Re: [R] Query regarding linking R with Matlab
Hi, It might be that R can't find Matlab; then you have to specify option 'matlab', see help(Matlab). Try also a different port. Try to add a line setVerbose(matlab, -2) to get more detailed output what is going on; matlab - Matlab(host=localhost, port=9998) setVerbose(matlab, -2) if (!open(matlab)) throw(Matlab server is not running: waited 30 seconds.) If you can't get it to work, send the output of the above. /Henrik On 12/27/06, Bhanu Kalyan.K [EMAIL PROTECTED] wrote: Respected Sir, I thank you for your concern. I have worked with the code that you have provided. But it has generated errors like: if (!open(matlab)) + throw(Matlab server is not running: waited 30 seconds.) //This command is not responding even after 30 seconds. res - evaluate(matlab, swissroll) Error in writeBin(con = con, as.integer(b), size = 1) : invalid connection vars - getVariable(matlab, c(Y, X, K, d)) Error in writeBin(con = con, as.integer(b), size = 1) : invalid connection Kindly help me with this. Regards Bhanu Kalyan K Bhanu Kalyan K BTech CSE Final Year [EMAIL PROTECTED] Tel :+91-9885238228 __ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] fitting all models in log-lm
Hello,
I am trying to use R to carry out loglinear analysis. So far, starting from a
previous post in this list
(http://finzi.psych.upenn.edu/R/Rhelp02a/archive/33698.html), I have been
able to use loglm() to generate likelihood ratio.
I now try to find a function that generates all the models and another one to
compare them.
Thanks for any help.
--
Nicolas Mazziotta
The contents of this e-mail, including any attachments, are ...{{dropped}}
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to transform string to variable name in a fuction?
I believe the split function should work in this case.
From the help file:
split(x, f, drop = FALSE, ...)
split(x, f, drop = FALSE, ...) - value
unsplit(value, f, drop = FALSE)
Arguments
|x| vector or data frame containing values to be divided into groups.
|f| a “factor” in the sense that |as.factor factor.html(f)| defines
the grouping, or a list of such factors in which case their interaction
is used for the grouping.
|drop| logical indicating if levels that do not occur should be dropped
(if |f| is a |factor| or a list).
|value| a list of vectors or data frames compatible with a splitting of
|x|. Recycling applies if the lengths do not match.
|...| further potential arguments passed to methods.
Abhijit Dasgupta, Ph.D.
Assistant Professor | Division of Biostatistics
Department of Pharmacology and Experimental Therapeutics | Thomas
Jefferson University
1015 Chestnut St | Suite M100 | Philadelphia, PA 19107
Ph: (215) 503-9201 | Fax: (215) 503-3804
jingjiangyan wrote:
there is a data frame, like this:
df
aa bb
1 a 20.27802
2 b 22.10664
3 c 21.33470
4 a 22.32898
5 b 19.73760
6 c 20.38979
.(suppressed)
what I want to do is to copy the data frame's rows into different data frames
according to the levels of 'aa' column,
df.a - df[df[,1]=='a',] ; df.b - df[df[,1]=='b',] ;
df.a
aa bb
1 a 20.27802
4 a 22.32898
...
So, when completed, there should be df.a, df.b,df.c, etc.
If we could do this by hand, it is pretty fine. But could I write a loop to
do this ?
when I tried this using a funciton, there is a problem.
for ( i in levels(df[,1])) {
+ name = paste('df',i,sep='')
+ name - df[df[,1]==i,]
+ }
name
aa bb
3 c 21.33470
6 c 20.38979
ls()
[1] df iname
i
[1] c
there is not data frames df.a, df.b,etc.
Could you please give me some suggestion?
I have found that write a function in R for a beginner is difficult. Is there
any tutorial on writing the functions in R?
Furthermore, someone also said that loop is not used as frequently as in
other script language (e.g. bash, perl). So, If you have any other smart
means do this more efficiently, please let me know, I would appreciate your
kindness.
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] counties in different colours using map()
Hi,
I would like to plot a map of US counties using different colors. map()
seems to be the function to use, e.g.
library(maps); map('usa'); map('county', 'colorado', add=T,fill = T,
col=c(1:5))
plots Colorado counties using colours 1 to 5.
However, I want each color to represent a certain value - a value to be
picked from a data frame.
This code should show a correspoding map at the level of states:
state.names - system('tr [A-Z] [a-z]', state.name)
map.states - unix('sed s/:.*//', map(names=T,plot=F))
state.to.map - match(map.states, state.names)
color- votes.repub[state.to.map, votes.year = 1900] / 100
map('state', fill=T, col=color); map('state', add=T)
It is copied from page 6 in
Richard A. Becker, and Allan R. Wilks, Maps in S, ATT Bell
Laboratories Statistics Research Report [93.2], 1993.
http://public.research.att.com/areas/stat/doc/93.2.ps
I also wonder whether the county names are available in the database
used by map(), and, if yes, how to extract or utilize them.
Thanks!
Tord
--
Tord Snäll
Department of Conservation Biology
Swedish University of Agricultural Sciences (SLU)
P.O. 7002, SE-750 07 Uppsala, Sweden
Office/Mobile/Fax
+46-18-672612/+46-730-891356/+46-18-673537
E-mail: [EMAIL PROTECTED]
www.nvb.slu.se/staff_tordsnall
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] axis and times() problem
Try: plot(x, y, bty = n, xaxs = i, yaxs = i) Actually I think there may be a bug here since the axes do not intersect. On 12/27/06, Knut Krueger [EMAIL PROTECTED] wrote: Dear R-Group, the first example is working as expected, but I need the plot without the box, normally no problem, but I am not able to get the x-axis formatted as times with the axis, command. I tried a lot of things, nothing was working so I used the most easy axis command in the second example here # working library(chron) # for times() library(graphics)# for axis par(cex=1.2,lwd=1) x - c(times(12:15:00),times(15:30:00)) y - c(1,5) plot(x, y, type=n,adj=0, asp=0, xlab=, ylab=,font.axis=2,yaxt='n') # axis() command problem: par(cex=1.2,lwd=1) x - c(times(12:15:00),times(15:30:00)) y - c(1,5) plot(x, y,axes=FALSE, type=n,adj=0, asp=0, xlab=, ylab=,font.axis=2,yaxt='n') axis(1) Maybe anybody could help me to disable the box around the plot and get the x-axis formatted as times Regards Knut __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] slightly inconsistent behavior
What do you think is `slightly inconsistent behavior' here? (You seem to be quite consistent in not telling us such relevant facts, including your OS and version of R!) If you think that the memory usage of R should be monotone in the size of the problem, your expectations are unfounded. Here it is likely you are seeing memory fragmentation on a 32-bit OS: see help(Memory-limits). But setting nrows= and 'extend nrows by a few more 9's' seems to me suggesting something like nrows=99 which is invalid and gives a warning message you did not mention. On Tue, 26 Dec 2006, ivo welch wrote: dear R experts: This is just a minor, minor nuisance, but I thought I would point it out: dataset - read.table(file=pipe(cmdline), header =T, + na.strings=c(NaN, C,I,M, E), sep=,, as.is=T, nrows=); Error: cannot allocate vector of size 781249 Kb If I extend nrows by a few more 9's, the error goes away. Similarly, if I use much fewer observations, the error goes away. regards, /iaw __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. PLEASE do, at long last. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] building R-package under windows - error - in options(deafultPackages) was not found
Given the lines after Description: have been wrapped during e-mail
transfer and are in one line in your DESCRIPTION file, here are two guesses:
- You have some additional library added that contains a broken
installation of some base package?
- You have declared some non-existing default packages in some Rprofile
or whereever that are not existing?
