Re: [R] help with R
li li-13 wrote:
Hi all,
I want to solve the following equation for x with rho - 0.5
pnorm(-x)*pnorm((rho*dnorm(x)/pnorm(x)-x)/sqrt(1-rho^2))==0.05
Is there a function in R to do this?
Or if you wish to try different values of rho
f - function(x, rho) {
pnorm(-x)*pnorm((rho*dnorm(x)/pnorm(x)-x)/sqrt(1-rho^2))-0.05
}
uniroot(f,c(-3,3),rho=.5)
uniroot(f,c(-3,3),rho=.3)
/Berend
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[R] How to output text to sink without initial line number [1], and data frame line by line without column names
I want to output a text file assembeld from various soruces within r (
actually as a genepop file)
the output should be formatted line 1 text comment
line 2:n selected
column names from data frame
line n+1on lines of
selected columns from data frame one row at a time
I have the following code, but cannot see how to remove the line numbers
and omit column names form the line by line data frame output
col1-c(2,45,67)
col2-c(a,B,C)
col3-c(234,44,566)
mydf-as.data.frame(cbind(col1,col2,col3))
n-ncol(mydf)
nr-nrow(mydf)
sink(test.txt)
print(I will be including text of various sorts in this file so cannot
use print table or similar command)
for (i in 1:n){
print(colnames(mydf[i]),quote=F) }
for (j in 1:nr){
print(mydf[j,c(2:n)],quote=F,row.names=F)}
sink()
The test.txt contains:
[1] I will be including text of various sorts in this file so cannot
use print table or similar command
[1] col1
[1] col2
[1] col3
col2 col3
a 234
col2 col3
B 44
col2 col3
C 566
what I would like in test.txt is:
I will be including text of various sorts in this file so cannot use
print table or similar command
col1
col2
col3
a 234
B 44
C 566
Many thanks
Nevil Amos
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Re: [R] How to output text to sink from data frame line by line without column names
OK I see how to remove the line numbers[1] etc by using cat instead of
print, but cannot work out how to remove the column names from the data
frame output
On 13/06/2010 4:21 PM, Nevil Amos wrote:
I want to output a text file assembeld from various soruces within r (
actually as a genepop file)
the output should be formatted line 1 text comment
line 2:n selected
column names from data frame
line n+1on lines
of selected columns from data frame one row at a time
I have the following code, but cannot see how to remove the line
numbers and omit column names form the line by line data frame output
col1-c(2,45,67)
col2-c(a,B,C)
col3-c(234,44,566)
mydf-as.data.frame(cbind(col1,col2,col3))
n-ncol(mydf)
nr-nrow(mydf)
sink(test.txt)
print(I will be including text of various sorts in this file so
cannot use print table or similar command)
for (i in 1:n){
print(colnames(mydf[i]),quote=F) }
for (j in 1:nr){
print(mydf[j,c(2:n)],quote=F,row.names=F)}
sink()
The test.txt contains:
[1] I will be including text of various sorts in this file so cannot
use print table or similar command
[1] col1
[1] col2
[1] col3
col2 col3
a 234
col2 col3
B 44
col2 col3
C 566
what I would like in test.txt is:
I will be including text of various sorts in this file so cannot use
print table or similar command
col1
col2
col3
a 234
B 44
C 566
Many thanks
Nevil Amos
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Re: [R] HOW to install RSQLite database
Yes i am asking how to install RSQLite packagein windows.Please help on this
regards
On 6/11/10, david.jessop [via R]
ml-node+2251498-1601505055-288...@n4.nabble.comml-node%2b2251498-1601505055-288...@n4.nabble.com
wrote:
Are you asking how to install the RSQLite package or how to create a
SQLite database? The two are somewhat distinct questions. RSQLite is
just a package of functions for R to be able to access data in an SQLite
database. There isn't a separate SQLite program - just a library that is
compiled into RSQLite.
Regards
David
-Original Message-
From: [hidden
email]http://user/SendEmail.jtp?type=nodenode=2251498i=0[mailto:[hidden
email] http://user/SendEmail.jtp?type=nodenode=2251498i=1]
On Behalf Of vijaysheegi
Sent: 10 June 2010 16:22
To: [hidden email] http://user/SendEmail.jtp?type=nodenode=2251498i=2
Subject: [R] HOW to install RSQLite database
Please let me know where i have to type below thing to RSQLite database
get installed.Please let me know the solution.Thanks in advance
RSQLite -- Embedding the SQLite engine in R
(The RSQLite package includes a recent copy of the SQLite distribution
from http://www.sqlite.org http://www.sqlite.org/?by-user=t.)
Installation
There are 3 alternatives for installation:
1. Simple installation:
R CMD INSTALL RSQLite-.tar.gz
the installation automatically detects whether SQLite is
available in any of your system directories; if it's not
available, it installs the SQLite engine and the R-SQLite
interface under the package directory $R_PACKAGE_DIR/sqlite.
2. If you have SQLite installed in a non-system directory (e.g,
in $HOME/sqlite),
a) You can use
export PKG_LIBS=-L$HOME/sqlite/lib -lsqlite
export PKG_CPPFLAGS=-I$HOME/sqlite/include
R CMD INSTALL RSQLite-.tar.gz
b) or you can use the --with-sqlite-dir configuration argument
R CMD INSTALL --configure-args=--with-sqlite-dir=$HOME/sqlite \
RSQLite-.tar.gz
3. If you don't have SQLite but you rather install the version we
provide
into a directory different than the RSQLite package, for instance,
$HOME/app/sqlite, use
R CMD INSTALL --configure-args=--enable-sqlite=$HOME/app/sqlite \
RSQLite-.tar.gz
Usage
-
Note that if you use an *existing* SQLite library that resides in a
non-system directory (e.g., other than /lib, /usr/lib, /usr/local/lib)
you may need to include it in our LD_LIBRARY_PATH, prior to invoking R.
For instance
export LD_LIBRARY_PATH=$HOME/sqlite/lib:$LD_LIBRARY_PATH
R
library(help=RSQLite)
library(RSQLite)
(if you use the --enable-sqlite=DIR configuration argument, the SQLite
library is statically linked to the RSQLite R package, and you need not
worry about setting LD_LIBRARY_PATH.)
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[R] Weighted Average application on Summary Dataset
Hi,
I have 2 huge datasets - May and Jun - a miniscule sample of one is given
below. I am trying to do 2 things with these datasets. I need to verify if
the weighted average of variable A for a Reason in Jun is same/different
from the same for May. To do this I am first computing the weighted average
for each SubReason using a function I wrote.
Where I need help is applying the function on both the datasets to arrive at
weighted averages for each SubReason. Then, I would like to know what the
best way would be, to compare the weighted average for a sub reason across 2
datasets to be able to state that there is a difference - t-test,ANOVA?
Would greatly appreciate any help!! The function I wrote for weighted
average computation is given below the dataset.
One of the datasets:
Reason SubReasonA N
A SR11115 29
B SR2734 24
B SR21054 31
A Sr1600 43
A SR31033 60
A Sr11163 30
B SR4732 43
B SR4988 70
A SR3569 25
B SR41073 65
Output I require:
R SR WA_A N (Sum of N)
A SR1912.0098 102
SR3896.5294118 85
B SR2914.3636364 55
SR4957.1966292 178
(Weighted Average
of A for N weights)
# FUNCTION TO CALCULATE THE WEIGHTED AVERAGE FOR A WEIGHTED BY N
WA-function(A,N) {
sp_A-c(A %*% N)
sum_N-sum(N)
WA-sp_A/sum_N
return(WA)
}
Thanks in advance!
