I think due to the fact that the lme spec works on variance inside the group. I need to show it that 1/n scales the variance between groups. Is that possible? I cannot disclose my dataset, but could post plots if that is possible somehow, let me know. Anova shows that everything in random=~poly(age,2) is significant. Thank you all very much. Stephen [[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] image() generates many border lines in pdf, not on screen (quartz) - R 2.9.1 GUI 1.28 Tiger build 32-bit (5444) - OS X 10.5.8
Your code is not reproducible, we do not have rfc, y, zVals nor NoCols. It's much easier to reproduce: just type in the first example from the image help page x - y - seq(-4*pi, 4*pi, len=27) r - sqrt(outer(x^2, y^2, +)) image(z = z - cos(r^2)*exp(-r/6), col=gray((0:32)/32)) then save from the quartz() display Well, not possible for me on Windows . Anyway, please use the pdf() device directly and see if it works: Works just as well as dev.copy2pdf() ... it really appears to be a bug in the Quartz framework. Unfortunately, I don't know why the OP doesn't just use the pdf() device. (I used the menu) and view with Adobe Reader 9 (I seem to have 9.0.0). You should upgrade (even just for security reasons and many bugfixes), 9.1.3 is recent on Windows. The newest I could get from Adobe is 9.1.0. Strangely enough, even though I have configured it to check for updates on startup, Adobe Reader never suggested to me that an update may be available. Thanks for mentioning this. Cheers, Stefan __ 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] Import/export ENVI files
2009/8/24 Lucas Sevilla García luckocto...@hotmail.com: Hi! I'm a beginner with this webpage so, I don't know if I'm sending my question to the correct site. Anyway, I'm working with R and I need to import and export ENVI files, (*.HDR files). A colleague told me that there is a package to import/export envi files but I haven't found that package, so does anyone know something about this? thank you so much :) . Ciaooo I'm guessing they are geographic data? Gridded, raster data perhaps? This information might have helped... Remember there are 36^3 = 46656 possible alpha-numeric three-character file endings and I'm not sure I know all of them. Lucky for you this one rang a bell... So, you probably want the rgdal package. If you've not got it already, then do this in R: install.packages(rgdal) when that's done, you need to load it: library(rgdal) and then try and read your file: map = readGDAL(file.hdr) and then try: summary(map) image(map) That's a start. Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to specify between group variance in lme
Clarification: Lm is much better than the base forecast from lme level=0, Level=1 produces a much tighter fit than lm. I was expecting that level=0 would produce something very close to lm, but it does not. [[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] error in creating gantt chart.
On Aug 24, 2009, at 8:22 AM, rajclinasia wrote: hi every one, i have a excel sheet like this labels starts ends 1 first task 1-Jan-04 3-Mar-04 2 second task 2-Feb-04 5-May-04 3 third task 3-Mar-04 6-Jun-04 4 fourth task 4-Apr-04 8-Aug-04 5 fifth task 5-May-04 9-Sep-04 now i converted this excel sheet into csv file and i read the csv file into R with the below code. my.gantt.info-read.csv(C:/Documents and Settings/balakrishna/Desktop/one.csv). my.gantt.info-read.csv(textConnection( labels starts ends + 1 first task 1-Jan-04 3-Mar-04 + 2 second task 2-Feb-04 5-May-04 + 3 third task 3-Mar-04 6-Jun-04 + 4 fourth task 4-Apr-04 8-Aug-04 + 5 fifth task 5-May-04 9-Sep-04)) my.gantt.info labels...starts.ends 1 1 first task 1-Jan-04 3-Mar-04 2 2 second task 2-Feb-04 5-May-04 3 3 third task 3-Mar-04 6-Jun-04 4 4 fourth task 4-Apr-04 8-Aug-04 5 5 fifth task 5-May-04 9-Sep-04 So that may look successful to you but that data.frame contains all of that data in a single (character) column. Why? Because a function was expecting commas on a file that did not have any. You will probably get further along if you use read.table with header=TRUE. Maybe you did something different or the file did have commas. With such a small file, you really should present the results of dput(my.gantt.info) That will contain all the values and attributes of the R object ... no more guessing, which is what we are doing now. and for create gantt chart i used below code. gantt.chart(my.gantt.info). That looks wrong. my.gantt.info is an R object. Is gantt.chart (from whatever unspecified package) really expecting to have its arguments quoted? I would bet against that possibility. I would also guess that, even if the data input issues are not a problem and the quotes are removed, you still have not converted those character values that you think look like dates into objects that R will interpret as dates. ?as.Date if i run this above code i am getting the error like this Error in x$starts : $ operator is invalid for atomic vectors. can anybody help in this aspect it would be very appreciable. Thanks in Advance. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R survival package error message - bug?!
Dear all, I have encountered a weird behaviour in R survival package which seems to me to be a bug. The weird behaviour happens when I am using 100 variables in the ridge function when calling coxph with following formula Surv(time = futime, event = fustat, type = right) ~ ridge(X1, X2, X3, X4, X5, X6, X7, X8, X9, X10, X11, X12, X13, X14, X15, X16, X17, X18, X19, X20, X21, X22, X23, X24, X25, X26, X27, X28, X29, X30, X31, X32, X33, X34, X35, X36, X37, X38, X39, X40, X41, X42, X43, X44, X45, X46, X47, X48, X49, X50, X51, X52, X53, X54, X55, X56, X57, X58, X59, X60, X61, X62, X63, X64, X65, X66, X67, X68, X69, X70, X71, X72, X73, X74, X75, X76, X77, X78, X79, X80, X81, X82, X83, X84, X85, X86, X87, X88, X89, X90, X91, X92, X93, X94, X95, X96, X97, X98, X99, X100, theta = lambda, scale = FALSE) and get the error message Error in model.frame.default(formula = form, data = sdf) : invalid variable names Calls: coxph - eval - eval - model.frame - model.frame.default Execution halted I am using R 2.8.0 and the latest version of the survival package 2.35-4. Here is how you can re-create the error message library(survival) Loading required package: splines x-as.data.frame(matrix(rnorm(100*500),ncol=100)) x$y-1:500 x$status-1 lambda-1.0 ff-as.formula(paste(Surv(y,status)~ridge(, paste(names(x)[1:100],collapse=,), ,theta = lambda, scale = FALSE))) coxph(ff,x) Error in model.frame.default(formula = ff, data = x) : invalid variable names print (ff) Surv(y, status) ~ ridge(V1, V2, V3, V4, V5, V6, V7, V8, V9, V10, V11, V12, V13, V14, V15, V16, V17, V18, V19, V20, V21, V22, V23, V24, V25, V26, V27, V28, V29, V30, V31, V32, V33, V34, V35, V36, V37, V38, V39, V40, V41, V42, V43, V44, V45, V46, V47, V48, V49, V50, V51, V52, V53, V54, V55, V56, V57, V58, V59, V60, V61, V62, V63, V64, V65, V66, V67, V68, V69, V70, V71, V72, V73, V74, V75, V76, V77, V78, V79, V80, V81, V82, V83, V84, V85, V86, V87, V88, V89, V90, V91, V92, V93, V94, V95, V96, V97, V98, V99, V100, theta = lambda, scale = FALSE) Also I have found that the code breaks with number of variables greater than 97. For 97 and less it works fine. I have found that it breaks at the following line of the coxph function if (is.R()) m - eval(temp, parent.frame()) I have been trying to understand why it breaks there and have not progressed much so far. With kind regards DK _ icons. [[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] R with MPI
On Mon, Aug 24, 2009 at 2:20 PM, polemon pole...@gmail.com wrote: Hello, I plan to use R with my cluster with OpenMPI. I need the packaged 'snow' and 'Rmpi' for that, however, I get an error while downloading and installing them: When I do a: install.packages(Rmpi, dependencies=T) I get this error: checking for mpi.h... no Try to find libmpi.so or libmpich.a checking for main in -lmpi... no libmpi not found. exiting... However, mpi.h is present via the openmpi-devel package on my RHEL 5.3. Some of those packages need sprng 2.0 (rsprng, for instance, which is a dependency for another MPI-related package). Sprng 2.0, however, isn't in developement for years, I wonder how I am supposed to keep my software up to date... Any ideas on how to workaround that mpi.h problem? Please help, --polemon I did as described here: http://www.cybaea.net/Blogs/Data/R-tips-Installing-Rmpi-on-Fedora-Linux.html Since Fedora and RHEL are pretty equal, I gave that installation a shot, and from what I can tell, I got pretty far. The package installed well, but when I try to load it with library(Rmpi): library(Rmpi) Error in dyn.load(file, DLLpath = DLLpath, ...) : unable to load shared library '/opt/R/lib64/R/library/Rmpi/libs/Rmpi.so': libmpi.so.0: cannot open shared object file: No such file or directory Error in library(Rmpi) : .First.lib failed for 'Rmpi' Error in dyn.unload(file.path(libpath, libs, paste(Rmpi, .Platform$dynlib.ext, : dynamic/shared library '/opt/R/lib64/R/library/Rmpi/libs/Rmpi.so' was not loaded As you can see, R is installed in /opt/R, libmpi.so.0 is available: /usr/lib/lam/lib/libmpi.so.0 /usr/lib/lam/lib/libmpi.so.0.0.0 /usr/lib/openmpi/1.2.7-gcc/lib/libmpi.so /usr/lib/openmpi/1.2.7-gcc/lib/libmpi.so.0 /usr/lib/openmpi/1.2.7-gcc/lib/libmpi.so.0.0.0 /usr/lib64/lam/lib/libmpi.so.0 /usr/lib64/lam/lib/libmpi.so.0.0.0 /usr/lib64/openmpi/1.2.7-gcc/lib/libmpi.so /usr/lib64/openmpi/1.2.7-gcc/lib/libmpi.so.0 /usr/lib64/openmpi/1.2.7-gcc/lib/libmpi.so.0.0.0 What should I do, to make Rmpi available in R? Cheers, --polemon [[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] Convert list to data frame while controlling column types
On 8/24/2009 2:06 AM, Petr PIKAL wrote:
Hi
r-help-boun...@r-project.org napsal dne 23.08.2009 17:29:48:
On 8/23/2009 9:58 AM, David Winsemius wrote:
I still have problems with this statement. As I understand R, this
should be impossible. I have looked at both you postings and neither of
them clarify the issues. How can you have blanks or spaces in an R
numeric vector?
Just because I search numeric columns doesn't mean that my regex matches
them! I posted some info on my data frame in an earlier email:
str(final_dataf)
'data.frame': 1127 obs. of 43 variables:
$ block : Factor w/ 1 level 2: 1 1 1 1 1 1 1 1 1 1 ...
$ treatment : Factor w/ 4 levels I,M,N,T: 1 1 1 1 1 1 ...
$ transect : Factor w/ 1 level 4: 1 1 1 1 1 1 1 1 1 1 ...
$ tag: chr NA 121AL 122AL 123AL ...
...
$ h1 : num NA NA NA NA NA NA NA NA NA NA ...
...
You can see that I do indeed have some numeric columns. And while I
Well, AFAICS you have a data frame with 3 columns which are factors and 1
which is character. I do not see any numeric column. If you want to change
block and transect to numeric you can use
df$block - as.numeric(as.character(df$block))
If you take a closer look at my data frame listing, you'll see that it
is 1127 obs. of 43 variables. I edited the column listing for
readability, and you'll see even in my editing listing I do indeed have
one numeric column - h1. And as I mentioned earlier, I use
colClasses, so no need to change anything to numeric here.
search them for spaces, I only do so because my dataset isn't so large
as to require me to exclude them from the search. If my dataset grows
too big at some point, I will exclude numeric columns, and other columns
which cannot hold blanks or spaces.
To clarify further with an example:
df = data.frame(a=c(1,2,3,4,5),b=c(a,,c,d, ))
df = as.data.frame(lapply(df, function(x){ is.na(x) -
+ grep('^\\s*$',x); return(x) }), stringsAsFactors = FALSE)
df
ab
1 1a
2 2 NA
3 3c
4 4d
5 5 NA
which can be done also by
df[,2] - levels(df[,2])[1:2]-NA
but maybe with less generalization
Yes - my point was to show how I looped through an entire data frame
looking for \\s*, even when some of the columns were numeric. I gave
this simple example with a 2-column data frame to illustrate that point.
str(df)
'data.frame': 5 obs. of 2 variables:
$ a: num 1 2 3 4 5
$ b: Factor w/ 5 levels , ,a,c,..: 3 NA 4 5 NA
And one final clarification: I left out as.data.frame in my previous
solution. So it now becomes:
final_dataf = as.data.frame(lapply(final_dataf, function(x){ is.na(x)
+ - grep('^\\s*$',x); return(x) }), stringsAsFactors = FALSE)
Again not too much of clarification, in your first data frame second
column is a factor with some levels you want to convert to NA, which can
be done by different approaches.
This clarification was to show the code that worked (for posterity), as
my previous post left out an argument. It seems that perhaps you missed
the previous emails.
Your final_dataf is same as df.
