[R] R: interpreting weight in meta-analysis of proportion
Fine ! Many many thanks! Mario -Messaggio originale- Da: Bernd Weiss [mailto:bernd.we...@uni-koeln.de] Inviato: venerdì 18 aprile 2014 15.58 A: petre...@unina.it; R-help@r-project.org Oggetto: Re: [R] interpreting weight in meta-analysis of proportion On 18.04.2014 13:02, petre...@unina.it wrote: Prof. Dewey, sorry for the trivial question and many thank for the replay. Using which package? In this case I used the meta package, but I know that for all but the DerSimonian-Laird method the R function rma.uni of R package metafor is called internally. What did you expect the weights to sum to, I wonder. I think that, to better explain the influence of single study in pooling the effect size, the weight are presented as percentage of the sum of total weight of each study, but I ask for a confirm. Nevertheless, I ask if it is possible to obtain for each study the value of the absolute weight, other than the relative weight, or at least the absolute value of the sum of the weights. ## This is an example from the examples-section res - metaprop(4:1, c(10, 20, 30, 40), comb.fixed=FALSE, comb.random=TRUE) ## Object res contains a lot of interesting information ## Open ?metaprop and read the section on values str(res) ## Obtaining the random-effects weights res$w.random ## Calculating relative weights manually res$w.random/(sum(res$w.random)) HTH, Bernd __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 Example scripts
On 18/04/2014, 6:07 PM, Gene Leynes wrote:
A few years ago R changed the way help was handled so that the HTML
files are no longer available in the library directory. Around that
time the R example files that used to be in some of the libraries also
vanished.
I'm wondering where the r-ex folder went. Is it totally unsupported
and gone? Is it hidden in the mysterious rdx or rdb files?
The source for the examples will be in the \examples{} section of the
.Rd source file -- that didn't change. What changed was how the source
is processed. The files are now parsed into a binary format that is
stored in the database files.
You can extract the examples from the source file using the tools::Rd2ex
function. You can extract them from the binary database using this
function on an Rd object, which is obtainable using the internal
function .getHelpFile.
So for example, to see the code for the example for rwishart, you could do
tools::Rd2ex(.../bayesm/man/rwishart.Rd)
if you have the source file installed there, or
library(bayesm)
Rd - utils:::.getHelpFile(?rwishart)
tools::Rd2ex(Rd)
if you have the package installed.
Duncan Murdoch
In particular I'm looking for the latest example scripts in the baysem
package. I don't see them in the source code or the installed library.
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Re: [R] Read.table mucks up headers
Hello Milan It had worked perfectly before, but now I am trying on a different text file but using the trick you showed I just get the headers in the output and that too as row.names and X. *code:* corr - read.table(E:/temp/corrosion data.txt,header=T,fileEncoding=UTF-8-BOM) dput(corr) structure(list(ï...Weight. = c(13.74, 12.97, 10.78, 10.53, 10.16, 9.38, 9.23, 9.2, 17.24, 18, 15.12, 16.08, 13.71, 13.81, 12.61, 14.03), Piece = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), Time = c(96L, 84L, 72L, 60L, 48L, 36L, 24L, 12L, 96L, 96L, 72L, 60L, 48L, 36L, 24L, 12L)), .Names = c(ï...Weight., Piece, Time), class = data.frame, row.names = c(NA, -16L )) str(corr) 'data.frame': 16 obs. of 3 variables: $ ï...Weight.: num 13.7 13 10.8 10.5 10.2 ... $ Piece : int 1 1 1 1 1 1 1 1 2 2 ... $ Time : int 96 84 72 60 48 36 24 12 96 96 ... Please tell me what I am doing wrong. -- View this message in context: http://r.789695.n4.nabble.com/Read-table-mucks-up-headers-tp4688742p4689099.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] Read.table mucks up headers