Raoul
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Re: [R] calling a function with new inputs every 1 minute
How about Sys.sleep(60) in a loop from 1 to 500? ?Sys.sleep On Saturday 12 June 2010 08:45:32 pm KstuS wrote: I have inputs to a function which are changing all the time - I pull these values from the internet. I then apply a function to the values. What I'd like to do is automate the process so it runs every one minute and adds the output of the function as a new element of a vector. Pseudo code: at Start time: input1_t0, input2_t0, input3_t0 function(input1_t0, input2_t0, input3_t0) function_result_0 at Time T2 (say one minute later) input1_t1, input2_t1, input3_t1 function(input1_t1, input2_t1, input3_t1 function_result_1 ... end_result - c(function_result_0, function_result_1, function_result_2, ..., n) Ideally I'd want to do this every 1 minute for the next 500 minutes. -- Friedrich Schuster Dompfaffenweg 6 69123 Heidelberg __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Get a list of installed commands
I'm pretty new to R, but have experience with other languages, both OO and scripting. I'm trying to add support for R to my text editor of choice and to do this I need a list of installed commands I can markup with XML. I'd then simply feed in the marked up list into my text editor's library and I'm off. I've done this in Stata before using the following command: getcmds using ~/Desktop/StataCommands.txt Does anyone know of a way to do this in R? Any pointers much appreciated. Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to write a customized hclust algorithm in R?
Hello dear R-help mailing list members, I wish to create an hclust object which will be based on a customized hierarchical clustering algorithm, programmed in R. After looking into the hclust function, I noticed that the algorithms themselves are implemented in Fortran. In order for me to implement my algorithm in R, it would be very helpful if there was some example of implementing (even one of) these algorithms in R. Which leads me to my question: *Is there an R implementation of hclust internal Fortran algorithmic functions?* Thanks in advance for any help, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[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] Scope and sapply
Worik R wrote:
...
N - 10
## x simulate a return series
x - runif(N)-.5
## Build an array of cumulative returns of a portfolio starting with $1 as
it changes over time
y - rep(0, length(x))
y[1] - 1+1*x[1]
for(i in 2:N){
y[i] - y[i-1]+y[i-1]*x[i]
}
## y is that return series. Use
test.1 - function(r.in){
v - rep(0, length(r.in))
foo - function(i, r){
if(i == 1){
s - 1
}else{
s - v[i-1]
}
v[i] - s + s*r[i]
return(v[i])
}
return(sapply(1:length(r.in), foo, r.in))
}
How about
cumprod(1+x)
/Berend
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Re: [R] Break in the y-axis
On 06/13/2010 01:48 PM, beloitstudent wrote:
Hello all,
I have been having trouble getting a break in my y-axis. All of my data
points are up around 100-200, but the graph has to start at zero, so i would
like to remove all the white space using a break symbol. I have been able
to get the break and labels to be correct, however, I can't seem to get the
data to match the axis anymore. I must be using the axis.break() in plotrix
incorrectly, but I cannot see where my issue is. This is what I have so
far.
##
library(plotrix)
par(mar=c(6,8,4,4))
###Data
Saline- structure(list(Time = c(-20L, 0, 30L, 45L, 60L, 80L,
110L,140L,200L, 260L, 320L), Average =
c(119.250,118.750,117.500,132.75,151.875,159.75,142.75,160,168,167.125,143),SEM=c(2.211,2.569,2.665,5.435146394,6.208741369,8.363550657,8.51349469,14.30284687,15.93865792,16.76541326,13.796)),
.Names = c(Time (min), Arterial Plasma Glucose (µg/mL), SEM), class =
data.frame, row.names = c(1, 2,3, 4, 5, 6, 7, 8, 9,
10, 11))
Ex- structure(list(Time = c(-20L, 0, 30L, 45L, 60L, 80L, 110L,140L,200L,
260L, 320L), Average =
c(117.500,117.625,117.375,134.5,166.25,173.5,164.25,162.5,160.375,150.25,139.875),SEM
=
c(1.484614978,1.748906364,1.761,5.613395058,9.642063459,9.493284415,8.220804866,8.967059901,11.91626825,11.27169111,10.92915498)),
.Names = c(Time (min), Arterial Plasma Glucose (µg/mL), SEM), class =
data.frame, row.names = c(1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11))
plotted data with error bars
plotCI(x=Saline [,1],y=Saline [,2], uiw=Saline [,3], err=y,
pt.bg=par(bg),pch=19, cex=2.5 ,gap=0, sfrac=0.005,
xlim=c(-20,340),xaxp=c(-20,320,12), xlab=Time (min),
ylim=c(0,200),yaxp=c(0,200,10), ylab=Arterial Plasma\nGlucose (µg/mL),
las=1, axes=FALSE, font.lab=2.2,cex.lab=1.6)
plotCI(x=Ex [,1],y=Ex [,2], uiw=Ex [,3], err=y,pt.bg=white,pch=21,
col=black,cex=2.5 ,gap=0, sfrac=0.005, xlim=c(-20,340),xaxp=c(-20,320,12),
xlab=Time (min), ylim=c(0,200), yaxp=c(0,200,10), ylab=Arterial
Plasma\nGlucose (µg/mL), las=1, font.lab=2.2, axes=FALSE, add=TRUE,
cex.lab=1.9)
##x-axis
axis(1, at=c(-20, 0, 30, 45, 60, 80, 110, 140, 200, 260,
320), lwd=2, font=2, pos=0,cex.axis=.9)
y-axis
axis(2, las=1, at=c(0,40,60,80,100,120, 140), labels=c(0, 100, 120,
140, 160, 180, 200), lwd=2, font=2, pos=-20, cex.axis=1.7)
#axis break
axis.break(2, 20, style=slash)
As you can see, my data does not fit my axis anymore. Any help with this
problem would be fantastic. Thanks!
Hi beloitstudent,
You have two problems here. First, if you want the plotted values to
match the offset labels on the y-axis, you are going to have to apply
the same offset to the y values.
plotCI(x=Saline [,1],y=Saline [,2]-60,...
plotCI(x=Ex [,1],y=Ex [,2]-60,...
The other problem is mine, I guess. I hadn't thought of offset axes when
writing axis.break, so I've added an argument for pos as in the axis
function. The modified function is attached, and will be in the next
version of plotrix. Thanks for bringing my attention to this.
Jim
# axis.break places a break marker at the position breakpos
# in user coordinates on the axis nominated - see axis().
axis.break-function(axis=1,breakpos=NULL,pos=NA,bgcol=white,breakcol=black,
style=slash,brw=0.02) {
# get the coordinates of the outside of the plot
figxy-par(usr)
# flag if either axis is logarithmic
xaxl-par(xlog)
yaxl-par(ylog)
# calculate the x and y offsets for the break
xw-(figxy[2]-figxy[1])*brw
yw-(figxy[4]-figxy[3])*brw
if(!is.na(pos)) figxy-rep(pos,4)
# if no break position was given, put it just off the plot origin
if(is.null(breakpos))
breakpos-ifelse(axis%%2,figxy[1]+xw*2,figxy[3]+yw*2)
if(xaxl (axis == 1 || axis == 3)) breakpos-log10(breakpos)
if(yaxl (axis == 2 || axis == 4)) breakpos-log10(breakpos)
# set up the blank rectangle (left, bottom, right, top)
switch(axis,
br-c(breakpos-xw/2,figxy[3]-yw/2,breakpos+xw/2,figxy[3]+yw/2),
br-c(figxy[1]-xw/2,breakpos-yw/2,figxy[1]+xw/2,breakpos+yw/2),
br-c(breakpos-xw/2,figxy[4]-yw/2,breakpos+xw/2,figxy[4]+yw/2),
br-c(figxy[2]-xw/2,breakpos-yw/2,figxy[2]+xw/2,breakpos+yw/2),
stop(Improper axis specification.))