Yes, that is the point. As I mentioned in the first email of this
thread, I was trying to get around as.data.frame's automatic conversion
routines, in order to retain the original column types. And it turned
out that gsub() was more of the problem than as.data.frame() was.
Please refer to the earlier emails for more information on that.
Columns which shall be numeric and are read as factor/character by
read.table likely contain some values which strictly can not be considered
numeric. You can see them quite often in Excel like programs and some
examples are
1..2, o.5, 12.o5 and or spaces, - e.t.c.
and you usually need handle them by hand.
Regards
Petr
Hope that clarifies things, and thanks for your help.
Thanks,
Allie
On 8/23/2009 9:58 AM, David Winsemius wrote:
On Aug 23, 2009, at 2:47 AM, Alexander Shenkin wrote:
On 8/21/2009 3:04 PM, David Winsemius wrote:
On Aug 21, 2009, at 3:41 PM, Alexander Shenkin wrote:
Thanks everyone for their replies, both on- and off-list. I should
clarify, since I left out some important information. My original
dataframe has some numeric columns, which get changed to character
by
gsub when I replace spaces with NAs.
If you used is.na() - that would not happen to a true _numeric_
vector
(but, of course, a numeric vector in a data.frame could not have
spaces,
so you are probably not using precise terminology).
I do have true numeric columns, but I loop through my entire
dataframe
looking for blanks and spaces for convenience.
I still have problems with this statement. As I understand R, this
should be impossible. I have looked at both you postings and neither
of
them clarify the issues. How can you have blanks or spaces in an R
numeric vector?
You would be well
advised to include the actual code rather than applying loose
Re: [R] Selecting groups with R
To drop empty factor levels from a subset, I use the following: a.subset - subset(dataset, Color!='BLUE') ifac - sapply(a.subset,is.factor) a.subset[ifac] - lapply(a.subset[ifac],factor) Mike dataset Color Score 1 RED10 2 RED13 3 RED12 4 WHITE22 5 WHITE27 6 WHITE25 7 BLUE18 8 BLUE17 9 BLUE16 table(dataset) Score Color 10 12 13 16 17 18 22 25 27 BLUE 0 0 0 1 1 1 0 0 0 RED1 1 1 0 0 0 0 0 0 WHITE 0 0 0 0 0 0 1 1 1 a.subset - subset(dataset, Color!='BLUE') a.subset Color Score 1 RED10 2 RED13 3 RED12 4 WHITE22 5 WHITE27 6 WHITE25 table(a.subset) Score Color 10 12 13 22 25 27 BLUE 0 0 0 0 0 0 RED1 1 1 0 0 0 WHITE 0 0 0 1 1 1 ifac - sapply(a.subset,is.factor) a.subset[ifac] - lapply(a.subset[ifac],factor) table(a.subset) Score Color 10 12 13 22 25 27 RED1 1 1 0 0 0 WHITE 0 0 0 1 1 1 __ 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] error in creating gantt chart.
On Mon, 24 Aug 2009 05:22:17 -0700 (PDT) rajclinasia r...@clinasia.com wrote: R my.gantt.info-read.csv(C:/Documents and R Settings/balakrishna/Desktop/one.csv). R R and for create gantt chart i used below code. R R gantt.chart(my.gantt.info). This again is why others have pointed you first to have a look at the basics of R which can be read at: http://cran.r-project.org/manuals.html or more extensively in: http://cran.r-project.org/other-docs.html You start to program without knowing the very basics of R. Which is in this case the data structures. my.gantt.info is a data frame. But with the quotation marks you don't even give this as an object to gantt.chart because with the quotation mark you have created a single variable with a single value which is the characters between the quotation marks that have nothing to do with the data - hence the atomic stuff error message... If you would have taken the real data you need the objects name which is: gantt.chart(my.gantt.info) but this would not work either since as I have already pointed out in an earlier mail lase week: you need a list but not a data.frame. So please also consider the documentation which you can do with ?gantt.chart which directly points you towards this. So what you need is a list. What that is you can see either in the example of example(gantt.chart) what is a list? thats basics. Its an object type (there is data.frame, matrix, list and so on) ?list ?as.list Please learn the basics before asking such questions because you save a lot of time for yourself and we do as well because we would not need to answer such questions. Stefan __ 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] CRAN (and crantastic) updates this week
CRAN (and crantastic) updates this week New packages Updated packages New reviews --- This email provided as a service for the R community by http://crantastic.org. Like it? Hate it? Please let us know: crana...@gmail.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] CRAN (and crantastic) updates this week
Sorry, we had some problems with the initial sending of our weekly digest which resulted in a rather empty email. Here is the correct version: CRAN (and crantastic) updates this week New packages * atm (0.1.0) Charlotte Maia http://crantastic.org/packages/atm An R package for creating additive models with semiparametric predictors, emphasizing term objects, especially (1) implementation of a term class hierarchy, and (2) interpretation and evaluation of term estimates as functions of explanatories. * gnumeric (0.5-1) Karoly Antal http://crantastic.org/packages/gnumeric Read data files readable into R. Can read whole sheet or a range, from several file formats, including the native format of gnumeric. Reading is done by using ssconvert (a file converter utility included in the gnumeric distribution) to convert the requested part to CSV. * KFAS (0.3.1) Jouni Lehtonen http://crantastic.org/packages/KFAS Fast multivariate Kalman filter, smoother, simulation smoother and forecasting. Uses exact diffuse initialisation when distributions of some or all elements of initial state vector are unknown. * munsell (0.1) Charlotte Wickham http://crantastic.org/packages/munsell Functions for exploring and using the Munsell colour system * oosp (0.1.0) Charlotte Maia http://crantastic.org/packages/oosp An R package designed to support object oriented statistical programming, especially by extending S3 capabilities, providing pointer and component objects, and providing basic support for symbolic-numeric statistical programming. * PLIS (1.0) Zhi Wei http://crantastic.org/packages/PLIS PLIS is a multiple testing procedure for testing several groups of hypotheses. Linear dependency is expected from the hypotheses within the same group and is modeled by hidden Markov Models. It is noted that, for PLIS, a smaller p value does not necessarily imply more significance because of dependency among the hypotheses. A typical applicaiton of PLIS is to analyze genome wide association studies datasets, where SNPs from the same chromosome are treated as a group and exhibit strong linear genomic dependency. * rrv (0.0.1) Charlotte Maia http://crantastic.org/packages/rrv An incomplete R package for working with random return variables. The current package provides limited support for formatting money. More features will be added in the future. * ttrTests (1.0) David St John http://crantastic.org/packages/ttrTests Four core functions evaluate the efficacy of a technical trading rule. - Conditional return statistics - Bootstrap resampling statistics - Reality Check for data snooping bias among parameter choices - Robustness, or Persistence, of parameter choices Updated packages AdMit (1-01.03), alr3 (1.1.9), alr3 (1.1.10), cmprskContin (1.1), dataframes2xls (0.4.3), digeR (1.2), doBy (4.0.1), DoE.base (0.7), FrF2 (0.97-1), FrF2 (0.97-3), hdrcde (2.10), HH (2.1-30), JM (0.4-0), memisc (0.95-21), minet (2.0.0), MLDS (0.2-0), monomvn (1.7-3), mrt (0.3), nsRFA (0.6-9), PBSmodelling (2.21), plink (1.2-0), plotrix (2.7), RaschSampler (0.8-2), RCurl (1.0-0), rgdal (0.6-14), RSiena (1.0.5), rtv (0.3.0), sdef (1.1), SIS (0.2), spdep (0.4-36) New reviews --- * sqldf, by m.e.driscoll http://crantastic.org/reviews/26 * SensoMineR, by padmanabhan.vijayan http://crantastic.org/reviews/25 * plyr, by eamani http://crantastic.org/reviews/24 This email provided as a service for the R community by http://crantastic.org. Like it? Hate it? Please let us know: crana...@gmail.com. On Mon, Aug 24, 2009 at 4:18 PM, Hadley Wickham crana...@gmail.com wrote: CRAN (and crantastic) updates this week New packages Updated packages New reviews --- This email provided as a service for the R community by http://crantastic.org. Like it? Hate it? Please let us know: crana...@gmail.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] CRAN (and crantastic) updates this week
I like it. Thanks. 2009/8/24 hadley wickham crana...@gmail.com: Sorry, we had some problems with the initial sending of our weekly digest which resulted in a rather empty email. Here is the correct version: CRAN (and crantastic) updates this week New packages * atm (0.1.0) Charlotte Maia http://crantastic.org/packages/atm An R package for creating additive models with semiparametric predictors, emphasizing term objects, especially (1) implementation of a term class hierarchy, and (2) interpretation and evaluation of term estimates as functions of explanatories. * gnumeric (0.5-1) Karoly Antal http://crantastic.org/packages/gnumeric Read data files readable into R. Can read whole sheet or a range, from several file formats, including the native format of gnumeric. Reading is done by using ssconvert (a file converter utility included in the gnumeric distribution) to convert the requested part to CSV. * KFAS (0.3.1) Jouni Lehtonen http://crantastic.org/packages/KFAS Fast multivariate Kalman filter, smoother, simulation smoother and forecasting. Uses exact diffuse initialisation when distributions of some or all elements of initial state vector are unknown. * munsell (0.1) Charlotte Wickham http://crantastic.org/packages/munsell Functions for exploring and using the Munsell colour system * oosp (0.1.0) Charlotte Maia http://crantastic.org/packages/oosp An R package designed to support object oriented statistical programming, especially by extending S3 capabilities, providing pointer and component objects, and providing basic support for symbolic-numeric statistical programming. * PLIS (1.0) Zhi Wei http://crantastic.org/packages/PLIS PLIS is a multiple testing procedure for testing several groups of hypotheses. Linear dependency is expected from the hypotheses within the same group and is modeled by hidden Markov Models. It is noted that, for PLIS, a smaller p value does not necessarily imply more significance because of dependency among the hypotheses. A typical applicaiton of PLIS is to analyze genome wide association studies datasets, where SNPs from the same chromosome are treated as a group and exhibit strong linear genomic dependency. * rrv (0.0.1) Charlotte Maia http://crantastic.org/packages/rrv An incomplete R package for working with random return variables. The current package provides limited support for formatting money. More features will be added in the future. * ttrTests (1.0) David St John http://crantastic.org/packages/ttrTests Four core functions evaluate the efficacy of a technical trading rule. - Conditional return statistics - Bootstrap resampling statistics - Reality Check for data snooping bias among parameter choices - Robustness, or Persistence, of parameter choices Updated packages AdMit (1-01.03), alr3 (1.1.9), alr3 (1.1.10), cmprskContin (1.1), dataframes2xls (0.4.3), digeR (1.2), doBy (4.0.1), DoE.base (0.7), FrF2 (0.97-1), FrF2 (0.97-3), hdrcde (2.10), HH (2.1-30), JM (0.4-0), memisc (0.95-21), minet (2.0.0), MLDS (0.2-0), monomvn (1.7-3), mrt (0.3), nsRFA (0.6-9), PBSmodelling (2.21), plink (1.2-0), plotrix (2.7), RaschSampler (0.8-2), RCurl (1.0-0), rgdal (0.6-14), RSiena (1.0.5), rtv (0.3.0), sdef (1.1), SIS (0.2), spdep (0.4-36) New reviews --- * sqldf, by m.e.driscoll http://crantastic.org/reviews/26 * SensoMineR, by padmanabhan.vijayan http://crantastic.org/reviews/25 * plyr, by eamani http://crantastic.org/reviews/24 This email provided as a service for the R community by http://crantastic.org. Like it? Hate it? Please let us know: crana...@gmail.com. On Mon, Aug 24, 2009 at 4:18 PM, Hadley Wickham crana...@gmail.com wrote: CRAN (and crantastic) updates this week New packages Updated packages New reviews --- This email provided as a service for the R community by http://crantastic.org. Like it? Hate it? Please let us know: crana...@gmail.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- HUANG Ronggui, Wincent PhD Candidate Dept of Public and Social Administration City University of Hong Kong Home page: http://asrr.r-forge.r-project.org/rghuang.html __ 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 current neural network (RNN) package?