Anyway, you might want to send the whole package in a private message.
Uwe Ligges
Daniel Berg wrote:
I have tried to uninstall R-2.4.1 and reinstall it but I still get the same
error message. I have also tried reinstalling both mingw and perl, both
with
no effect.
My DESCRIPTION file:
Package: copulaGOF
Type: Package
Title: Copula simulation, estimation and goodness-of-fit
Version: 1.5
Date: 2006-12-15
Author: Daniel Berg, Henrik Bakken (Norwegian Computing Center)
Depends: adapt, copula, fBasics, mnormt, mvtnorm, scatterplot3d, sn
Maintainer: Daniel Berg [EMAIL PROTECTED]
URL: www.danielberg.no
Description: Functions for copula simulation, estimation and
goodness-of-fit
testing. Includes functions for simulating and estimating higher
dimensional
distributions.
License: GPL Version 2 or later, GSL-1.6
and the error message read:
$ Rcmd build --binary copulaGOF
* checking for file 'copulaGOF/DESCRIPTION' ... OK
* preparing 'copulaGOF' :
* checking DESCRIPTION meta-information ... ERROR
During startup - Warning messages:
' in: library(package, lib.loc = lib.loc, character.only = TRUE, logical =
TRUE,
in options(defaultPackages) was not found
I don't have access to binaries of older versions of R now but will try
with
older version after the holidays to see if this has an effect.
Regards,
Daniel
On 12/23/06, Brian Ripley [EMAIL PROTECTED] wrote:
On Sat, 23 Dec 2006, Uwe Ligges wrote:
Daniel Berg wrote:
Dear all,
I have been building R packages under windows on my old pc,
successfully.
Now I have bought a new pc, still running windows, and I am trying to
build
the same R packages as before, but now without the same success. I
have
installed the Rtools, perl, mingw and added them to the environment
variables.
I am running Windows XP Professional on a Thinkpad T60. I have
installed
R-2.4.1, ActivePerl 5.8.8 Build 819, MinGW 5.1.2, and I downloaded
tools.zipfrom
http://www.murdoch-sutherland.com/Rtools.
I receive the following error message:
$ Rcmd build --binary copulaGOF/
* checking for file 'mypackage/DESCRIPTION' ... OK
* preparing 'copulaGOF' :
* checking DESCRIPTION meta-information ... ERROR
Looks like something is wrong with your DESCRIPTION file. Can you send
us the contents of that file?
I think rather with his R: the message below says 'During startup', and
indicates that one of the default packages is missing. That would mean
that it has not got to running the code to look at DESCRIPTION.
Uwe Ligges
During startup - Warning messages :
' in: library(package, lib.loc = lob.loc, character.only = TRUE,
logical =
TRUE,
in options(defaultPackages) was not found
In my package I have included a zzz.r file that contains the
following,
perhaps this is the cause?
.First.lib -function (lib, pkg) {
library(adapt)
library(copula)
library(fBasics)
library(mvtnorm)
runif(1)
library.dynam(mypackage, package=mypackage)
}
Any help or comments is most welcome. Thank you.
Best wishes,
Daniel Berg
-
danielberg.no
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
http://www.stats.ox.ac.uk/%7Eripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] plotting time series with zoo pckg
Dear Gabor, sorry for not posting the code. below I have a piece of code that generates a multivariate zoo data (3 columns) and graphs it using the axis commands to generate the labels. This does not work. However, if one extracts one column the labelling works using the same commands! I cannot figure out what I am missing here. Thanks for any suggestion. AA. I am using R version 2.4.0 (windows XP) and zoo package version 1.2-1(2006-09-20). #-Begin-- # Let's take the previous example to create a simple zoo data. z - structure(c(21,34,33,41,39,38,37,28,33,40), index = structure(c(8044,8051,8058,8065,8072, 8079,8086,8093,8100,8107), class=Date), class = zoo) # generating 3 random vectors with the same length as z. jx1 - rnorm(10); jx2 - rnorm(10); jx3 - rnorm(10) # create a zoo class data using the random vectors. jx - cbind(jx1,jx2,jx3) z1 - zoo(jx, index(z)) # now we just repete the previous example. plot(z1, xaxt = n) axis(1, time(z1), lab = FALSE) jd - time(z1)[seq(1, dim(z1)[[1]], 3)] axis(1,jd, as.character(jd),cex.axis = 0.8, tcl = -0.7, las = 2) # now we extract one column of z1 and graph it with the same axis commands. z2 - z1[,1, drop = F] windows() plot(z2, xaxt = n) axis(1, time(z2), lab = FALSE) jd - time(z2)[seq(1, dim(z2)[[1]], 3)] axis(1,jd, as.character(jd),cex.axis = 0.8, tcl = -0.7, las = 2) #---End--- - Original Message From: Gabor Grothendieck [EMAIL PROTECTED] To: ahmad ajakh [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Tuesday, December 26, 2006 11:17:44 PM Subject: Re: [R] plotting time series with zoo pckg Please read the last line of every message to r-help and follow that. On 12/26/06, ahmad ajakh [EMAIL PROTECTED] wrote: Dear Gabor, Thank you for your quick reply. This solution works for my univariate zoo class time series. I first tried it for a timeseries with 4 columns of data, it did not plot the labels nor the ticks, I tried it on a one dim timeseries (one column zoo class data as the example in the question) and it worked! is there something that I am missing? Thanks again. AA. - Original Message From: Gabor Grothendieck [EMAIL PROTECTED] To: ahmad ajakh [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Sent: Tuesday, December 26, 2006 8:31:07 PM Subject: Re: [R] plotting time series with zoo pckg Try this: # test data library(zoo) z - structure(c(21, 34, 33, 41, 39, 38, 37, 28, 33, 40), index = structure(c(8044, 8051, 8058, 8065, 8072, 8079, 8086, 8093, 8100, 8107), class = Date), class = zoo) z # plot without X axis plot(z, xaxt = n) # unlabelled tick at each point axis(1, time(z), lab = FALSE) # labelled tick every third point dd - time(z)[seq(1, length(z), 3)] axis(1, dd, as.character(dd), cex.axis = 0.7, tcl = -0.7) On 12/26/06, ahmad ajakh [EMAIL PROTECTED] wrote: Hi all, I am using the zoo package to plot time series. I have a problem with formatting the axes. my zoo object (z) looks like the following. c1 1992-01-10 21 1992-01-17 34 1992-01-24 33 1992-01-31 41 1992-02-07 39 1992-02-14 38 1992-02-21 37 1992-02-28 28 1992-03-06 33 1992-03-13 40 plot.zoo(z) produces a plot with the labels on the x-axis that I cannot control. I want a an xtick every 10 data points with corresponding date labels. I have tried different combination of axis command without success any idea? Thanks __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ Do You Yahoo!? http://mail.yahoo.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to suppress a loading required package: ... message
Hi, how to suppress a loading required package:... message? Kind regards Jaci -- __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to suppress a loading required package: ... message
On 12/27/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Hi, how to suppress a loading required package:... message? require(package, quiet=TRUE) jab -- John Bollinger, CFA, CMT www.BollingerBands.com If you advance far enough, you arrive at the beginning. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Formatting an arry to typeset as a table
I am writting some functions that return an array of coefficients along with confidence intetervals for each coefficient. My intent is to eventually typeset the coefficients and intervals into a table (or tables) in a document. I would like to use existing tools such as the 'latex' function in the Hmisc package, the 'odfTable' function in odfWeave package, or 'HTML' in the R2HTML package (or any others that work similarly). This is fairly straight forward for 2 dimensional arrays (matrices) without the confidence intervals, just add the proper dimnames and call latex or whatever. In some cases the array is 3 dimensional and could possibly have 4 dimensions, which complicates things a bit. If nothing better presents itself I can use apply to produce several 2 dimensional tables from the higher dimentsional array. The big complication is that I would like each cell of the final table to have the coefficient value with the confidence limits below it, e.g.: 5.3 (4.2,6.4) My current thinking is to have a function (possibly a summary method) that takes the output from my functions and returns something that could be passed directly to the latex, odfTable, etc. functions: myobj - myfunc(myargs) latex( summary(myobj), file='mytempfile' ) I prefer to use LaTeX, but many of the people I work with are stuck with MS products so the odf or html options are attractive there. Does anyone have any suggestions on how best to format the output so it can be passed to latex and friends? or will I need to write my own methods for these functions? (I know enough LaTeX to probably do a method for latex, but don't have enough knowledge of html or odf to do the others myself) Thanks for any suggestions, [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plotting time series with zoo pckg
To do it with plot.zoo one has to create a custom panel.
