Re: [R] Unusual separators
On Aug 17, 2011, at 05:57 , Jim Holtman wrote: just read in the file using the tab as the separator. if this is a problem because a tab might appear by itself, then use readLines to read in the file, gsub to replace the blank/tab with a new separator, writeLines to write out to a temporary and then read in from the temporary file. You can skip the write and read back step by reading from a text connection. In R 2.14-to-be, there's a text= argument to read.table (and scan too), so you'll be able to do the whole thing on the fly: read.table( text=gsub(readLines(.))) Sent from my iPad On Aug 16, 2011, at 11:02, Matt Curcio matt.curcio...@gmail.com wrote: Hi all, I have a list that I got from a web page that I would like to crunch. Unfortunately, the list has some unusual separators in it. I believe the columns are separated by 1 space and 1 tab. I tried to insert this into the read.table( ..., sep= \t, ...) but got an error that said something like 'only one byte separators can be used. I have thought about using a gsub to 'swap out' the space + tab and replace it with commas, etc but thought there might be another way. Any suggestions? M -- Matt Curcio M: 401-316-5358 E: matt.curcio...@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. __ 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. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com Døden skal tape! --- Nordahl Grieg __ 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] chisq.test(): standardized (adjusted) Pearson residuals
On Aug 19, 2011, at 20:40 , David Winsemius wrote: On Aug 19, 2011, at 1:28 PM, Stephen Davies wrote: I'm using chisq.test() on a matrix of categorical data, and I see that the residuals attribute of the returned object will give me the Pearson residuals. That's cool. However, what I'd really like is the standardized (adjusted) Pearson residuals, which have a N(0,1) distribution. Is there a way to do that in R (other than by me programming it myself?) ?scale chisq.test(...)$stdres, more likely. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com Døden skal tape! --- Nordahl Grieg __ 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] Raw epoch time from XTS
Hi, I have a very large data set stored as an xts object. xts is very nice about showing row labels as human readable dates and times. I want the actual epoch values that are stored internally. The only way I can find to access them is one-at-a-time using the internal function: xcoredata() Calling this in an entire column, the R way doesn't work. It will only return a single value. Calling it in a loop for each row works but is painfully slow. Since the epoch is stored internally, there must be some way to just grab it as a vector. Does anyone know how? Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 __ 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] Raw epoch time from XTS
Hi Noah, This is one of those cases where following the posting guide (particularly the minimal, reproducible example part) would have really helped. Are you saying that calling: xcoredata(your_xts_object) does not give you the internal representation of time that you want? data(sample_matrix) sample.xts - as.xts(sample_matrix, descr='my new xts object') # returns a list, the index is the numeric representation of time displayed in the rows xcoredata(sample.xts) You could also try the more direct: attr(xts_object, index) If this is not what you want or is not working for you, providing us the output of dput() from the first few rows of your dataset and an example of what you do want would be spectacular. Cheers, Josh On Sat, Aug 20, 2011 at 12:44 AM, Noah Silverman noahsilver...@ucla.edu wrote: Hi, I have a very large data set stored as an xts object. xts is very nice about showing row labels as human readable dates and times. I want the actual epoch values that are stored internally. The only way I can find to access them is one-at-a-time using the internal function: xcoredata() Calling this in an entire column, the R way doesn't work. It will only return a single value. Calling it in a loop for each row works but is painfully slow. Since the epoch is stored internally, there must be some way to just grab it as a vector. Does anyone know how? Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building Los Angeles, CA 90095 __ 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. -- Joshua Wiley Ph.D. Student, Health Psychology Programmer Analyst II, ATS Statistical Consulting Group University of California, Los Angeles https://joshuawiley.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] hetlp_type text doen't work on Ubuntu
Hi, I' using Ubuntu 10.10/11.04 and on both of the systems help doesn't work unless I set options(help_type=html). I hate to change context and help_type='html' distracts me. I want text based help inside console. Where should I look to fix this problem? -- Amol __ 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] Unusual separators
In the current version (2.13.1) textConnection is much slower if you have a large file (1 lines) than using a temporary output file. Try timing a script using the two different approachs to get an appreciation for the difference. Sent from my iPad On Aug 20, 2011, at 3:39, peter dalgaard pda...@gmail.com wrote: On Aug 17, 2011, at 05:57 , Jim Holtman wrote: just read in the file using the tab as the separator. if this is a problem because a tab might appear by itself, then use readLines to read in the file, gsub to replace the blank/tab with a new separator, writeLines to write out to a temporary and then read in from the temporary file. You can skip the write and read back step by reading from a text connection. In R 2.14-to-be, there's a text= argument to read.table (and scan too), so you'll be able to do the whole thing on the fly: read.table( text=gsub(readLines(.))) Sent from my iPad On Aug 16, 2011, at 11:02, Matt Curcio matt.curcio...@gmail.com wrote: Hi all, I have a list that I got from a web page that I would like to crunch. Unfortunately, the list has some unusual separators in it. I believe the columns are separated by 1 space and 1 tab. I tried to insert this into the read.table( ..., sep= \t, ...) but got an error that said something like 'only one byte separators can be used. I have thought about using a gsub to 'swap out' the space + tab and replace it with commas, etc but thought there might be another way. Any suggestions? M -- Matt Curcio M: 401-316-5358 E: matt.curcio...@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. __ 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. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com Døden skal tape! --- Nordahl Grieg __ 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] Groups and bwplot
Dear R-users,
A while ago, Deepayan Sarkar suggested some code that uses the group
argument in bwplot to create some 'side-by-side' boxplots
(https://stat.ethz.ch/pipermail/r-help/2010-February/230065.html). The
example he gave was relatively specific and I wanted to generalize his
approach into a function. Unfortunately, I seem to have some issues
passing the correct arguments to the panel function, and would greatly
appreciate any suggestions to solve these issues:
require(lattice)
mybwplot - function(x,y,data,groups){
if (missing(groups)||is.null(groups)) {
groups - NULL
ngroups - 1
} else {
data[[groups]] - as.factor(data[[groups]])
ngroups - nlevels(data[[groups]])
}
mywidth - 1/(ngroups+1)
mypanel - function(x,y,groups,...){
if (missing(groups)||is.null(groups)) {
panel.bwplot(x,y,...)
} else {
panel.superpose(x,y,...)
}
}
mypanel.groups - function(x,y,groups,ngroups,...){
if (missing(groups)||is.null(groups)){
NULL
} else {
function(x, y, ..., group.number) {
panel.bwplot(x+(group.number-0.5*(ngroups+1))/(ngroups+1),y, ...)}
}
}
bwplot(formula(paste(y,x,sep=' ~ ')),
data = data,
groups = 'Plant',
ngroups=ngroups,
pch = |,
box.width = mywidth,
panel = mypanel,
panel.groups = mypanel.groups)
}
myCO2 - CO2
myCO2$year - 2011
summary(myCO2)
mybwplot('Type','uptake',myCO2) # works
mybwplot('Type','uptake',myCO2,'Treatment') # Error using packet 1,
'x' is missing
mybwplot('Type','uptake',myCO2,'year') # Error using packet 1, 'x' is missing
# Deepayan Sarkar suggested code (adapted to myC02)
# bwplot(uptake ~ Plant, data = myCO2, groups = Treatment,
#pch = |, box.width = 1/3,
#panel = panel.superpose,
#panel.groups = function(x, y, ..., group.number) {
#panel.bwplot(x + (group.number-1.5)/3, y, ...)
#})
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Re: [R] How to use PC1 of PCA and dim1 of MCA as a predictor in logistic regression model for data reduction
Dear Mark,
Thank you very much for your advice.
I will try it.
I really appreciate your all kind advice.
Thanks a lot again.
Best regards,
Kohkichi
(11/08/19 22:28), Mark Difford wrote:
On Aug 19, 2011 khosoda wrote:
I used x10.homals4$objscores[, 1] as a predictor for logistic regression
as in the same way as PC1 in PCA.
Am I going the right way?
Hi Kohkichi,
Yes, but maybe explore the sets= argument (set Response as the target
variable and the others as the predictor variables). Then use Dim1 scores.
Also think about fitting a rank-1 restricted model, combined with the sets=
option.
See the vignette to the package and look at
@ARTICLE{MIC98,
author = {Michailides, G. and de Leeuw, J.},
title = {The {G}ifi system of descriptive multivariate analysis},
journal = {Statistical Science},
year = {1998},
volume = {13},
pages = {307--336},
abstract = {}
}
Regards, Mark.
-
Mark Difford (Ph.D.)
Research Associate
Botany Department
Nelson Mandela Metropolitan University
Port Elizabeth, South Africa
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-use-PC1-of-PCA-and-dim1-of-MCA-as-a-predictor-in-logistic-regression-model-for-data-reduction-tp3750251p3755163.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] Build a package - check error
On 20.08.2011 00:41, Eduardo Mendes wrote:
Hi
I have modified the path to
dyn.load(paste(Sys.getenv(R_LIBS_USER),/fortran/src/fortran.so,sep=))
Hmmm, looks like you never took a look into the relevant manual Writing
R Extensions nor in the help files I cited below.
?.First.lib has an example:
## Suppose a package needs to call a DLL named 'fooEXT',
## where 'EXT' is the system-specific extension. Then you should use
.First.lib - function(lib, pkg)
library.dynam(foo, pkg, lib)
You cannot know that the package is in Sys.getenv(R_LIBS_USER)
Note also that the .so file won't be in the path /src/... but in
.../lib/... once you install the package in the recommended way using R
CMD INSTALL.
Really, please do read manual!!!
and the package could installed, loaded and the lines with dyn.load worked.
It does not look like a pretty solution but works on my linux (I am not sure
if it works on my mac or windows).
I am not sure if this is what you meant but as I have no clue what
.First.lib does or NAMESPACES means this is the best I come up with.
So again time to read the manual and the help pages.
Best,
Uwe Ligges
Please correct me if I am wrong.
Many thanks
Ed
On Fri, Aug 19, 2011 at 6:03 PM, Uwe Liggeslig...@statistik.tu-dortmund.de
wrote:
On 19.08.2011 22:53, Eduardo Mendes wrote:
Dear R-users
I am slowly migrating my mex files (MATLAB - Fortran and C) to R. To get
my
own functions available on R section I have decided to learn how to build
a
R package. I choose a simple example with a few Fortran and R functions
(wrapper).
The fortran sources are located at src and the R functions at R (as
recommended). The building process went ok but R CMD check did not. The
error mgs was
Error in dyn.load(fortran.so) :
unable to load shared object
'/home/eduardo/R_packages/**test.Rcheck/fortran.so':
Although I can see that R cannot find the compiled fortran code I do not
know what to do. I believe it is something to do with the following
lines
in the R-wrapper file
if (!is.loaded('calnpr'))
dyn.load(fortran.so)
1. If the package is called calnpr, the shared library is also called that
way.
2. you have to provide the path to the shared library.
See ?.First.lib for how to do it in a package without NAMESPACE (and note
that NAMESPACES are forced for the next R release).
Best,
Uwe Ligges
How to add the path so that once the package is installed the compiled
fortran code can be found?