Hello, I'm trying to tackle a problem that would require the implementation of a recurrent NN. However, even though the CRAN is very big, I can’t seem to find a package for this. Does anybody here know if one exits? BR, John -- View this message in context: http://www.nabble.com/Recurrent-neural-network-%28RNN%29-package--tp25116968p25116968.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] CRAN (and crantastic) updates this week
Great idea - thx! On Mon, Aug 24, 2009 at 9:30 AM, hadley wickham crana...@gmail.com wrote: Sorry, we had some problems with the initial sending of our weekly digest which resulted in a rather empty email. Here is the correct version: CRAN (and crantastic) updates this week New packages * atm (0.1.0) Charlotte Maia http://crantastic.org/packages/atm An R package for creating additive models with semiparametric predictors, emphasizing term objects, especially (1) implementation of a term class hierarchy, and (2) interpretation and evaluation of term estimates as functions of explanatories. * gnumeric (0.5-1) Karoly Antal http://crantastic.org/packages/gnumeric Read data files readable into R. Can read whole sheet or a range, from several file formats, including the native format of gnumeric. Reading is done by using ssconvert (a file converter utility included in the gnumeric distribution) to convert the requested part to CSV. * KFAS (0.3.1) Jouni Lehtonen http://crantastic.org/packages/KFAS Fast multivariate Kalman filter, smoother, simulation smoother and forecasting. Uses exact diffuse initialisation when distributions of some or all elements of initial state vector are unknown. * munsell (0.1) Charlotte Wickham http://crantastic.org/packages/munsell Functions for exploring and using the Munsell colour system * oosp (0.1.0) Charlotte Maia http://crantastic.org/packages/oosp An R package designed to support object oriented statistical programming, especially by extending S3 capabilities, providing pointer and component objects, and providing basic support for symbolic-numeric statistical programming. * PLIS (1.0) Zhi Wei http://crantastic.org/packages/PLIS PLIS is a multiple testing procedure for testing several groups of hypotheses. Linear dependency is expected from the hypotheses within the same group and is modeled by hidden Markov Models. It is noted that, for PLIS, a smaller p value does not necessarily imply more significance because of dependency among the hypotheses. A typical applicaiton of PLIS is to analyze genome wide association studies datasets, where SNPs from the same chromosome are treated as a group and exhibit strong linear genomic dependency. * rrv (0.0.1) Charlotte Maia http://crantastic.org/packages/rrv An incomplete R package for working with random return variables. The current package provides limited support for formatting money. More features will be added in the future. * ttrTests (1.0) David St John http://crantastic.org/packages/ttrTests Four core functions evaluate the efficacy of a technical trading rule. - Conditional return statistics - Bootstrap resampling statistics - Reality Check for data snooping bias among parameter choices - Robustness, or Persistence, of parameter choices Updated packages AdMit (1-01.03), alr3 (1.1.9), alr3 (1.1.10), cmprskContin (1.1), dataframes2xls (0.4.3), digeR (1.2), doBy (4.0.1), DoE.base (0.7), FrF2 (0.97-1), FrF2 (0.97-3), hdrcde (2.10), HH (2.1-30), JM (0.4-0), memisc (0.95-21), minet (2.0.0), MLDS (0.2-0), monomvn (1.7-3), mrt (0.3), nsRFA (0.6-9), PBSmodelling (2.21), plink (1.2-0), plotrix (2.7), RaschSampler (0.8-2), RCurl (1.0-0), rgdal (0.6-14), RSiena (1.0.5), rtv (0.3.0), sdef (1.1), SIS (0.2), spdep (0.4-36) New reviews --- * sqldf, by m.e.driscoll http://crantastic.org/reviews/26 * SensoMineR, by padmanabhan.vijayan http://crantastic.org/reviews/25 * plyr, by eamani http://crantastic.org/reviews/24 This email provided as a service for the R community by http://crantastic.org. Like it? Hate it? Please let us know: crana...@gmail.com. On Mon, Aug 24, 2009 at 4:18 PM, Hadley Wickham crana...@gmail.com wrote: CRAN (and crantastic) updates this week New packages Updated packages New reviews --- This email provided as a service for the R community by http://crantastic.org. Like it? Hate it? Please let us know: crana...@gmail.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Two lines, two scales, one graph
First of all, thanks to everyone who answers these questions - it's most helpful. I'm new to R and despite searching have not found an example of what I want to do (there are some good beginner's guides and a lot of complex plots, but I haven't found this). I would like to plot two variables against the same abscissa values. They have different scales. I've found how to make a second axis on the right for labeling, but not how to plot two lines at different scales. Thanks, - Rick r...@ece.pdx.edu __ 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] robust method to obtain a correlation coeff?
Hi, Being a R-newbie I am wondering how to calculate a correlation coefficient (preferably with an associated p-value) for data like: d[,1] [1] 25.5 25.3 25.1 NA 23.3 21.5 23.8 23.2 24.2 22.7 27.6 24.2 ... d[,2] [1] 0.0 11.1 0.0 NA 0.0 10.1 10.6 9.5 0.0 57.9 0.0 0.0 ... Apparently corr(d) from the boot-library fails with NAs in the data, also cor.test cannot cope with a different number of NAs. Is there a solution to this problem (calculating a correlation coefficient and ignoring different number of NAs), e.g. Pearson's corr coeff? If so, please point me to the relevant piece of documentation. TIA Christian __ 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] robust method to obtain a correlation coeff?
On 24-Aug-09 14:47:02, Christian Meesters wrote: Hi, Being a R-newbie I am wondering how to calculate a correlation coefficient (preferably with an associated p-value) for data like: d[,1] [1] 25.5 25.3 25.1 NA 23.3 21.5 23.8 23.2 24.2 22.7 27.6 24.2 ... d[,2] [1] 0.0 11.1 0.0 NA 0.0 10.1 10.6 9.5 0.0 57.9 0.0 0.0 ... Apparently corr(d) from the boot-library fails with NAs in the data, Yes, apparently corr() has no option for dealing with NAs. also cor.test cannot cope with a different number of NAs. On the other hand, cor.test() does have an option na.action which, by default, is the same as what is in getOption(na.action). In my R installation, this, by default, is na.omit. This has the effect that, for any pair in (x,y) where at least one of the pair is NA, that pair will be omitted from the calculation. For example, basing two vectors x,y on your data above, and a third z which is y with an extra NA: x-c(25.5,25.3,25.1,NA,23.3,21.5,23.8,23.2,24.2,22.7,27.6,24.2) y-c( 0.0,11.1, 0.0,NA, 0.0,10.1,10.6, 9.5, 0.0,57.9, 0.0, 0.0) z-y; z[8]-NA I get cor.test(x,y) # Pearson's product-moment correlation # data: x and y # t = -1.3986, df = 9, p-value = 0.1954 # alternative hypothesis: true correlation is not equal to 0 # 95 percent confidence interval: # -0.8156678 0.2375438 # sample estimates: # cor # -0.422542 # cor.test(x,z) # Pearson's product-moment correlation # data: x and z # t = -1.3466, df = 8, p-value = 0.215 # alternative hypothesis: true correlation is not equal to 0 # 95 percent confidence interval: # -0.8338184 0.2738824 # sample estimates: #cor # -0.4298726 So it has worked in both cases (see the difference in 'df'), despite the different numbers of NAs in x and z. For functions such as corr() which do not have provision for omitting NAs, you can fix it up for yourself before calling the function. In the case of your two series d[,1], d[,2] you could use an index variable to select cases: ix - (!is.na(d[,1]))(!is.na(d[,2])) corr(d[ix,]) With my variables x,y,z I get ix.1 - (!is.na(x))(!is.na(y)) ix.2 - (!is.na(x))(!is.na(z)) d.1 -cbind(x,y) corr(d.1[ix.1,]) # [1] -0.422542 ## (and -0.422542 from cor.test above as well) d.2 - cbind(x,z) corr(d.2[ix.2,]) # [1] -0.4298726 ## (and -0.4298726 from cor.test above as well) Hoping this helps, Ted. Is there a solution to this problem (calculating a correlation coefficient and ignoring different number of NAs), e.g. Pearson's corr coeff? If so, please point me to the relevant piece of documentation. TIA Christian __ 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. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 24-Aug-09 Time: 16:26:53 -- XFMail -- __ 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] robust method to obtain a correlation coeff?
On Aug 24, 2009, at 11:26 AM, (Ted Harding) wrote: On 24-Aug-09 14:47:02, Christian Meesters wrote: Hi, Being a R-newbie I am wondering how to calculate a correlation coefficient (preferably with an associated p-value) for data like: d[,1] [1] 25.5 25.3 25.1 NA 23.3 21.5 23.8 23.2 24.2 22.7 27.6 24.2 ... d[,2] [1] 0.0 11.1 0.0 NA 0.0 10.1 10.6 9.5 0.0 57.9 0.0 0.0 ... Apparently corr(d) from the boot-library fails with NAs in the data, Yes, apparently corr() has no option for dealing with NAs. also cor.test cannot cope with a different number of NAs. On the other hand, cor.test() does have an option na.action which, by default, is the same as what is in getOption(na.action). In my R installation, this, by default, is na.omit. This has the effect that, for any pair in (x,y) where at least one of the pair is NA, that pair will be omitted from the calculation. For example, basing two vectors x,y on your data above, and a third z which is y with an extra NA: x-c(25.5,25.3,25.1,NA,23.3,21.5,23.8,23.2,24.2,22.7,27.6,24.2) y-c( 0.0,11.1, 0.0,NA, 0.0,10.1,10.6, 9.5, 0.0,57.9, 0.0, 0.0) z-y; z[8]-NA I get cor.test(x,y) snipped unneeded output # sample estimates: #cor # -0.4298726 So it has worked in both cases (see the difference in 'df'), despite the different numbers of NAs in x and z. You may not need to go through the material that follows. There are already a set of functions to handle such concerns: ?na.omit will bring a help page describing: na.fail(object, ...) na.omit(object, ...) na.exclude(object, ...) na.pass(object, ...) It reminded me that: na.action: the name of a function for treating missing values (NA's) for certain situations. ... but I do not know what those certain situations really are. For functions such as corr() which do not have provision for omitting NAs, you can fix it up for yourself before calling the function. In the case of your two series d[,1], d[,2] you could use an index variable to select cases: ix - (!is.na(d[,1]))(!is.na(d[,2])) corr(d[ix,]) With my variables x,y,z I get ix.1 - (!is.na(x))(!is.na(y)) ix.2 - (!is.na(x))(!is.na(z)) d.1 -cbind(x,y) corr(d.1[ix.1,]) # [1] -0.422542 ## (and -0.422542 from cor.test above as well) d.2 - cbind(x,z) corr(d.2[ix.2,]) # [1] -0.4298726 ## (and -0.4298726 from cor.test above as well) Hoping this helps, Ted. Is there a solution to this problem (calculating a correlation coefficient and ignoring different number of NAs), e.g. Pearson's corr coeff? If so, please point me to the relevant piece of documentation. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] robust method to obtain a correlation coeff?
On Aug 24, 2009, at 11:38 AM, David Winsemius wrote: On Aug 24, 2009, at 11:26 AM, (Ted Harding) wrote: On 24-Aug-09 14:47:02, Christian Meesters wrote: Hi, Being a R-newbie I am wondering how to calculate a correlation coefficient (preferably with an associated p-value) for data like: d[,1] [1] 25.5 25.3 25.1 NA 23.3 21.5 23.8 23.2 24.2 22.7 27.6 24.2 ... d[,2] [1] 0.0 11.1 0.0 NA 0.0 10.1 10.6 9.5 0.0 57.9 0.0 0.0 ... Apparently corr(d) from the boot-library fails with NAs in the data, Yes, apparently corr() has no option for dealing with NAs. also cor.test cannot cope with a different number of NAs. On the other hand, cor.test() does have an option na.action which, by default, is the same as what is in getOption(na.action). In my R installation, this, by default, is na.omit. This has the effect that, for any pair in (x,y) where at least one of the pair is NA, that pair will be omitted from the calculation. For example, basing two vectors x,y on your data above, and a third z which is y with an extra NA: x-c(25.5,25.3,25.1,NA,23.3,21.5,23.8,23.2,24.2,22.7,27.6,24.2) y-c( 0.0,11.1, 0.0,NA, 0.0,10.1,10.6, 9.5, 0.0,57.9, 0.0, 0.0) z-y; z[8]-NA I get cor.test(x,y) snipped unneeded output # sample estimates: #cor # -0.4298726 So it has worked in both cases (see the difference in 'df'), despite the different numbers of NAs in x and z. You may not need to go through the material that follows. There are already a set of functions to handle such concerns: ?na.omit will bring a help page describing: na.fail(object, ...) na.omit(object, ...) na.exclude(object, ...) na.pass(object, ...) Apologies; this was a bit hastily constructed. What I was quoting in what follows was from the Options help page and Options set in package stats section of that help page. na.action: the name of a function for treating missing values (NA's) for certain situations. ... but I do not know what those certain situations really are. So there are some function that may be affected by settings of options(na.action) but I cannot tell you where to find a list of such functions. For functions such as corr() which do not have provision for omitting NAs, you can fix it up for yourself before calling the function. In the case of your two series d[,1], d[,2] you could use an index variable to select cases: ix - (!is.na(d[,1]))(!is.na(d[,2])) corr(d[ix,]) With my variables x,y,z I get ix.1 - (!is.na(x))(!is.na(y)) ix.2 - (!is.na(x))(!is.na(z)) d.1 -cbind(x,y) corr(d.1[ix.1,]) # [1] -0.422542 ## (and -0.422542 from cor.test above as well) d.2 - cbind(x,z) corr(d.2[ix.2,]) # [1] -0.4298726 ## (and -0.4298726 from cor.test above as well) Hoping this helps, Ted. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Multiply List by a Numeric
I apologize for what seems like it should be a straighforward query.
I am trying to multiply a list by a numeric and thought there would be a
straightforward way to do this, but the best solution I found so far has a
for loop.