Another approach is to use xyplot.zoo since that supports custom scales
directly. Note that the ?xyplot.zoo examples contain code that is along the
lines of the xyplot.zoo solution.
Here are examples of both approaches:
library(zoo)
# test data
z - structure(c(21,34,33,41,39,38,37,28,33,40),
index = structure(c(8044,8051,8058,8065,8072,
8079,8086,8093,8100,8107), class=Date), class = zoo)
set.seed(1) # needed to make it reproducible
jx1 - rnorm(10); jx2 - rnorm(10); jx3 - rnorm(10)
# create a zoo class data using the random vectors.
jx - cbind(jx1,jx2,jx3)
z1 - zoo(jx, index(z))
# 1. plot.zoo solution using custom panel function, my.panel
my.panel - function(...) {
lines(...)
if (parent.frame()$j == ncol(z1)) {
# following line only if non-labelled ticks wanted for each point
axis(1, at = time(z1), lab = FALSE)
ix - seq(1, length(z1), 3)
labs - format(time(z1)[ix], %b-%d)
axis(1, at = time(z1)[ix], lab = labs, tcl = -0.7, cex.axis = 0.7)
}
}
plot(z1, panel = my.panel, xaxt = n)
# 2. xyplot.zoo solution
# z1 is from above
library(lattice)
ix - seq(1, length(z1), 3)
labs - format(time(z1)[ix], %b-%d)
xyplot(z1, scales = list(x = list(at = time(z1)[ix], labels = labs)))
# only need following if non-labelled ticks are to be added for each point
trellis.focus(panel, 1, 1, clip.off = TRUE)
panel.axis(bottom, check.overlap = TRUE, outside = TRUE, labels = FALSE,
tck = .7, at = time(z1))
trellis.unfocus()
On 12/27/06, ahmad ajakh [EMAIL PROTECTED] wrote:
Dear Gabor,
sorry for not posting the code. below I have a piece of code that generates
a multivariate zoo data (3 columns) and graphs it using the axis
commands to generate the labels. This does not work. However, if one
extracts one column the labelling works using the same commands!
I cannot figure out what I am missing here.
Thanks for any suggestion.
AA.
I am using R version 2.4.0 (windows XP) and zoo package version
1.2-1(2006-09-20).
#-Begin--
# Let's take the previous example to create a simple zoo data.
z - structure(c(21,34,33,41,39,38,37,28,33,40),
index = structure(c(8044,8051,8058,8065,8072,
8079,8086,8093,8100,8107), class=Date), class = zoo)
# generating 3 random vectors with the same length as z.
jx1 - rnorm(10); jx2 - rnorm(10); jx3 - rnorm(10)
# create a zoo class data using the random vectors.
jx - cbind(jx1,jx2,jx3)
z1 - zoo(jx, index(z))
# now we just repete the previous example.
plot(z1, xaxt = n)
axis(1, time(z1), lab = FALSE)
jd - time(z1)[seq(1, dim(z1)[[1]], 3)]
axis(1,jd, as.character(jd),cex.axis = 0.8, tcl = -0.7, las = 2)
# now we extract one column of z1 and graph it with the same axis commands.
z2 - z1[,1, drop = F]
windows()
plot(z2, xaxt = n)
axis(1, time(z2), lab = FALSE)
jd - time(z2)[seq(1, dim(z2)[[1]], 3)]
axis(1,jd, as.character(jd),cex.axis = 0.8, tcl = -0.7, las = 2)
#---End---
- Original Message
From: Gabor Grothendieck [EMAIL PROTECTED]
To: ahmad ajakh [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Tuesday, December 26, 2006 11:17:44 PM
Subject: Re: [R] plotting time series with zoo pckg
Please read the last line of every message to r-help and follow that.
On 12/26/06, ahmad ajakh [EMAIL PROTECTED] wrote:
Dear Gabor,
Thank you for your quick reply.
This solution works for my univariate zoo class time series. I first tried
it for a timeseries with 4 columns of data, it did not plot the labels nor
the
ticks, I tried it on a one dim timeseries (one column zoo class data as the
example
in the question) and it worked! is there something that I am missing?
Thanks again.
AA.
- Original Message
From: Gabor Grothendieck [EMAIL PROTECTED]
To: ahmad ajakh [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Tuesday, December 26, 2006 8:31:07 PM
Subject: Re: [R] plotting time series with zoo pckg
Try this:
# test data
library(zoo)
z - structure(c(21, 34, 33, 41, 39, 38, 37, 28, 33, 40),
index = structure(c(8044, 8051, 8058, 8065, 8072, 8079, 8086,
8093, 8100, 8107), class = Date), class = zoo)
z
# plot without X axis
plot(z, xaxt = n)
# unlabelled tick at each point
axis(1, time(z), lab = FALSE)
# labelled tick every third point
dd - time(z)[seq(1, length(z), 3)]
axis(1, dd, as.character(dd), cex.axis = 0.7, tcl = -0.7)
On 12/26/06, ahmad ajakh [EMAIL PROTECTED] wrote:
Hi all,
I am using the zoo package to plot time series. I have a problem with
formatting the axes.
my zoo object (z) looks like the following.
c1
1992-01-10 21
1992-01-17 34
1992-01-24 33
1992-01-31 41
1992-02-07 39
1992-02-14 38
1992-02-21 37
Re: [R] stacked plots
If this is time series data try library(zoo) example(plot.zoo) example(xyplot.zoo) to see if any of those fit your requirements. On 12/27/06, BBands [EMAIL PROTECTED] wrote: Dear helpeRs, Is there a better method of producing stacked charts than par(mfrow(3,1)), plot(x), plot(y), plot(z)? What I would like to do is produce a chart of several panes stacked vertically with no space between them so they appeared to be a single figure. I've attached a small example, though it is not clear that it will make it, as the posting guide doesn't say which sort of images are allowed--it is a gif. My data will be in zoo objects like those from get.hist.quote() with the data for the extra panes in additional columns. Thanks in advance, jab -- John Bollinger, CFA, CMT www.BollingerBands.com If you advance far enough, you arrive at the beginning. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] stacked plots
John, On 27 December 2006 at 08:36, BBands wrote: | Dear helpeRs, | | Is there a better method of producing stacked charts than | par(mfrow(3,1)), plot(x), plot(y), plot(z)? What I would like to do is | produce a chart of several panes stacked vertically with no space | between them so they appeared to be a single figure. I've attached a | small example, though it is not clear that it will make it, as the | posting guide doesn't say which sort of images are allowed--it is a | gif. My data will be in zoo objects like those from get.hist.quote() | with the data for the extra panes in additional columns. Do you remember the bollingerBands example we worked on a few years ago and that is still at Romain's incredible R Graph Gallery at http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=65 It uses layout, you can also use the simpler par(mfrow=...) approach *if* you also reduce bottom and top spacing accordingly as e.g. in the plot functions in the script referenced above. Cheers, Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] counties in different colours using map()
since dr. bivand seems to be away, i will try to help a little :-)
i did not work with the database that comes with maps, but with shapefiles from
http://www.census.gov/geo/www/cob/co2000.html#shp. you need to merge the
dataframe that you have with the shapefile.
do you have access to gis software? the merging can be done with r but it is
quite cumbersome.