Many thanks
Cheers
Ed
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Re: [R] hetlp_type text doen't work on Ubuntu
On 20.08.2011 08:25, Amol Jadhav wrote: Hi, I' using Ubuntu 10.10/11.04 and on both of the systems help doesn't work unless I set options(help_type=html). I hate to change context and help_type='html' distracts me. I want text based help inside console. Where should I look to fix this problem? What's wrong with the text help? What does doesn't work mean? What is the output? Error message? R version? . PLEASE do read the posting guide http://www.R-project.org/posting-guide.html Uwe Ligges -- Amol __ 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 add horizontal lines above bar graph to display p-values?
On 19.08.2011 22:27, Sébastien Vigneau wrote: Hi, I would like to draw horizontal lines above a bar graph, in order to display the p-values of a Fisher test. Here is an examplehttp://thejns.org/action/showPopup?citid=citart1id=f3-1060501doi=10.3171%2Fped.2007.106.6.501of the type of display I would like to have. Is there a way to draw the horizontal lines See ?abline, ?lines, ?segments, and write their associated p-values in R? See ?text Uwe Ligges Thanks for you help! Sebastien Vigneau [[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] adding text to a plot created with strat.plot() from package rioja
On 19.08.2011 18:40, Jason Paul Joines wrote: I have a plot created with strat.plot() from package rioja. When the plot is created with scale.percent=FALSE, each x axes is labeled at 0 and its maximum. However, when scale.percent=TRUE, the x axes are not labeled. I need to use scale.percent=TRUE and I need labels for the x axes. I have been able to add labels to the x axes with mtext but it is very tedious to find the correct position. Is there a better way to do this or a better way to find the desired coordinates than trial and error? Yes: You can change the code and suggest your improvements to the package maintainer, for example. The relevant line in that function obviously is: axis(side = 1, at = seq(0, colM[i], by = 10), labels = FALSE) Uwe Ligges Jason === __ 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] questions about metafor package
At 16:21 17/08/2011, Emilie MAILLARD wrote: Hello,  I would like to do a meta-analysis with the package « metafor ». Ideally I would like to use a mixed model because Iâm interested to see the effect of some moderators. But the data set I managed to collect from literature presents two limits.  -        Firstly, for each observation, I have means for a treatment and for a control, but I donât always have corresponding standard deviations (52 of a total of 93 observations donât have standard deviations). Nevertheless I have the sample sizes for all observations so I wonder if it was possible to weight observations by sample size in the package « metafor ». -        Secondly, some observations are probably not independent as I have sometimes several relevant observations for a same design. More precisely, for these cases, the control mean is identical but treatment means varied. Ideally, I would not like to do a weighted average for these non-independent observations because these observations represent levels of a moderator. I know that the package « metafor » is not designed for the analysis of correlated outcomes. What are the dangers of using the package even if observations are not really independent ?  Emilie, I am not sure whether this is the answer to your problem of observations which are not independent but you might also look at the metaSEM package http://courses.nus.edu.sg/course/psycwlm/internet/metaSEM/ I am still trying to understand his paper on this (see link for reference) but he is trying to embed meta-analysis within the structural equation framework and it may be possible to cope with lack of independence in that way. But as I say I am still trying to come to grips with the paper.  Thank you for your help,  Ãmilie. [[alternative HTML version deleted]] Michael Dewey i...@aghmed.fsnet.co.uk http://www.aghmed.fsnet.co.uk/home.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] R training course
Dear Sir/Madam Hi. I am a general paediatrician who works in Middle-East. I will be in UK on September, and October 2011. Would you mind writing for me, please, if you suggest a good and appropriate training course for *new R* users in UK? -- Kind regards Mehrshad Koleini, MD General Paediatrician [[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 extract smoothed plot data?
Dear R, I have this script which from my data points creates smoothed plot: require(graphics) require(stats) A=read.table(some.txt,header =FALSE,sep = , dec = ,,fileEncoding = , encoding = unknown, skip=18,nrows=400) attach(A) plot(A$V1, A$V2,col=white) lines(smooth.spline(A$V1, A$V2,spar=0.4),col=green,lwd=3) I would like to export points coordinates from smoothed plot to file, so that I can then use new data for further analysis. In other words to pipe defined sequence of x coordinates and corresponding y values to file. -- View this message in context: http://r.789695.n4.nabble.com/How-to-extract-smoothed-plot-data-tp3756894p3756894.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] hetlp_type text doen't work on Ubuntu
Don't know the internals behind, but when I did this:
help.search('qplot', rebuild=T)
the text help started working. It would be interesting to know why it
worked that way. Also if it is happens always, it should have been
documented somewhere -- so that new users (who needs help) will not
get disappointed [I was using R without help until i found about
option(help_type='html')].
Anyway, i'm glad that it started working.
--
Amol
On Sat, Aug 20, 2011 at 11:55 AM, Amol Jadhav amolj.1...@gmail.com wrote:
Hi,
I' using Ubuntu 10.10/11.04 and on both of the systems help doesn't
work unless I set options(help_type=html). I hate to change context
and help_type='html' distracts me. I want text based help inside
console. Where should I look to fix this problem?
--
Amol
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to extract smoothed plot data?
On 20.08.2011 12:57, derek wrote: Dear R, I have this script which from my data points creates smoothed plot: require(graphics) require(stats) A=read.table(some.txt,header =FALSE,sep = , dec = ,,fileEncoding = , encoding = unknown, skip=18,nrows=400) attach(A) plot(A$V1, A$V2,col=white) lines(smooth.spline(A$V1, A$V2,spar=0.4),col=green,lwd=3) sspline - smooth.spline(A$V1, A$V2, spar=0.4) write.table(data.frame(sspline[c(x, y)]), file = ...) Uwe Ligges I would like to export points coordinates from smoothed plot to file, so that I can then use new data for further analysis. In other words to pipe defined sequence of x coordinates and corresponding y values to file. -- View this message in context: http://r.789695.n4.nabble.com/How-to-extract-smoothed-plot-data-tp3756894p3756894.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.
[R] Pattern names matching
Dear R magic guys.. I have two tables (actually will be dataframes), both with names to be matched. The names on the first dataframe are from a study with antenatal visits on some health centers here. It happens that we need the delivery info. And half and some thing else of the women decided to delivery some where else our health units. We managed to get the names from some other places but now we have to match our 4000 original names with over 2 other names. To make thing more bitter some names have badly written. So I need some algorithm like Levenstein or sondex or phonix or something better already on R. Can you help me? Orvalho [[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] Build a package - check error
Hi there
Many thanks.
Just to be clear I did read the manual (section 5 and all) and have even
looked at the package expm which has a src dir with codes in fortran and C
to understand what it is going on.
I have also read the help for .First.lib.
Cheers
Ed
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Saturday, August 20, 2011 10:38 AM
To: Eduardo Mendes
Cc: r-help@r-project.org
Subject: Re: [R] Build a package - check error
On 20.08.2011 00:41, Eduardo Mendes wrote:
Hi
I have modified the path to
dyn.load(paste(Sys.getenv(R_LIBS_USER),/fortran/src/fortran.so,sep
=))
Hmmm, looks like you never took a look into the relevant manual Writing R
Extensions nor in the help files I cited below.
?.First.lib has an example:
## Suppose a package needs to call a DLL named 'fooEXT', ## where 'EXT' is
the system-specific extension. Then you should use .First.lib -
function(lib, pkg)
library.dynam(foo, pkg, lib)
You cannot know that the package is in Sys.getenv(R_LIBS_USER)
Note also that the .so file won't be in the path /src/... but in
.../lib/... once you install the package in the recommended way using R CMD
INSTALL.
Really, please do read manual!!!
and the package could installed, loaded and the lines with dyn.load
worked.
It does not look like a pretty solution but works on my linux (I am
not sure if it works on my mac or windows).
I am not sure if this is what you meant but as I have no clue what
.First.lib does or NAMESPACES means this is the best I come up with.
So again time to read the manual and the help pages.
Best,
Uwe Ligges
Please correct me if I am wrong.
Many thanks
Ed
On Fri, Aug 19, 2011 at 6:03 PM, Uwe
Liggeslig...@statistik.tu-dortmund.de
wrote:
On 19.08.2011 22:53, Eduardo Mendes wrote:
Dear R-users
I am slowly migrating my mex files (MATLAB - Fortran and C) to R.
To get my own functions available on R section I have decided to
learn how to build
a
R package. I choose a simple example with a few Fortran and R
functions (wrapper).
The fortran sources are located at src and the R functions at R (as
recommended). The building process went ok but R CMD check did not.
The error mgs was
Error in dyn.load(fortran.so) :
unable to load shared object
'/home/eduardo/R_packages/**test.Rcheck/fortran.so':
Although I can see that R cannot find the compiled fortran code I do not
know what to do. I believe it is something to do with the following
lines
in the R-wrapper file
if (!is.loaded('calnpr'))
dyn.load(fortran.so)
1. If the package is called calnpr, the shared library is also called
that way.
2. you have to provide the path to the shared library.
See ?.First.lib for how to do it in a package without NAMESPACE (and
note that NAMESPACES are forced for the next R release).
Best,
Uwe Ligges
How to add the path so that once the package is installed the
compiled
fortran code can be found?