Everything else I try seems to throw an error non-numeric argument to
binary operator
Consider the example:
a - 1
b - 1:2
c - 1:3
abc - list(a,b,c)
To multiply every element of abc by a numeric, say 3, I wrote a for-loop:
for (i in 1:length(abc))
{
abc[[i]] - 3*abc[[i]]
}
Is this really the simplest way or am I missing something?
Thanks!
[[alternative HTML version deleted]]
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Re: [R] transforming data glm
Mcdonald, Grant wrote: Dear sir, I am fitting a glm with default identity link: model-glm(timetoacceptsecs~maleage*maletub*relweight*malemobtrue*femmobtrue) the model is overdisperesed and plot model shows a low level of linearity of the residuals. I don't see how the model can be *over*dispersed unless you are using a family with a fixed scale parameter (binomial/Poisson/etc.) ? The overdispersion and linearity of residulas on the normal Q-Q plot is corrected well by using: model-glm(log(timetoacceptsecs)~maleage*maletub*relweight*malemobtrue*femmobtrue)) Boxcox of my model also suggests that the log transformation is what i should do. I ask how i am able to do this by changing the link function or error family of my glm and not diretly taking the log of the response variable. For instance: model-glm(log(timetoacceptsecs)~maleage*maletub*relweight*malemobtrue*femmobtrue, family=poisson)) does not improve my model in terms of overdispersion etc as much as taking the log. I don't see why you are using a Poisson family for data that are (apparently, based on their name time to accept in seconds) -- unless you have some particular reason to believe that in your system they should follow a Poisson (it seems unlikely -- some form of waiting time distribution [exponential, gamma, Weibull) seems more plausible)) the difference between glm(y~x,family=gaussian(link=log)) and glm(log(y)~x, family=gaussian(link=identity)) (which is essentially equivalent to glm(log(y)~x) or lm(log(y)~x)) is in whether the error is assumed to be normal with a constant variance on the original scale (the first method) or on the log-transformed scale (the second method) note that you have to be careful about model comparisons between continuous data transformed to different scales. Bottom line: I don't see what's wrong with your second model. Why not just use it? -- View this message in context: http://www.nabble.com/transforming-data-glm-tp25115147p25118604.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] table function
hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes I don t find a clear explnation in examples of ?table !!! example x y z 1 0 100 5 1 1500 6 1 1200 2 2 500 1 1 3500 5 2 2000 8 5 4500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] robust method to obtain a correlation coeff?
Inline below. Bert Gunter Genentech Nonclinical Biostatisics -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of David Winsemius Sent: Monday, August 24, 2009 8:53 AM To: David Winsemius Cc: r-help@r-project.org Help; ted.hard...@manchester.ac.uk Subject: Re: [R] robust method to obtain a correlation coeff? On Aug 24, 2009, at 11:38 AM, David Winsemius wrote: ... but I do not know what those certain situations really are. So there are some function that may be affected by settings of options(na.action) but I cannot tell you where to find a list of such functions. --- Because this is up to the whim of package developers, there is no such list. In general, most of the modeling functions of base R and recommended packages, e.g. lm, glm, rlm in MASS, lme in nlme,... have an na.action argument. But not universally: e.g. loess() has an na.action argument, but lowess() does not. A real gotcha with missing values is that some of R's core arithmetic functions like mean(), median(), sum() etc. use a logical argument, na.rm = TRUE or FALSE to control the handling of missings (as I'm sure you know). -- Bert __ 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] Multiply List by a Numeric
On Aug 24, 2009, at 10:58 AM, Brigid Mooney wrote:
I apologize for what seems like it should be a straighforward query.
I am trying to multiply a list by a numeric and thought there would
be a
straightforward way to do this, but the best solution I found so far
has a
for loop.
Everything else I try seems to throw an error non-numeric argument to
binary operator
Consider the example:
a - 1
b - 1:2
c - 1:3
abc - list(a,b,c)
To multiply every element of abc by a numeric, say 3, I wrote a for-
loop:
for (i in 1:length(abc))
{
abc[[i]] - 3*abc[[i]]
}
Is this really the simplest way or am I missing something?
Thanks!
Try:
abc
[[1]]
[1] 1
[[2]]
[1] 1 2
[[3]]
[1] 1 2 3
lapply(abc, *, 3)
[[1]]
[1] 3
[[2]]
[1] 3 6
[[3]]
[1] 3 6 9
See ?lapply for more information.
HTH,
Marc Schwartz
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Re: [R] Multiply List by a Numeric
Try
lapply(abc, function(x) x*3)
Peter Ehlers
Brigid Mooney wrote:
I apologize for what seems like it should be a straighforward query.
I am trying to multiply a list by a numeric and thought there would be a
straightforward way to do this, but the best solution I found so far has a
for loop.
Everything else I try seems to throw an error non-numeric argument to
binary operator
Consider the example:
a - 1
b - 1:2
c - 1:3
abc - list(a,b,c)
To multiply every element of abc by a numeric, say 3, I wrote a for-loop:
for (i in 1:length(abc))
{
abc[[i]] - 3*abc[[i]]
}
Is this really the simplest way or am I missing something?
Thanks!
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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[R] Combining matrices
If I have two matrices like x - matrix(rep(c(1,2,3),3),3) y - matrix(rep(c(4,5,6),3),3) How can I combine them to get ? 1 1 1 4 4 4 1 1 1 5 5 5 1 1 1 6 6 6 2 2 2 4 4 4 2 2 2 5 5 5 2 2 2 6 6 6 3 3 3 4 4 4 3 3 3 5 5 5 3 3 3 6 6 6 The number of rows and the actual numbers above are unimportant, they are given so as to illustrate how I want to combine the matrices. I.e., I am looking for a general way to combine the first row of x with each row of y, then the second row of x with y, Thanks, Dan Daniel Nordlund Bothell, WA USA Thanks for __ 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] problem with BRugs
Hello, I am sorry, I have this problem before and Uwe send me the answer but I
misplaced it and can not find it.
writing a model for BRugs
library(BRugs)
Loading required package: coda
Loading required package: lattice
Welcome to BRugs running on OpenBUGS version 3.0.3
setwd(c:/tmp)
Error in setwd(c:/tmp) : cannot change working directory
mo - function(){
+ for (k in 1:p){
+ delta[1,k] ~ dnorm(0,0.1)I(,delta[2,k])
Error: unexpected symbol in:
for (k in 1:p){
delta[1,k] ~ dnorm(0,0.1)I
delta[2,k] ~ dnorm(0,0.1)I(delta[1,k],delta[3,k])
Error: unexpected symbol in delta[2,k] ~ dnorm(0,0.1)I
delta[3,k] ~ dnorm(0,0.1)I(delta[2,k],)}
Error: unexpected symbol in delta[3,k] ~ dnorm(0,0.1)I
}
Error: unexpected '}' in }
so R parser does not like the I(,) construct, What is the alternative way of
propgramming the
constrain I(lower,upper)
Thanks
Heberto Ghezzo
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Re: [R] table function
You need to create a factor that indicates which group the values in 'z' belong to. The easiest way to do that based on your situation is to use the 'cut' function to construct the factor, and then call 'table' using the result created by 'cut'. See ?cut and ?factor -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Inchallah Yarab Sent: Monday, August 24, 2009 10:59 AM To: r-help@r-project.org Subject: [R] table function hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes I don t find a clear explnation in examples of ?table !!! example x y z 1 0 100 5 1 1500 6 1 1200 2 2 500 1 1 3500 5 2 2000 8 5 4500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] table function
On Aug 24, 2009, at 10:59 AM, Inchallah Yarab wrote: hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes I don t find a clear explnation in examples of ?table !!! example x yz 10 100 51 1500 61 1200 22 500 11 3500 5 2 2000 8 5 4500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help See ?cut, which bins a continuous variable. DF x yz 1 1 0 100 2 5 1 1500 3 6 1 1200 4 2 2 500 5 1 1 3500 6 5 2 2000 7 8 5 4500 table(cut(DF$z, breaks = c(-Inf, 1000, 3000, Inf), labels = c(0 - 1000, 1000 - 3000, 3000))) 0 - 1000 1000 - 30003000 232 HTH, Marc Schwartz __ 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] table function
Inchallah Yarab wrote: i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] You can use cut to create a new vector of labels and tabulate the result. Options control closed/open endpoints (see ?cut): z - c(100,1500,1200,500,3500,2000,4500) table(cut(z,c(0,1000,3000,max(z (0,1e+03] (1e+03,3e+03] (3e+03,4.5e+03] 2 3 2 -- View this message in context: http://www.nabble.com/table-function-tp25118909p25119226.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] [Rd] Formulas in gam function of mgcv package
[Note R-Devel is the wrong list for such questions. R-Help is where this should have been directed - redirected there now] On Mon, 2009-08-24 at 17:02 +0100, Corrado wrote: Dear R-experts, I have a question on the formulas used in the gam function of the mgcv package. I am trying to understand the relationships between: y~s(x1)+s(x2)+s(x3)+s(x4) and y~s(x1,x2,x3,x4) Does the latter contain the former? what about the smoothers of all interaction terms? I'm not 100% certain how this scales to smooths of more than 2 variables, but Sections 4.10.2 and 5.2.2 of Simon Wood's book GAM: An Introduction with R (2006, Chapman Hall/CRC) discuss this for smooths of 2 variables. Strictly y ~ s(x1) + s(x2) is not nested in y ~ s(x1, x2) as the bases used to produce the smoothers in the two models may not be the same in both models. One option to ensure nestedness is to fit the more complicated model as something like this: ## if simpler model were: y ~ s(x1, k=20) + s(x2, k = 20) y ~ s(x1, k=20) + s(x2, k = 20) + s(x1, x2, k = 60) ^ where the last term (^^^ above) has the same k as used in s(x1, x2) Note that these are isotropic smooths; are x1 and x2 measured in the same units etc.? Tensor product smooths may be more appropriate if not, and if we specify the bases when fitting models s(x1) + s(x2) *is* strictly nested in te(x1, x2), eg. y ~ s(x1, bs = cr, k = 10) + s(x2, bs = cr, k = 10) is strictly nested within y ~ te(x1, x2, k = 10) ## is the same as y ~ te(x1, x2, bs = cr, k = 10) [Note that bs = cr is the default basis in te() smooths, hence we don't need to specify it, and k = 10 refers to each individual smooth in the te().] HTH G I have (tried to) read the manual pages of gam, formula.gam, smooth.terms, linear.functional.terms but could not understand properly. Regards -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ 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] Combining matrices
Try this; do.call(rbind, lapply(split(x, seq(nrow(x))), cbind, y)) On Mon, Aug 24, 2009 at 1:16 PM, Daniel Nordlund djnordl...@verizon.netwrote: If I have two matrices like x - matrix(rep(c(1,2,3),3),3) y - matrix(rep(c(4,5,6),3),3) How can I combine them to get ? 1 1 1 4 4 4 1 1 1 5 5 5 1 1 1 6 6 6 2 2 2 4 4 4 2 2 2 5 5 5 2 2 2 6 6 6 3 3 3 4 4 4 3 3 3 5 5 5 3 3 3 6 6 6 The number of rows and the actual numbers above are unimportant, they are given so as to illustrate how I want to combine the matrices. I.e., I am looking for a general way to combine the first row of x with each row of y, then the second row of x with y, Thanks, Dan Daniel Nordlund Bothell, WA USA Thanks for __ 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] table function
Your question is a little vague. Do you just want to know how often z falls in one the three classes? If so, you could either code an indicator variable (e.g. z.cat) that expresses the three categories and then do table(z.cat). Alternatively, you could just do sum(z0z1000) sum(z1z3000) sum(z3000) It's unclear to me whether you want to summarize the other variables (x and y) for each of the categories of z. If you want to do that, use tapply (see ?tapply). Daniel - cuncta stricte discussurus - -Ursprüngliche Nachricht- Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im Auftrag von Inchallah Yarab Gesendet: Monday, August 24, 2009 11:59 AM An: r-help@r-project.org Betreff: [R] table function hi, i want to use the function table to build a table not of frequence (number of time the vareable is repeated in a list or a data frame!!) but in function of classes I don t find a clear explnation in examples of ?table !!! example x yz 10 100 51 1500 61 1200 22 500 11 3500 5 2 2000 8 5 4500 i want to do a table summerizing the number of variable where z is in [0-1000],],[1000-3000], [ 3000] thank you very much for your help [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] image plot
ogbos okike schreef: Hi, I am trying to use the image function to do a color plot. My matrix columns are labeled y and x. I tried image(y, x) but I had error message (Error in image.default(y, x) : increasing 'x' and 'y' values expected). Could anybody please tell me how to add these increasing 'x' and 'y' values. Thanks [[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. Hi, Please provide a reproducible example. An example that works with image: x = 1:10 y = seq(1,100,by =10) z = matrix(runif(100), 10, 10) image(x,y,z) x = sort(runif(10)) y = sort(runif(10)) image(x,y,z) So z is a matrix with the values and x and y tell the dimensions of each cell, often with equal interval. cheers, Paul __ 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 a fast way to do several hundred thousand ANOVA tests?
have you tried:
fits - lm(a~b)
fstat - sapply(summary(fits), function(x) x[[fstatistic]][[value]])
it takes 3secs for 100K columns on my machine (running on batt)
b
On Aug 23, 2009, at 9:55 PM, big permie wrote:
Dear R users,
I have a matrix a and a classification vector b such that
str(a)
num [1:50, 1:80]
and
str(b)
Factor w/ 3 levels cond1,cond2,cond3
I'd like to do an anova on all 80 columns and record the F
statistic for
each test; I currently do this using
f.stat.vec - numeric(length(a[1,])
for (i in 1:length(a[1,]) {
f.test.frame - data.frame(nums = a[,i], cond = b)
aov.vox - aov(nums ~ cond, data = f.test.frame)
f.stat - summary(aov.vox)[[1]][1,4]
f.stat.vec[i] - f.stat
}
The problem is that this code takes about 70 minutes to run.