The road to plot the map could be:
_
require(maptools)
a=read.shape(your_merged_file.shp, dbf.data=TRUE, verbose=TRUE)
attach(a)
win.graph(width=5,height=3, pointsize=8)
plot.Map(a, auxvar=the_variable_to_plot_from_dbf_file, xlab= , ylab= ,
ol=NA)
__
more about this at: http://cran.r-project.org/src/contrib/Views/Spatial.html
hth,
Mihai Nica
170 East Griffith St. G5
Jackson, MS 39201
601-914-0361
- Original Message
From: Tord Snäll [EMAIL PROTECTED]
To: R-help@stat.math.ethz.ch
Sent: Wednesday, December 27, 2006 5:12:04 AM
Subject: [R] counties in different colours using map()
Hi,
I would like to plot a map of US counties using different colors. map()
seems to be the function to use, e.g.
library(maps); map('usa'); map('county', 'colorado', add=T,fill = T,
col=c(1:5))
plots Colorado counties using colours 1 to 5.
However, I want each color to represent a certain value - a value to be
picked from a data frame.
This code should show a correspoding map at the level of states:
state.names - system('tr [A-Z] [a-z]', state.name)
map.states - unix('sed s/:.*//', map(names=T,plot=F))
state.to.map - match(map.states, state.names)
color- votes.repub[state.to.map, votes.year = 1900] / 100
map('state', fill=T, col=color); map('state', add=T)
It is copied from page 6 in
Richard A. Becker, and Allan R. Wilks, Maps in S, ATT Bell
Laboratories Statistics Research Report [93.2], 1993.
http://public.research.att.com/areas/stat/doc/93.2.ps
I also wonder whether the county names are available in the database
used by map(), and, if yes, how to extract or utilize them.
Thanks!
Tord
--
Tord Snäll
Department of Conservation Biology
Swedish University of Agricultural Sciences (SLU)
P.O. 7002, SE-750 07 Uppsala, Sweden
Office/Mobile/Fax
+46-18-672612/+46-730-891356/+46-18-673537
E-mail: [EMAIL PROTECTED]
www.nvb.slu.se/staff_tordsnall
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to suppress a loading required package: ... message
On 27 December 2006 at 08:52, BBands wrote: | On 12/27/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: | Hi, | | how to suppress a loading required package:... message? | | require(package, quiet=TRUE) Some packages insist on talking even when they are asked to be quiet, in which case I have also resorted to wrapping sink() around the loading: sink(/dev/null) library(Hmisc) sink() cat(Hi again\n) Hi again Hth, Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to suppress a loading required package: ... message
Or try: invisible(capture.output(library(Hmisc))) On 12/27/06, Dirk Eddelbuettel [EMAIL PROTECTED] wrote: On 27 December 2006 at 08:52, BBands wrote: | On 12/27/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: | Hi, | | how to suppress a loading required package:... message? | | require(package, quiet=TRUE) Some packages insist on talking even when they are asked to be quiet, in which case I have also resorted to wrapping sink() around the loading: sink(/dev/null) library(Hmisc) sink() cat(Hi again\n) Hi again Hth, Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] stacked plots
On 12/27/06, Dirk Eddelbuettel [EMAIL PROTECTED] wrote: Do you remember the bollingerBands example we worked on a few years ago and that is still at Romain's incredible R Graph Gallery at http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=65 It uses layout, you can also use the simpler par(mfrow=...) approach *if* you also reduce bottom and top spacing accordingly as e.g. in the plot functions in the script referenced above. I do indeed remember that, it was a nice piece of work that I learned a lot from. At the time you mentioned there were some newer methods in the works that might serve better, which prompted my question after an appropriate delay. ;-) jab -- John Bollinger, CFA, CMT www.BollingerBands.com If you advance far enough, you arrive at the beginning. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] stacked plots
Hi John, I cannot see the attached file but if you read the vignette of the zoo package there is an example with Lucent stock price (High Low Open Close) doing what you want. the command plot(z) (z being the zoo multivariate object) produces the graph that you want I guess. Also there are some posts today on how to label them. good luck AA. - Original Message From: BBands [EMAIL PROTECTED] To: R-Help r-help@stat.math.ethz.ch Sent: Wednesday, December 27, 2006 11:36:00 AM Subject: [R] stacked plots Dear helpeRs, Is there a better method of producing stacked charts than par(mfrow(3,1)), plot(x), plot(y), plot(z)? What I would like to do is produce a chart of several panes stacked vertically with no space between them so they appeared to be a single figure. I've attached a small example, though it is not clear that it will make it, as the posting guide doesn't say which sort of images are allowed--it is a gif. My data will be in zoo objects like those from get.hist.quote() with the data for the extra panes in additional columns. Thanks in advance, jab -- John Bollinger, CFA, CMT www.BollingerBands.com If you advance far enough, you arrive at the beginning. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to suppress a loading required package: ... message
Dirk Eddelbuettel wrote: On 27 December 2006 at 08:52, BBands wrote: | On 12/27/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: | Hi, | | how to suppress a loading required package:... message? | | require(package, quiet=TRUE) Some packages insist on talking even when they are asked to be quiet, in which case I have also resorted to wrapping sink() around the loading: sink(/dev/null) library(Hmisc) For that one, do options(Hverbose=FALSE) before library(Hmisc) Frank sink() cat(Hi again\n) Hi again Hth, Dirk -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to write string dynamicly?
Hi everyone: I'm trying to compose a string dynamicly for the parameter input of some function. For example: In package MASS, function lda() require to input the name of predictor variable. Let's say the 16th column is the predictor variable. Then we call the function like this: lda(V16~., data=mydata). I don't want to hard-code the call, instead, I would like to use a dynamic expression for this parameter so that I can use my program on different set of data. I guess there are some function that can do this, but I didn't find it in Introduction to R so far, could someone please tell me this kind of function? Thank you! Best, Feng __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Consensus Fork Index
Dear All, I am trying to automate series of dendrograms for binary data in R as well as calculate the Consensus Fork Index (Colless' index). I'll appreciate any assistance with regards to this. Thank you and best regards, Taiwo __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to debug R program?