On 19.04.2014 14:30, starter wrote: Hello Milan It had worked perfectly before, but now I am trying on a different text file but using the trick you showed I just get the headers in the output and that too as row.names and X. *code:* corr - read.table(E:/temp/corrosion data.txt,header=T,fileEncoding=UTF-8-BOM) Which R version? Are you sure this is UTF-8-BOM and not UCS-2LE or UTF-16LE? Best, Uwe Ligges dput(corr) structure(list(ï...Weight. = c(13.74, 12.97, 10.78, 10.53, 10.16, 9.38, 9.23, 9.2, 17.24, 18, 15.12, 16.08, 13.71, 13.81, 12.61, 14.03), Piece = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), Time = c(96L, 84L, 72L, 60L, 48L, 36L, 24L, 12L, 96L, 96L, 72L, 60L, 48L, 36L, 24L, 12L)), .Names = c(ï...Weight., Piece, Time), class = data.frame, row.names = c(NA, -16L )) str(corr) 'data.frame': 16 obs. of 3 variables: $ ï...Weight.: num 13.7 13 10.8 10.5 10.2 ... $ Piece : int 1 1 1 1 1 1 1 1 2 2 ... $ Time : int 96 84 72 60 48 36 24 12 96 96 ... Please tell me what I am doing wrong. -- View this message in context: http://r.789695.n4.nabble.com/Read-table-mucks-up-headers-tp4688742p4689099.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with pip2d() from ptinpoly
Thank you for catching that, Boris. I'm surprised, given that there is no mention of sense of direction in the package documentation. Scott Waichler -Original Message- From: Boris Steipe [mailto:boris.ste...@utoronto.ca] Sent: Friday, April 18, 2014 3:49 PM To: Waichler, Scott R; r-help@r-project.org Subject: Re: [R] problem with pip2d() from ptinpoly Apparently it matters whether your polygon is defined clockwise or counterclockwise. A point outside your triangle is recognized ... q2 - matrix(c(594893.0,115435.0), ncol=2, byrow=T) pip2d(Vertices = verts, Queries = q2) [1] 1 ... and defining the triangle in counterclockwise sense gives the expected behaviour. v2 - matrix(c(594891,115309,594891,117201,59,117201), ncol=2, byrow=T) pip2d(Vertices = v2, Queries = query) [1] 1 Cheers, B. On 2014-04-18, at 6:00 PM, Waichler, Scott R wrote: Hi, pip2d() doesn't seem to work correctly for me. I have a plot of a triangle that a query point fits inside, but the point is defined as outside the polygon by pip2d. library(ptinpoly) verts - matrix(c(594891,115309,59,117201,594891,117201), ncol=2, byrow=T) query - matrix(c(594885.0,115435.0), ncol=2, byrow=T) pip2d(Vertices = verts, Queries = query) # result = -1 # contrary to -1 output of pip2d, plot shows point lies within triangle plot(c(594400, 595000), c(115000, 117500), type=n) polygon(verts, border=red) points(x=query[,1], y=query[,2], col=blue) Scott Waichler Pacific Northwest National Laboratory Richland, WA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] XML getNodeSet
SOLVED: (mostly)
So I'll post this here in case it helps someone in the future.
fileName - '001.xml'
doc - xmlTreeParse(fileName,
handlers=list(comment=function(x,...){NULL}), asTree = TRUE)
root - xmlRoot(doc)
Connected_To - xmlToDataFrame(getNodeSet(root,
'//ccd:el-e3657392-8e26-42a4-a996-258f9e645c46',ccd))
Must use xmlTreeParse.
Connected_To is a dataframe with the element names as colnames and the rows
are for each instance in the file.
I still get an warning:
Namespace prefix ccd on el-e3657392-8e26-42a4-a996-258f9e645c46 is not defined
I haven't quite found the solution for this namespace problem. I have tried
a few functions/examples from the XML package but they usually return a
list instead of a character vector. But the result dataframe from the XML
is correct and that is my initial goal.
Cheers,
Tim
On Thu, Apr 17, 2014 at 3:54 PM, Timothy W. Cook t...@mlhim.org wrote:
Apologies, I forgot to add details:
platform x86_64-pc-linux-gnu
arch x86_64
os linux-gnu
system x86_64, linux-gnu
status
major 3
minor 0.1
year 2013
month 05
day16
svn rev62743
language R
version.string R version 3.0.1 (2013-05-16)
nickname Good Sport
Executed inside R Studio Version 0.98.501
On Thu, Apr 17, 2014 at 12:35 PM, Timothy W. Cook t...@mlhim.org wrote:
R newbie, experienced software developer.
I have a bit of confusion regarding using this function. See the XML
fragment at the end of the post.
This works as far as retrieving the nodeset:
fileName - '/home/tim/MLHIM/git/EpiS3/test_ccd/inst/examples/001.xml'
doc - xmlInternalTreeParse(fileName) nodes -
getNodeSet(doc,'//ccd:el-e3657392-8e26-42a4-a996-258f9e645c46')
Connected_To=xmlToDataFrame(nodes)
The result is:
str(Connected_To)'data.frame': 1 obs. of 7 variables:
$ comment : Factor w/ 1 level DvURI : 1
$ data-name : Factor w/ 1 level Connected To: 1
$ valid-time-begin: Factor w/ 1 level Use any subtype of ExceptionalValue
here when a value is missing: 1
$ valid-time-end : Factor w/ 1 level 2020-06-07T01:31:49Z: 1
$ DvURI-dv: Factor w/ 1 level 2009-01-20T08:40:53Z: 1
$ relation: Factor w/ 1 level http://www.ccdgen.com: 1
$ NA : Factor w/ 1 level connected to: 1
The line:
$ valid-time-begin: Factor w/ 1 level Use any subtype of ExceptionalValue
here when a value is missing: 1
has a value that is a comment that occurs before the element. Then each
value is shifted by one place with an extra one at the end.