Is there a faster way to do an anova record the F stat for each
column?
Any help would be appreciated.
Thanks
Heath
[[alternative HTML version deleted]]
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Re: [R] Combining matrices
On Aug 24, 2009, at 11:16 AM, Daniel Nordlund wrote: If I have two matrices like x - matrix(rep(c(1,2,3),3),3) y - matrix(rep(c(4,5,6),3),3) How can I combine them to get ? 1 1 1 4 4 4 1 1 1 5 5 5 1 1 1 6 6 6 2 2 2 4 4 4 2 2 2 5 5 5 2 2 2 6 6 6 3 3 3 4 4 4 3 3 3 5 5 5 3 3 3 6 6 6 The number of rows and the actual numbers above are unimportant, they are given so as to illustrate how I want to combine the matrices. I.e., I am looking for a general way to combine the first row of x with each row of y, then the second row of x with y, Thanks, Dan nr.x - nrow(x) nr.y - nrow(y) cbind(x[rep(1:nr.x, each = nr.x), ], y[rep(1:nr.y, nr.y), ]) [,1] [,2] [,3] [,4] [,5] [,6] [1,]111444 [2,]111555 [3,]111666 [4,]222444 [5,]222555 [6,]222666 [7,]333444 [8,]333555 [9,]333666 HTH, Marc Schwartz __ 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] Selecting groups with R
jlwoodard wrote: Each of the above lines successfully excludes the BLUE subjects, but the BLUE category is still present in my data set; that is, if I try table(Color) I get RED WHITE BLUE 82 151 0 How can I eliminate the BLUE category completely so I can do a t-test using Color (with just the RED and WHITE subjects)? A simpler example. See details in the help file for factor() for an explanation. #Factor with 3 levels x - rep(c(blue,red,white),c(1,1,2)) x - factor(x) table(x) x blue red white 1 1 2 #Subset is still a factor with 3 levels y - x[x!=blue] table(y) y blue red white 0 1 2 #Drops unused levels; result a factor with 2 levels table(factor(y)) red white 1 2 -- View this message in context: http://www.nabble.com/Selecting-groups-with-R-tp25088073p25119474.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 recalling data points in a specific region of the plot
Hi all, Is there a quick way to display or recall data points from a specific region on the plot? For example I want the points from x5 and y5? Thank you very much! -- Edward Chen [[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] Combining matrices
On Aug 24, 2009, at 11:46 AM, Marc Schwartz wrote: On Aug 24, 2009, at 11:16 AM, Daniel Nordlund wrote: If I have two matrices like x - matrix(rep(c(1,2,3),3),3) y - matrix(rep(c(4,5,6),3),3) How can I combine them to get ? 1 1 1 4 4 4 1 1 1 5 5 5 1 1 1 6 6 6 2 2 2 4 4 4 2 2 2 5 5 5 2 2 2 6 6 6 3 3 3 4 4 4 3 3 3 5 5 5 3 3 3 6 6 6 The number of rows and the actual numbers above are unimportant, they are given so as to illustrate how I want to combine the matrices. I.e., I am looking for a general way to combine the first row of x with each row of y, then the second row of x with y, Thanks, Dan nr.x - nrow(x) nr.y - nrow(y) cbind(x[rep(1:nr.x, each = nr.x), ], y[rep(1:nr.y, nr.y), ]) [,1] [,2] [,3] [,4] [,5] [,6] [1,]111444 [2,]111555 [3,]111666 [4,]222444 [5,]222555 [6,]222666 [7,]333444 [8,]333555 [9,]333666 Actually, correction...that will work in this case, but in the general case, I believe that it needs to be: x - matrix(rep(c(1,2,3),3),3) y - matrix(rep(c(4,5,6,7),3),4) x [,1] [,2] [,3] [1,]111 [2,]222 [3,]333 y [,1] [,2] [,3] [1,]444 [2,]555 [3,]666 [4,]777 nr.x - nrow(x) nr.y - nrow(y) cbind(x[rep(1:nr.x, each = nr.y), ], y[rep(1:nr.y, nr.x), ]) [,1] [,2] [,3] [,4] [,5] [,6] [1,]111444 [2,]111555 [3,]111666 [4,]111777 [5,]222444 [6,]222555 [7,]222666 [8,]222777 [9,]333444 [10,]333555 [11,]333666 [12,]333777 We need to replicate each row by the number of rows in the other matrix. HTH, Marc __ 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] robust method to obtain a correlation coeff?
I may be misunderstanding the question but would cor(d1, use='complete.obs') or some other variant of use help? --- On Mon, 8/24/09, Christian Meesters meest...@imbie.uni-bonn.de wrote: From: Christian Meesters meest...@imbie.uni-bonn.de Subject: [R] robust method to obtain a correlation coeff? To: r-help@r-project.org Help r-help@r-project.org Received: Monday, August 24, 2009, 10:47 AM Hi, Being a R-newbie I am wondering how to calculate a correlation coefficient (preferably with an associated p-value) for data like: d[,1] [1] 25.5 25.3 25.1 NA 23.3 21.5 23.8 23.2 24.2 22.7 27.6 24.2 ... d[,2] [1] 0.0 11.1 0.0 NA 0.0 10.1 10.6 9.5 0.0 57.9 0.0 0.0 ... Apparently corr(d) from the boot-library fails with NAs in the data, also cor.test cannot cope with a different number of NAs. Is there a solution to this problem (calculating a correlation coefficient and ignoring different number of NAs), e.g. Pearson's corr coeff? If so, please point me to the relevant piece of documentation. TIA Christian __ 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. __ Be smarter than spam. See how smart SpamGuard is ons in Mail and switch to New Mail today or register for free at http://mail.yahoo.ca __ 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] Combining matrices
Try this: kronecker(cbind(x, y), rep(1, 3)) On Mon, Aug 24, 2009 at 12:16 PM, Daniel Nordlunddjnordl...@verizon.net wrote: If I have two matrices like x - matrix(rep(c(1,2,3),3),3) y - matrix(rep(c(4,5,6),3),3) How can I combine them to get ? 1 1 1 4 4 4 1 1 1 5 5 5 1 1 1 6 6 6 2 2 2 4 4 4 2 2 2 5 5 5 2 2 2 6 6 6 3 3 3 4 4 4 3 3 3 5 5 5 3 3 3 6 6 6 The number of rows and the actual numbers above are unimportant, they are given so as to illustrate how I want to combine the matrices. I.e., I am looking for a general way to combine the first row of x with each row of y, then the second row of x with y, Thanks, Dan Daniel Nordlund Bothell, WA USA Thanks for __ 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] Combining matrices
-Original Message- From: Marc Schwartz [mailto:marc_schwa...@me.com] Sent: Monday, August 24, 2009 9:57 AM To: Daniel Nordlund Cc: r help Subject: Re: [R] Combining matrices On Aug 24, 2009, at 11:46 AM, Marc Schwartz wrote: On Aug 24, 2009, at 11:16 AM, Daniel Nordlund wrote: If I have two matrices like x - matrix(rep(c(1,2,3),3),3) y - matrix(rep(c(4,5,6),3),3) How can I combine them to get ? 1 1 1 4 4 4 1 1 1 5 5 5 1 1 1 6 6 6 2 2 2 4 4 4 2 2 2 5 5 5 2 2 2 6 6 6 3 3 3 4 4 4 3 3 3 5 5 5 3 3 3 6 6 6 The number of rows and the actual numbers above are unimportant, they are given so as to illustrate how I want to combine the matrices. I.e., I am looking for a general way to combine the first row of x with each row of y, then the second row of x with y, Thanks, Dan nr.x - nrow(x) nr.y - nrow(y) cbind(x[rep(1:nr.x, each = nr.x), ], y[rep(1:nr.y, nr.y), ]) [,1] [,2] [,3] [,4] [,5] [,6] [1,]111444 [2,]111555 [3,]111666 [4,]222444 [5,]222555 [6,]222666 [7,]333444 [8,]333555 [9,]333666 Actually, correction...that will work in this case, but in the general case, I believe that it needs to be: x - matrix(rep(c(1,2,3),3),3) y - matrix(rep(c(4,5,6,7),3),4) x [,1] [,2] [,3] [1,]111 [2,]222 [3,]333 y [,1] [,2] [,3] [1,]444 [2,]555 [3,]666 [4,]777 nr.x - nrow(x) nr.y - nrow(y) cbind(x[rep(1:nr.x, each = nr.y), ], y[rep(1:nr.y, nr.x), ]) [,1] [,2] [,3] [,4] [,5] [,6] [1,]111444 [2,]111555 [3,]111666 [4,]111777 [5,]222444 [6,]222555 [7,]222666 [8,]222777 [9,]333444 [10,]333555 [11,]333666 [12,]333777 We need to replicate each row by the number of rows in the other matrix. HTH, Marc Thanks to all who responded (including those off-list). I now have options to apply to solving my programming task. Thanks, Dan __ 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] help with recalling data points in a specific region of the plot
I am assuming that you want to desplay all the data and highlight the subset. Set up a vector to indicate the breakdown of the data and you can do it fairly easily in ggplot2 if you treat the vector as a factor. library(ggplot) mydata - data.frame(x=1:21, y= -10:10) z - ifelse(mydata[,1]5 mydata[,2] 5, 1, 0) # identify points 5 mydata - data.frame(mydata, z) ggplot(mydata, aes(x=x, y=y, colour= factor(z))) + geom_point() --- On Mon, 8/24/09, Edward Chen edche...@gmail.com wrote: From: Edward Chen edche...@gmail.com Subject: [R] help with recalling data points in a specific region of the plot To: r-help@r-project.org Received: Monday, August 24, 2009, 12:55 PM Hi all, Is there a quick way to display or recall data points from a specific region on the plot? For example I want the points from x5 and y5? Thank you very much! -- Edward Chen [[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. __ Looking for the perfect gift? Give the gift of Flickr! http://www.flickr.com/gift/ __ 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] RMySQL - overwrite record, not table
whizvast wrote:
Hi, Adrian-
If you use overwrite=T parameter, you will overwrite the entire table,
not each record. this is the essence of my problem and i still haven't
found out right solution. i am thinking of writing my own MySQLwriteTable
function...
Thank you for your answer anyway!
Sorry for the late reply (I'm on my vacation). If you want to replace a
variable instead of the whole dataframe, I wrote a function about a year ago
and I used it succesfully a few times.
Try this:
dbUpdateVars -
function(conn, dbtable, dataframe=NULL, primary, vars) {
if (!dbExistsTable(conn, dbtable)) {
stop(The target table \, dbtable, \ doesn't exist in the
database \, dbGetInfo(conn)$dbname, \\n\n, call. = FALSE)
}
if (is.null(dataframe)) {
stop(The source dataframe is missing, with no default\n\n, call. =
FALSE)
}
if (!(toupper(primary) %in% toupper(names(dataframe {
stop(The primary key variable doesn't exist in the source
dataframe\n\n, call. = FALSE)
}
if (!all(toupper(vars) %in% toupper(names(dataframe {
stop(One or more variables don't exist in the source
dataframe\n\n, call. = FALSE)
}
if (!(toupper(primary) %in% toupper(dbListFields(con, dbtable {
stop(The primary key variable doesn't exist in the target
table\n\n, call. = FALSE)
}
if (!all(toupper(vars) %in% toupper(dbListFields(con, dbtable {
stop(One or more variables don't exist in the target table\n\n,
call. = FALSE)
}
if(length(vars) 1) {
pastedvars - paste(', apply(dataframe[, vars], 1, paste,
collapse=', '), ', sep=)
}
else {
pastedvars - paste(', dataframe[, vars], ', sep=)
}
varlist - paste(dbtable, (, paste(c(primary, vars), collapse=, ),
), sep=)
datastring - paste((, paste(paste(dataframe[, primary], pastedvars,
sep=, ), collapse=), (), ), sep=)
toupdate - paste(paste(vars, =VALUES(, vars, ), sep=),
collapse=, )
sqlstring - paste(INSERT INTO, varlist, VALUES, datastring, ON
DUPLICATE KEY UPDATE, toupdate)
dbSendQuery(conn, sqlstring)
}
I hopw it helps you,
Adrian
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Re: [R] hdf5 package segfault when processing large data
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Budi Mulyono
Sent: Monday, August 24, 2009 3:38 AM
To: r-help@r-project.org
Subject: [R] hdf5 package segfault when processing large data
Hi there,
I am currently working on something that uses hdf5 library. I think
hdf5 is a great data format, I've used it somewhat extensively in
python via PyTables. I was looking for something similar to that in R.