Hi everyone: I wrote a R program which has loops. When I run the program, it crashed. I would like to identify in which loop the pragram crashed, how can I debug ? I'm new to R, could somebody please give me a general idea about debugging in R.(I'm a C/C++ programmer and have general knowledge about program debugging.) Thank you! Best, Feng __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] model comparison + use of offset in glmmPQL
Hi, I have 2 questions. First - is it possible to use the offset term in a glmmPQL formula rather than transforming the variables in the dataset beforehand? Second - how do you compare the output/fit of 2 models produced through glmmPQL if you can't use the Anova(model1, model2, test=) method? __ [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Question about predict function
I am working with a non-parametic smoothing operation using a Generalized Additive Model. It is a bivariate data set. I know how to do the smooth, and out comes a nice smooth curve. Now I want to find the value of the smoothed curve for several values of x (the abscissa). This can be done (please correct me if I am wrong) by using the predict.gam function. You feed the predict.gam function a data frame, telling it what you want. Let's start with the predict.gam function. Are you supposed to be able to look up how to use it? E.g., what goes into the various columns of the data frame? I do have a working function call for predict. It says: pred_out - predict (mod, data.frame (x = x), type = response) (mod is the model) Now, if you tell me that x = x, I will believe you. But what is meant by data.frame (x = x), I know not. Or would it better to call the class, names, and str functions, using some well chosen objects? Tom Jones __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R-help Digest, Vol 46, Issue 27
On Wednesday 27 December 2006 06:00, [EMAIL PROTECTED] wrote:
jingjiangyan
I agree, you can use 'assign'. To be more explicit, you could use the
following function.
jingjiangyan -
function(formula, data)
{
m - match.call()
%,% - function(x,y)paste(x,y,sep=)
d.nm - as.character(m$data)
y.nm - as.character(formula[[2]])
x.nm - as.character(formula[[3]])
for(i in levels(data[[x.nm]])){
var.name - d.nm %,% . %,% i
var.val - data[[y.nm]][data[[x.nm]]==i]
cmd - var.name %,% - %,% var.val
eval(cmd)
assign(var.name, var.val, globalenv())
}
}
Next, assuming the data.frame listed in the previous posting, 'df'
exists in your workspace, the call
jingjiangyan(bb ~ aa, data=df)
would produce the desired results.
Cheers,
Grant Izmirlian
--
Հրանդ Իզմիրլյան
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to write string dynamicly?
Try: lda(iris[-5], iris[,5]) On 12/26/06, Feng Qiu [EMAIL PROTECTED] wrote: Hi everyone: I'm trying to compose a string dynamicly for the parameter input of some function. For example: In package MASS, function lda() require to input the name of predictor variable. Let's say the 16th column is the predictor variable. Then we call the function like this: lda(V16~., data=mydata). I don't want to hard-code the call, instead, I would like to use a dynamic expression for this parameter so that I can use my program on different set of data. I guess there,- are some function that can do this, but I didn't find it in Introduction to R so far, could someone please tell me this kind of function? Thank you! Best, Feng __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] vectorizing an iterative process.
Geoffrey == Geoffrey Zhu [EMAIL PROTECTED] on Tue, 26 Dec 2006 10:05:55 -0600 writes: Geoffrey I meant x[i] - x[i-1] + y[i-1] and Y[i] - y[i-1] + x[i] below. Geoffrey The mailing list software just keep adding 3D's. Sorry. It's not the mailing list software - not per se at least. Rather it's your own e-mail software's behavior (maybe in conjunction with our (i.e. r-help's) mail server software configuration) If I write x[i] = x[i-i] then I'm pretty sure you won't get an extra 3D when you'll read it. Regards Martin Maechler ETH Zurich (where the R-help list is running from) __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plotting time series with zoo pckg
Thank you Gabor. This is very helpful.
AA.
- Original Message
From: Gabor Grothendieck [EMAIL PROTECTED]
To: ahmad ajakh [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Wednesday, December 27, 2006 12:08:41 PM
Subject: Re: [R] plotting time series with zoo pckg
To do it with plot.zoo one has to create a custom panel.
Another approach is to use xyplot.zoo since that supports custom scales
directly. Note that the ?xyplot.zoo examples contain code that is along the
lines of the xyplot.zoo solution.
Here are examples of both approaches:
library(zoo)
# test data
z - structure(c(21,34,33,41,39,38,37,28,33,40),
index = structure(c(8044,8051,8058,8065,8072,
8079,8086,8093,8100,8107), class=Date), class = zoo)
set.seed(1) # needed to make it reproducible
jx1 - rnorm(10); jx2 - rnorm(10); jx3 - rnorm(10)
# create a zoo class data using the random vectors.
jx - cbind(jx1,jx2,jx3)
z1 - zoo(jx, index(z))
# 1. plot.zoo solution using custom panel function, my.panel
my.panel - function(...) {
lines(...)
if (parent.frame()$j == ncol(z1)) {
# following line only if non-labelled ticks wanted for each point
axis(1, at = time(z1), lab = FALSE)
ix - seq(1, length(z1), 3)
labs - format(time(z1)[ix], %b-%d)
axis(1, at = time(z1)[ix], lab = labs, tcl = -0.7, cex.axis = 0.7)
}
}
plot(z1, panel = my.panel, xaxt = n)
# 2. xyplot.zoo solution
# z1 is from above
library(lattice)
ix - seq(1, length(z1), 3)
labs - format(time(z1)[ix], %b-%d)
xyplot(z1, scales = list(x = list(at = time(z1)[ix], labels = labs)))
# only need following if non-labelled ticks are to be added for each point
trellis.focus(panel, 1, 1, clip.off = TRUE)
panel.axis(bottom, check.overlap = TRUE, outside = TRUE, labels = FALSE,
tck = .7, at = time(z1))
trellis.unfocus()
On 12/27/06, ahmad ajakh [EMAIL PROTECTED] wrote:
Dear Gabor,
sorry for not posting the code. below I have a piece of code that generates
a multivariate zoo data (3 columns) and graphs it using the axis
commands to generate the labels. This does not work. However, if one
extracts one column the labelling works using the same commands!
I cannot figure out what I am missing here.
Thanks for any suggestion.
AA.
I am using R version 2.4.0 (windows XP) and zoo package version
1.2-1(2006-09-20).
#-Begin--
# Let's take the previous example to create a simple zoo data.
z - structure(c(21,34,33,41,39,38,37,28,33,40),
index = structure(c(8044,8051,8058,8065,8072,
8079,8086,8093,8100,8107), class=Date), class = zoo)
# generating 3 random vectors with the same length as z.
jx1 - rnorm(10); jx2 - rnorm(10); jx3 - rnorm(10)
# create a zoo class data using the random vectors.
jx - cbind(jx1,jx2,jx3)
z1 - zoo(jx, index(z))
# now we just repete the previous example.
plot(z1, xaxt = n)
axis(1, time(z1), lab = FALSE)
jd - time(z1)[seq(1, dim(z1)[[1]], 3)]
axis(1,jd, as.character(jd),cex.axis = 0.8, tcl = -0.7, las = 2)
# now we extract one column of z1 and graph it with the same axis commands.
z2 - z1[,1, drop = F]
windows()
plot(z2, xaxt = n)
axis(1, time(z2), lab = FALSE)
jd - time(z2)[seq(1, dim(z2)[[1]], 3)]
axis(1,jd, as.character(jd),cex.axis = 0.8, tcl = -0.7, las = 2)
#---End---
- Original Message
From: Gabor Grothendieck [EMAIL PROTECTED]
To: ahmad ajakh [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Tuesday, December 26, 2006 11:17:44 PM
Subject: Re: [R] plotting time series with zoo pckg
Please read the last line of every message to r-help and follow that.
On 12/26/06, ahmad ajakh [EMAIL PROTECTED] wrote:
Dear Gabor,
Thank you for your quick reply.
This solution works for my univariate zoo class time series. I first tried
it for a timeseries with 4 columns of data, it did not plot the labels nor
the
ticks, I tried it on a one dim timeseries (one column zoo class data as the
example
in the question) and it worked! is there something that I am missing?
Thanks again.
AA.