My desired solution would be to remove commenteds from the parsed tree.
So I did this:
doc - xmlInternalTreeParse(fileName,
handlers=list(comment=function(x,...){NULL}), useInternalNodes = TRUE)
But now when I attempt to get the nodes I get an error.
nodes -
getNodeSet(doc,'//ccd:el-e3657392-8e26-42a4-a996-258f9e645c46')Error in
UseMethod(xpathApply) :
no applicable method for 'xpathApply' applied to an object of class list
I am not sure what the error is telling me nor how to fix it.
All suggestions are welcome and appreciated. Maybe I should be using a
different approach to solving the issue with extracting the data.frame?
Regads,
Tim
XML fragment:
ccd:el-e3657392-8e26-42a4-a996-258f9e645c46
!-- DvURI --
data-nameConnected To/data-name
!-- Use any subtype of ExceptionalValue here when a value is
missing--
valid-time-begin2020-06-07T01:31:49Z/valid-time-begin
valid-time-end2009-01-20T08:40:53Z/valid-time-end
DvURI-dvhttp://www.ccdgen.com/DvURI-dv
relationconnected to/relation
/ccd:el-e3657392-8e26-42a4-a996-258f9e645c46
--
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Timothy Cook, MSc +55 21 994711995
MLHIM http://www.mlhim.org
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--
MLHIM VIP Signup: http://goo.gl/22B0U
Timothy Cook, MSc +55 21 994711995
MLHIM http://www.mlhim.org
Like Us on FB: https://www.facebook.com/mlhim2
Circle us on G+: http://goo.gl/44EV5
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LinkedIn Profile:http://www.linkedin.com/in/timothywaynecook
MLHIM http://www.mlhim.org
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[R] Foreign function interface
I read the Foreign Function Interface for R. The documentation does not prescribe the calling convention though. Does that mean __stdcall on W32 and __cdecl for Linux? Regards, Alex van der Spek __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] cbind with row names to serveral files in R
Hi,
The rownames part is not clear as your expected output and input files didn't
show them as rownames.
##Suppose you have all the files in a folder
##here I am creating the names of those files
files1 - paste0(sample, rep(1:777, each=29),chr,1:29,.txt)
length(files1)
#[1] 22533
lst1 - split(files1,as.numeric(gsub(.*chr(\\d+)\\..*,\\1,files1)))
##use list.files()
##in your case, ##not tested
lst1 - split(list.files(pattern=sample\\d+),
as.numeric(gsub(.*chr(\\d+)\\..*,\\1,list.files(pattern=sample\\d+
##in case other files are also in the folder
library(plyr)
lst2 - lapply(lst1,function(x) {x1 -join_all(lapply(x,function(y)
read.table(y,header=TRUE,stringsAsFactors=FALSE,sep=)),c(Name,Chr,Position))
})
lapply(seq_along(lst2),function(i)
write.table(lst2[[i]],paste0(LRRrawallchr,i,.txt),row.names=FALSE,quote=FALSE))
###if you need to create rownames using the first three columns:
lapply(seq_along(lst2),function(i) {x1 - lst2[[i]]; row.names(x1) -
as.character(interaction(x1[,1:3],sep=_)); x2 - x1[,-(1:3)]; write.table(x2,
paste0(LRRrawallchr,i,.txt), quote=FALSE)})
A.K.
I would like to use the `cbind` in a list of files. However each file are
splited in a specific chromosome (chr) `(k in 1:29)`, and specific sample `(i
in 1:777)`. The files are like:
sample1chr1.txt, sample1chr2.txt ... sample1chr29.txt, sample2chr1.txt ...
sample777chr29.txt
All files have exactly the same rows names (3 first collumns represent my row
names). I would like to get a final file to each chr merging to all sample
files, with and do not repeat the row names in the final file (the first 3
collumns representing my rows names).