The closest I can get is this library: hdf5. While it does not work
the same way as PyTables did, but it's good enough to let them
exchange data via hdf5 file.
There is just 1 problem, I keep getting Segfault error when trying to
process large files (10MB), although this is by no mean large when we
talk about hdf5 capabilities. I have included the example code and
data below. I have tried with different OS (WinXP and Ubuntu 8.04),
architecture (32 and 64bit) and R versions (2.7.1, 2.72, and 2.9.1),
but all of them present the same problem. I was wondering if anyone
have any clue as to what's going on here and maybe can advice me to
handle it.
This sort of problem should be sent to the package's maintainer.
packageDescription(hdf5)
Package: hdf5
Version: 1.6.9
Title: HDF5
Author: Marcus G. Daniels mdani...@lanl.gov
Maintainer: Marcus G. Daniels mdani...@lanl.gov
Description: Interface to the NCSA HDF5 library
...
This is probably due to the code in hdf5.c allocating a huge
matrix, buf, on the stack with
883 unsigned char buf[rowcount][size];
It dies with the segmentatio fault (stack overflow, in particular)
at line 898, where it tries to access this buf.
885 for (ri = 0; ri rowcount; ri++)
886 for (pos = 0; pos colcount; pos++)
887 {
888 SEXP item = VECTOR_ELT (val, pos);
889 SEXPTYPE type = TYPEOF (item);
890 void *ptr = buf[ri][offsets[pos]];
891
892 switch (type)
893 {
894 case REALSXP:
895 memcpy (ptr, REAL (item)[ri], sizeof
(double));
896 break;
897 case INTSXP:
898 memcpy (ptr, INTEGER (item)[ri], sizeof
(int));
899 break;
The code should use one of the allocators in the R API instead
of putting the big memory block on the stack.
Bill Dunlap
TIBCO Software Inc - Spotfire Division
wdunlap tibco.com
Thank you, appreciate any help i can get.
Cheers,
Budi
The example script
library(hdf5)
fileName - sample.txt
myTable - read.table(fileName,header=TRUE,sep=\t,as.is=TRUE)
hdf5save(test.hdf, myTable)
The data example, the list continue for more than 250,000
rows: sample.txt
Date Timef1 f2 f3 f4 f5
20070328 07:56 463 463.07 462.9 463.01 1100
20070328 07:57 463.01 463.01 463.01 463.01 200
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Re: [R] trouble with Vista reading files
Dear Mike:
I don't know.
1. What specific error message do you get?
2. Your example is too long for me to parse, especially with
the color being stripped before I saw it.
3. Have you tried using debug(cro.etest.grab), then
walking through the code line by line until you get to the offending
line. Then you can study the error message and try different things
quickly. This may not work, but typically provides access to more
information on the problem AND often usually substantially the test
cycle for potential fixes.
Hope this helps.
Spencer
Mike Williamson wrote:
All,
I am having trouble with a read.table() function that is inside of
another function. But if I call the function by itself, it works fine.
Moreover, if I run the script on a Mac OS X (with the default Mac OS X
version of R installed, rev 2.8), it works fine. But it does not work if I
run it on windows vista (also default Windows version of R, rev. 2.8).
Again, both calls shown below work fine in Mac, but only the call by
itself works in Vista. The other call embedded in a function does not.
Thanks in advance for all the help!!
Regards, Mike
Below are the calls:
#
Below is the call which DOES work, as long as it is called by itself.
##
* eTestData -
read.table(C:/Users/userID/Documents/R/eTestDataDir/EtestExample.csv,header
= TRUE,
as.is = TRUE)
*
#
Below is the call which does NOT work. The problem function call
highlighted in *red*
Especially strange with this is that there is a call below to ask for all
the files in the directory, which I
have highlighted in *purple*, and that call works fine. So it is some sort
of permissions thing.
##
cro.etest.grab - function(dataDir=raw.etest.data, header=hdr,
dataHeaders=datasets/eTestDataHeaders.txt,
slotCol=Wafer, dateFormat=%m/%d/%Y %H:%M:%S,
lotCol=eTestLotID) {
### Function: grab data in its raw form from SVTC's HP electrical tester and
### munge it into a format more friendly for analysis in R.
### Requires: dataDir -- the directory where the raw SVTC data set is
stored
### header -- the differentiation between the names of the
data
### files and the header files. E.g., if data file
### is CORR682..18524 and header file is
### CORR682.hdr.18524, then the header is hdr.
### dataHeaders -- Sometimes the labels for the data is missing,
but
### they are NEARLY always the same. If the labels
### are ever missing, this fills them in with the
### vector of headers given here. E.g., c(Wafer,
### Site,R2_ET1_M1,etc.)
### slotCol -- In the data file, typically column Wafer is
### actually the slot ID. This renames it to Slot.
### So, SlotCol is the name IN THE RAW DATA.
### lotCol -- The data files have no lot ID column, the lot
ID
### is grabbed from the file name. This provides a
### column header name for the lot ID.
### dateFormat -- The test data header file has eTest time,
written
### in the format month/day/year hour:minute:sec.
### If another format is being read, it can be
### altered here.
dataHeaders - read.table(dataHeaders, stringsAsFactors = FALSE)[,1]
print(paste(dataDir:,dataDir,header:,header,
slotCol:,slotCol,
lotCol:,lotCol))
*allFiles - list.files(path = dataDir)*
tmp - grep(hdr,allFiles,ignore.case = TRUE)
dataFiles - allFiles[-tmp]
hdrFiles - sub(\\.(.*)\\.,\\.hdr\\1\\.,dataFiles)
* eTestData - read.table(paste(dataDir,/,dataFiles[1],sep=),header =
TRUE,
as.is = TRUE)
* eTestData[,slotCol] - as.character(eTestData[,slotCol])
eTestData[,lotCol] - rep(dataFiles[1],length(eTestData[,1]))
tmp - try(scan(paste(dataDir,/, hdrFiles[1], sep=), what =
character,
sep=\n, quiet=TRUE), silent=TRUE)
if (is.null(attr(tmp,class))) {
dateCols - grep([0-9][0-9]/[0-9][0-9]/20[01][0-9],tmp)
hdrDF -
data.frame(tmp[(dateCols-1)],tmp[dateCols],stringsAsFactors=FALSE)
hdrDF$LotDate - rep(hdrDF[1,2],length(hdrDF[,1])) ; hdrDF - hdrDF[-1,]
hdrDF[1,1] - tmp[(dateCols-2)][1]
names(hdrDF) - c(slotCol,Date,LotDate)
hdrDF[,slotCol] - substring(hdrDF[,slotCol],
(regexpr(=,hdrDF[,slotCol])+2),
Re: [R] help with recalling data points in a specific region of the plot
On Aug 24, 2009, at 12:55 PM, Edward Chen wrote: Hi all, Is there a quick way to display or recall data points from a specific region on the plot? For example I want the points from x5 and y5? Thank you very much! Your question is pretty light on specifics but assuming that you have columns x and y in a dataframe, df1, then in ordinary graphics you could just execute this after another plotting function has been performed: with(subset(df1, x5 y5), points(x,y, col=red) ) # I generally make my points smaller with cex=0.2 or cex=0.1 subset(f1, x5 y5)[ , c(x,y)] # should recall the values, if my wetware R interpreter is working properly. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lme, lmer, gls, and spatial autocorrelation
Hello folks, I have some data where spatial autocorrelation seems to be a serious problem, and I'm unclear on how to deal with it in R. I've tried to do my homework - read through 'The R Book,' use the online help in R, search the internet, etc. - and I still have some unanswered questions. I'd greatly appreciate any help you could offer. The super-super short explanation is that I'd like to draw a straight line through my data, accounting for spatial autocorrelation and using Poisson errors (I have count data). There's a longer explanation at the end of this e-mail, I just didn't want to overdo it at the start. There are three R functions that do at least some of what I would like, but I'm unclear on some of their specifics. 1. lme - Maybe models spatial autocorrelation, but doesn't allow for Poisson errors. I get mixed messages from The R Book. On p. 647, there's an example that uses lme with temporal autocorrelation, so it seems that you can specify a correlation structure. On the other hand, on p.778, The R Book says, the great advantage of the gls function is that the errors are allowed to be correlated. This suggests that only gls (not lme or lmer) allows specification of a corStruct class. Though it may also suggest that I have an incomplete understanding of these functions. 2. lmer - Allows specification of a Poisson error structure. However, it seems that lmer does not yet handle correlated errors. 3. gls - Surely works with spatial autocorrelation, but doesn't allow for Poisson errors. Does allow the spatial autocorrelation to be assessed independently for different groups (I have two groups, one at each of two different spatial scales). Since gls is what The R Book uses in the example of spatial autocorrelation, this seems like the best option. I'd rather have Poisson errors, but Gaussian would be OK. However, I'm still somewhat confused by these three functions. In particular, I'm unclear on the difference between lme and gls. I'd feel more confident in my results if I had a better understanding of these choices. I'd greatly appreciate advice on the matter More detailed explanation of the data/problem is below: The data: [1] A count of the number of plant species present on each of 96 plots that are 1m^2 in area. [2] A count of the number of plant species present on each of 24 plots that are 100m^2 in area. [3] X,Y coordinates for the centroid of all plots (both sizes). Goal: 1. A best fit straight-line relating log10(area) to #species. 2. The slope of that line, and the standard error of that slope. (I want to compare the slope of this line with the slope of another line) The problem: Spatial autocorrelation. Across our range of plot-separation-distances, Moran's I ranges from -.5 to +.25. Depending on the size of the distance-bins, about 1 out of 10 of these I values are statistically significant. Thus, there seems to be a significant degree of spatial autocorrelation. if I want 'good' values for my line parameters, I need to account for this somehow. Tim Handley Fire Effects Monitor Santa Monica Mountains National Recreation Area 401 W. Hillcrest Dr. Thousand Oaks, CA 91360 805-370-2347 __ 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] Unique command not deleting all duplicate rows
Hello everyone, when I run the unique command on my data frame, it deletes the majority of duplicate rows, but not all of them. Here is a sample of my data. How do I get it to delete all the rows? 6 -115.38 32.894 195 162.94 D 8419 D 7 -115.432 32.864 115 208.91 D 8419 D 8 -115.447 32.773 1170 264.57 D 8419 D 9 -115.447 32.773 1170 264.57 D 8419 D 10 -115.447 32.773 1170 264.57 D 8419 D 11 -115.447 32.773 1170 264.57 D 8419 D 12 -115.447 32.773 149 186.21 D 8419 D 13 -115.466 32.855 114 205.63 D 8419 D 14 -115.473 32.8 1121 207.469 D 8419 D Thanks a bunch! Mehdi Khan [[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] Unique command not deleting all duplicate rows
I really don't think this is the issue. I think the issue is that some columns of the data.frame, specifically V1, V2, and V4 should be checked versus R FAQ 7.31. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Don McKenzie Sent: Monday, August 24, 2009 1:35 PM To: Mehdi Khan Cc: r-help@r-project.org Subject: Re: [R] Unique command not deleting all duplicate rows duplicated() test.df V1 V2 V3 V4 V5 V6 V7 1 -115.380 32.894 195 162.940 D 8419 D 2 -115.432 32.864 115 208.910 D 8419 D 3 -115.447 32.773 1170 264.570 D 8419 D 4 -115.447 32.773 1170 264.570 D 8419 D 5 -115.447 32.773 1170 264.570 D 8419 D 6 -115.447 32.773 1170 264.570 D 8419 D 7 -115.447 32.773 149 186.210 D 8419 D 8 -115.466 32.855 114 205.630 D 8419 D 9 -115.473 32.800 1121 207.469 D 8419 D test.df[!duplicated(test.df),] V1 V2 V3 V4 V5 V6 V7 1 -115.380 32.894 195 162.940 D 8419 D 2 -115.432 32.864 115 208.910 D 8419 D 3 -115.447 32.773 1170 264.570 D 8419 D 7 -115.447 32.773 149 186.210 D 8419 D 8 -115.466 32.855 114 205.630 D 8419 D 9 -115.473 32.800 1121 207.469 D 8419 D On 24-Aug-09, at 11:23 AM, Mehdi Khan wrote: Hello everyone, when I run the unique command on my data frame, it deletes the majority of duplicate rows, but not all of them. Here is a sample of my data. How do I get it to delete all the rows? 6 -115.38 32.894 195 162.94 D 8419 D 7 -115.432 32.864 115 208.91 D 8419 D 8 -115.447 32.773 1170 264.57 D 8419 D 9 -115.447 32.773 1170 264.57 D 8419 D 10 -115.447 32.773 1170 264.57 D 8419 D 11 -115.447 32.773 1170 264.57 D 8419 D 12 -115.447 32.773 149 186.21 D 8419 D 13 -115.466 32.855 114 205.63 D 8419 D 14 -115.473 32.8 1121 207.469 D 8419 D Thanks a bunch! Mehdi Khan [[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. Don McKenzie, Research Ecologist Pacific WIldland Fire Sciences Lab US Forest Service Affiliate Professor School of Forest Resources, College of the Environment CSES Climate Impacts Group University of Washington desk: 206-732-7824 cell: 206-321-5966 d...@u.washington.edu donaldmcken...@fs.fed.us __ 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] lme, lmer, gls, and spatial autocorrelation