- Original Message
From: Gabor Grothendieck [EMAIL PROTECTED]
To: ahmad ajakh [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Tuesday, December 26, 2006 8:31:07 PM
Subject: Re: [R] plotting time series with zoo pckg
Try this:
# test data
library(zoo)
z - structure(c(21, 34, 33, 41, 39, 38, 37, 28, 33, 40),
index = structure(c(8044, 8051, 8058, 8065, 8072, 8079, 8086,
8093, 8100, 8107), class = Date), class = zoo)
z
# plot without X axis
plot(z, xaxt = n)
# unlabelled tick at each point
axis(1, time(z), lab = FALSE)
# labelled tick every third point
dd - time(z)[seq(1, length(z), 3)]
axis(1, dd, as.character(dd), cex.axis = 0.7, tcl = -0.7)
On 12/26/06, ahmad ajakh [EMAIL PROTECTED] wrote:
Hi all,
I am using the zoo package to plot time
Re: [R] Question about predict function
Thomas L Jones wrote: I am working with a non-parametic smoothing operation using a Generalized Additive Model. It is a bivariate data set. I know how to do the smooth, and out comes a nice smooth curve. Now I want to find the value of the smoothed curve for several values of x (the abscissa). This can be done (please correct me if I am wrong) by using the predict.gam function. You feed the predict.gam function a data frame, telling it what you want. Let's start with the predict.gam function. Are you supposed to be able to look up how to use it? E.g., what goes into the various columns of the data frame? I do have a working function call for predict. It says: pred_out - predict (mod, data.frame (x = x), type = response) (mod is the model) Now, if you tell me that x = x, I will believe you. But what is meant by data.frame (x = x), I know not. Or would it better to call the class, names, and str functions, using some well chosen objects? Does this help? library(gam) data(kyphosis) kyphosis.gam - gam(Kyphosis ~ s(Age,4) + Number, family = binomial, data=kyphosis) predict(kyphosis.gam, data.frame(Age = c(10,100,200), Number = 3), type=response) 1 2 3 0.019098701 0.248043366 0.005807921 The data frame should contain the predictor variables in the model and the values for the variables should be those at which you want a prediction. Tom Jones __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] counties in different colours using map()
The following example shows how to get/display the county names:
library(maps)
# Get County Data
m - map('county', 'colorado', plot=FALSE)
names(m)
m$names # State,County names
# The names appear to be in alphabetical order by state, e.g.:
m$names[1:3]
[1] colorado,adamscolorado,alamosa colorado,arapahoe
# Show county names on map
map.text('county', 'colorado', proj='bonne', param=45)
# Show county indices on map
map.text('county', 'colorado', proj='bonne', param=45,
labels=paste(1:length(m$names)))
#or perhaps
map.text('county', 'colorado', proj='bonne', param=45,
labels=paste(1:length(m$names)), col=1:length(m$names))
You can use your own labels vector above to show county abbreviations
instead of full names, or other info, if desired.
Once you get the mapping of the counties, you can connect to other sources
of information.
efg
Tord Snäll [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Hi,
I would like to plot a map of US counties using different colors. map()
seems to be the function to use, e.g.
library(maps); map('usa'); map('county', 'colorado', add=T,fill = T,
col=c(1:5))
plots Colorado counties using colours 1 to 5.
However, I want each color to represent a certain value - a value to be
picked from a data frame.
This code should show a correspoding map at the level of states:
state.names - system('tr [A-Z] [a-z]', state.name)
map.states - unix('sed s/:.*//', map(names=T,plot=F))
state.to.map - match(map.states, state.names)
color- votes.repub[state.to.map, votes.year = 1900] / 100
map('state', fill=T, col=color); map('state', add=T)
It is copied from page 6 in
Richard A. Becker, and Allan R. Wilks, Maps in S, ATT Bell
Laboratories Statistics Research Report [93.2], 1993.
http://public.research.att.com/areas/stat/doc/93.2.ps
I also wonder whether the county names are available in the database
used by map(), and, if yes, how to extract or utilize them.
Thanks!
Tord
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Google Desktop Search and R script files
I want to be able to search my saved R script files on my hard drive. Thankfully the files are all saved with the .R filename extension which means that filetype:R in the Google Desktop Search (GDS) box limits the search to those files. Unfortunately if I put any other term in the search box (for example, hist to find scripts where I have created a histogram) then GDS does not find it. It only seems to index based on the filename and not on the contents of the file. So for instance, if I had a file called HistAnalysis.R then the search filetype:R hist would find it. Google web based search is working correctly (As somebody pointed out previously). I am only having the problem with GDS. Has anyone succeeded with what I am attempting to do. -- Farrel Buchinsky [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Google Desktop Search and R script files
If you're on Windows switch to http://www.copernic.com/en/products/desktop-search/index.html , last time I looked it was quite a lot better than Google Desktop Search. -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Farrel Buchinsky Sent: Wednesday, December 27, 2006 4:16 PM To: r-help@stat.math.ethz.ch Subject: [R] Google Desktop Search and R script files I want to be able to search my saved R script files on my hard drive. Thankfully the files are all saved with the .R filename extension which means that filetype:R in the Google Desktop Search (GDS) box limits the search to those files. Unfortunately if I put any other term in the search box (for example, hist to find scripts where I have created a histogram) then GDS does not find it. It only seems to index based on the filename and not on the contents of the file. So for instance, if I had a file called HistAnalysis.R then the search filetype:R hist would find it. Google web based search is working correctly (As somebody pointed out previously). I am only having the problem with GDS. Has anyone succeeded with what I am attempting to do. -- Farrel Buchinsky [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with histograms
I would like to make one histogram combining two value vectors. One vector has more than 50 entries and the other only has 16. Both vectors contain dates in the POSIXct format. I need to be able to highlight the 16 entries from the smaller data set. Is it possible to do this with the hist function? Which function should I use? Sloan -- ___ Surf the Web in a faster, safer and easier way: Download Opera 9 at http://www.opera.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Google Desktop Search and R script files
Go to the gooogle desktop preferences page and install Larry's Any Text File Indexer 1.00 by Larry Gadea (Index any file extension specified as plaintext) then tell it to search the .R and other extensions as plaintext. Google will then index all indicated files on your harddisk and will find them very fast. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Google Desktop Search and R script files
Wonderful. I did it and it works perfectly. Thanks a lot. Richard M. Heiberger [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] Go to the gooogle desktop preferences page and install Larry's Any Text File Indexer 1.00 by Larry Gadea (Index any file extension specified as plaintext) then tell it to search the .R and other extensions as plaintext. Google will then index all indicated files on your harddisk and will find them very fast. -- Farrel Buchinsky, MD Pediatric Otolaryngologist Allegheny General Hospital Pittsburgh, PA __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Google Desktop Search and R script files
Thank you for your advice. I have read quite a bit about Copernic Desktop Search. Nevertheless, I chose not to download yet another indexing program. I am very happy with Google Desktop Search and particularly how well it integrates with many other features. Therefore, when I saw the reply to my original posting that allowed me to simply download a GDS add-on, I decided to go with that. bogdan romocea [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] If you're on Windows switch to http://www.copernic.com/en/products/desktop-search/index.html , last time I looked it was quite a lot better than Google Desktop Search. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with histograms
If you look at the help for the tkBrush function in the TeachingDemos package, the examples show the definition of a function called colhist. That function may do what you want. Hope this helps, -Original Message- From: [EMAIL PROTECTED] on behalf of sloan jones Sent: Wed 12/27/2006 2:49 PM To: r-help@stat.math.ethz.ch Subject: [R] Help with histograms I would like to make one histogram combining two value vectors. One vector has more than 50 entries and the other only has 16. Both vectors contain dates in the POSIXct format. I need to be able to highlight the 16 entries from the smaller data set. Is it possible to do this with the hist function? Which function should I use? Sloan -- ___ Surf the Web in a faster, safer and easier way: Download Opera 9 at http://www.opera.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to test difference in my case?