# get the current setting of xpd
old.xpd-par(xpd)
# don't cut the break off at the edge of the plot
par(xpd=TRUE)
# correct for logarithmic axes
if(xaxl) br[c(1,3)]-10^br[c(1,3)]
if(yaxl) br[c(2,4)]-10^br[c(2,4)]
if(style == gap) {
if(xaxl) {
figxy[1]-10^figxy[1]
figxy[2]-10^figxy[2]
}
if(yaxl) {
figxy[3]-10^figxy[3]
figxy[4]-10^figxy[4]
}
# blank out the gap area and calculate the line segments
if(axis == 1 || axis == 3) {
rect(breakpos,figxy[3],breakpos+xw,figxy[4],col=bgcol,border=bgcol)
xbegin-c(breakpos,breakpos+xw)
ybegin-c(figxy[3],figxy[3])
xend-c(breakpos,breakpos+xw)
yend-c(figxy[4],figxy[4])
if(xaxl) {
xbegin-10^xbegin
xend-10^xend
}
}
else {
rect(figxy[1],breakpos,figxy[2],breakpos+yw,col=bgcol,border=bgcol)
xbegin-c(figxy[1],figxy[1])
Re: [R] Get a list of installed commands
Hi, Take a look at any of the R-editors, like Tinn-R, Emacs-ESS, Eclipse with StatET,... They contain lists you can use. Also the listings package of LaTeX contains a wordlist for R. Getting all installed commands out of R is not doable with a single command as far as I know. R works completely different than Stata; R is a fullblown programming language, not a statistical program. Try to find a list of all installed Perl commands for example... Attached is the recognition file of Tinn-R. Reserved 1 are special keywords, Reserved 2 is a list of the most commonly used commands in the pre-installed packages, and Reserved 3 is a list of common parameters of those functions. It's submitted with Tinn-R under the GNU license, so keep that in mind when using it. But instead of re-inventing the wheel and constructing your own editor, you could take a look at one of those mentioned above. On Windows, I recommend Tinn-R for daily scripting, and Eclipse for developing packages and the likes. Both offer the advantage of direct communication with the R console. Cheers Joris On Sun, Jun 13, 2010 at 9:31 AM, Data Monkey coco.datamon...@gmail.com wrote: I'm pretty new to R, but have experience with other languages, both OO and scripting. I'm trying to add support for R to my text editor of choice and to do this I need a list of installed commands I can markup with XML. I'd then simply feed in the marked up list into my text editor's library and I'm off. I've done this in Stata before using the following command: getcmds using ~/Desktop/StataCommands.txt Does anyone know of a way to do this in R? Any pointers much appreciated. Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joris Meys Statistical consultant Ghent University Faculty of Bioscience Engineering Department of Applied mathematics, biometrics and process control tel : +32 9 264 59 87 joris.m...@ugent.be --- Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Scope and sapply
On Sat, Jun 12, 2010 at 11:40 PM, Worik R wor...@gmail.com wrote: I was careless. Here is a better example of what I am trying to. With the '-' you offered. ?- That was exactly what I needed, thankyou. Just as an aside I am assuming these are just examples to illustrate scope. In reality you would rarely use - and instead would just return the computed value as the value of the function. Also see cumsum and cumprod and check out this page: http://cran.r-project.org/web/views/Finance.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.
Re: [R] Get a list of installed commands
On Sun, Jun 13, 2010 at 3:31 AM, Data Monkey coco.datamon...@gmail.com wrote: I'm pretty new to R, but have experience with other languages, both OO and scripting. I'm trying to add support for R to my text editor of choice and to do this I need a list of installed commands I can markup with XML. I'd then simply feed in the marked up list into my text editor's library and I'm off. I've done this in Stata before using the following command: getcmds using ~/Desktop/StataCommands.txt Does anyone know of a way to do this in R? Any pointers much appreciated. If installed commands means reserved words in R's parser then see: ?Reserved __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Get a list of installed commands
Thanks Joris. Very helpful. I had thought of that, just curious to see if it was possible to get a fresh list in R. After reading your email I think perhaps my wording was a bit loose. I meant commands in the pre-installed packages. So basically, out of the box what commands will R recognize. I had found the full reference manual for R online and parsed the table of contents into a nice XML file I could use as a possible solution. However, your attached file looks a bit more complete (haven't really given much thought to the reason yet). Thanks again for sending it. I'm using Coda on OS X, which is really designed for Web dev, but I like it so much I've added support for Stata and now want R. It's got Terminal bult-in, so I can invoke R through there. Anyway, my ultimate goal is to not only get syntax highlighting, but also autocompletion. If possible in the prediction, also get the method (where applicable) signature. Coda allows all this. I use this a lot when coding in PHP in Coda and C# in Visual Studio - I think Microsoft call it Intellisense or something. Cheers for the help. On 13/06/2010, at 9:47 PM, Joris Meys wrote: Hi, Take a look at any of the R-editors, like Tinn-R, Emacs-ESS, Eclipse with StatET,... They contain lists you can use. Also the listings package of LaTeX contains a wordlist for R. Getting all installed commands out of R is not doable with a single command as far as I know. R works completely different than Stata; R is a fullblown programming language, not a statistical program. Try to find a list of all installed Perl commands for example... Attached is the recognition file of Tinn-R. Reserved 1 are special keywords, Reserved 2 is a list of the most commonly used commands in the pre-installed packages, and Reserved 3 is a list of common parameters of those functions. It's submitted with Tinn-R under the GNU license, so keep that in mind when using it. But instead of re-inventing the wheel and constructing your own editor, you could take a look at one of those mentioned above. On Windows, I recommend Tinn-R for daily scripting, and Eclipse for developing packages and the likes. Both offer the advantage of direct communication with the R console. Cheers Joris On Sun, Jun 13, 2010 at 9:31 AM, Data Monkey coco.datamon...@gmail.com wrote: I'm pretty new to R, but have experience with other languages, both OO and scripting. I'm trying to add support for R to my text editor of choice and to do this I need a list of installed commands I can markup with XML. I'd then simply feed in the marked up list into my text editor's library and I'm off. I've done this in Stata before using the following command: getcmds using ~/Desktop/StataCommands.txt Does anyone know of a way to do this in R? Any pointers much appreciated. Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joris Meys Statistical consultant Ghent University Faculty of Bioscience Engineering Department of Applied mathematics, biometrics and process control tel : +32 9 264 59 87 joris.m...@ugent.be --- Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php Tinn-R_recognized words.r __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] setting the current working directory to the location of the source file
Charles' hint was what I was looking for. Thanks! mg. On Fri, Jun 11, 2010 at 7:40 AM, Charles C. Berry cbe...@tajo.ucsd.eduwrote: cat(print(eval(sys.calls()[[1]][[2]])),file='test.R') See ?sys.calls [[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] setting the current working directory to the location of the source file
On Fri, Jun 11, 2010 at 11:30 AM, Henrik Bengtsson h...@stat.berkeley.eduwrote: Isn't this what source(..., chdir=TRUE) is for? See help(source). not really. Imagine you give someone a script, but you have no control over where and how they run it. They shouldn't be required to put in the chdir parameter. Say, the script outputs a plot to a file (pdf,png,...) and you want the file to be created next to the source script, as the user would expect it. mg. [[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] setting the current working directory to the location of the source file
On Thu, Jun 10, 2010 at 8:33 PM, Marcin Gomulka mrgo...@gmail.com wrote: AFAIK a script run through source() does not have any legit way to learn about it's own location. I need this to make sure that the script will find its datafiles after I move the whole directory. (The datafiles are in the same directory.) Here is a hack I invented to work around it: print(getwd()) source_pathname = get(ofile,envir = parent.frame()) source_dirname = dirname(source_pathname ) setwd(source_dirname) print(getwd()) This variation is nearly the same: https://stat.ethz.ch/pipermail/r-help/2010-May/238804.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.