Have you looked at the Spatial task view on CRAN? That would seem to me the logical first place to go. Bert Gunter Genentech Nonclinical Biostatisics -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of timothy_hand...@nps.gov Sent: Monday, August 24, 2009 11:12 AM To: r-help@r-project.org Subject: [R] lme, lmer, gls, and spatial autocorrelation Hello folks, I have some data where spatial autocorrelation seems to be a serious problem, and I'm unclear on how to deal with it in R. I've tried to do my homework - read through 'The R Book,' use the online help in R, search the internet, etc. - and I still have some unanswered questions. I'd greatly appreciate any help you could offer. The super-super short explanation is that I'd like to draw a straight line through my data, accounting for spatial autocorrelation and using Poisson errors (I have count data). There's a longer explanation at the end of this e-mail, I just didn't want to overdo it at the start. There are three R functions that do at least some of what I would like, but I'm unclear on some of their specifics. 1. lme - Maybe models spatial autocorrelation, but doesn't allow for Poisson errors. I get mixed messages from The R Book. On p. 647, there's an example that uses lme with temporal autocorrelation, so it seems that you can specify a correlation structure. On the other hand, on p.778, The R Book says, the great advantage of the gls function is that the errors are allowed to be correlated. This suggests that only gls (not lme or lmer) allows specification of a corStruct class. Though it may also suggest that I have an incomplete understanding of these functions. 2. lmer - Allows specification of a Poisson error structure. However, it seems that lmer does not yet handle correlated errors. 3. gls - Surely works with spatial autocorrelation, but doesn't allow for Poisson errors. Does allow the spatial autocorrelation to be assessed independently for different groups (I have two groups, one at each of two different spatial scales). Since gls is what The R Book uses in the example of spatial autocorrelation, this seems like the best option. I'd rather have Poisson errors, but Gaussian would be OK. However, I'm still somewhat confused by these three functions. In particular, I'm unclear on the difference between lme and gls. I'd feel more confident in my results if I had a better understanding of these choices. I'd greatly appreciate advice on the matter More detailed explanation of the data/problem is below: The data: [1] A count of the number of plant species present on each of 96 plots that are 1m^2 in area. [2] A count of the number of plant species present on each of 24 plots that are 100m^2 in area. [3] X,Y coordinates for the centroid of all plots (both sizes). Goal: 1. A best fit straight-line relating log10(area) to #species. 2. The slope of that line, and the standard error of that slope. (I want to compare the slope of this line with the slope of another line) The problem: Spatial autocorrelation. Across our range of plot-separation-distances, Moran's I ranges from -.5 to +.25. Depending on the size of the distance-bins, about 1 out of 10 of these I values are statistically significant. Thus, there seems to be a significant degree of spatial autocorrelation. if I want 'good' values for my line parameters, I need to account for this somehow. Tim Handley Fire Effects Monitor Santa Monica Mountains National Recreation Area 401 W. Hillcrest Dr. Thousand Oaks, CA 91360 805-370-2347 __ 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] Unique command not deleting all duplicate rows
Duplicated did not work, I agree with Erik. Is there any way I can specify a tolerance limit and then delete? On Mon, Aug 24, 2009 at 11:41 AM, Erik Iverson eiver...@nmdp.org wrote: I really don't think this is the issue. I think the issue is that some columns of the data.frame, specifically V1, V2, and V4 should be checked versus R FAQ 7.31. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Don McKenzie Sent: Monday, August 24, 2009 1:35 PM To: Mehdi Khan Cc: r-help@r-project.org Subject: Re: [R] Unique command not deleting all duplicate rows duplicated() test.df V1 V2 V3 V4 V5 V6 V7 1 -115.380 32.894 195 162.940 D 8419 D 2 -115.432 32.864 115 208.910 D 8419 D 3 -115.447 32.773 1170 264.570 D 8419 D 4 -115.447 32.773 1170 264.570 D 8419 D 5 -115.447 32.773 1170 264.570 D 8419 D 6 -115.447 32.773 1170 264.570 D 8419 D 7 -115.447 32.773 149 186.210 D 8419 D 8 -115.466 32.855 114 205.630 D 8419 D 9 -115.473 32.800 1121 207.469 D 8419 D test.df[!duplicated(test.df),] V1 V2 V3 V4 V5 V6 V7 1 -115.380 32.894 195 162.940 D 8419 D 2 -115.432 32.864 115 208.910 D 8419 D 3 -115.447 32.773 1170 264.570 D 8419 D 7 -115.447 32.773 149 186.210 D 8419 D 8 -115.466 32.855 114 205.630 D 8419 D 9 -115.473 32.800 1121 207.469 D 8419 D On 24-Aug-09, at 11:23 AM, Mehdi Khan wrote: Hello everyone, when I run the unique command on my data frame, it deletes the majority of duplicate rows, but not all of them. Here is a sample of my data. How do I get it to delete all the rows? 6 -115.38 32.894 195 162.94 D 8419 D 7 -115.432 32.864 115 208.91 D 8419 D 8 -115.447 32.773 1170 264.57 D 8419 D 9 -115.447 32.773 1170 264.57 D 8419 D 10 -115.447 32.773 1170 264.57 D 8419 D 11 -115.447 32.773 1170 264.57 D 8419 D 12 -115.447 32.773 149 186.21 D 8419 D 13 -115.466 32.855 114 205.63 D 8419 D 14 -115.473 32.8 1121 207.469 D 8419 D Thanks a bunch! Mehdi Khan [[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. Don McKenzie, Research Ecologist Pacific WIldland Fire Sciences Lab US Forest Service Affiliate Professor School of Forest Resources, College of the Environment CSES Climate Impacts Group University of Washington desk: 206-732-7824 cell: 206-321-5966 d...@u.washington.edu donaldmcken...@fs.fed.us __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Between-group variance from ANOVA
Hi Emma, R gives you the tools to work this out. ## Example set.seed(7) TDat - data.frame(response = c(rnorm(100, 5, 2), rnorm(100, 20, 2))) TDat$group - gl(2, 100, labels=c(A,B)) with(TDat, boxplot(split(response, group))) summary(aov(response ~ group, data=TDat)) Regards, Mark. emj83 wrote: can anyone advise me please? emj83 wrote: I have done some ANOVA tables for some data that I have, from this I can read the within-group variance. can anyone tell me how i may find out the between-group variance? Thanks Emma -- View this message in context: http://www.nabble.com/Between-group-variance-from-ANOVA-tp24954045p25121532.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] create list entry from variable
Hi; assume i-10 how can i create a list having key=10 and value=11 list(i=11) generates a list with 'i' [1] 11 and not 10 [1] 11 any help? Thanks _ Facebook. :ON:WL:en-US:SI_SB_facebook:082009 [[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] Unique command not deleting all duplicate rows
?round Bert Gunter Genentech Nonclinical Biostatisics -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Mehdi Khan Sent: Monday, August 24, 2009 11:52 AM To: Erik Iverson Cc: r-help@r-project.org Subject: Re: [R] Unique command not deleting all duplicate rows Duplicated did not work, I agree with Erik. Is there any way I can specify a tolerance limit and then delete? On Mon, Aug 24, 2009 at 11:41 AM, Erik Iverson eiver...@nmdp.org wrote: I really don't think this is the issue. I think the issue is that some columns of the data.frame, specifically V1, V2, and V4 should be checked versus R FAQ 7.31. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Don McKenzie Sent: Monday, August 24, 2009 1:35 PM To: Mehdi Khan Cc: r-help@r-project.org Subject: Re: [R] Unique command not deleting all duplicate rows duplicated() test.df V1 V2 V3 V4 V5 V6 V7 1 -115.380 32.894 195 162.940 D 8419 D 2 -115.432 32.864 115 208.910 D 8419 D 3 -115.447 32.773 1170 264.570 D 8419 D 4 -115.447 32.773 1170 264.570 D 8419 D 5 -115.447 32.773 1170 264.570 D 8419 D 6 -115.447 32.773 1170 264.570 D 8419 D 7 -115.447 32.773 149 186.210 D 8419 D 8 -115.466 32.855 114 205.630 D 8419 D 9 -115.473 32.800 1121 207.469 D 8419 D test.df[!duplicated(test.df),] V1 V2 V3 V4 V5 V6 V7 1 -115.380 32.894 195 162.940 D 8419 D 2 -115.432 32.864 115 208.910 D 8419 D 3 -115.447 32.773 1170 264.570 D 8419 D 7 -115.447 32.773 149 186.210 D 8419 D 8 -115.466 32.855 114 205.630 D 8419 D 9 -115.473 32.800 1121 207.469 D 8419 D On 24-Aug-09, at 11:23 AM, Mehdi Khan wrote: Hello everyone, when I run the unique command on my data frame, it deletes the majority of duplicate rows, but not all of them. Here is a sample of my data. How do I get it to delete all the rows? 6 -115.38 32.894 195 162.94 D 8419 D 7 -115.432 32.864 115 208.91 D 8419 D 8 -115.447 32.773 1170 264.57 D 8419 D 9 -115.447 32.773 1170 264.57 D 8419 D 10 -115.447 32.773 1170 264.57 D 8419 D 11 -115.447 32.773 1170 264.57 D 8419 D 12 -115.447 32.773 149 186.21 D 8419 D 13 -115.466 32.855 114 205.63 D 8419 D 14 -115.473 32.8 1121 207.469 D 8419 D Thanks a bunch! Mehdi Khan [[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. Don McKenzie, Research Ecologist Pacific WIldland Fire Sciences Lab US Forest Service Affiliate Professor School of Forest Resources, College of the Environment CSES Climate Impacts Group University of Washington desk: 206-732-7824 cell: 206-321-5966 d...@u.washington.edu donaldmcken...@fs.fed.us __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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] Unique command not deleting all duplicate rows
duplicated() test.df V1 V2 V3 V4 V5 V6 V7 1 -115.380 32.894 195 162.940 D 8419 D 2 -115.432 32.864 115 208.910 D 8419 D 3 -115.447 32.773 1170 264.570 D 8419 D 4 -115.447 32.773 1170 264.570 D 8419 D 5 -115.447 32.773 1170 264.570 D 8419 D 6 -115.447 32.773 1170 264.570 D 8419 D 7 -115.447 32.773 149 186.210 D 8419 D 8 -115.466 32.855 114 205.630 D 8419 D 9 -115.473 32.800 1121 207.469 D 8419 D test.df[!duplicated(test.df),] V1 V2 V3 V4 V5 V6 V7 1 -115.380 32.894 195 162.940 D 8419 D 2 -115.432 32.864 115 208.910 D 8419 D 3 -115.447 32.773 1170 264.570 D 8419 D 7 -115.447 32.773 149 186.210 D 8419 D 8 -115.466 32.855 114 205.630 D 8419 D 9 -115.473 32.800 1121 207.469 D 8419 D On 24-Aug-09, at 11:23 AM, Mehdi Khan wrote: Hello everyone, when I run the unique command on my data frame, it deletes the majority of duplicate rows, but not all of them. Here is a sample of my data. How do I get it to delete all the rows? 6 -115.38 32.894 195 162.94 D 8419 D 7 -115.432 32.864 115 208.91 D 8419 D 8 -115.447 32.773 1170 264.57 D 8419 D 9 -115.447 32.773 1170 264.57 D 8419 D 10 -115.447 32.773 1170 264.57 D 8419 D 11 -115.447 32.773 1170 264.57 D 8419 D 12 -115.447 32.773 149 186.21 D 8419 D 13 -115.466 32.855 114 205.63 D 8419 D 14 -115.473 32.8 1121 207.469 D 8419 D Thanks a bunch! Mehdi Khan [[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. Don McKenzie, Research Ecologist Pacific WIldland Fire Sciences Lab US Forest Service Affiliate Professor School of Forest Resources, College of the Environment CSES Climate Impacts Group University of Washington desk: 206-732-7824 cell: 206-321-5966 d...@u.washington.edu donaldmcken...@fs.fed.us __ 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] lme, lmer, gls, and spatial autocorrelation