hello all, I wonder if anyone could give me a hint on which statistical technique I should use and how to carry it out in R in my case. Thanks in advance. My data is composed of two columns, the same numerical variable (continuous) from actual measurement and model prediction. My objective is to compare the data agreement (if there is significant difference) and make conclusions about the model efficiency. Since the measured and predicted variable was based on the same unit, the first test came into my mind was paired t-test. However, the paired difference is not normal (p-value = 0.0048 from SAS proc univariate). In this case, I can either do a wilcoxon signed-rank test or do transformations about the data. I was told that wilcoxon signed-rank test is not as widely recognized as paired t-test in the literature, so I prefer to do transformation. My question is: do I need to do transformations on both columns of original data, or just the paired difference? What transformation is appropriate? I thought about log transformation, but if I find significant (or no significant) difference between the logged data (measured and predicted), can I say there is significant (or no significant) difference between the original data? After this step of analysis, I will convert the continuous numerical data into qualitative categorical ranking (value=1, 2, 3 and 4). Which statistical test and R command should I use to compare the ranking agreement between the actual measurement and prediction? Thank you very much for helping me out. I haven't slept since a long time ago and this is kind of emergency. If there is any confusion about my description, please let me know. Regards, XY __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Query regarding linking R with Matlab
Hi. From what you tell me you manage to start Matlab in the background by calling: Matlab$startServer() but that R fails to connect to Matlab by: matlab - Matlab(host=localhost, port=9998) if (!open(matlab)) throw(Matlab server is not running: waited 30 seconds.) Sorry for not being explicit enough; if no port is given, Matlab$startServer() will setup up Matlab to listen to port (as explained in the help), but then you try to communicate with it via port 9998. I realize that the example might be a bit confusing since it is using port 9998 (for the purpose of illustrating the fact that you can choose another port). Either try the above with Matlab$startServer(port=9998), *or*, maybe simpler: Matlab$startServer() matlab - Matlab(host=localhost) if (!open(matlab)) throw(Matlab server is not running: waited 30 seconds.) Does this work for you? Henrik On 12/27/06, Bhanu Kalyan.K [EMAIL PROTECTED] wrote: library(R.matlab) Loading required package: R.oo R.oo v1.2.3 (2006-09-07) successfully loaded. See ?R.oo for help. R.matlab v1.1.2 (2006-05-08) successfully loaded. See ?R.matlab for help. help(Matlab) Matlab$startServer()Loading required package: R.utils R.utils v0.8.0 (2006-08-21) successfully loaded. See ?R.utils for help. [1] 0 // Here a Matlab window opened but that window couldnot be maximized. matlab - Matlab(host=localhost, port=9998) matlab [1] Matlab: The Matlab host is 'localhost' and communication goes via port 9998. Objects are passed via the local file system (remote=FALSE). The connection to the Matlab server is closed (not opened). setVerbose(matlab, -2) open(matlab) Opens a blocked connection to host 'localhost' (port 9998)... Try #0. Try #1. Try #2. Try #3. Try #4. Try #5. Try #6. Try #7. Try #8. Try #9. Try #10. Try #11. Try #12. There were 12 warnings (use warnings() to see them) Opens a blocked connection to host 'localhost' (port 9998)...done if (!open(matlab)) + throw(Matlab server is not running: waited 30 seconds.) Opens a blocked connection to host 'localhost' (port 9998)... Try #0. Try #1. // I 'stopped' the computation here Warning message: localhost:9998 cannot be opened Opens a blocked connection to host 'localhost' (port 9998)...done This is the output obtained after running the commands. Kindly go through the above commands and help me identify the problem. Regards, Bhanu Kalyan K Bhanu Kalyan K BTech CSE Final Year [EMAIL PROTECTED] Tel :+91-9885238228 __ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R.matlab question
Hi, a follow up after realizing that you might not have started the Matlab application to listen on port 9998. Try: Matlab$startServer(port=9998) and then matlab - Matlab(host=localhost, port=9998) if (!open(matlab)) throw(Matlab server is not running: waited 30 seconds.) Does this help? Henrik On 12/20/06, Henrik Bengtsson [EMAIL PROTECTED] wrote: Hi. On 12/20/06, Aimin Yan [EMAIL PROTECTED] wrote: Does anyone know how to solve this question about R.matlab? I am in windowsXP, my matlab is matlab 7.0.0 19920(R14) thanks, Aimin matlab - Matlab(host=localhost, port=9998) if (!open(matlab)) throw(Matlab server is not running: waited 30 seconds.) Error in list(throw(Matlab server is not running: waited 30 seconds.) = environment, : [2006-12-17 22:26:03] Exception: Matlab server is not running: waited 30 seconds. at throw(Exception(...)) at throw.default(Matlab server is not running: waited 30 seconds.) at throw(Matlab server is not running: waited 30 seconds.) In addition: There were 30 warnings (use warnings() to see them) warnings function (...) UseMethod(warnings) warnings() Warning messages: 1: localhost:9998 cannot be opened 2: localhost:9998 cannot be opened [snip] 30: localhost:9998 cannot be opened This could be because your firewall is blocking R from connecting to Matlab. Try a few different port numbers. I recently learned that the current default port in R.matlab might not be the best one; different port intervals are reserved for different purposes, cf. http://www.iana.org/assignments/port-numbers. That document indicates that a port number in [49152, 65535] might be better. See if this helps. Does someone else knowof a port interval that is more likely to work in general? You can also tell the Matlab object to report more details what it is trying to do by setting the verbosity threshold, i.e. setVerbose(matlab, threshold=-1); the lower the threshold the more details you'll see. Cheers Henrik __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to write string dynamicly?
Generally, you can create formula from string: lda(formula(paste(names(iris)[5],~.)),iris) On 12/28/06, Gabor Grothendieck [EMAIL PROTECTED] wrote: Try: lda(iris[-5], iris[,5]) On 12/26/06, Feng Qiu [EMAIL PROTECTED] wrote: Hi everyone: I'm trying to compose a string dynamicly for the parameter input of some function. For example: In package MASS, function lda() require to input the name of predictor variable. Let's say the 16th column is the predictor variable. Then we call the function like this: lda(V16~., data=mydata). I don't want to hard-code the call, instead, I would like to use a dynamic expression for this parameter so that I can use my program on different set of data. I guess there,- are some function that can do this, but I didn't find it in Introduction to R so far, could someone please tell me this kind of function? Thank you! Best, Feng __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to suppress a loading required package: ... message
Frank, On 27 December 2006 at 12:30, Frank E Harrell Jr wrote: | Dirk Eddelbuettel wrote: | On 27 December 2006 at 08:52, BBands wrote: | | On 12/27/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: | | how to suppress a loading required package:... message? | | | | require(package, quiet=TRUE) | | Some packages insist on talking even when they are asked to be quiet, in | which case I have also resorted to wrapping sink() around the loading: | | sink(/dev/null) | library(Hmisc) | | For that one, do options(Hverbose=FALSE) before library(Hmisc) With all due respect, I think you are misguided here. Per-package options for verbosity strike me as suboptimal. IMHO, if options(verbose) is FALSE, or if the quiet argument to require() has been given, Hmisc should simply be quiet. In any event, Gabor's one-liner is preferable here as it is generic. Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Dates in R
Hello all, Can somebody point me to references or provide some code on dealing with this date issue. Basically, I have two vectors of values that represent dates. I want to convert these values into a date format and subtract the differences to show elapsed time in days. More specifically, here is the example: Date1 Date2 032398061585 032398061585 111694101994 111694101994 062695021595 051898111597 072495040195 072495040195 The dates are in the mmddyy format, but when I attempt to format these in R with the function, date.mmddyy(Date1), I get very odd results. Any help on this matter would be greatly appreciated! Thanks in advance, Brian [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dates in R
Try this:
x - 'Date1 Date2
+ 032398061585
+ 032398061585
+ 111694101994
+ 111694101994
+ 062695021595
+ 051898111597
+ 072495040195
+ 072495040195'
data.in - read.table(textConnection(x), header=TRUE,
colClasses=c('character', 'character'))
data.in$Date1 - as.POSIXct(strptime(data.in$Date1, %m%d%y))
data.in$Date2 - as.POSIXct(strptime(data.in$Date2, %m%d%y))
data.in$diff - difftime(data.in$Date1, data.in$Date2, units='days')
data.in
Date1 Date2 diff
1 1998-03-23 1985-06-15 4664
2 1998-03-23 1985-06-15 4664
3 1994-11-16 1994-10-19 28
4 1994-11-16 1994-10-19 28
5 1995-06-26 1995-02-15 131
6 1998-05-18 1997-11-15 184
7 1995-07-24 1995-04-01 114
8 1995-07-24 1995-04-01 114
On 12/27/06, Brian Edward [EMAIL PROTECTED] wrote:
Hello all,
Can somebody point me to references or provide some code on dealing with
this date issue. Basically, I have two vectors of values that represent
dates. I want to convert these values into a date format and subtract the
differences to show elapsed time in days. More specifically, here is the
example:
Date1 Date2
032398061585
032398061585
111694101994
111694101994
062695021595
051898111597
072495040195
072495040195
The dates are in the mmddyy format, but when I attempt to format these in
R
with the function, date.mmddyy(Date1), I get very odd results. Any help
on
this matter would be greatly appreciated!