Re: [R] How to output text to sink from data frame line by line without column names
col1-c(2,45,67)
col2-c(a,B,C)
col3-c(234,44,566)
mydf-as.data.frame(cbind(col1,col2,col3),stringsAsFactors=F)
n-ncol(mydf)
nr-nrow(mydf)
#sink(test.txt)
cat(I will be including text of various sorts in this file so cannot
use print table or similar command)
for (i in 1:n){
cat(colnames(mydf[i]), )
cat(\n) }
for (j in 1:nr){
cat(unlist(mydf[j,c(2:n)]), )
cat(\n)}
#sink()
Cheers
Joris
On Sun, Jun 13, 2010 at 8:41 AM, Nevil Amos nevil.a...@gmail.com wrote:
OK I see how to remove the line numbers[1] etc by using cat instead of
print, but cannot work out how to remove the column names from the data
frame output
On 13/06/2010 4:21 PM, Nevil Amos wrote:
I want to output a text file assembeld from various soruces within r (
actually as a genepop file)
the output should be formatted line 1 text comment
line 2:n selected
column names from data frame
line n+1on lines of
selected columns from data frame one row at a time
I have the following code, but cannot see how to remove the line numbers
and omit column names form the line by line data frame output
col1-c(2,45,67)
col2-c(a,B,C)
col3-c(234,44,566)
mydf-as.data.frame(cbind(col1,col2,col3))
n-ncol(mydf)
nr-nrow(mydf)
sink(test.txt)
print(I will be including text of various sorts in this file so cannot
use print table or similar command)
for (i in 1:n){
print(colnames(mydf[i]),quote=F) }
for (j in 1:nr){
print(mydf[j,c(2:n)],quote=F,row.names=F)}
sink()
The test.txt contains:
[1] I will be including text of various sorts in this file so cannot use
print table or similar command
[1] col1
[1] col2
[1] col3
col2 col3
a 234
col2 col3
B 44
col2 col3
C 566
what I would like in test.txt is:
I will be including text of various sorts in this file so cannot use
print table or similar command
col1
col2
col3
a 234
B 44
C 566
Many thanks
Nevil Amos
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--
Joris Meys
Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
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Re: [R] Can one get a list of recommended packages?
On 13.06.2010 01:09, Dr. David Kirkby wrote: On 06/12/10 05:27 PM, Douglas Bates wrote: On Sat, Jun 12, 2010 at 8:37 AM, Dr. David Kirkby david.kir...@onetel.net wrote: R 2.10.1 is used in the Sage maths project. Several recommended packages (Matrix, class, mgcv, nnet, rpart, spatial, and survival) are failing to build on Solaris 10 (SPARC). Have you checked the dependencies for those packages? Some require GNU make. We used GNU make. We would like to be able to get a list of the recommended packages for R 2.10.1, but ideally via a call to R, so it is not necessary to update that list every time a new version of R is released. We do not want to access the Internet to get this information. Is there a way in R to list the recommended packages? I'm not sure I understand the logic of this. If you are going to build R then presumably you have the tar.gz file which contains the sources for the recommended packages in the subdirectory src/library/Recommended/. Why not get the list from there? The reason is when the version of R gets updated in Sage, then someone will have to check that list again, and more than likely fail to do so, with the result tests will fail since packages do not exist, or worst still we will be unaware they have failed to build properly. Therefore, being able to get them from a command would be useful, but can understand if that is not possible. $ cd ~/src/R-devel/src/library/Recommended/ $ ls *.tgz boot.tgz codetools.tgz lattice.tgz mgcv.tgz rpart.tgz class.tgz foreign.tgz MASS.tgz nlme.tgz spatial.tgz cluster.tgz KernSmooth.tgz Matrix.tgz nnet.tgz survival.tgz OK, thank you for that list. Better still, is there a way to list the recommended packages which have not been installed, so getting a list of any failures? Again, this seems to be a rather convoluted approach. Why not check why the packages don't install properly? R had built, and the failure of the packages to build was not very obvious, since it did not cause make to exit with a non-zero exit code. Nobody had noticed until very recently that there was a problem. Therefore I proposed to make a test of the packages that should have been installed, and ensure they actually all had. You need to be aware that R is just one part of Sage. Building the whole of Sage takes a long time (24 hours on some computers) so needless to say, people will not view every line of error messages. The fact that 'make' succeeded left us a false sense of security, when later it was realsed there were problems when R run its self-tests. Dave But if you really want to sense some security, you should really run make check-all after the installation, particularly since you are on a platform that is not really mainstream any more. Uwe Ligges __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ERROR need finite 'ylim' values
Hello: I use R with MAC I have a simple data table, numeric and text columns, named dt. The table is imported through read.csv from a csv file. Row numbers are automatically assigned, header is set to TRUE. there are 599 rows and several columns. I am trying to plot using the stripchart command: one numeric variable (say dt$fnatg) vs a text column (say dt$pat). dt$pat contains one of 3 values: pos, neg, So I issue the following command: stripchart (dt$fnatg~dt$pat) and works well. it works well also with several options and nuances: stripchart (dt$fnatg ~ dt$pat, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) Now I want my graph to exclude values for which dt$pat == I tried: stripchart (dt$fnatg ~ dt$pat, method =jitter, subset (dt, dt$pat!=),jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) there is no effect: the plot contains the same values as before the I tried first subsetting the table: patonly-(dt, dt$pat!=) which works well in creating a new table excluding the unwanted rows. I have noticed that the new table keeps the same row numbers assigned in the previous table. So row numbers now go 1 to 599 but with some intervals (for example there is no row 475 etc.). then I use: stripchart (patonly$fnatg ~ patonly$pat, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) and I get the following error: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In min(x) : no non-missing arguments to min; returning Inf 2: In max(x) : no non-missing arguments to max; returning -Inf 3: In min(x) : no non-missing arguments to min; returning Inf 4: In max(x) : no non-missing arguments to max; returning -Inf I f I try the same command but I use another text variable (for example patonly$gr) in the same table to split the plot, it now works: stripchart (patonly$fnatg ~ patonly$gr, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) My question is two fold: Why does not the subset command work within the stripchart command? Why the subsetted table cannot be used in the stripchart command, when the plotting variable is the same previously used in the subsetting process? Thank you Giuseppe -- View this message in context: http://r.789695.n4.nabble.com/ERROR-need-finite-ylim-values-tp2253388p2253388.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] Simulating a Poisson Process in R by calling C Code over .Call
Hi
I want to write a C function for the R Code below and call it with .Call:
SimPoisson - function(lambda,tgrid,T2M)
#Simulation eines Poissonprozesses
{
NT - 0
Ni - rep(0,length(tgrid))
tau - 0
sign - 0
if(lambda != 0)
{
i=1
j=1
while (1)
{
EVar - rexp(1,lambda)
sign - sign + EVar
if (sign T2M)
{
break
}
tau[i] - sign
i = i+1
for (j in j:length(tgrid))
{
if (tgrid[j] sign)
{
Ni[j] - NT
}else
{
break
}
}
NT - NT + 1
}
for (j in j:length(tgrid))
{
Ni[j] - NT
}
}
return(list(NT=NT,Ni=Ni,tau=tau))
}
I read the manual writing R extensions over and over again, but i have
no idea, how to solve the problem with tau because i dont no the length
of tau at the begining of the function
Fabian
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[R] Boxplot intervals combining names