I tried this:
`#Creating file with row names (3 first collumns) to each Chr
{
{for(k in 1:29){
infile - paste0(sample1chr,k,.txt)
outfile - paste0(LRRrawallchr,k,.txt)
rows - read.table(infile, header=TRUE, sep=\t)
rows - rows[, -grep(Log.R.Ratio, colnames(rows))]
write.table(rows, outfile, sep=;)}}`
`#Cbind in one file per Chr
{ for(i in 1:777)
for(k in 1:29){
base - paste0(LRRrawallchr,k,.txt)
chr - read.table(base, header=TRUE, sep=;)
infile - paste0(sample,i,chr,k,.txt)
chr2 - read.table(infile, header=TRUE, sep=\t)
outfile - paste0(LRRrawallchr,k,.txt)
chr2 - chr2[, -grep(Name, colnames(chr2))]
chr2 - chr2[, -grep(Chr, colnames(chr2))]
chr2 - chr2[, -grep(Position, colnames(chr2))]
chr - cbind(chr, chr2)
write.table(chr, outfile, sep=;, row.names=FALSE, col.names=FALSE)}
}`
Input example (sample1chr1.txt):
Name Chr Position sample1value
BAC-11034 1 128 0.302
BAC-11044 1 129 -0.56
BAC-11057 1 134 0.0840
Input example (sample2chr1.txt):
Name Chr Position sample2value
BAC-11034 1 128 0.25
BAC-11044 1 129 0.41
BAC-11057 1 134 -0.14
Expected output (LRRrawallchr1):
Name Chr Position sample1value sample2value
BAC-11034 1 128 0.302 0.25
BAC-11044 1 129 -0.56 0.41
BAC-11057 1 134 0.0840 -0.14
I have 22553 diferent .txt files (29 files (one per chr) to each of 777
samples). All 22553 files (sample1chr1.txt, sample1chr2.txt ...
sample1chr29.txt, sample2chr1.txt ... sample777chr29.txt) are like above
example.
I wanna 29 files like (LRRrawallchr1), one per Chr. The LRRrawallchr,k, files
have to be with 777+3 (800 collumns). The 3 row names and one collumn per
sample.
Cheers!
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[R] cex values being ignored in the curve function
Dear R-Community, The cex values in curve seem to be being ignored. I have searched previous help questions and also the web generally, and cannot find this being a major problem so I am suspicious of something odd happening but I am at a loss to work out why. I am trying to plot this: b1 - -0.858 pow- 0.8 ratio - 1 sig - 0.05 curve((qnorm(1-sig/2)+qnorm(pow))^2/b1^2/x/(ratio/(1+ratio))/(1/(1+ratio)),from=0.005,to=0.08,main=This needs to be bigger,xlab=And this bigger too,ylab=And this too, cex=1,cex.lab=3.5, cex.axis=3.5, cex.main=3.5, cex.sub=3.5) But no matter what value of cex I make it, or whether I break up the arguments into : title(main=something,cex.main=3) axis(cex.axis=2) or parse this before plotting : par(cex.lab=1.5, cex.axis=1.5, cex.main=3.5, cex.sub=1.5) I cannot get the main, axis titles or axis numbers to change in size, whatever I do. I am using a macbook pro, with mavericks, R-studio and R 3.1.0 spring-dance, and I am initiating the plot with: X11( width=width , height=height , type=cairo). Any help would be greatly appreciated. thanks Philip __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Difference between times
Thank you for your reply. I discovered the OlsonNames() function to get the time-zone names in my system. Rui get a warning message when using a not recognized tz. On my system this doesn't succed. I solved as follow: dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = Europe/Rome) dt2 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = America/Los_Angeles) dt1[1] 2014-04-18 09:00:00 CESTdt2[1] 2014-04-18 09:00:00 PDTdt1-dt2Time difference of -9 hours Thank you, Nicola 2014-04-18 21:59 GMT+02:00 Prof Brian Ripley rip...@stats.ox.ac.uk: On 18/04/2014 19:46, Rui Barradas wrote: Hello, The reason why is that you've misspelled CET (not CEST) Neither CET nor CEST are portable time-zone names. We have not been given the 'at a minimum' information required by the posting guide, so please read ?Sys.timezone on your system. dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = CEST) Warning messages: 1: In strptime(x, format, tz = tz) : unknown timezone 'CEST' 2: In as.POSIXct.POSIXlt(as.POSIXlt(x, tz, ...), tz, ...) : unknown timezone 'CEST' dt2 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = GMT) dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = CET) dt1-dt2 Time difference of -2 hours Hope this helps, Rui Barradas Em 18-04-2014 17:13, Nicola Sturaro Sommacal escreveu: Hi. I am new to POSIX and I'd like to understand the reason of this difference. dt1 = as.POSIXct(2014-03-29 09.00, format=%Y-%m-%d %H.%M) dt2 = as.POSIXct(2014-03-30 09.00, format=%Y-%m-%d %H.%M) dt2-dt1 dt1[1] 2014-03-29 09:00:00 CET dt2[1] 2014-03-30 09:00:00 CEST dt2-dt1 Time difference of 23 hours This is right, because on Mar 31 at 2 PM we jump directly to 3PM, DST. On the contrary, I don't understand the following: dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = CEST) dt2 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = GMT) dt1[1] 2014-04-18 09:00:00 CEST dt2[1] 2014-04-18 09:00:00 GMT dt1-dt2Time difference of 0 secs I should expected a time difference of 2 hours, as CEST is GMT+2. Anyone can help me? Thank you, Nicola [[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. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 [[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] Need help to convert data frame to transaction set.