Many thanks
Cheers
Ed
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[R] RMysql on Windows 7
Hello all, good day I have installed R 1.13.0 on windows 7. then configured R + latex + sweave to use eclipse indigo.Then would like to be able to work with R + mysql 5.5(full package + connectors) which I have already installed.My problem is to build RMYSQL binary package, using Rtools.Should I remove existing R and install RTools?. thank you for your help Udd [[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] Pattern names matching
See the stringMatch function in the MiscPsycho package for an implementation of Levenshtein From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of Orvalho Augusto [orvaq...@gmail.com] Sent: Saturday, August 20, 2011 11:08 AM To: r-help@r-project.org Subject: [R] Pattern names matching Dear R magic guys.. I have two tables (actually will be dataframes), both with names to be matched. The names on the first dataframe are from a study with antenatal visits on some health centers here. It happens that we need the delivery info. And half and some thing else of the women decided to delivery some where else our health units. We managed to get the names from some other places but now we have to match our 4000 original names with over 2 other names. To make thing more bitter some names have badly written. So I need some algorithm like Levenstein or sondex or phonix or something better already on R. Can you help me? Orvalho [[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] RMysql on Windows 7
On 20.08.2011 17:22, Twaha Mlwilo wrote: Hello all, good day I have installed R 1.13.0 on windows 7. then configured R + latex + sweave to use eclipse indigo.Then would like to be able to work with R + mysql 5.5(full package + connectors) which I have already installed.My problem is to build RMYSQL binary package, using Rtools.Should I remove existing R and install RTools?. You need the Rtools *in addition* to R. See the manual R Installation and Administration for details. Uwe ligges thank you for your help Udd [[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] AFT model time-dependent with weibull distribution
On Sat, Aug 20, 2011 at 4:19 AM, JPF xpfen...@gmail.com wrote: JPF wrote: weibullaft-aftreg(Surv(sta,time,S) ~ TDC1 + TIC1, dist=weibull, data.frame=Data) ## aftreg gives an error when I add an ID argument... That should be used for controlling for time-varying variables. Error in aftreg.fit(X, Y, dist, strats, offset, init, shape, id, control, : *Overlapping intervals for id 2 * From help(aftreg): id If there are more than one spell per individual, it is essential to keep spells together by the id argument. This allows for time-varying covariates. This is solved. It gave the overlapping intervals error, but now it is solved: /id=ID/, just as before. Good. Do you still need answers to your other questions? G. -- View this message in context: http://r.789695.n4.nabble.com/AFT-model-time-dependent-with-weibull-distribution-tp3755079p3756502.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. -- Göran Broström __ 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 nls fitting
Dear all, I'm trying to fit the following function slope_pp3_mrna = ( (k3 * v3_K_d *p1^v3_h) / ( (v3_Kd^v3_h) + p2^v3_h ) ) * ( 1/(1 + (p2/v4_Kd)^v4_h) ) - pp3_mrna to this experimental data in the datafraeme Data_pp3_mrna (see it at the end of this e-mail) I'm using the nls function in the following code. IN the last step of the fit fm_pp3_mrna_4, when I add to the funziont the paramter v4_Kd something goes wrong, and I reeive this message Error in numericDeriv(form[[3L]], names(ind), env) : Missing value or an infinity produced when evaluating the model What could be the error? I tried with differnet intial values of v4_Kd, but I did not fix the problem. Here below the code. Thanks in advance, Paola. Data_pp3_mrna - data.frame( p1 = protein_1, p2 = protein_2, pp3_mrna = protein_3_mrna, slope_pp3_mrna = stinemanSlopes(times, protein_3_mrna) ) one_par - nls(slope_pp3_mrna ~ ( (k3 * p1) / ( (1) + p1 ) ) * ( 1/(1 + (p2)) ) - pp3_mrna, data=Data_pp3_mrna, start=list( k3=0.1 )) summary(one_par) fm_pp3_mrna_1 - nls(slope_pp3_mrna ~ ((k3 * v3_Kd *p1) / ( (v3_Kd) + p1 ) ) * ( 1/(1 + (p2)) ) - pp3_mrna, data=Data_pp3_mrna, start=list(k3=24, v3_Kd=1 )) summary(fm_pp3_mrna_1) fm_pp3_mrna_2 - nls(slope_pp3_mrna ~ ((k3 * v3_Kd *p1^v3_h) / ( (v3_Kd)^v3_h + p1^v3_h ) ) * ( 1/(1 + (p2)) ) - pp3_mrna, control = list(maxiter = 500), data=Data_pp3_mrna, start=list(k3=69, v3_Kd=0.3238, v3_h=1 )) summary(fm_pp3_mrna_2) fm_pp3_mrna_3 - nls(slope_pp3_mrna ~ ((k3 * v3_Kd *p1^v3_h) / ( (v3_Kd)^v3_h + p1^v3_h ) ) * ( 1/(1 + (p2)^v4_h) ) - pp3_mrna, control = list(maxiter = 500), data=Data_pp3_mrna, start=list(k3=37.451, v3_Kd=0.59, v3_h=2.013, v4_h=0.01 )) summary(fm_pp3_mrna_3) fm_pp3_mrna_4 - nls(slope_pp3_mrna ~ ((k3 * v3_Kd *p1^v3_h) / ( (v3_Kd)^v3_h + p1^v3_h ) ) * ( 1/(1 + (p2/v4_Kd)^v4_h) ) - pp3_mrna, control = list(maxiter = 500), data=Data_pp3_mrna, start=list(k3=56.2823, v3_Kd=0.3366, v3_h=1.8040, v4_Kd=0.03, v4_h=0.7693 )) Here the data. Data_pp3_mrna p1 p2 pp3_mrna slope_pp3_mrna 1 1.006 0.921 0.041 8.63741887 2 2.235 2.047 2.9069031 2.82619343 3 3.744 3.937 4.052 0.84354113 4 4.222 9.340 4.3237353 0.47577213 5 9.022 14.609 4.531-0.03940131 6 11.326 22.765 3.510-2.0420 7 6.899 17.852 2.489-1.86822481 8 10.709 27.777 1.6222048-1.55625973 9 14.084 27.785 0.911-0.48800514 10 14.922 23.613 0.826-0.1700 11 14.340 18.422 0.741-0.22560156 12 13.066 24.085 0.599-0.2840 13 17.553 18.594 0.457-0.13847372 14 14.803 16.831 0.4550965 0.03588624 15 11.945 14.495 0.493 0.09536674 16 11.427 12.458 0.5505361 0.15549062 17 11.556 9.082 0.649 0.49638596 18 20.107 9.987 1.2486525 1.36871828 19 15.999 10.305 2.059 1.87868197 20 16.094 5.793 3.2285000 2.3390 21 11.752 6.944 4.398 0.80395869 22 15.841 5.575 4.651 0.5060 23 12.601 5.221 4.904 0.80270128 24 13.598 2.872 5.819 1.8300 25 13.879 2.883 6.734-0.03571884 26 16.270 2.213 6.4135000-0.6410 27 17.176 3.381 6.093-0.33913760 28 12.332 2.781 6.032-0.1220 29 12.373 3.073 5.971-0.41224459 30 14.781 2.948 5.489-0.9640 31 17.578 3.953 5.007-0.68258311 32 18.865 2.279 4.7568901-0.39425557 33 14.735 2.806 4.606 0.13264481 34 16.160 1.676 4.987 0.7620 35 13.416 2.478 5.368 0.63914248 36 13.864 1.394 5.6374239 0.45419096 37 16.219 2.299 5.827-0.12540453 38 13.249 1.457 5.2505000-1.1530 39 14.445 2.325 4.674-0.09880849 40 13.210 2.230 5.5576966 2.17200142 41 12.358 2.116 8.94711.38521228 -- *Paola Lecca, PhD* *The Microsoft Research - University of Trento* *Centre for Computational and Systems Biology* *Piazza Manci 17 38123 Povo/Trento, Italy* *Phome: +39 0461282843* *Fax: +39 0461282814* [[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] Groups and bwplot
You may want to consult a recent post by Felix
(https://stat.ethz.ch/pipermail/r-help/2011-August/286707.html) on how
to pass group parameter.
Weidong Gu
On Sat, Aug 20, 2011 at 6:59 AM, Sébastien Bihorel pomc...@free.fr wrote:
Dear R-users,
A while ago, Deepayan Sarkar suggested some code that uses the group
argument in bwplot to create some 'side-by-side' boxplots
(https://stat.ethz.ch/pipermail/r-help/2010-February/230065.html). The
example he gave was relatively specific and I wanted to generalize his
approach into a function. Unfortunately, I seem to have some issues
passing the correct arguments to the panel function, and would greatly
appreciate any suggestions to solve these issues:
require(lattice)
mybwplot - function(x,y,data,groups){
if (missing(groups)||is.null(groups)) {
groups - NULL
ngroups - 1
} else {
data[[groups]] - as.factor(data[[groups]])
ngroups - nlevels(data[[groups]])
}
mywidth - 1/(ngroups+1)
mypanel - function(x,y,groups,...){
if (missing(groups)||is.null(groups)) {
panel.bwplot(x,y,...)
} else {
panel.superpose(x,y,...)
}
}
mypanel.groups - function(x,y,groups,ngroups,...){
if (missing(groups)||is.null(groups)){
NULL
} else {
function(x, y, ..., group.number) {
panel.bwplot(x+(group.number-0.5*(ngroups+1))/(ngroups+1),y, ...)}
}
}
bwplot(formula(paste(y,x,sep=' ~ ')),
data = data,
groups = 'Plant',
ngroups=ngroups,
pch = |,
box.width = mywidth,
panel = mypanel,
panel.groups = mypanel.groups)
}
myCO2 - CO2
myCO2$year - 2011
summary(myCO2)
mybwplot('Type','uptake',myCO2) # works
mybwplot('Type','uptake',myCO2,'Treatment') # Error using packet 1,
'x' is missing
mybwplot('Type','uptake',myCO2,'year') # Error using packet 1, 'x' is missing
# Deepayan Sarkar suggested code (adapted to myC02)
# bwplot(uptake ~ Plant, data = myCO2, groups = Treatment,
# pch = |, box.width = 1/3,
# panel = panel.superpose,
# panel.groups = function(x, y, ..., group.number) {
# panel.bwplot(x + (group.number-1.5)/3, y, ...)
# })
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Re: [R] Extract p value from coxme object
Cleaning up some old messages, I found this one: The p-values are not stored in the coxme object. They are calculated on the fly when you ask to rpint the object. See coxme:::print.coxme for the code. Uwe Ligges On 02.08.2011 17:26, Catarina Miranda wrote: Dear R experts; I am trying to extract the p values from a coxme object (package coxme). I can see the value in the model output, but I wanted to have the result with a higher number of decimal places. I have searched the mailing list and followed equivalent suggestions for nlme/lme objects, but I wasn't successful. Thanks; Catarina [[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] Problem Installing/Uninstalling Rattle
Cleaning up some old messages, I found this one On 02.08.2011 14:51, adarwish wrote: Rattle won't install properly on my Windows 7 64 bit laptop. Here is what I've tried: I've followed the instructions here: http://rattle.togaware.com/rattle-install-mswindows.html I had R installed already. I downloaded the GTK+ packages, unzipped the 32 bit one into c:\gtkwin32. See the repository's ReadMe in http://cran.r-project.org/bin/windows/contrib/2.13/ReadMe that tells you: - Package RGtk2 requires an an installation of Gtk+ aka Gtk2 = 2.20. For 32-bit R, version 2.20 or later from http://www.gtk.org/download/win32.php, e.g. http://ftp.gnome.org/pub/gnome/binaries/win32/gtk+/2.22/gtk+-bundle_2.22.0-20101016_win32.zip For 64-bit R, version 2.20 or later from http://www.gtk.org/download/win64.php, e.g. http://ftp.gnome.org/pub/gnome/binaries/win64/gtk+/2.22/gtk+-bundle_2.22.0-20101016_win64.zip In each case, unpack the zip file in a suitable empty directory and put the 'bin' directory in your path. NB: the 32-bit and 64-bit distributions contain DLLs of the same names, and so you must ensure that you have the 32-bit version in your path when running 32-bit R and the 64-bit version when running 64-bit R - and the error messages you get with the wrong version are confusing. I put c:\gtkwin32\bin in the system variables PATH. I launched R, installed the rattle package, called the rattle library, called rattle(). It told me RGtk2 could not be found and asked to install it. I let it download it to install, but still nothing. Restarting/resintalling R has not helped. And when I try remove.packages(rattle) I get the error: Removing package(s) from ‘C:/Users/darwish/Documents/R/win-library/2.13’ (as ‘lib’ is unspecified) Error in match(x, table, nomatch = 0L) : 'match' requires vector arguments The package name has to be quotes (i.e. specified as character rather than symbol). Uwe Ligges I've restarted R before trying anything multiple times. From what I understand, I need to clean everything off and start anew. How do I remove rattle so I can start fresh? What did I do wrong in my steps? Thanks in advance. -- View this message in context: http://r.789695.n4.nabble.com/Problem-Installing-Uninstalling-Rattle-tp3712221p3712221.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] chisq.test(): standardized (adjusted) Pearson residuals
On Aug 20, 2011, at 3:43 AM, peter dalgaard wrote: On Aug 19, 2011, at 20:40 , David Winsemius wrote: On Aug 19, 2011, at 1:28 PM, Stephen Davies wrote: I'm using chisq.test() on a matrix of categorical data, and I see that the residuals attribute of the returned object will give me the Pearson residuals. Actually they are not an attribute in the R sense, but rather a list value. That's cool. However, what I'd really like is the standardized (adjusted) Pearson residuals, which have a N(0,1) distribution. Is there a way to do that in R (other than by me programming it myself?) ?scale chisq.test(...)$stdres, more likely. Agree that does have a much greater chance of keeping the questioner in the mainstream of statistics terminology and is most likely what he was looking for, but do not think the result will in general have an N(1,0) distribution. I believe the correct statement is that standardized residuals would (in the statistical asymptotic sense) have an N(1,0) distribution if and when the null hypothesis of marginal homogeneity were true, but should not be N(1,0) in any case when an alternate hypothesis holds. My error was in taking the questioner's request at face value. -- David Winsemius, MD 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] Pattern names matching
On Aug 20, 2011, at 11:25 AM, Doran, Harold wrote: See the stringMatch function in the MiscPsycho package for an implementation of Levenshtein The agrep function in base R also returns a Levenshtein distance. -- David. From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of Orvalho Augusto [orvaq...@gmail.com] Sent: Saturday, August 20, 2011 11:08 AM To: r-help@r-project.org Subject: [R] Pattern names matching Dear R magic guys.. I have two tables (actually will be dataframes), both with names to be matched. The names on the first dataframe are from a study with antenatal visits on some health centers here. It happens that we need the delivery info. And half and some thing else of the women decided to delivery some where else our health units. We managed to get the names from some other places but now we have to match our 4000 original names with over 2 other names. To make thing more bitter some names have badly written. So I need some algorithm like Levenstein or sondex or phonix or something better already on R. Can you help me? Orvalho David Winsemius, MD 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] reshape a matrix
Hi all,
I have a data.frame like following
A-c('d0','d0','d1','d1','d2','d2')
B-rep(c('control','sample'),3)
C-c(rep(10,2),200,300,400,500)
dataframe-data.frame(A,B,C)
I want to reshape the matrix, so the matrix with 'd0', 'd1' and 'd2' in rows
and 'control' and 'sample' in columns. Is there a function for doing this
easily?