Hi R users, This seems like a simple problem but I have searched nabble for the answer and can't seem to find it. All I want to do is produce a boxplot where I have two boxes for one Individual but on the xaxis I only have one tick mark centred between the boxes so I can add the Individuals' name. I have 30 IDs and have shown the code I use below for a couple of IDs, I figure the data is not important here so it's not included. boxplot (ID1[,8],ID1[,9],ID2[,8],ID2[,9],xaxt='n') I have put all the ID names in as 'names1' and I have tried numerous variations on axis, e.g. axis(1,at=1:30,labels=names1) but nothing works: the boxplot appears to 'know' that there are 60 tick marks (data) and therefore only puts ticks half way up the graph, and using: axis(1,at=1:30,labels=names1) complains that there is a difference of length, which of course there is! I must be missing something simple here, but any suggestions would be gratefully received, Ross -- View this message in context: http://r.789695.n4.nabble.com/Boxplot-intervals-combining-names-tp2253442p2253442.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] Clustering algorithms don't find obvious clusters
Henrik, the methods you use are NOT applicable to directed graphs, in the contrary even. They will split up what you want to put together. In your data, an author never cites himself. Hence, A and B are far more different than B and D according to the techniques you use. Please check out Etiennes solution, that is what you want. Cheers Joris On Sat, Jun 12, 2010 at 8:43 PM, Henrik Aldberg henrik.aldb...@gmail.com wrote: Dave, I used daisy with the default settings (daisy(M) where M is the matrix). Henrik On 11 June 2010 21:57, Dave Roberts dvr...@ecology.msu.montana.edu wrote: Henrik, The clustering algorithms you refer to (and almost all others) expect the matrix to be symmetric. They do not seek a graph-theoretic solution, but rather proximity in geometric or topological space. How did you convert y9oru matrix to a dissimilarity? Dave Roberts Henrik Aldberg wrote: I have a directed graph which is represented as a matrix on the form 0 4 0 1 6 0 0 0 0 1 0 5 0 0 4 0 Each row correspond to an author (A, B, C, D) and the values says how many times this author have cited the other authors. Hence the first row says that author A have cited author B four times and author D one time. Thus the matrix represents two groups of authors: (A,B) and (C,D) who cites each other. But there is also a weak link between the groups. In reality this matrix is much bigger and very sparce but it still consists of distinct groups of authors. My problem is that when I cluster the matrix using pam, clara or agnes the algorithms does not find the obvious clusters. I have tried to turn it into a dissimilarity matrix before clustering but that did not help either. The layout of the clustering is not that important to me, my primary interest is the to get the right nodes into the right clusters. Sincerely Henrik [[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. - [[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. -- Joris Meys Statistical consultant Ghent University Faculty of Bioscience Engineering Department of Applied mathematics, biometrics and process control tel : +32 9 264 59 87 joris.m...@ugent.be --- Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Boxplot intervals combining names
Hi Ross, You are indeed missing something simple. If there are 60 bars, the axis runs from 1 to 60, and you want a label at every second one, rather than at every one. Like this (with 10 instead of 60): testdata - matrix(runif(100), ncol=10) boxplot(testdata, xaxt=n) axis(1, at=seq(1.5, 9.5, by=2), labels=letters[1:5]) Sarah On Sun, Jun 13, 2010 at 9:40 AM, RCulloch ross.cull...@dur.ac.uk wrote: Hi R users, This seems like a simple problem but I have searched nabble for the answer and can't seem to find it. All I want to do is produce a boxplot where I have two boxes for one Individual but on the xaxis I only have one tick mark centred between the boxes so I can add the Individuals' name. I have 30 IDs and have shown the code I use below for a couple of IDs, I figure the data is not important here so it's not included. boxplot (ID1[,8],ID1[,9],ID2[,8],ID2[,9],xaxt='n') I have put all the ID names in as 'names1' and I have tried numerous variations on axis, e.g. axis(1,at=1:30,labels=names1) but nothing works: the boxplot appears to 'know' that there are 60 tick marks (data) and therefore only puts ticks half way up the graph, and using: axis(1,at=1:30,labels=names1) complains that there is a difference of length, which of course there is! I must be missing something simple here, but any suggestions would be gratefully received, Ross -- View this message in context: http://r.789695.n4.nabble.com/Boxplot-intervals-combining-names-tp2253442p2253442.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. -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Finding an order for an hclust (dendrogram) object without intersections
Hello all, I manually created an hclust object. Now I am looking to reorder the leafs so they won't intersect with each other, and would be happy for advises on how to do that. Here is an example code: #- a - list() # initialize empty object # define merging pattern: #negative numbers are leaves, #positive are merged clusters (defined by row number in $merge) a$merge - matrix(c(-1, -2, -3, -4, 1, 2, -5,-6, 3,4), nc=2, byrow=TRUE ) a$height - c(1, 1.5, 3,4,4.5)# define merge heights a$order - c(1,4,2,3,6,5) # order of leaves(trivial if hand-entered) a$labels - 1:6# LETTERS[1:4]# labels of leaves class(a) - hclust# make it an hclust object plot(a) # look at the result #- A working order solution in this example would be 1:6. The question is how can I find it. Thanks, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[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] add only selected labels using thigmophobe
hi, I am trying to label data points within a scatter plot using thigmophobe. While the data set consists of about 3000 points I only would like to label a subset of these points. I read the x and y coordinates in from a txt file and define them as: MLPM1-log2(Ratio.M.L.G2.PM) HLG2-log2(Ratio.H.L..G2.PM) the names and a corresponding ID are read in from the same txt file as: GENE-(Gene.Names) ID-(id) plotting is: plot(x=MLPM1,y=HLG2,main='G2 vs PM',xlab='Fold change M/L PM',ylab='Fold change H/L G2)',xlim=c(-7,7),ylim=c(-7,7)) labelling of the whole dataset I can do with: thigmophobe.labels(MLPM1,HLG2,labels=c(GENE),col=c(2:6,8:12),font=2) but how can I label only a few points from the whole dataset with names or IDs ?? Thanks a lot for help and sorry for the simple question. Just started to use R. -- View this message in context: http://r.789695.n4.nabble.com/add-only-selected-labels-using-thigmophobe-tp2253497p2253497.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] Finding an order for an hclust (dendrogram) object without intersections
On Sun, 13 Jun 2010, Tal Galili wrote: Hello all, I manually created an hclust object. Now I am looking to reorder the leafs so they won't intersect with each other, and would be happy for advises on how to do that. Have a look at the code for hclust(). If you can instead create an object like 'hcl', then pass the results to .Fortran(hcass2,...), you should be done. HTH, Chuck Here is an example code: #- a - list() # initialize empty object # define merging pattern: #negative numbers are leaves, #positive are merged clusters (defined by row number in $merge) a$merge - matrix(c(-1, -2, -3, -4, 1, 2, -5,-6, 3,4), nc=2, byrow=TRUE ) a$height - c(1, 1.5, 3,4,4.5)# define merge heights a$order - c(1,4,2,3,6,5) # order of leaves(trivial if hand-entered) a$labels - 1:6# LETTERS[1:4]# labels of leaves class(a) - hclust# make it an hclust object plot(a) # look at the result #- A working order solution in this example would be 1:6. The question is how can I find it. Thanks, Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:cbe...@tajo.ucsd.edu UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Finding an order for an hclust (dendrogram) object without intersections
Thanks Charles.