Bert - I took a look at that page just now, and I'd classify my problem as spatial regression. Unfortunately, I don't think the spdep library fits my needs. Or at least, I can't figure out how to use it for this problem. The examples I have seen all use spdep with networks. They build a graph, connecting each location to something like the nearest N neighbors, attach some set of weights, and then do an analysis. The plots in my data have a very irregular, semi-random, yet somewhat clumped (several isolated islands), spatial distribution. Honestly, it's quite weird looking. I don't know how to cleanly turn this into a network, and even if I did, I don't know that I ought to. To me (and please feel free to disagree) it seems more natural to use a matrix of distances and associated correlations, which is what the gls function appears to do. In the ecological analysis section, it looks like both 'ade4' and 'vegan' may have helpful tools. I'll explore that some more. However, I still think that one of lme or gls already has the functionality I need, and I just need to learn how to use them properly. Tim Handley Fire Effects Monitor Santa Monica Mountains National Recreation Area 401 W. Hillcrest Dr. Thousand Oaks, CA 91360 805-370-2347 Bert Gunter gunter.ber...@ge ne.comTo timothy_hand...@nps.gov, 08/24/2009 11:43 r-help@r-project.org AM cc Subject RE: [R] lme, lmer, gls, and spatial autocorrelation Have you looked at the Spatial task view on CRAN? That would seem to me the logical first place to go. Bert Gunter Genentech Nonclinical Biostatisics -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of timothy_hand...@nps.gov Sent: Monday, August 24, 2009 11:12 AM To: r-help@r-project.org Subject: [R] lme, lmer, gls, and spatial autocorrelation Hello folks, I have some data where spatial autocorrelation seems to be a serious problem, and I'm unclear on how to deal with it in R. I've tried to do my homework - read through 'The R Book,' use the online help in R, search the internet, etc. - and I still have some unanswered questions. I'd greatly appreciate any help you could offer. The super-super short explanation is that I'd like to draw a straight line through my data, accounting for spatial autocorrelation and using Poisson errors (I have count data). There's a longer explanation at the end of this e-mail, I just didn't want to overdo it at the start. There are three R functions that do at least some of what I would like, but I'm unclear on some of their specifics. 1. lme - Maybe models spatial autocorrelation, but doesn't allow for Poisson errors. I get mixed messages from The R Book. On p. 647, there's an example that uses lme with temporal autocorrelation, so it seems that you can specify a correlation structure. On the other hand, on p.778, The R Book says, the great advantage of the gls function is that the errors are allowed to be correlated. This suggests that only gls (not lme or lmer) allows specification of a corStruct class. Though it may also suggest that I have an incomplete understanding of these functions. 2. lmer - Allows specification of a Poisson error structure. However, it seems that lmer does not yet handle correlated errors. 3. gls - Surely works with spatial autocorrelation, but doesn't allow for Poisson errors. Does allow the spatial autocorrelation to be assessed independently for different groups (I have two groups, one at each of two different spatial scales). Since gls is what The R Book uses in the example of spatial autocorrelation, this seems like the best option. I'd rather have Poisson errors, but Gaussian would be OK. However, I'm still somewhat confused by these three functions. In particular, I'm unclear on the difference between lme and gls. I'd feel more confident in my results if I had a better
Re: [R] create list entry from variable
Try this: l - list(i + 1) names(l) - i On Mon, Aug 24, 2009 at 4:01 PM, rami jiossy sra...@hotmail.com wrote: Hi; assume i-10 how can i create a list having key=10 and value=11 list(i=11) generates a list with 'i' [1] 11 and not 10 [1] 11 any help? Thanks _ Facebook. :ON:WL:en-US:SI_SB_facebook:082009 [[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. -- 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] create list entry from variable
Yep; great thanks :) Date: Mon, 24 Aug 2009 16:11:20 -0300 Subject: Re: [R] create list entry from variable From: www...@gmail.com To: sra...@hotmail.com CC: r-help@r-project.org Try this: l - list(i + 1) names(l) - i On Mon, Aug 24, 2009 at 4:01 PM, rami jiossy sra...@hotmail.com wrote: Hi; assume i-10 how can i create a list having key=10 and value=11 list(i=11) generates a list with 'i' [1] 11 and not 10 [1] 11 any help? Thanks _ Facebook. :ON:WL:en-US:SI_SB_facebook:082009 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list 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] create list entry from variable
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of rami jiossy Sent: Monday, August 24, 2009 12:01 PM To: R-Help Subject: [R] create list entry from variable Hi; assume i-10 how can i create a list having key=10 and value=11 list(i=11) generates a list with 'i' [1] 11 and not 10 [1] 11 any help? You can use [[- with a character argument for the key myList-list() i-10 myList[[i]] - 11 myList $`10` [1] 11 If you use an integer argument, as in, myList[[10]] - 11 then 11 becomes the 10'th element of myList, not the element named '10'. myList[[i]] - NULL will remove the element. Bill Dunlap TIBCO Software Inc - Spotfire Division wdunlap tibco.com Thanks _ Facebook. :ON:WL:en-US:SI_SB_facebook:082009 [[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] Two lines, two scales, one graph
On Mon, 24-Aug-2009 at 08:00AM -0700, Rick wrote:
First of all, thanks to everyone who answers these questions - it's
most helpful.
I'm new to R and despite searching have not found an example of what I
want to do (there are some good beginner's guides and a lot of complex
plots, but I haven't found this).
I would like to plot two variables against the same abscissa values. They
have different scales. I've found how to make a second axis on the right
for labeling, but not how to plot two lines at different scales.
The idea is that you rescale the second lot of y values to fit into
the same range as the first lot.
If your first ones range from 0 to 10 and your second ones from 0 to
1000, you do the second line (using the lines() function) by dividing
every value by 100 and I think you will have found how to use axis
with side = 3 to do the axis. If the zeros don't coincide, you need
to make further adjustments which should become obvious.
HTH
--
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
___Patrick Connolly
{~._.~} Great minds discuss ideas
_( Y )_ Average minds discuss events
(:_~*~_:) Small minds discuss people
(_)-(_) . Eleanor Roosevelt
~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.~.
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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] Help on comparing two matrices
David: Well, e.g. the first row has 2 ones in your output while there were no rows with 2 ones in the original matrix. Since the row and column sums can't be changed by sorting them, the output matrix can't be equivalent to the original one. But that means nothing, maybe it's intended and just for comparison reasons? :) But I don't get how the ones can get lost by making a string out of the row values... Steve: The two matrices I want to compare really are graph matrices, just not adjacency but incidence matrices. There should be a way to get an adjacency matrix of a graph out of its incidence matrix but I don't know it... David Winsemius schrieb: On Aug 23, 2009, at 4:14 PM, Michael Kogan wrote: Thanks for all the replies! Steve: I don't know whether my suggestion is a good one. I'm quite new to programming, have absolutely no experience and this was the only one I could think of. :-) I'm not sure whether I'm able to put your tips into practice, unfortunately I had no time for much reading today but I'll dive into it tomorrow. David: To be honest I don't understand your code yet, but the result is not equivalent to the original matrix since the row sums don't match, isn't it? Or is it intended like this? I'll try to ask Google (or rather RSeek) tomorrow to understand your code. :-) Not sure what you mean by the row sums don't match. All I did (working from the inside of that function outward) was: a) concatenate the row values into a string: Reduce(paste, sm[1,]) [1] 0 0 0 0 1 1 0 0 # the innermost function applied to the first row. b) do it for every row sapply(1:7, function(x) Reduce(paste, sm[x,])) [1] 0 0 0 0 1 1 0 0 1 1 1 1 0 1 1 0 1 1 1 1 0 0 1 0 1 0 0 1 0 0 0 0 0 0 1 1 1 0 0 1 [6] 1 0 0 0 0 0 0 1 1 1 0 0 1 1 0 1 c) create a sorting vector from that vector (of characters): order(sapply(1:7, function(x) Reduce(paste, sm[x,])) ) [1] 1 5 6 4 7 3 2 d) use that sort vector to order the rows: sm[order(sapply(1:7, function(x) Reduce(paste, sm[x,])) ), ] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,]00001100 [2,]00111001 [3,]10000001 [4,]10010000 [5,]11001101 [6,]11110010 [7,]11110110 All of the original vectors are output, just in a different order, so I am therefore confused. why you think the rowSums don't match don't match what? I assumed you would take this process and apply it to _each_ of the two matrices in question and then see if you got a TRUE result with the identical function or perhaps the == function. __ 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] plotting a grid with grid() ?
I am trying to come up with a way of shading-in a grid for a simple pattern So far I can draw a square where I want but I cannot seem to draw a complete grid. I am just drawing them along the diagonal!! Clearly I am missing something simple but what? Any suggestions gratefully accepted. Example # op - par(xaxs=i, yaxs=i) plot(c(1, 11), c(1,11), type =n, xlab=, ylab=) x1 - rep(1:10, each=10) x2 - rep(2:11, each=10) y1 - rep(1:10, each=10) y2 - rep(2:11, each=10) # no grid :( rect(x1,y1,x2,y2, border=blue) rect(2,2,3,3, col=red) x1 - rep(1:10,10) x2 - rep(2:11, 10) y1 - rep(1:10, 10) y2 - rep(2:11, 10) # no grid again :( rect(x1,y1,x2,y2, border=blue) par - op #= __ The new Internet Explorer® 8 - Faster, safer, easier. Optimized for Yahoo! Get it Now for Free! at http://downloads.yahoo.com/ca/internetexplorer/ __ 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 lines, two scales, one graph
On Aug 24, 2009, at 11:00 AM, Rick wrote:
First of all, thanks to everyone who answers these questions - it's
most helpful.
I'm new to R and despite searching have not found an example of what I
want to do (there are some good beginner's guides and a lot of complex
plots, but I haven't found this).
I would like to plot two variables against the same abscissa values.
They
have different scales. I've found how to make a second axis on the
right
for labeling, but not how to plot two lines at different scales.
After using R site search I would ask: have you looked at as.layer in
{latticeExtra}?
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on comparing two matrices
On Aug 24, 2009, at 4:01 PM, Michael Kogan wrote: David: Well, e.g. the first row has 2 ones in your output while there were no rows with 2 ones in the original matrix. Since the row and column sums can't be changed by sorting them, the output matrix can't be equivalent to the original one. But that means nothing, maybe it's intended and just for comparison reasons? :) But I don't get how the ones can get lost by making a string out of the row values... OK, so shoot me. I screwed up and forgot to use byrow=TRUE in my scan operation. So I ended up with a different starting matrix than you. This is what it should have looked like: sm - matrix(scan(textConnection( + 01110110 + 11000101 + 10100011 + 11001000 + 10111000 + 01011000 + 00000111)), 7, 8, byrow=TRUE) Read 56 items sm [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,]01110110 [2,]11000101 [3,]10100011 [4,]11001000 [5,]10111000 [6,]01011000 [7,]00000111 order(sapply(1:7, function(x) Reduce(paste, sm[x,])) ) [1] 7 6 1 3 5 2 4 sm[order(sapply(1:7, function(x) Reduce(paste, sm[x,])) ), ] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,]00000111 [2,]01011000 [3,]01110110 [4,]10100011 [5,]10111000 [6,]11000101 [7,]11001000 The process creates a sorted index and then just outputs rows from the original matrix, so there cannot be any row that was not there at the start. Gabor's solution will do the same operation and certainly looks more elegant than mine. (His input operation did the same mutilation on your input string as did mine.) -- David Steve: The two matrices I want to compare really are graph matrices, just not adjacency but incidence matrices. There should be a way to get an adjacency matrix of a graph out of its incidence matrix but I don't know it... David Winsemius schrieb: On Aug 23, 2009, at 4:14 PM, Michael Kogan wrote: Thanks for all the replies! Steve: I don't know whether my suggestion is a good one. I'm quite new to programming, have absolutely no experience and this was the only one I could think of. :-) I'm not sure whether I'm able to put your tips into practice, unfortunately I had no time for much reading today but I'll dive into it tomorrow. David: To be honest I don't understand your code yet, but the result is not equivalent to the original matrix since the row sums don't match, isn't it? Or is it intended like this? I'll try to ask Google (or rather RSeek) tomorrow to understand your code. :-) Not sure what you mean by the row sums don't match. All I did (working from the inside of that function outward) was: a) concatenate the row values into a string: Reduce(paste, sm[1,]) [1] 0 0 0 0 1 1 0 0 # the innermost function applied to the first row. b) do it for every row sapply(1:7, function(x) Reduce(paste, sm[x,])) [1] 0 0 0 0 1 1 0 0 1 1 1 1 0 1 1 0 1 1 1 1 0 0 1 0 1 0 0 1 0 0 0 0 0 0 1 1 1 0 0 1 [6] 1 0 0 0 0 0 0 1 1 1 0 0 1 1 0 1 c) create a sorting vector from that vector (of characters): order(sapply(1:7, function(x) Reduce(paste, sm[x,])) ) [1] 1 5 6 4 7 3 2 d) use that sort vector to order the rows: sm[order(sapply(1:7, function(x) Reduce(paste, sm[x,])) ), ] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,]00001100 [2,]00111001 [3,]10000001 [4,]10010000 [5,]11001101 [6,]11110010 [7,]11110110 All of the original vectors are output, just in a different order, so I am therefore confused. why you think the rowSums don't match don't match what? I assumed you would take this process and apply it to _each_ of the two matrices in question and then see if you got a TRUE result with the identical function or perhaps the == function. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