Thank you in advance.
Wendy
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Re: [R] chisq.test(): standardized (adjusted) Pearson residuals
I'm using chisq.test() on a matrix of categorical data, and I see that the residuals attribute of the returned object will give me the Pearson residuals. Actually they are not an attribute in the R sense, but rather a list value. Oh. I was just going by: attributes(my.chisq.test) $names [1] statistic parameter p.value methoddata.name observed [7] expected residuals $class [1] htest which I interpreted as this object has 8 attributes, called 'statistic', 'parameter', ..., 'residuals'. Is that not the right terminology? That's cool. However, what I'd really like is the standardized (adjusted) Pearson residuals, which have a N(0,1) distribution. Is there a way to do that in R (other than by me programming it myself?) ?scale chisq.test(...)$stdres, more likely. scale is not what I want. As for $stdres, that would be wonderful, but as you can see from the above list of attributes, it's not one of the 8 returned. What am I missing? - Stephen __ 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] reshape a matrix
On 20.08.2011 17:04, Wendy wrote:
Hi all,
I have a data.frame like following
A-c('d0','d0','d1','d1','d2','d2')
B-rep(c('control','sample'),3)
C-c(rep(10,2),200,300,400,500)
dataframe-data.frame(A,B,C)
I want to reshape the matrix, so the matrix with 'd0', 'd1' and 'd2' in rows
and 'control' and 'sample' in columns. Is there a function for doing this
easily?
See ?reshape
reshape(data=dataframe, idvar=A, timevar=B, direction=wide)
Uwe Ligges
Thank you in advance.
Wendy
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] chisq.test(): standardized (adjusted) Pearson residuals
On Aug 20, 2011, at 12:04 PM, Stephen Davies wrote: I'm using chisq.test() on a matrix of categorical data, and I see that the residuals attribute of the returned object will give me the Pearson residuals. Actually they are not an attribute in the R sense, but rather a list value. Oh. I was just going by: attributes(my.chisq.test) $names [1] statistic parameter p.value methoddata.name observed [7] expected residuals $class [1] htest which I interpreted as this object has 8 attributes, called 'statistic', 'parameter', ..., 'residuals'. Is that not the right terminology? The names attribute let's you know what characters to use if you want to access values in a list. Unless you are doing programming attributes is not a particular useful function. It is much more common to access the names attribute with the `names` function: names(Xsq) [1] statistic parameter p.value methoddata.name observed expected [8] residuals stdres So stdres is not an attribute but rather one value in a particular attribute called names. You would get (much) more information by using str on the htest object as below: str(Xsq) List of 9 $ statistic: Named num 30.1 ..- attr(*, names)= chr X-squared $ parameter: Named num 2 ..- attr(*, names)= chr df $ p.value : num 2.95e-07 $ method : chr Pearson's Chi-squared test $ data.name: chr M $ observed : table [1:2, 1:3] 762 484 327 239 468 477 ..- attr(*, dimnames)=List of 2 .. ..$ gender: chr [1:2] M F .. ..$ party : chr [1:3] Democrat Independent Republican $ expected : num [1:2, 1:3] 704 542 320 246 534 ... ..- attr(*, dimnames)=List of 2 .. ..$ gender: chr [1:2] M F .. ..$ party : chr [1:3] Democrat Independent Republican $ residuals: table [1:2, 1:3] 2.199 -2.505 0.411 -0.469 -2.843 ... ..- attr(*, dimnames)=List of 2 .. ..$ gender: chr [1:2] M F .. ..$ party : chr [1:3] Democrat Independent Republican $ stdres : table [1:2, 1:3] 4.502 -4.502 0.699 -0.699 -5.316 ... ..- attr(*, dimnames)=List of 2 .. ..$ gender: chr [1:2] M F .. ..$ party : chr [1:3] Democrat Independent Republican - attr(*, class)= chr htest Now you can see that the values in the stdres object are really a list element and are in a table with particular row and column names. You get that object one of two ways. you ca use the $ method as Dalgaard suggested or you can use [[ with the name of the object: Xsq[[stdres]] That's cool. However, what I'd really like is the standardized (adjusted) Pearson residuals, which have a N(0,1) distribution. Is there a way to do that in R (other than by me programming it myself?) ?scale chisq.test(...)$stdres, more likely. scale is not what I want. As for $stdres, that would be wonderful, but as you can see from the above list of attributes, it's not one of the 8 returned. What am I missing? David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] t() prior to data rotation
Well, did you try it with a simple test case to see what comes out? Do you have a sample mydata.txt file so you know how it's ordered in the first place? Carl quote From: christopher stratton cfstratton_at_gmail.com Date: Fri, 19 Aug 2011 23:10:07 -0400 Dear All, I have come upon an R-mode PCA protocol that uses the following arguments, where mydata.txt is an nxm matrix of n objects and m variables: a - read.table(mydata.txt) b - t(a) c - prcomp(b) c$rotation The user then plots the coordinates given by c$rotation (PC1 and PC2) as the scores of their PCA plot. This doesn't make sense to me as the user transposed the matrix prior to rotating the data, so they have solved for the eigenvectors of the objects and by plotting the values of c$rotation the user is in effect plotting the loading matrix and not the scores. If anything, this looks like a Q-mode PCA where the rotation matrix should be multiplied by the original data matrix to give scores for the variables. Am I missing something or does this procedure look incorrect? Thank you for your time, Chris /quote -- - Sent from my Cray XK6 __ 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] hetlp_type text doen't work on Ubuntu
Amol Jadhav amolj.1306 at gmail.com writes:
Don't know the internals behind, but when I did this:
help.search('qplot', rebuild=T)
the text help started working. It would be interesting to know why it
worked that way. Also if it is happens always, it should have been
documented somewhere -- so that new users (who needs help) will not
get disappointed [I was using R without help until i found about
option(help_type='html')].
Anyway, i'm glad that it started working.
I'm glad it started working too, but it seems extremely likely
that your problem was specific to your system. R's help system does
normally work fine out of the box -- there are thousands of Ubuntu
users, and none that I'm aware of has ever reported this particular
problem before.
Ben Bolker
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Re: [R] chisq.test(): standardized (adjusted) Pearson residuals
On Aug 20, 2011, at 18:04 , Stephen Davies wrote: As for $stdres, that would be wonderful, but as you can see from the above list of attributes, it's not one of the 8 returned. What am I missing? An upgrade, most likely. -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com Døden skal tape! --- Nordahl Grieg __ 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] reshape a matrix
On Aug 20, 2011, at 12:32 PM, Uwe Ligges wrote:
On 20.08.2011 17:04, Wendy wrote:
Hi all,
I have a data.frame like following
A-c('d0','d0','d1','d1','d2','d2')
B-rep(c('control','sample'),3)
C-c(rep(10,2),200,300,400,500)
dataframe-data.frame(A,B,C)
I want to reshape the matrix, so the matrix with 'd0', 'd1' and
'd2' in rows
and 'control' and 'sample' in columns. Is there a function for
doing this
easily?
See ?reshape
reshape(data=dataframe, idvar=A, timevar=B, direction=wide)
Uwe Ligges
Many people have experienced problems understanding the mechanics of
the base function 'reshape'. If you do not and if do continue to use
it, you would be doing the world a great service by writing a tutorial
manual with a bunch of worked examples. I have never found a tutorial
that clarified how I should use it in the variety of situations where
I have needed it.
So Hadley wrote an alternate facility ... the reshape package that
does not have a reshape function in it but rather two functions 'melt'
and 'cast'. Your data is all ready molten, i.e. it is in the long
format (in the terminology of the base reshape function) with
identifier values in each row and a single column of values.
library(reshape)
cast(dataframe,A~B)
Using C as value column. Use the value argument to cast to override
this choice
A control sample
1 d0 1e+05 1e+05
2 d1 2e+02 3e+02
3 d2 4e+02 5e+02
Basically the cast formula keeps the LHS variables in the rows and hte
RHD variables get arranges in columns. (For reasons that are unclear
to me the dataframe argument was placed first when using positional
argument passing, unlike most other formula methods in R.)