In the meantime, I found out the following code does the trick.
But I am wondering if:
1) I might have made a mistake in it somewhere
2) If there are other (smarter) ways of going about this.
Here is the solution I wrote:
# -
order.a.tree - function(tree)
{
num.of.leafs - length(tree$order)
for(i in 2:(num.of.leafs-1))
{
tree$order - order( cutree(tree, k = i))
}
return(tree)
}
#Example:
a - list() # initialize empty object
# define merging pattern:
#negative numbers are leaves,
#positive are merged clusters (defined by row number in $merge)
a$merge - matrix(c(-1, -2,
-3, -4,
1, 2,
-5,-6,
3,4), nc=2, byrow=TRUE )
a$height - c(1, 1.5, 3,4,4.5)# define merge heights
a$order - c(1,4,2,3,6,5) # order of leaves(trivial if
hand-entered)
a$labels - 1:6# LETTERS[1:4]# labels of leaves
class(a) - hclust# make it an hclust object
par(mfrow = c(1,2))
plot(a) # look at the result
plot(order.a.tree(a))
# -
Any comments, or other solutions will be very welcomed.
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--
On Sun, Jun 13, 2010 at 7:08 PM, Charles C. Berry cbe...@tajo.ucsd.eduwrote:
On Sun, 13 Jun 2010, Tal Galili wrote:
Hello all,
I manually created an hclust object.
Now I am looking to reorder the leafs so they won't intersect with each
other, and would be happy for advises on how to do that.
Have a look at the code for hclust().
If you can instead create an object like 'hcl', then pass the results to
.Fortran(hcass2,...), you should be done.
HTH,
Chuck
Here is an example code:
#-
a - list() # initialize empty object
# define merging pattern:
#negative numbers are leaves,
#positive are merged clusters (defined by row number in $merge)
a$merge - matrix(c(-1, -2,
-3, -4,
1, 2,
-5,-6,
3,4), nc=2, byrow=TRUE )
a$height - c(1, 1.5, 3,4,4.5)# define merge heights
a$order - c(1,4,2,3,6,5) # order of leaves(trivial if
hand-entered)
a$labels - 1:6# LETTERS[1:4]# labels of leaves
class(a) - hclust# make it an hclust object
plot(a) # look at the result
#-
A working order solution in this example would be 1:6.
The question is how can I find it.
Thanks,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--
[[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.
Charles C. Berry(858) 534-2098
Dept of Family/Preventive
Medicine
E mailto:cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mid-P value for a chi-squared test
On Jun 1, 2010, at 4:17 AM, Wilson, Andrew wrote: Can anyone tell me how to calculate a mid-p value for a chi-squared test in R? I cannot see that this has been answered. It has a date from 12 days ago but I cannot see a reply in the archives. So, what is a mid-p value and which chi-square test are you asking about? A simple data setup in R code with expected output would speed this discussion along. 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] Simulating a Poisson Process in R by calling C Code over .Call
On Sun, 13 Jun 2010, Fabian Zäpernick wrote: Hi I want to write a C function for the R Code below and call it with .Call: SimPoisson - function(lambda,tgrid,T2M) #Simulation eines Poissonprozesses snip return(list(NT=NT,Ni=Ni,tau=tau)) } I read the manual writing R extensions over and over again, but i have no idea, how to solve the problem with tau because i dont no the length of tau at the begining of the function The standard approach is to start off with some reasonable guess at the length of tau and then if the vector fills up, allocate one that is twice as large and copy the current values of tau into the new vector. In this case you can do better, since you have a very good initial guess for how long tau will be. If you make the vector of length qpois(0., lambda), then there is a 99.99% chance that it is long enough. Using lambda+4*sqrt(lambda) is almost as good. Yet another approach is to take advantage of the memoryless property of the Poisson process: set tau to some reasonable length, then if it fill up, return and call the function again to handle the remainder of the duration. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Finding an order for an hclust (dendrogram) object without intersections
o.k,
I found an example where my algorithm can't fix the tree order.
But I don't know how to resolve it.
Here is the code to reproduce the problem:
#
order.a.tree - function(tree)
{
num.of.leafs - length(tree$order)
for(i in 1:(num.of.leafs))
{
tree$order - order( cutree(tree, k = i))
}
return(tree)
}
problematic.tree -
structure(list(merge = structure(c(-3, -24, 1, -25, 4, -27, 5,
-22, 8, 3, 9, -5, 10, 11, 12, 6, 14, 7, 18, 19, 20, 21, 22, 23,
24, 25, -13, -23, -14, -21, -20, -26, -19, -1, 2, -15, -7, -4,
-9, -18, -6, -17, -12, 17, -2, -8, -10, -16, 13, 15, 16, -11), .Dim =
c(26L,
2L)), height = c(0.0833, 0.0867, 0.117,
0.136507936507937, 0.220634920634921, 0.622,
0.674603174603175,
0.823, 1.06349206349206, 1.27698412698413, 1.37,
2.00952380952381, 2.2975, 2.39, 2.686667,
2.9, 3.14736842105263, 3.55634920634921, 3.7921768707483, 3.84183673469388,
3.93817373103087, 4.54464285714286, 4.81438464274599, 5.10895156778615,
5.36142237562854, 6.19122779197967), order = c(1L, 2L, 3L, 4L,
5L, 6L, 7L, 8L, 9L, 10L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L,
20L, 21L, 22L, 23L, 24L, 25L, 26L, 27L, 11L), labels = 1:27), .Names =
c(merge,
height, order, labels), class = hclust)
plot(order.a.tree(problematic.tree ))
#
Thanks,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--
On Sun, Jun 13, 2010 at 7:39 PM, Tal Galili tal.gal...@gmail.com wrote:
Thanks Charles.
In the meantime, I found out the following code does the trick.
But I am wondering if:
1) I might have made a mistake in it somewhere
2) If there are other (smarter) ways of going about this.