Hi, To convert coerce the data set to transaction data set I used the code trans4 - as(split(a[,Cust_ID], a[,Parts]), transactions) but I am getting the following error- Error in as(split(a[, Cust_ID], a[, Parts]), transactions) : nomethod or default for coercing “list” to “transactions” Then I tried first converting the data set to matrix structure using the code c_m-as.matrix(c_df) c_m then entered the following code trans2 - as(c_m, transactions) but got the following error Error in as(c_m, transactions) : no method or default for coercing “matrix” to “transactions” Please let me know how to correct the problem. Thanks Sathish __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 Statistical Computing Language Preference Help
Dear R Developers and Help Specialists; I downloaded and installed R to learn and use it during solutions of my scientific studies. At first, I got in trouble with it with the menu languages. It is in Turkish and I have to change it to English but I can not find where the language preferences menu or options menu is. Could you please let me know how I can select the English Language as the main language of the software (for menus, help etc.) Your kindness and help will be very much appriciated. Please do not publish on web my e-mail. Have a nice day. Best Regards Dr. Burak Omer Saracoglu PhD in Graduate School of Science Engineering and Technology MSc in Industrial Engineering BSc in Naval Architecture and Marine Engineering [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to use rainbow function without the gamma argument
Hi all, I am using an old code (probably written for R 2.5) and it stops when calling rainbow() with gamma argument. I saw that gamma argument is not present in newer version of R rainbow function. How can I translate this line of code: rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75, gamma = 1.5) ? It fails with: Error in rainbow(100, s = 1, v = 0.75, start = 0, end = 0.75, gamma = 1.5) : unused argument (gamma = 1.5) Calls: nmfconsensus ... matrix.abs.plot - image - image.default - rainbow Execution halted Thanks ~ Francesco Brundu [[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] Difference between times
I forgot: sysname release Linux 3.5.0-48-generic version #72-Ubuntu SMP Mon Mar 10 23:18:29 UTC 2014 2014-04-19 14:03 GMT+02:00 Nicola Sturaro Sommacal mailingl...@nicolasturaro.com: Thank you for your reply. I discovered the OlsonNames() function to get the time-zone names in my system. Rui get a warning message when using a not recognized tz. On my system this doesn't succed. I solved as follow: dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = Europe/Rome) dt2 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = America/Los_Angeles) dt1[1] 2014-04-18 09:00:00 CESTdt2 [1] 2014-04-18 09:00:00 PDTdt1-dt2Time difference of -9 hours Thank you, Nicola 2014-04-18 21:59 GMT+02:00 Prof Brian Ripley rip...@stats.ox.ac.uk: On 18/04/2014 19:46, Rui Barradas wrote: Hello, The reason why is that you've misspelled CET (not CEST) Neither CET nor CEST are portable time-zone names. We have not been given the 'at a minimum' information required by the posting guide, so please read ?Sys.timezone on your system. dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = CEST) Warning messages: 1: In strptime(x, format, tz = tz) : unknown timezone 'CEST' 2: In as.POSIXct.POSIXlt(as.POSIXlt(x, tz, ...), tz, ...) : unknown timezone 'CEST' dt2 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = GMT) dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = CET) dt1-dt2 Time difference of -2 hours Hope this helps, Rui Barradas Em 18-04-2014 17:13, Nicola Sturaro Sommacal escreveu: Hi. I am new to POSIX and I'd like to understand the reason of this difference. dt1 = as.POSIXct(2014-03-29 09.00, format=%Y-%m-%d %H.%M) dt2 = as.POSIXct(2014-03-30 09.00, format=%Y-%m-%d %H.%M) dt2-dt1 dt1[1] 2014-03-29 09:00:00 CET dt2[1] 2014-03-30 09:00:00 CEST dt2-dt1 Time difference of 23 hours This is right, because on Mar 31 at 2 PM we jump directly to 3PM, DST. On the contrary, I don't understand the following: dt1 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = CEST) dt2 = as.POSIXct(2014-04-18 09.00, format=%Y-%m-%d %H.%M, tz = GMT) dt1[1] 2014-04-18 09:00:00 CEST dt2[1] 2014-04-18 09:00:00 GMT dt1-dt2Time difference of 0 secs I should expected a time difference of 2 hours, as CEST is GMT+2. Anyone can help me? Thank you, Nicola [[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. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 [[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] Extracting the names of coefficients of random effects