--
David Winsemius, MD
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] chisq.test(): standardized (adjusted) Pearson residuals
On Aug 20, 2011, at 12:57 PM, peter dalgaard wrote: On Aug 20, 2011, at 18:04 , Stephen Davies wrote: As for $stdres, that would be wonderful, but as you can see from the above list of attributes, it's not one of the 8 returned. What am I missing? An upgrade, most likely. Whoosh. Sometimes I am simply clueless. I didn't notice that 'stdres' was missing from the names in Stephen's output. Laura Thompson has a very nice R/S accompaniment to Agresti's Categorical Data Analysis text and she shows how to adjust the Pearson residuals to make them standardized. What follows is directly from pages 37-38 of her work: #--# resid.pear - residuals(fit.glm, type = pearson) Note that the sum of the squared Pearson residuals equals the Pearson chi-squared statistic: sum(resid.pear^2) [1] 69.11429 To get the standardized residuals, just modify resid.pear according to the formula on p. 81 of Agresti. ni-rowSums(table.3.2.array) # row sums nj-colSums(table.3.2.array) # column sums n-sum(table.3.2.array) # total sample size resid.pear.mat-matrix(resid.pear, nc=3, byrow=T, dimnames=list(c(HS,HS or JH, Bachelor or Grad),c(Fund, Mod, Lib))) n*resid.pear.mat/sqrt(outer(n-ni,n-nj,*) ) # standardized Pearson residuals FundMod Lib HS 4.534062 -2.5520482 -1.941537 HS or JH 2.552988 1.2859745 -3.994669 Bachelor or Grad -6.806638 0.7007539 6.250329 #--# You can also look at the code (once you upgrade) and the method in R is quite similar, although the R codes calcualtes the stdres values separately rather than adjusting the Pearson residuals -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School David Winsemius, MD 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] reshape a matrix
On 20.08.2011 19:04, David Winsemius wrote:
On Aug 20, 2011, at 12:32 PM, Uwe Ligges wrote:
On 20.08.2011 17:04, Wendy wrote:
Hi all,
I have a data.frame like following
A-c('d0','d0','d1','d1','d2','d2')
B-rep(c('control','sample'),3)
C-c(rep(10,2),200,300,400,500)
dataframe-data.frame(A,B,C)
I want to reshape the matrix, so the matrix with 'd0', 'd1' and 'd2'
in rows
and 'control' and 'sample' in columns. Is there a function for doing
this
easily?
See ?reshape
reshape(data=dataframe, idvar=A, timevar=B, direction=wide)
Uwe Ligges
Many people have experienced problems understanding the mechanics of the
base function 'reshape'. If you do not and if do continue to use it, you
would be doing the world a great service by writing a tutorial manual
with a bunch of worked examples. I have never found a tutorial that
clarified how I should use it in the variety of situations where I have
needed it.
David,
I think there are some good examples on the help page. What is missing?
What is not clearly explained? If a longer tutorial is needed, that may
be an article for the R Help Desk in The R Journal. Anybody volunteering?
Best,
Uwe
So Hadley wrote an alternate facility ... the reshape package that does
not have a reshape function in it but rather two functions 'melt' and
'cast'.
Your data is all ready molten, i.e. it is in the long format
(in the terminology of the base reshape function) with identifier values
in each row and a single column of values.
library(reshape)
cast(dataframe,A~B)
Using C as value column. Use the value argument to cast to override this
choice
A control sample
1 d0 1e+05 1e+05
2 d1 2e+02 3e+02
3 d2 4e+02 5e+02
Basically the cast formula keeps the LHS variables in the rows and hte
RHD variables get arranges in columns. (For reasons that are unclear to
me the dataframe argument was placed first when using positional
argument passing, unlike most other formula methods in R.)
__
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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] Groups and bwplot
Thanks for your input and this link. I realize that there was a typo
in my example code that impacted the group argument... That's king of
stupid.
However, even with the implementation of Felix's syntax, the Error
using packet 1, 'x' is missing error message is still displayed, even
if the call appears correct. So I believe that group argument in not
the issue but rather my panel functions. (Plus, Felix's notation
creates side-issues such as the calculation of ngroups, which I have
hard-coded in the following modified example).
require(lattice)
mybwplot - function(x,data,groups){
if (missing(groups)) {
ngroups - 1
} else {
ngroups - 2
}
mywidth - 1/(ngroups+1)
mypanel - function(x,y,groups,...){
if (missing(groups)||is.null(groups)) {
panel.bwplot(x,y,...)
} else {
panel.superpose(x,y,...)
}
}
mypanel.groups - function(x,y,groups,ngroups,group.number,...){
if (missing(groups)||is.null(groups)){
NULL
} else {
panel.bwplot(x+(group.number-0.5*(ngroups+1))/(ngroups+1),y, ...)
}
}
ccall - quote(bwplot(x,
data = data,
ngroups=ngroups,
pch = |,
box.width = mywidth,
panel = mypanel,
panel.groups = mypanel.groups))
ccall$groups - substitute(groups)
str(ccall)
eval(ccall)
}
myCO2 - CO2
myCO2$year - 2011
mybwplot(uptake~Type,myCO2) # works
mybwplot(uptake~Type,myCO2,groups=Treatment) # Error using packet 1,
'x' is missing
#mybwplot(uptake~Type,myCO2,groups=year) # Error using packet 1, 'x' is missing
# Deepayan Sarkar suggested code (adapted to myC02)
# bwplot(uptake ~ Type, data = myCO2, groups = Treatment,
#pch = |, box.width = 1/3,
#panel = panel.superpose,
#panel.groups = function(x, y, ..., group.number) {
#panel.bwplot(x + (group.number-1.5)/3, y, ...)
#})
On Sat, Aug 20, 2011 at 11:38 AM, Weidong Gu anopheles...@gmail.com wrote:
You may want to consult a recent post by Felix
(https://stat.ethz.ch/pipermail/r-help/2011-August/286707.html) on how
to pass group parameter.
Weidong Gu
On Sat, Aug 20, 2011 at 6:59 AM, Sébastien Bihorel pomc...@free.fr wrote:
Dear R-users,
A while ago, Deepayan Sarkar suggested some code that uses the group
argument in bwplot to create some 'side-by-side' boxplots
(https://stat.ethz.ch/pipermail/r-help/2010-February/230065.html). The
example he gave was relatively specific and I wanted to generalize his
approach into a function. Unfortunately, I seem to have some issues
passing the correct arguments to the panel function, and would greatly
appreciate any suggestions to solve these issues:
require(lattice)
mybwplot - function(x,y,data,groups){
if (missing(groups)||is.null(groups)) {
groups - NULL
ngroups - 1
} else {
data[[groups]] - as.factor(data[[groups]])
ngroups - nlevels(data[[groups]])
}
mywidth - 1/(ngroups+1)
mypanel - function(x,y,groups,...){
if (missing(groups)||is.null(groups)) {
panel.bwplot(x,y,...)
} else {
panel.superpose(x,y,...)
}
}
mypanel.groups - function(x,y,groups,ngroups,...){
if (missing(groups)||is.null(groups)){
NULL
} else {
function(x, y, ..., group.number) {
panel.bwplot(x+(group.number-0.5*(ngroups+1))/(ngroups+1),y, ...)}
}
}
bwplot(formula(paste(y,x,sep=' ~ ')),
data = data,
groups = 'Plant',
ngroups=ngroups,
pch = |,
box.width = mywidth,
panel = mypanel,
panel.groups = mypanel.groups)
}
myCO2 - CO2
myCO2$year - 2011
summary(myCO2)
mybwplot('Type','uptake',myCO2) # works
mybwplot('Type','uptake',myCO2,'Treatment') # Error using packet 1,
'x' is missing
mybwplot('Type','uptake',myCO2,'year') # Error using packet 1, 'x' is missing
# Deepayan Sarkar suggested code (adapted to myC02)
# bwplot(uptake ~ Plant, data = myCO2, groups = Treatment,
# pch = |, box.width = 1/3,
# panel = panel.superpose,
# panel.groups = function(x, y, ..., group.number) {
# panel.bwplot(x + (group.number-1.5)/3, y, ...)
# })
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] reshape a matrix
On Aug 20, 2011, at 1:21 PM, Uwe Ligges wrote:
On 20.08.2011 19:04, David Winsemius wrote:
On Aug 20, 2011, at 12:32 PM, Uwe Ligges wrote:
On 20.08.2011 17:04, Wendy wrote:
Hi all,
I have a data.frame like following
A-c('d0','d0','d1','d1','d2','d2')
B-rep(c('control','sample'),3)
C-c(rep(10,2),200,300,400,500)
dataframe-data.frame(A,B,C)
I want to reshape the matrix, so the matrix with 'd0', 'd1' and
'd2'
in rows
and 'control' and 'sample' in columns. Is there a function for
doing
this
easily?
See ?reshape
reshape(data=dataframe, idvar=A, timevar=B, direction=wide)
Uwe Ligges
Many people have experienced problems understanding the mechanics
of the
base function 'reshape'. If you do not and if do continue to use
it, you
would be doing the world a great service by writing a tutorial manual
with a bunch of worked examples. I have never found a tutorial that
clarified how I should use it in the variety of situations where I
have
needed it.
David,
I think there are some good examples on the help page. What is
missing? What is not clearly explained?
The stumbling blocks I have encountered are in trying to figure out
which of the multiple arguments are needed a) in going from wide to
long and b) in going from long to wide, c) and what are the reasons
for the various error messages I provoke . I am almost never able to
do it correctly on the first try and rarely able to do it even on the
fourth try. I bought Spector's book in hopes of understanding it
better, but his efforts did not take root in my brain.
In the instance above, how would I have know how to apply the help
page description of timevar (below) to this problem?
timevar
the variable in long format that differentiates multiple records from
the same group or individual.
To my reading that does not distinguish the purpose of 'timevar' from
the purpose of 'idvar' and then reading the 'idvar' definition just
below it does not help at all.
--
David.
If a longer tutorial is needed, that may be an article for the R
Help Desk in The R Journal. Anybody volunteering?
Best,
Uwe
So Hadley wrote an alternate facility ... the reshape package that
does
not have a reshape function in it but rather two functions 'melt' and
'cast'.
Your data is all ready molten, i.e. it is in the long format
(in the terminology of the base reshape function) with identifier
values
in each row and a single column of values.
library(reshape)
cast(dataframe,A~B)
Using C as value column. Use the value argument to cast to override
this
choice
A control sample
1 d0 1e+05 1e+05
2 d1 2e+02 3e+02
3 d2 4e+02 5e+02
Basically the cast formula keeps the LHS variables in the rows and
hte
RHD variables get arranges in columns. (For reasons that are
unclear to
me the dataframe argument was placed first when using positional
argument passing, unlike most other formula methods in R.)
David Winsemius, MD
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] 'install package(s) from local zip files': what is the syntax for that?
Dear all, if I want to install a package in windows system then generally I use the 'install package(s) from local zip files' from the 'package' menu. However I am interested to know that whether there is any syntax which I can use in the R console instead. I have tried with install.packages(__package_name, repos = 'f:/), however could not. Any suggestion will be highly appreciated. 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.
Re: [R] 'install package(s) from local zip files': what is the syntax for that?