Here is the solution I wrote:
# -
order.a.tree - function(tree)
{
num.of.leafs - length(tree$order)
for(i in 2:(num.of.leafs-1))
{
tree$order - order( cutree(tree, k = i))
}
return(tree)
}
#Example:
a - list() # initialize empty object
# define merging pattern:
#negative numbers are leaves,
#positive are merged clusters (defined by row number in $merge)
a$merge - matrix(c(-1, -2,
-3, -4,
1, 2,
-5,-6,
3,4), nc=2, byrow=TRUE )
a$height - c(1, 1.5, 3,4,4.5)# define merge heights
a$order - c(1,4,2,3,6,5) # order of leaves(trivial if
hand-entered)
a$labels - 1:6# LETTERS[1:4]# labels of leaves
class(a) - hclust# make it an hclust object
par(mfrow = c(1,2))
plot(a) # look at the result
plot(order.a.tree(a))
# -
Any comments, or other solutions will be very welcomed.
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--
On Sun, Jun 13, 2010 at 7:08 PM, Charles C. Berry cbe...@tajo.ucsd.eduwrote:
On Sun, 13 Jun 2010, Tal Galili wrote:
Hello all,
I manually created an hclust object.
Now I am looking to reorder the leafs so they won't intersect with each
other, and would be happy for advises on how to do that.
Have a look at the code for hclust().
If you can instead create an object like 'hcl', then pass the results to
.Fortran(hcass2,...), you should be done.
HTH,
Chuck
Here is an example code:
#-
a - list() # initialize empty object
# define merging pattern:
#negative numbers are leaves,
#positive are merged clusters (defined by row number in $merge)
a$merge - matrix(c(-1, -2,
-3, -4,
1, 2,
-5,-6,
3,4), nc=2, byrow=TRUE )
a$height - c(1, 1.5, 3,4,4.5)# define merge heights
a$order - c(1,4,2,3,6,5) # order of leaves(trivial if
hand-entered)
a$labels - 1:6# LETTERS[1:4]# labels of leaves
class(a) - hclust# make it an hclust object
plot(a) # look at the result
#-
A working order solution in this example would be 1:6.
The question is how can I find it.
Thanks,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
[R] Pairwise cross correlation from data set
Dear list,
Following up on an earlier post, I would like to reorder a dataset and
compute pairwise correlations. But I'm having some real problems
getting this done.
My data looks something like:
Participant Stimulus Measurement
p1 s`15
p1 s`26.1
p1 s`37
p2 s`14.8
p2 s`26
p2 s`36.5
p3 s`14
p3 s`27
p3 s`36
As a first step I would imagine that I have to rearrange my data into
a frame more like this
Stimulus p1 p2 p3
s1 5 4.8 4
s2 6.1 67
s3 7 6.5 6
And then do the pairwise correlations between {p1,p2},{p2,p3}.{p2,p3}
I can do all of this manually, i.e., using some messy case specific
code, but can anyone please point out the best way to do this in a
more generalizable way.
Thanks for any help you can give a novice!
Claus
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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] add only selected labels using thigmophobe
joerg wrote:
I am trying to label data points within a scatter plot using thigmophobe.
While the data set consists of about 3000 points I only would like to
label a subset of these points.
I read the x and y coordinates in from a txt file and define them as:
MLPM1-log2(Ratio.M.L.G2.PM)
HLG2-log2(Ratio.H.L..G2.PM)
the names and a corresponding ID are read in from the same txt file as:
GENE-(Gene.Names)
ID-(id)
plotting is:
plot(x=MLPM1,y=HLG2,main='G2 vs PM',xlab='Fold change M/L PM',ylab='Fold
change H/L G2)',xlim=c(-7,7),ylim=c(-7,7))
labelling of the whole dataset I can do with:
thigmophobe.labels(MLPM1,HLG2,labels=c(GENE),col=c(2:6,8:12),font=2)
but how can I label only a few points from the whole dataset with names or
IDs ??
require(plotrix)
#Loading required package: plotrix
?thigmophobe.labels
#starting httpd help server ... done
dta - data.frame(a=rlnorm(20), b=rlnorm(20))
with( dta, plot(a,b))
with( dta[c(2,7,10), ], thigmophobe.labels(a, b, labels=c('A', 'B', 'C') )
)
--
David.
--
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Sent from the R help mailing list archive at Nabble.com.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Count of unique factors within another factor
I have a data frame with two factors (sampling 'unit', 'species'). I want to calculate the number of unique 'species' per 'unit.' I can calculate the number of unique values for each variable separately, but can't get a count for each ‘unit’. data=read.csv(C:/Desktop/sr_sort_practice.csv) attach(data) data[1:10,] unit species 1 123ACMA 2 123LIDE 3 123LIDE 4 123SESE 5 123SESE 6 123SESE 7 345HEAR 8 345LOHI 9 345QUAG 10 345TODI….. sr.unique- lapply (data, unique) $unit [1] 123 345 216 $species [1] ACMA LIDE SESE HEAR LOHI QUAG TODI UMCA ARSP LIDE sapply (sr.unique,length) unit species 3 10 Then, I get stuck here because this unique species count is not given for each ‘unit’. What I'd like to get is: unit species 1233 3454 216-- Thanks-- -- View this message in context: http://r.789695.n4.nabble.com/Count-of-unique-factors-within-another-factor-tp2253545p2253545.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ERROR need finite 'ylim' values
Giuseppe wrote: Hello: I use R with MAC I have a simple data table, numeric and text columns, named dt. The table is imported through read.csv from a csv file. Row numbers are automatically assigned, header is set to TRUE. there are 599 rows and several columns. I am trying to plot using the stripchart command: one numeric variable (say dt$fnatg) vs a text column (say dt$pat). dt$pat contains one of 3 values: pos, neg, So I issue the following command: stripchart (dt$fnatg~dt$pat) and works well. it works well also with several options and nuances: stripchart (dt$fnatg ~ dt$pat, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) Now I want my graph to exclude values for which dt$pat == I tried: stripchart (dt$fnatg ~ dt$pat, method =jitter, subset (dt, dt$pat!=),jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) there is no effect: the plot contains the same values as before the I tried first subsetting the table: patonly-(dt, dt$pat!=) which works well in creating a new table excluding the unwanted rows. I have noticed that the new table keeps the same row numbers assigned in the previous table. So row numbers now go 1 to 599 but with some intervals (for example there is no row 475 etc.). then I use: stripchart (patonly$fnatg ~ patonly$pat, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) and I get the following error: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In min(x) : no non-missing arguments to min; returning Inf 2: In max(x) : no non-missing arguments to max; returning -Inf 3: In min(x) : no non-missing arguments to min; returning Inf 4: In max(x) : no non-missing arguments to max; returning -Inf I f I try the same command but I use another text variable (for example patonly$gr) in the same table to split the plot, it now works: stripchart (patonly$fnatg ~ patonly$gr, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) My question is two fold: Why does not the subset command work within the stripchart command? Why the subsetted table cannot be used in the stripchart command, when the plotting variable is the same previously used in the subsetting process? You appear to have adopted a strategy of using positional matching. Naming your arguments will often result in more informative error messages. Looking at the help page for stripchart, it appears that there is no subset parameter to set in any of its methods and only the formula method has a data argument. It should work with: stripchart(formula1 , data=subset(dta, subset=criteria), rest of arguments preferably named ) Your other option might be to use the with() function: with( subset(patonly, pat!=), stripchart(fnatg ~ gr, ... named arguments) ) HTH. and if it doesn't, then submit a reproducible data example to work with. -- David. -- View this message in context: http://r.789695.n4.nabble.com/ERROR-need-finite-ylim-values-tp2253388p2253560.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] Count of unique factors within another factor