Hi All, I need to be able to manipulate the names of the coefficients from *ranef()*. If there is any missing data when fitting a mixed model using lmer, no estimate is returned for the associated level for that random effect. Thus if the data input for regions had levels *Region* Bolton Bradford Cambridge Durham and there was missing data on Bradford then * ranef(model)* gives (Intercept) Bolton: -0.0981763413 Cambridge 0.0151102347 Durham 0.1837142259 This becomes a problem if I want to use *predict( )* on new data where there is no missing data on Bradford. In such an instance *predict (model, newdata = newInput) * gives the following error message ‘Error in (function (x, n) : new levels detected in newdata’ I could get round this by checking the Region field of the new data ‘newInput’ against the names of the levels of the intercept coefficients from* ranef().* However, I’m not sure how to access these since if *x- ranef(model) x * This gives the same output above, but *x[1,1]* only returns the numeric value -0.0981763413 x is a dataframe with only one field (column) with numeric values and the names of the levels are not present or identified. There must be a variable which defines ‘Bolton’, Bradford’ etc since if I use the write.table function *write.table (x, file=/desktop/Dummy.csv, sep = ,, col.names = NA, qmethod = double)* This outputs both the names (‘Bolton’, Bradford’..) and their corresponding numeric values to a spreadsheet. Does anyone know how to do this without resorting to outputting to a spreadsheet? Regards, Brian H Willis Health and Population Sciences University of Birmingham -- View this message in context: http://r.789695.n4.nabble.com/Extracting-the-names-of-coefficients-of-random-effects-tp4689109.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] cex values being ignored in the curve function
Don't know if this helps, but works for me on a PC under Windows 7 with the default graphics device. Try a different device, maybe (e.g. pdf) ?? -- Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374(650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch Call Send SMS Add to Skype You'll need Skype CreditFree via Skype __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 Statistical Computing Language Preference Help
Search! Googling on How to set language for R produced this: https://stat.ethz.ch/pipermail/r-help/2008-October/178503.html If this is not what you need, a little more searching may be required. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Sat, Apr 19, 2014 at 2:10 AM, burak ömer saraçoglu burakomersaraco...@hotmail.com wrote: Dear R Developers and Help Specialists; I downloaded and installed R to learn and use it during solutions of my scientific studies. At first, I got in trouble with it with the menu languages. It is in Turkish and I have to change it to English but I can not find where the language preferences menu or options menu is. Could you please let me know how I can select the English Language as the main language of the software (for menus, help etc.) Your kindness and help will be very much appriciated. Please do not publish on web my e-mail. Have a nice day. Best Regards Dr. Burak Omer Saracoglu PhD in Graduate School of Science Engineering and Technology MSc in Industrial Engineering BSc in Naval Architecture and Marine Engineering [[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] How to use rainbow function without the gamma argument
Have you looked at ?rainbow ? Is there a reason why you don't simply leave the gamma parameter away? Try: pie(rep(1,100), col=rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75)) Cheers, B. On 2014-04-19, at 6:05 AM, Francesco Brundu wrote: Hi all, I am using an old code (probably written for R 2.5) and it stops when calling rainbow() with gamma argument. I saw that gamma argument is not present in newer version of R rainbow function. How can I translate this line of code: rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75, gamma = 1.5) ? It fails with: Error in rainbow(100, s = 1, v = 0.75, start = 0, end = 0.75, gamma = 1.5) : unused argument (gamma = 1.5) Calls: nmfconsensus ... matrix.abs.plot - image - image.default - rainbow Execution halted Thanks ~ Francesco Brundu [[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] inverse normal distribution function
You may want to read about generalized linear modelling and link functions for forming appropriate categorical variable/link function. See documentations in R: ?glm, ?family and ?inverse.gaussian. Also look at the original paper of Nelder, John; Wedderburn, Robert , it is available freely with the courtesy of JSTOR: http://www.jstor.org/discover/10.2307/2344614 On 18 April 2014 09:13, thanoon younis thanoon.youni...@gmail.com wrote: dear all members i want to use inverse normal distribution in R to show the value of variable Z when Z represent the ordered categorical variables. i hope anyone gives me an example on this distribution . thanks to all [[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] inverse normal distribution function
Not sure how would you do that but there is a package SEM on CRAN for structural equation models. On 20 April 2014 01:10, thanoon younis thanoon.youni...@gmail.com wrote: thank you so much Suzen i want to use bayesian analysis in structural equation models with ordered categorical data and i want to use inverse normal as a distribution of thresholds and i dont find any paper or documents in R or another program about inverse normal. best regards __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 use rainbow function without the gamma argument
If it MUST be parameter-compatible with the old call, you could just add ...