On 20.08.2011 20:07, Bogaso Christofer wrote: Dear all, if I want to install a package in windows system then generally I use the 'install package(s) from local zip files' from the 'package' menu. However I am interested to know that whether there is any syntax which I can use in the R console instead. I have tried with install.packages(__package_name, repos = 'f:/), however could not. Let me read ?install.packages for you: The description of the first argument tells us: If repos = NULL, a character vector of file paths of ‘.zip’ files containing binary builds of packages. Hence install.packages(f:/__package_name, repos = NULL) seems to be more appropriate here. Uwe Ligges Any suggestion will be highly appreciated. 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. __ 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] Pattern names matching
Thank you. Orvalho On Sat, Aug 20, 2011 at 6:02 PM, David Winsemius dwinsem...@comcast.netwrote: On Aug 20, 2011, at 11:25 AM, Doran, Harold wrote: See the stringMatch function in the MiscPsycho package for an implementation of Levenshtein The agrep function in base R also returns a Levenshtein distance. -- David. __**__ From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of Orvalho Augusto [orvaq...@gmail.com] Sent: Saturday, August 20, 2011 11:08 AM To: r-help@r-project.org Subject: [R] Pattern names matching Dear R magic guys.. I have two tables (actually will be dataframes), both with names to be matched. The names on the first dataframe are from a study with antenatal visits on some health centers here. It happens that we need the delivery info. And half and some thing else of the women decided to delivery some where else our health units. We managed to get the names from some other places but now we have to match our 4000 original names with over 2 other names. To make thing more bitter some names have badly written. So I need some algorithm like Levenstein or sondex or phonix or something better already on R. Can you help me? Orvalho David Winsemius, MD West Hartford, CT [[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] t() prior to data rotation
Yes. I know for certain that mydata.txt is ordered as an nxm matrix with n objects and m variables. Being that the data is transposed prior to performing the PCA, the rotation matrix ends up being given in terms of object loadings on the principal components, rather than variables as it's normally done. I tried a sample data set and if you plot the values of b$rotation from the transposed data, you end up with a similar (but, not equivalent) plot as if you processed the data without transposing it first, i.e.,: a - read.table(mydata.txt) b - prcomp(a, retx = TRUE) b$x Even though the results are similar, conceptually I am not sure why it's okay to transpose the data, perform PCA, and then plot the rotation matrix as though they were scores. Thanks again, Chris On Sat, Aug 20, 2011 at 12:45 PM, Carl Witthoft c...@witthoft.com wrote: Well, did you try it with a simple test case to see what comes out? Do you have a sample mydata.txt file so you know how it's ordered in the first place? Carl quote From: christopher stratton cfstratton_at_gmail.com Date: Fri, 19 Aug 2011 23:10:07 -0400 Dear All, I have come upon an R-mode PCA protocol that uses the following arguments, where mydata.txt is an nxm matrix of n objects and m variables: a - read.table(mydata.txt) b - t(a) c - prcomp(b) c$rotation The user then plots the coordinates given by c$rotation (PC1 and PC2) as the scores of their PCA plot. This doesn't make sense to me as the user transposed the matrix prior to rotating the data, so they have solved for the eigenvectors of the objects and by plotting the values of c$rotation the user is in effect plotting the loading matrix and not the scores. If anything, this looks like a Q-mode PCA where the rotation matrix should be multiplied by the original data matrix to give scores for the variables. Am I missing something or does this procedure look incorrect? Thank you for your time, Chris /quote -- - Sent from my Cray XK6 __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Christopher F. Stratton Tan Laboratory Memorial SloanKettering Cancer Center 1275 York Ave., Box 422, ZRC-2131, New York, NY 10065 Tel: 646-888-2229 - Fax: 646-422-0416 [[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] Build a package - check error
Hello
I managed to find what was wrong and it has nothing to do with .First.lib.
The clue was in NAMESPACE and in the way the fortran was called (don't use
dyn.load).
I thank the author of the package digest where I found the lines that help
me to crack what was wrong and Uwe Ligges for pushing me hard.
Cheers
Ed
-Original Message-
From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
Sent: Saturday, August 20, 2011 10:38 AM
To: Eduardo Mendes
Cc: r-help@r-project.org
Subject: Re: [R] Build a package - check error
On 20.08.2011 00:41, Eduardo Mendes wrote:
Hi
I have modified the path to
dyn.load(paste(Sys.getenv(R_LIBS_USER),/fortran/src/fortran.so,sep
=))
Hmmm, looks like you never took a look into the relevant manual Writing R
Extensions nor in the help files I cited below.
?.First.lib has an example:
## Suppose a package needs to call a DLL named 'fooEXT', ## where 'EXT' is
the system-specific extension. Then you should use .First.lib -
function(lib, pkg)
library.dynam(foo, pkg, lib)
You cannot know that the package is in Sys.getenv(R_LIBS_USER)
Note also that the .so file won't be in the path /src/... but in
.../lib/... once you install the package in the recommended way using R CMD
INSTALL.
Really, please do read manual!!!
and the package could installed, loaded and the lines with dyn.load
worked.
It does not look like a pretty solution but works on my linux (I am
not sure if it works on my mac or windows).
I am not sure if this is what you meant but as I have no clue what
.First.lib does or NAMESPACES means this is the best I come up with.
So again time to read the manual and the help pages.
Best,
Uwe Ligges
Please correct me if I am wrong.
Many thanks
Ed
On Fri, Aug 19, 2011 at 6:03 PM, Uwe
Liggeslig...@statistik.tu-dortmund.de
wrote:
On 19.08.2011 22:53, Eduardo Mendes wrote:
Dear R-users
I am slowly migrating my mex files (MATLAB - Fortran and C) to R.
To get my own functions available on R section I have decided to
learn how to build
a
R package. I choose a simple example with a few Fortran and R
functions (wrapper).
The fortran sources are located at src and the R functions at R (as
recommended). The building process went ok but R CMD check did not.
The error mgs was
Error in dyn.load(fortran.so) :
unable to load shared object
'/home/eduardo/R_packages/**test.Rcheck/fortran.so':
Although I can see that R cannot find the compiled fortran code I do not
know what to do. I believe it is something to do with the following
lines
in the R-wrapper file
if (!is.loaded('calnpr'))
dyn.load(fortran.so)
1. If the package is called calnpr, the shared library is also called
that way.
2. you have to provide the path to the shared library.
See ?.First.lib for how to do it in a package without NAMESPACE (and
note that NAMESPACES are forced for the next R release).
Best,
Uwe Ligges
How to add the path so that once the package is installed the
compiled
fortran code can be found?
Many thanks
Cheers
Ed
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Re: [R] adding text to a plot created with strat.plot() from package rioja
I'm pretty new to R and finding the responsible line would not have been obvious to me without your help. I downloaded the source for package Rioja and was surprised to see that each function was supplied in it's own file. That made it pretty straightforward to copy, modify, and use my own version of it. Still, it's going to take quite a while to get familiar with all of the graphical capabilities of R. Thanks, Jason === Original Message Subject: Re: [R] adding text to a plot created with strat.plot() from package rioja From: Uwe Ligges lig...@statistik.tu-dortmund.de To: Jason Paul Joines ja...@joines.org Date: 2011.08.20.Sat.8:43:30 On 19.08.2011 18:40, Jason Paul Joines wrote: I have a plot created with strat.plot() from package rioja. When the plot is created with scale.percent=FALSE, each x axes is labeled at 0 and its maximum. However, when scale.percent=TRUE, the x axes are not labeled. I need to use scale.percent=TRUE and I need labels for the x axes. I have been able to add labels to the x axes with mtext but it is very tedious to find the correct position. Is there a better way to do this or a better way to find the desired coordinates than trial and error? Yes: You can change the code and suggest your improvements to the package maintainer, for example. The relevant line in that function obviously is: axis(side = 1, at = seq(0, colM[i], by = 10), labels = FALSE) Uwe Ligges Jason === __ 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] How to abort function execution after x-seconds
Hello all, I'm running a parameter grid optimization ( ksvm, kernlab package) and the optimizer seems not to converge for certain parameters and stays in a infinity loop. Would it be possible to abort the execution after x-seconds and continue with the next parameter set? Which R function do need to use to accomplish this? I had a look at, try() etc. but they didn't seem so fit my needs. I would appreciate any suggestions.. 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 abort function execution after x-seconds
?proc.time
e.g. something like
time0 - proc.time()[2]
state - not converged
while(proc.time()[2] time0[2] + 600 ) ## 10 minutes
{
##... Do your thing...
if(get an answer) {state - converged; break}
}
Refinement required, but you get the idea.
Cheers,
Bert
On Sat, Aug 20, 2011 at 1:52 PM, Immanuel mane.d...@googlemail.com wrote:
Hello all,
I'm running a parameter grid optimization ( ksvm, kernlab package) and the
optimizer
seems not to converge for certain parameters and stays in a infinity loop.
Would it be possible to abort the execution after x-seconds and continue
with the next parameter set? Which R function do need to use to accomplish
this?
I had a look at, try() etc. but they didn't seem so fit my needs.
I would appreciate any suggestions..
best regards
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--
Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions.
-- Maimonides (1135-1204)
Bert Gunter
Genentech Nonclinical Biostatistics
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Re: [R] How to abort function execution after x-seconds
I should have said: The below only works if you can get it into the
function code. If you cannot, and the function has no options that
allow you to limit execution time or iterations, then I don't see how
you can do it.
-- Bert
On Sat, Aug 20, 2011 at 2:16 PM, Bert Gunter bgun...@gene.com wrote:
?proc.time
e.g. something like
time0 - proc.time()[2]
state - not converged
while(proc.time()[2] time0[2] + 600 ) ## 10 minutes
{
##... Do your thing...
if(get an answer) {state - converged; break}
}
Refinement required, but you get the idea.
Cheers,
Bert
On Sat, Aug 20, 2011 at 1:52 PM, Immanuel mane.d...@googlemail.com wrote:
Hello all,
I'm running a parameter grid optimization ( ksvm, kernlab package) and the
optimizer
seems not to converge for certain parameters and stays in a infinity loop.
Would it be possible to abort the execution after x-seconds and continue
with the next parameter set? Which R function do need to use to accomplish
this?
I had a look at, try() etc. but they didn't seem so fit my needs.
I would appreciate any suggestions..
best regards
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions.
-- Maimonides (1135-1204)
Bert Gunter
Genentech Nonclinical Biostatistics
--
Men by nature long to get on to the ultimate truths, and will often
be impatient with elementary studies or fight shy of them. If it were
possible to reach the ultimate truths without the elementary studies
usually prefixed to them, these would not be preparatory studies but
superfluous diversions.