Thanks in advance,
Brian
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem you are trying to solve?
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dates in R
Try: as.date(Date1)-as.date(Date2) On 12/28/06, Brian Edward [EMAIL PROTECTED] wrote: Hello all, Can somebody point me to references or provide some code on dealing with this date issue. Basically, I have two vectors of values that represent dates. I want to convert these values into a date format and subtract the differences to show elapsed time in days. More specifically, here is the example: Date1 Date2 032398061585 032398061585 111694101994 111694101994 062695021595 051898111597 072495040195 072495040195 The dates are in the mmddyy format, but when I attempt to format these in R with the function, date.mmddyy(Date1), I get very odd results. Any help on this matter would be greatly appreciated! Thanks in advance, Brian [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dates in R
See the documentation for as.Date Something like (untested): as.Date(Date1, '%m%d%y) - as.Date(Date2,%m%d%y) That's assuming Date1 and Date2 have already been loaded into R, and are character vectors. They would have to be character vectors in order to display the leading zero, as in 032398. -Don At 10:18 PM -0600 12/27/06, Brian Edward wrote: Hello all, Can somebody point me to references or provide some code on dealing with this date issue. Basically, I have two vectors of values that represent dates. I want to convert these values into a date format and subtract the differences to show elapsed time in days. More specifically, here is the example: Date1 Date2 032398061585 032398061585 111694101994 111694101994 062695021595 051898111597 072495040195 072495040195 The dates are in the mmddyy format, but when I attempt to format these in R with the function, date.mmddyy(Date1), I get very odd results. Any help on this matter would be greatly appreciated! Thanks in advance, Brian [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- - Don MacQueen Lawrence Livermore National Laboratory Livermore, CA, USA __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to suppress a loading required package: ... message
Dirk Eddelbuettel wrote: Frank, On 27 December 2006 at 12:30, Frank E Harrell Jr wrote: | Dirk Eddelbuettel wrote: | On 27 December 2006 at 08:52, BBands wrote: | | On 12/27/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: | | how to suppress a loading required package:... message? | | | | require(package, quiet=TRUE) | | Some packages insist on talking even when they are asked to be quiet, in | which case I have also resorted to wrapping sink() around the loading: | | sink(/dev/null) | library(Hmisc) | | For that one, do options(Hverbose=FALSE) before library(Hmisc) With all due respect, I think you are misguided here. Per-package options for verbosity strike me as suboptimal. IMHO, if options(verbose) is FALSE, or if the quiet argument to require() has been given, Hmisc should simply be quiet. Dirk, I spent a significant amount of time a year ago trying to get the more general approach to work but to no avail. As I recall there either was missing documentation or some failure of the system to pass a correct argument to .First.lib. Hence the creation of the workaround. Frank In any event, Gabor's one-liner is preferable here as it is generic. Dirk -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Colored Dendrogram
In your colLab function, you are assigning the label colors using the mycols 4-element vector, which you generated using the rainbow() function. So, to pick colors using the values in vector c, normalize c and then try the colorRamp function. I don't know how to use that function, myself, so instead I did: # Generate lots of colors, starting at blue (0.6) and ending at red (0) # We started at blue because you wanted blue for the lower-valued elements # and red for the higher-valued elements. pal - rainbow(4000,start=0.5,end=0.05) # Normalize vector c to the range 0-1 (well, not really) # and scale the range to a color index mycols - pal[length(pal)*c/max(c)] Then define your colLab function and run. On Tue, 26 Dec 2006 11:20:45 -0500, Om [EMAIL PROTECTED] wrote: In this dendrogram, each leaf is colored differently. But, I do not want the leaf colored on a random basis. I want to assign color for leaf on basis of some criterion. My actual problem: How to generate a dendrogram with the leaf colored according to the values in the second matrix (which is 4x1 dim, say matrix c)? Meaning, the leaf 1 and 2 should be colored in neighboring spectra of color (say different shade of red) and leaf 3 and 4 in a different color (say different shade of blue) Below is R code and data matrix __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] importing bitmap images to R
All - I'm creating some plots in R that I would like to overlay on images that are created outside of R. I've used image before to plot image-like data within R, and have added vector plots on top of them, but I can't find a way to read in an external bitmap file into R and use that. Currently the external files are .png's, but I could generate a few other types, if something else might be easier to import. Thanks, -Mike __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Re : how to test difference in my case?
How about fitted statistics of the model ? Justin BEM Elève Ingénieur Statisticien Economiste BP 294 Yaoundé. Tél (00237)9597295. - Message d'origine De : Xu Yuan [EMAIL PROTECTED] À : r-help@stat.math.ethz.ch Envoyé le : Jeudi, 28 Décembre 2006, 1h14mn 35s Objet : [R] how to test difference in my case? hello all, I wonder if anyone could give me a hint on which statistical technique I should use and how to carry it out in R in my case. Thanks in advance. My data is composed of two columns, the same numerical variable (continuous) from actual measurement and model prediction. My objective is to compare the data agreement (if there is significant difference) and make conclusions about the model efficiency. Since the measured and predicted variable was based on the same unit, the first test came into my mind was paired t-test. However, the paired difference is not normal (p-value = 0.0048 from SAS proc univariate). In this case, I can either do a wilcoxon signed-rank test or do transformations about the data. I was told that wilcoxon signed-rank test is not as widely recognized as paired t-test in the literature, so I prefer to do transformation. My question is: do I need to do transformations on both columns of original data, or just the paired difference? What transformation is appropriate? I thought about log transformation, but if I find significant (or no significant) difference between the logged data (measured and predicted), can I say there is significant (or no significant) difference between the original data? After this step of analysis, I will convert the continuous numerical data into qualitative categorical ranking (value=1, 2, 3 and 4). Which statistical test and R command should I use to compare the ranking agreement between the actual measurement and prediction? Thank you very much for helping me out. I haven't slept since a long time ago and this is kind of emergency. If there is any confusion about my description, please let me know. Regards, XY __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ___ interface révolutionnaire. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