to your local version of rainbow. The unused parameter will then be dropped.
Here's how:
# The original creates an error ...
rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75, gamma = 1.5)
Error in rainbow(100, s = 1, v = 0.75, start = 0, end = 0.75, gamma = 1.5) :
unused argument (gamma = 1.5)
# The code of the function is here:
rainbow
function (n, s = 1, v = 1, start = 0, end = max(1, n - 1)/n,
alpha = 1)
{
if ((n - as.integer(n[1L])) 0) {
if (start == end || any(c(start, end) 0) || any(c(start,
end) 1))
stop('start' and 'end' must be distinct and in [0, 1].)
hsv(h = seq.int(start, ifelse(start end, 1, 0) + end,
length.out = n)%%1, s, v, alpha)
}
else character()
}
bytecode: 0x101968950
environment: namespace:grDevices
# I add ... to the parameters and define a local version of rainbow
rainbow = function (n, s = 1, v = 1, start = 0, end = max(1, n - 1)/n,
alpha = 1, ...)
{
if ((n - as.integer(n[1L])) 0) {
if (start == end || any(c(start, end) 0) || any(c(start,
end) 1))
stop('start' and 'end' must be distinct and in [0, 1].)
hsv(h = seq.int(start, ifelse(start end, 1, 0) + end,
length.out = n)%%1, s, v, alpha)
}
else character()
}
# same code except for the dots...
# Now it accepts and discards unused arguments
rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75, gamma = 1.5,
gefingerpoken = TRUE)
[1] #BFFF #BF0900FF #BF1100FF #BF1A00FF #BF2300FF #BF2B00FF
#BF3400FF
[...]
[99] #5700BFFF #6000BFFF
# and if I want the original back, I just delete my local version ...
rm(rainbow)
rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75, gamma = 1.5)
Error in rainbow(100, s = 1, v = 0.75, start = 0, end = 0.75, gamma = 1.5) :
unused argument (gamma = 1.5)
rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75)
[1] #BFFF #BF0900FF #BF1100FF #BF1A00FF #BF2300FF #BF2B00FF
#BF3400FF
[...]
[99] #5700BFFF #6000BFFF
You can read about the '...' argument in the Introduction to R. Here we use it
not to pass variables on, but to have them not cause an error when present.
HOWEVER: I personally would consider this poor style. It's probably better to
review and update your old code. There may be other less obvious problems.
YMMV
B.
On 2014-04-19, at 7:19 PM, Francesco Brundu wrote:
Hi Boris,
yes I tried this way and it worked. The fact is that I wanted to be compliant
with the old code, I did not want to change anything. So I wanted to find a
new way to rewrite the code.
Thanks
~ Francesco Brundu
On 19 April 2014 23:18, Boris Steipe boris.ste...@utoronto.ca wrote:
Have you looked at ?rainbow ?
Is there a reason why you don't simply leave the gamma parameter away?
Try:
pie(rep(1,100), col=rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75))
Cheers,
B.
On 2014-04-19, at 6:05 AM, Francesco Brundu wrote:
Hi all,
I am using an old code (probably written for R 2.5) and it stops when
calling rainbow() with gamma argument. I saw that gamma argument is not
present in newer version of R rainbow function. How can I translate this
line of code:
rainbow(100, s = 1.0, v = 0.75, start = 0.0, end = 0.75, gamma = 1.5)
?
It fails with:
Error in rainbow(100, s = 1, v = 0.75, start = 0, end = 0.75, gamma = 1.5)
:
unused argument (gamma = 1.5)
Calls: nmfconsensus ... matrix.abs.plot - image - image.default - rainbow
Execution halted
Thanks
~ Francesco Brundu
[[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.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Need help to convert data frame to transaction set.