-- Maimonides (1135-1204)
Bert Gunter
Genentech Nonclinical Biostatistics
467-7374
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
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[R] val.surv
Dear R-users, I have two questions regarding validation and calibration of Survival regression models. 1. I am trying to calibrate and validate a cox model using val.surv. here is my code: f.1-cph(Surv(time,event)~age, x=T, y=T, data=train) test1-test[,age] val.surv(f.1, newdata=data.frame(test1), u=10) but I get an error message: Error in val.surv(f.1, newdata = data.frame(testi), u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length I ran the example in the r-documentation but couldn't extract dxy from result. What am I doing wrong? 2. In validate and calibrate cph functions. If it is frailty fit, does the the bootstrap resample clusters or just individuals regards, Salvo Mac [[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] AFT model time-dependent with weibull distribution
Göran Broström wrote: Good. Do you still need answers to your other questions? Yes. Could answer the following two questions: 1- Can I use phreg function to estimate a model with time-dependent covariates? In case of a positive answer, how? 2- I could not find any example that clearly explains how to interpret aftreg output. Specially, refering to the diference between survreg and aftreg output (intercept and sign of the estimates). I include below an example of output of a regression with coxph, survreg, phreg and aftreg and a time-independent variable. I would appreciate if you could explain it or provide an external example that explains how it works. n=26 events=25 time at risk=45 a/ coxph(Surv(time,s) ~ Z1, data=data.frame(data)) coef exp(coef) se(coef)z p Z1 0.0249 1.03 0.00907 2.75 0.006 b/ phreg(Surv(time,s) ~ Z1, data=data.frame(data), dist=weibull) Covariate W.mean Coef Exp(Coef) se(Coef)Wald p Z1 43.689 0.033 1.033 0.009 0.000 log(scale) 0.641 1.899 0.065 0.000 log(shape) 1.172 3.230 0.158 0.000 Max. log. likelihood -22.135 LR test statistic 13.1 Degrees of freedom1 Overall p-value 0.000302689 c/ aftreg(Surv(time,s) ~ Z1, data=data.frame(data), dist=weibull) Covariate W.mean Coef Exp(Coef) se(Coef)Wald p mas 43.689 0.0101.010 0.002 0.000 log(scale) 1.147 3.149 0.141 0.000 log(shape) 1.172 3.230 0.158 0.000 Max. log. likelihood -22.135 LR test statistic 13.1 Degrees of freedom1 Overall p-value 0.000302692 d/ survreg(Surv(time,s) ~ Z1, data=data.frame(data), dist=weibull) Value Std. Error zp (Intercept) 1.14760.13498 8.50 1.87e-17 mas -0.0101 0.00232-4.34 1.45e-05 Log(scale) -1.17240.15787-7.43 1.11e-13 Scale= 0.310 Weibull distribution Loglik(model)= -22.1 Loglik(intercept only)= -28.7 Chisq= 13.05 on 1 degrees of freedom, p= 3e-04 Number of Newton-Raphson Iterations: 5 Thank you very much, J -- View this message in context: http://r.789695.n4.nabble.com/AFT-model-time-dependent-with-weibull-distribution-tp3755079p3757387.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] I have a problem with R!!
Dear all
i´m working with a program i´ve made in R (using functions that others
created)
to run my program i need a sample. if i generate the sample using for
example, rnorm(n, mu, sigma) i have no problem
but if i obtain a sample from a column in excel and i copy it, the program
says that there is a mistake: it says Error en `[.data.frame`(data,
indices) : undefined columns selected
my program is:
d- read.delim(clipboard, header = T, dec = ,)
#Para determinar los valores de las componentes del vector de capacidad es
necesario definir primero las especificaciones y el valor objetivo, T, así
como el máximo valor admitido para la proporción de producción no
conforme, a cada lado de los límites de especificaciones#
# Ingrese ahora el valor del límite inferior de especificaciones#
LIE - 13
# Ingrese ahora el valor del límite superior de especificaciones#
LSE - 17
# Ingrese ahora el valor objetivo#
T - 14.5
# Ingrese ahora el máximo valor admitido para la proporción de producción
no conforme a cada lado de los límites de especificaciones#
MA- 0.00135
D- min ((LSE-T), (T-LIE))
compo1 - function(data, indices)
{
d- data[indices]
n = length (d)
desvio - sd(d)
y- rep(1:n)
y[x = mean(d)] - 1
y[x mean(d)] - 0
RI1- D/(3*desvio*2*mean(y))
RI2 - D/(3*desvio*2*(1-mean(y)))
return (min (RI1, RI2))
}
compo2- function(data, indices)
{
d - data[indices]
c2 - (abs(mean(d) - T))/D
return (1-c2)
}
compo3-function(data, indices)
{
d- data[indices]
n- length (d)
y- rep(1:n)
y[d LIE] - 1
y[d = LIE] - 0
INFE - mean (y);
y- rep(1:n)
y[d LSE] - 1
y[d = LSE] - 0
SUPE- mean (y);
PPI - (1 - INFE)/(1-MA)
PPS - (1 - SUPE)/(1-MA)
return (min (PPI, PPS))
}
save(file = compo1.RData)
save(file = compo2.RData)
save(file = compo3.RData)
compos- function(data, indices)
{
d - data[indices]
capacidad - c(compo1(d), compo2(d), compo3(d))
return(capacidad)
}
save(file = compos.RData)
require (boot)
vectorcapacidad - boot (d, compos, R = 3000)
ETC. ETC.
WHEN I START MY PROGRAM WRITING:
d- rnorm (n, mu, sigma)
I HAVE NO PROBLEM. BUT WHEN I READ A VECTOR FROM EXCEL, R TELLS ME
Error en `[.data.frame`(data, indices) : undefined columns selected
CAN YOU HELP ME THANK YOU VERY MUCH!
NOEMI FERRERI, ROSARIO, ARGENTINA
SCHOOL OF INDUSTRIAL ENGINEERING
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Re: [R] Calculating p-value for 1-tailed test in a linear model
My question isn't related to homework. It is a small part of an actual problem I'm trying to solve. I've been unable to find a solution in R help files and discussions, in statistics books, or from colleagues. The solution may not be overly complicated, but any assistance is appreciated. Thanks, Andy - Original Message - From: Rolf Turner rolf.tur...@xtra.co.nz To: Andrew Campomizzi acampomi...@tamu.edu Cc: r-help@r-project.org Sent: Friday, August 19, 2011 9:18:07 PM GMT -06:00 US/Canada Central Subject: Re: [R] Calculating p-value for 1-tailed test in a linear model The r-help mailing list is *not* for giving assistance with homework. cheers, Rolf Turner On 20/08/11 10:20, Andrew Campomizzi wrote: Hello, I'm having trouble figuring out how to calculate a p-value for a 1-tailed test of beta_1 in a linear model fit using command lm. My model has only 1 continuous, predictor variable. I want to test the null hypothesis beta_1 is= 0. I can calculate the p-value for a 2-tailed test using the code 2*pt(-abs(t-value), df=degrees.freedom), where t-value and degrees.freedom are values provided in the summary of the lm. The resulting p-value is the same as provided by the summary of the lm for beta_1. I'm unsure how to change my calculation of the p-value for a 1-tailed test. Thanks for your assistance, Andy __ 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 abort function execution after x-seconds
Hello Bert, thanks for the suggestion. I should have mentioned that, sadly I can NOT change the function that does the optimization. best, Immanuel __ 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] val.surv
On Aug 20, 2011, at 3:32 PM, Salvo Mac wrote: Dear R-users, I have two questions regarding validation and calibration of Survival regression models. 1. I am trying to calibrate and validate a cox model using val.surv. here is my code: f.1-cph(Surv(time,event)~age, x=T, y=T, data=train) test1-test[,age] val.surv(f.1, newdata=data.frame(test1), u=10) but I get an error message: Error in val.surv(f.1, newdata = data.frame(testi), u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length I ran the example in the r-documentation but couldn't extract dxy from result. What am I doing wrong? The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object. I would think that if test were a dataframe then wrapping data.frame around a vector might not get it named correctly as 'age'. You might try newdata= test['age']. Just a guess. 2. In validate and calibrate cph functions. If it is frailty fit, does the the bootstrap resample clusters or just individuals The code above appears to be dependent on the rms package. The frailty function is part of the underlying survival package and I do not see it mentioned in the index for Harrell's RMS text. You will probably need to wait until Frank comes across this. He is generally very good about correction my errors and knowledge gaps. -- David Winsemius, MD 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] adding text to a plot created with strat.plot() from package rioja
I completely missed that entering just the function name prints the code for that function making it even easier to copy and create a custom version. Jason === Original Message Subject: Re: [R] adding text to a plot created with strat.plot() from package rioja From: Jason Paul Joines ja...@joines.org To: r-help@r-project.org Date: 2011.08.20.Sat.14:48:35 I'm pretty new to R and finding the responsible line would not have been obvious to me without your help. I downloaded the source for package Rioja and was surprised to see that each function was supplied in it's own file. That made it pretty straightforward to copy, modify, and use my own version of it. Still, it's going to take quite a while to get familiar with all of the graphical capabilities of R. Thanks, Jason === Original Message Subject: Re: [R] adding text to a plot created with strat.plot() from package rioja From: Uwe Ligges lig...@statistik.tu-dortmund.de To: Jason Paul Joines ja...@joines.org Date: 2011.08.20.Sat.8:43:30 On 19.08.2011 18:40, Jason Paul Joines wrote: I have a plot created with strat.plot() from package rioja. When the plot is created with scale.percent=FALSE, each x axes is labeled at 0 and its maximum. However, when scale.percent=TRUE, the x axes are not labeled. I need to use scale.percent=TRUE and I need labels for the x axes. I have been able to add labels to the x axes with mtext but it is very tedious to find the correct position. Is there a better way to do this or a better way to find the desired coordinates than trial and error? Yes: You can change the code and suggest your improvements to the package maintainer, for example. The relevant line in that function obviously is: axis(side = 1, at = seq(0, colM[i], by = 10), labels = FALSE) Uwe Ligges Jason === __ 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 p-value for 1-tailed test in a linear model
On Aug 19, 2011, at 6:20 PM, Andrew Campomizzi wrote: Hello, I'm having trouble figuring out how to calculate a p-value for a 1- tailed test of beta_1 in a linear model fit using command lm. My model has only 1 continuous, predictor variable. I want to test the null hypothesis beta_1 is = 0. I can calculate the p-value for a 2-tailed test using the code 2*pt(-abs(t-value), df=degrees.freedom), where t-value and degrees.freedom are values provided in the summary of the lm. The resulting p-value is the same as provided by the summary of the lm for beta_1. I'm unsure how to change my calculation of the p-value for a 1-tailed test. You need to clearly state your hypothesis. Then using the output from the regression function should be straightforward. (Yes. this is a intentionally vague answer designed to elicit further information about your understanding of the statistical issues and how they relate to your domain knowledge. Many time peole already have the data and because they didn't get the answer they wanted, they search for other ways to game the system by ad-hoc changes in the statistical rules of the road.) -- David Winsemius, MD 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] Best way to setClass and setMethod for an R package?
Folks: I'm putting together an R package, and I was wondering where, specifically, I put both class definitions (via setClass) as well as setMethod calls? Is there a particular file name I need to use? Thanks! --j -- Jonathan A. Greenberg, PhD Assistant Project Scientist Center for Spatial Technologies and Remote Sensing (CSTARS) Department of Land, Air and Water Resources University of California, Davis One Shields Avenue Davis, CA 95616 Phone: 415-763-5476 AIM: jgrn307, MSN: jgrn...@hotmail.com, Gchat: jgrn307 __ 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] reshape a matrix
On Aug 20, 2011, at 1:39 PM, David Winsemius wrote:
On Aug 20, 2011, at 1:21 PM, Uwe Ligges wrote:
On 20.08.2011 19:04, David Winsemius wrote:
On Aug 20, 2011, at 12:32 PM, Uwe Ligges wrote:
On 20.08.2011 17:04, Wendy wrote:
Hi all,
I have a data.frame like following
A-c('d0','d0','d1','d1','d2','d2')
B-rep(c('control','sample'),3)
C-c(rep(10,2),200,300,400,500)
dataframe-data.frame(A,B,C)
I want to reshape the matrix, so the matrix with 'd0', 'd1' and
'd2'
in rows
and 'control' and 'sample' in columns. Is there a function for
doing
this
easily?
See ?reshape
reshape(data=dataframe, idvar=A, timevar=B, direction=wide)
Uwe Ligges
Many people have experienced problems understanding the mechanics
of the
base function 'reshape'. If you do not and if do continue to use
it, you
would be doing the world a great service by writing a tutorial
manual
with a bunch of worked examples. I have never found a tutorial that
clarified how I should use it in the variety of situations where I
have
needed it.