Hi, Without a reproducible example using ?dput() or the package name, it is a bit difficult to comment. Assuming that you used: library(arules) data(AdultUCI) AdultUCI$ID - 1:nrow(AdultUCI) lst1 - split(AdultUCI[,ID], AdultUCI[,marital-status]) as(lst1, transactions) #transactions in sparse format with # 7 transactions (rows) and # 48842 items (columns) A.K. On Saturday, April 19, 2014 5:01 PM, Sathish Kumar tellsath...@gmail.com wrote: Hi, To convert coerce the data set to transaction data set I used the code trans4 - as(split(a[,Cust_ID], a[,Parts]), transactions) but I am getting the following error- Error in as(split(a[, Cust_ID], a[, Parts]), transactions) : nomethod or default for coercing “list” to “transactions” Then I tried first converting the data set to matrix structure using the code c_m-as.matrix(c_df) c_m then entered the following code trans2 - as(c_m, transactions) but got the following error Error in as(c_m, transactions) : no method or default for coercing “matrix” to “transactions” Please let me know how to correct the problem. Thanks Sathish __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Need help to convert data frame to transaction set.
Hi Satish, Using your code: part-read.csv(spares.csv) str(part) #'data.frame': 838 obs. of 1 variable: # $ Cust_ID.Parts: Factor w/ 838 levels 100\tAIR\tFILTER\tHOUSING\t,..: 161 333 495 727 728 768 769 770 784 785 ... #I guess you have only two columns in the dataset lines1 - readLines(spares.csv) head(lines1) #[1] Cust_ID\tParts\t\t\t 1\tFENDERS\t\t\t #[3] 2\tFENDERS\t\t\t 3\tOIL\tFILTERS\t\t #[5] 4\tBMW\tSERVICE\tFLUIDS\t 4\tOIL\tFILTER\t\t part-read.csv(spares.csv,header=TRUE,sep=\t) str(part) #'data.frame': 838 obs. of 5 variables: # $ Cust_ID: int 1 2 3 4 4 5 5 5 6 6 ... # $ Parts : Factor w/ 26 levels AIR,ALTERNATOR,..: 9 9 16 4 16 6 9 12 6 9 ... # $ X : Factor w/ 14 levels ,BLADES,CAR,..: 1 1 6 14 5 1 1 1 1 1 ... # $ X.1 : Factor w/ 6 levels ,(AXLE,CARE,..: 1 1 1 4 1 1 1 1 1 1 ... # $ X.2 : Factor w/ 2 levels ,BOOT): 1 1 1 1 1 1 1 1 1 1 ... part$Parts - interaction(part[,2:5],sep= ,drop=TRUE) part - part[,1:2] str(part) #'data.frame': 838 obs. of 2 variables: # $ Cust_ID: int 1 2 3 4 4 5 5 5 6 6 ... # $ Parts : Factor w/ 31 levels ALTERNATOR ,..: 6 6 21 28 20 3 6 8 3 6 ... as(split(part[,Cust_ID],part[,Parts]),transactions) #transactions in sparse format with # 31 transactions (rows) and # 502 items (columns) A.K. On Saturday, April 19, 2014 11:08 PM, Sathish Kumar tellsath...@gmail.com wrote: Hi AK, I have attached the data set. And as you suggested, I coded as follows but still I am getting an error part-read.csv(spares.csv) part lst1-split(part[,Cust_ID],part[,Parts]) as(lst1, transactions) Error in as(lst1, transactions) : no method or default for coercing “list” to “transactions” On Sat, Apr 19, 2014 at 6:43 PM, arun smartpink...@yahoo.com wrote: Hi, Without a reproducible example using ?dput() or the package name, it is a bit difficult to comment. Assuming that you used: library(arules) data(AdultUCI) AdultUCI$ID - 1:nrow(AdultUCI) lst1 - split(AdultUCI[,ID], AdultUCI[,marital-status]) as(lst1, transactions) #transactions in sparse format with # 7 transactions (rows) and # 48842 items (columns) A.K. On Saturday, April 19, 2014 5:01 PM, Sathish Kumar tellsath...@gmail.com wrote: Hi, To convert coerce the data set to transaction data set I used the code trans4 - as(split(a[,Cust_ID], a[,Parts]), transactions) but I am getting the following error- Error in as(split(a[, Cust_ID], a[, Parts]), transactions) : nomethod or default for coercing “list” to “transactions” Then I tried first converting the data set to matrix structure using the code c_m-as.matrix(c_df) c_m then entered the following code trans2 - as(c_m, transactions) but got the following error Error in as(c_m, transactions) : no method or default for coercing “matrix” to “transactions” Please let me know how to correct the problem. Thanks Sathish __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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.