David,
I think there are some good examples on the help page. What is
missing? What is not clearly explained?
The stumbling blocks I have encountered are in trying to figure out
which of the multiple arguments are needed a) in going from wide to
long and b) in going from long to wide, c) and what are the reasons
for the various error messages I provoke . I am almost never able to
do it correctly on the first try and rarely able to do it even on
the fourth try. I bought Spector's book in hopes of understanding it
better, but his efforts did not take root in my brain.
In the instance above, how would I have know how to apply the help
page description of timevar (below) to this problem?
timevar
the variable in long format that differentiates multiple records
from the same group or individual.
I'm wondering it this would be clearer (if it is correct):
timevar
the variable in long format that order multiple records which vary
within the same groups or individual specified in idvar.
--
David
To my reading that does not distinguish the purpose of 'timevar'
from the purpose of 'idvar' and then reading the 'idvar' definition
just below it does not help at all.
--
David.
If a longer tutorial is needed, that may be an article for the R
Help Desk in The R Journal. Anybody volunteering?
Best,
Uwe
So Hadley wrote an alternate facility ... the reshape package that
does
not have a reshape function in it but rather two functions 'melt'
and
'cast'.
Your data is all ready molten, i.e. it is in the long format
(in the terminology of the base reshape function) with identifier
values
in each row and a single column of values.
library(reshape)
cast(dataframe,A~B)
Using C as value column. Use the value argument to cast to
override this
choice
A control sample
1 d0 1e+05 1e+05
2 d1 2e+02 3e+02
3 d2 4e+02 5e+02
Basically the cast formula keeps the LHS variables in the rows and
hte
RHD variables get arranges in columns. (For reasons that are
unclear to
me the dataframe argument was placed first when using positional
argument passing, unlike most other formula methods in R.)
David Winsemius, MD
West Hartford, CT
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Re: [R] val.surv
Thanks David However, I tried your trick on val.surv with newdata=test['age'] but still didn't work. Still gives the same error message: Error in val.surv(f.1, newdata = test1[age], u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length Salvo From: David Winsemius dwinsem...@comcast.net Cc: r-help@R-project.org r-help@r-project.org Sent: Sunday, August 21, 2011 12:55 AM Subject: Re: [R] val.surv On Aug 20, 2011, at 3:32 PM, Salvo Mac wrote: Dear R-users, I have two questions regarding validation and calibration of Survival regression models. 1. I am trying to calibrate and validate a cox model using val.surv. here is my code: f.1-cph(Surv(time,event)~age, x=T, y=T, data=train) test1-test[,age] val.surv(f.1, newdata=data.frame(test1), u=10) but I get an error message: Error in val.surv(f.1, newdata = data.frame(testi), u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length I ran the example in the r-documentation but couldn't extract dxy from result. What am I doing wrong? The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object. I would think that if test were a dataframe then wrapping data.frame around a vector might not get it named correctly as 'age'. You might try newdata= test['age']. Just a guess. 2. In validate and calibrate cph functions. If it is frailty fit, does the the bootstrap resample clusters or just individuals The code above appears to be dependent on the rms package. The frailty function is part of the underlying survival package and I do not see it mentioned in the index for Harrell's RMS text. You will probably need to wait until Frank comes across this. He is generally very good about correction my errors and knowledge gaps. -- David Winsemius, MD West Hartford, CT [[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] reshape a matrix
On 21/08/11 03:04, Wendy wrote:
Hi all,
I have a data.frame like following
A-c('d0','d0','d1','d1','d2','d2')
B-rep(c('control','sample'),3)
C-c(rep(10,2),200,300,400,500)
dataframe-data.frame(A,B,C)
I want to reshape the matrix, so the matrix with 'd0', 'd1' and 'd2' in rows
and 'control' and 'sample' in columns. Is there a function for doing this
easily?
Your question has been answered But if you remember that a *matrix*
and a *data frame* are ***NOT*** the same thing, you'll get into a lot less
trouble. Why do you think there are two *different* terms for these sorts
of object if they are the same thing? Psigh!!!
cheers,
Rolf Turner
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Re: [R] I have a problem with R!!
No, you don't have a problem with R. You have a problem with Excel. Solution: Don't use Excel. cheers, Rolf Turner __ 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] val.surv
On Aug 20, 2011, at 8:08 PM, Salvo Mac wrote: Thanks David However, I tried your trick on val.surv with newdata=test['age'] but still didn't work. Still gives the same error message: Error in val.surv(f.1, newdata = test1[age], u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length As I said (and you did not act upon): The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object. I said it was a guess. Now stop wasting our time and offer what is needed. -- david. Salvo From: David Winsemius dwinsem...@comcast.net Cc: r-help@R-project.org r-help@r-project.org Sent: Sunday, August 21, 2011 12:55 AM Subject: Re: [R] val.surv On Aug 20, 2011, at 3:32 PM, Salvo Mac wrote: Dear R-users, I have two questions regarding validation and calibration of Survival regression models. 1. I am trying to calibrate and validate a cox model using val.surv. here is my code: f.1-cph(Surv(time,event)~age, x=T, y=T, data=train) test1-test[,age] val.surv(f.1, newdata=data.frame(test1), u=10) but I get an error message: Error in val.surv(f.1, newdata = data.frame(testi), u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length I ran the example in the r-documentation but couldn't extract dxy from result. What am I doing wrong? The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object. I would think that if test were a dataframe then wrapping data.frame around a vector might not get it named correctly as 'age'. You might try newdata= test['age']. Just a guess. 2. In validate and calibrate cph functions. If it is frailty fit, does the the bootstrap resample clusters or just individuals The code above appears to be dependent on the rms package. The frailty function is part of the underlying survival package and I do not see it mentioned in the index for Harrell's RMS text. You will probably need to wait until Frank comes across this. He is generally very good about correction my errors and knowledge gaps. -- David Winsemius, MD West Hartford, CT [[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. David Winsemius, MD 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] val.surv
The test and train are like split data sets, contain similar variables but from different countries so the two sets are somehow independent. And yes it is a data frame. So I extracted age, time and event. So test is data frame,(age, time, event). does that suffice? From: David Winsemius dwinsem...@comcast.net Cc: r-help@R-project.org r-help@r-project.org Sent: Sunday, August 21, 2011 3:19 AM Subject: Re: [R] val.surv On Aug 20, 2011, at 8:08 PM, Salvo Mac wrote: Thanks David However, I tried your trick on val.surv with newdata=test['age'] but still didn't work. Still gives the same error message: Error in val.surv(f.1, newdata = test1[age], u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length As I said (and you did not act upon): The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object. I said it was a guess. Now stop wasting our time and offer what is needed. --david. Salvo From: David Winsemius dwinsem...@comcast.net Cc: r-help@R-project.org r-help@r-project.org Sent: Sunday, August 21, 2011 12:55 AM Subject: Re: [R] val.surv On Aug 20, 2011, at 3:32 PM, Salvo Mac wrote: Dear R-users, I have two questions regarding validation and calibration of Survival regression models. 1. I am trying to calibrate and validate a cox model using val.surv. here is my code: f.1-cph(Surv(time,event)~age, x=T, y=T, data=train) test1-test[,age] val.surv(f.1, newdata=data.frame(test1), u=10) but I get an error message: Error in val.surv(f.1, newdata = data.frame(testi), u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length I ran the example in the r-documentation but couldn't extract dxy from result. What am I doing wrong? The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object. I would think that if test were a dataframe then wrapping data.frame around a vector might not get it named correctly as 'age'. You might try newdata= test['age']. Just a guess. 2. In validate and calibrate cph functions. If it is frailty fit, does the the bootstrap resample clusters or just individuals The code above appears to be dependent on the rms package. The frailty function is part of the underlying survival package and I do not see it mentioned in the index for Harrell's RMS text. You will probably need to wait until Frank comes across this. He is generally very good about correction my errors and knowledge gaps. -- David Winsemius, MD West Hartford, CT [[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. David Winsemius, MD West Hartford, CT [[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] val.surv
On Aug 20, 2011, at 10:25 PM, Salvo Mac wrote: The test and train are like split data sets, contain similar variables but from different countries so the two sets are somehow independent. And yes it is a data frame. What is a data.frame? test and train may be dataframes, but test[, age] is not a dataframe. So I extracted age, time and event. Code? The code you offered before would have created a newdata object (yes, a data.frame) with a single column bearing the same name as the vector argument , test1. Not named age. Try it. do str on such an object: str(dataframe(test1)) So test is data frame,(age, time, event). does that suffice? It certainly does not allow me to reproduce the error you got (which I still think is probably related to the structure of your argument to newdata.) That's all I can say without data and code. -- David. From: David Winsemius dwinsem...@comcast.net Cc: r-help@R-project.org r-help@r-project.org Sent: Sunday, August 21, 2011 3:19 AM Subject: Re: [R] val.surv On Aug 20, 2011, at 8:08 PM, Salvo Mac wrote: Thanks David However, I tried your trick on val.surv with newdata=test['age'] but still didn't work. Still gives the same error message: Error in val.surv(f.1, newdata = test1[age], u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length As I said (and you did not act upon): The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object. I said it was a guess. Now stop wasting our time and offer what is needed. --david. Salvo From: David Winsemius dwinsem...@comcast.net Cc: r-help@R-project.org r-help@r-project.org Sent: Sunday, August 21, 2011 12:55 AM Subject: Re: [R] val.surv On Aug 20, 2011, at 3:32 PM, Salvo Mac wrote: Dear R-users, I have two questions regarding validation and calibration of Survival regression models. 1. I am trying to calibrate and validate a cox model using val.surv. here is my code: f.1-cph(Surv(time,event)~age, x=T, y=T, data=train) test1-test[,age] val.surv(f.1, newdata=data.frame(test1), u=10) but I get an error message: Error in val.surv(f.1, newdata = data.frame(testi), u = 10) : dims [product 1797] do not match the length of object [2496] In addition: Warning message: In est.surv + S[, 1] : longer object length is not a multiple of shorter object length I ran the example in the r-documentation but couldn't extract dxy from result. What am I doing wrong? The fundamental thing you are doing wrong for q1 is failing to unambiguously describe the test object. I would think that if test were a dataframe then wrapping data.frame around a vector might not get it named correctly as 'age'. You might try newdata= test['age']. Just a guess. 2. In validate and calibrate cph functions. If it is frailty fit, does the the bootstrap resample clusters or just individuals The code above appears to be dependent on the rms package. The frailty function is part of the underlying survival package and I do not see it mentioned in the index for Harrell's RMS text. You will probably need to wait until Frank comes across this. He is generally very good about correction my errors and knowledge gaps. David Winsemius, MD 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] AFT model time-dependent with weibull distribution
question 2
*aftreg vs. survreg*
for aftreg = S0 *{t/exp(b-BXi)]^a} a= shape and b= log(scale)
for survreg and stata S0 *{t*exp(intercept+BXi)]^1/p} p=shape
/intercept, log(scale) and estimates are equivalent with reversed sign./
*PH and AFT*
/phreg.Bhat= aftreg.Bhat * shape /
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