I think ?tapply will help here. But *please* read the posting guide and provide minimal, reproducible examples! Birdnerd wrote: I have a data frame with two factors (sampling 'unit', 'species'). I want to calculate the number of unique 'species' per 'unit.' I can calculate the number of unique values for each variable separately, but can't get a count for each ‘unit’. data=read.csv(C:/Desktop/sr_sort_practice.csv) attach(data) data[1:10,] unit species 1 123ACMA 2 123LIDE 3 123LIDE 4 123SESE 5 123SESE 6 123SESE 7 345HEAR 8 345LOHI 9 345QUAG 10 345TODI….. sr.unique- lapply (data, unique) $unit [1] 123 345 216 $species [1] ACMA LIDE SESE HEAR LOHI QUAG TODI UMCA ARSP LIDE sapply (sr.unique,length) unit species 3 10 Then, I get stuck here because this unique species count is not given for each ‘unit’. What I'd like to get is: unit species 1233 3454 216-- Thanks-- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Count of unique factors within another factor
Hi there, Try with(data, tapply(species, unit, function(x) length(unique(x HTH, Jorge On Sun, Jun 13, 2010 at 12:07 PM, Birdnerd wrote: I have a data frame with two factors (sampling 'unit', 'species'). I want to calculate the number of unique 'species' per 'unit.' I can calculate the number of unique values for each variable separately, but can't get a count for each unit. data=read.csv(C:/Desktop/sr_sort_practice.csv) attach(data) data[1:10,] unit species 1 123ACMA 2 123LIDE 3 123LIDE 4 123SESE 5 123SESE 6 123SESE 7 345HEAR 8 345LOHI 9 345QUAG 10 345TODI .. sr.unique- lapply (data, unique) $unit [1] 123 345 216 $species [1] ACMA LIDE SESE HEAR LOHI QUAG TODI UMCA ARSP LIDE sapply (sr.unique,length) unit species 3 10 Then, I get stuck here because this unique species count is not given for each unit. What I'd like to get is: unit species 1233 3454 216-- Thanks-- -- View this message in context: http://r.789695.n4.nabble.com/Count-of-unique-factors-within-another-factor-tp2253545p2253545.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. [[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] Are any values in one list contained within a second list
Silly question, but, can I test to see if any value of list a is contained in list b without doing a loop? A loop is easy enough, but wanted to see if there was a cleaner way. By way of example: List 1: a, b, c, d, e, f, g List 2: z, y, x, w, v, u, b Return true, since both lists contain b List 1: a, b, c, d, e, f, g List 2: z, y, x, w, v, u, t Return false, since the lists have no mutual values -- View this message in context: http://r.789695.n4.nabble.com/Are-any-values-in-one-list-contained-within-a-second-list-tp2253637p2253637.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] Mid-P value for a chi-squared test
On 13-Jun-10 17:12:45, David Winsemius wrote: On Jun 1, 2010, at 4:17 AM, Wilson, Andrew wrote: Can anyone tell me how to calculate a mid-p value for a chi-squared test in R? I cannot see that this has been answered. It has a date from 12 days ago but I cannot see a reply in the archives. So, what is a mid-p value and which chi-square test are you asking about? A simple data setup in R code with expected output would speed this discussion along. David Winsemius, MD The mid-p value is a device for improving the accuracy of a continuous approximation to a distribution which in reality is discrete. Intuitively, the idea is to treat the discrete probabilities of the discrete distribution as if they were proportions in a histogram. Then imagine fitting a continuous curve (e.g. a chi-squared density) to the histogram. The fit (agreement between the proportion in one histogram bar, and the probability below that portion of the curve which lies in the same range) will be better if the curve goes through the midpoint of the top of the bar. This leads to the formal definition: mid-P = Prob(X Xobs) + 0.5*Prob(X = Xobs) A number of R functions use this idea. Check out what you get by going to http://finzi.psych.upenn.edu/nmz.html and entering mid-p into the search box, and see whether any of them match (or come close to) your particular case. In the case of the chi-squared test, the idea is related to (but not the same as) the Yates correction for continuity. chisq.test() has an option correct=TRUE to force this, but only for 2x2 tables. Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 13-Jun-10 Time: 19:19:44 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Pairwise cross correlation from data set
On Jun 13, 2010, at 1:47 PM, Claus O'Rourke wrote:
Dear list,
Following up on an earlier post, I would like to reorder a dataset and
compute pairwise correlations. But I'm having some real problems
getting this done.
My data looks something like:
Participant Stimulus Measurement
p1 s`15
p1 s`26.1
p1 s`37
p2 s`14.8
p2 s`26
p2 s`36.5
p3 s`14
p3 s`27
p3 s`36
As a first step I would imagine that I have to rearrange my data into
a frame more like this
It is not particularly natural to do the re-ordering to my mind as
long asyou preserve the ordering of the Stimulus variable. Personally,
I would avoid using back-quotes in values as they may have special
syntactic implications.
Stimulus p1 p2 p3
s1 5 4.8 4
s2 6.1 67
s3 7 6.5 6
And then do the pairwise correlations between {p1,p2},{p2,p3}.{p2,p3}
I can do all of this manually, i.e., using some messy case specific
code, but can anyone please point out the best way to do this in a
more generalizable way.
rd.txt
function(txt, header=TRUE, ...) {
rd - read.table(textConnection(txt), header=header, ...)
closeAllConnections()
rd }
dtat - rd.txt(Participant Stimulus Measurement
+ p1 s`15
+ p1 s`26.1
+ p1 s`37
+ p2 s`14.8
+ p2 s`26
+ p2 s`36.5
+ p3 s`14
+ p3 s`27
+ p3 s`36, stringsAsFactors=F)
str(dtat)
'data.frame': 9 obs. of 3 variables:
$ Participant: chr p1 p1 p1 p2 ...
$ Stimulus : chr s`1 s`2 s`3 s`1 ...
$ Measurement: num 5 6.1 7 4.8 6 6.5 4 7 6
combn(unique(dtat$Participant), 2)
[,1] [,2] [,3]
[1,] p1 p1 p2
[2,] p2 p3 p3
apply( combn(unique(dtat$Participant), 2), 2,
# read combinations by columns
function(x) {with(subset(dtat, Participant %in% x),
# used only the desired combo's
cor(Measurement, as.numeric(factor(Stimulus
# needed to turn Stimulus into factor to get an ordering
} )
# [1] 0.9696635 0.7627701 0.7424791
David Winsemius, MD
West Hartford, CT
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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] Are any values in one list contained within a second list
On Jun 13, 2010, at 2:17 PM, GL wrote: Silly question, but, can I test to see if any value of list a is contained in list b without doing a loop? A loop is easy enough, but wanted to see if there was a cleaner way. By way of example: List 1: a, b, c, d, e, f, g List 2: z, y, x, w, v, u, b Return true, since both lists contain b List 1: a, b, c, d, e, f, g List 2: z, y, x, w, v, u, t Return false, since the lists have no mutual values length(intersect(List_1, List_2)) 0 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] Are any values in one list contained within a second list
Hi GL,
Tr this:
# example 1
list1 - list(letters[1:7])
list1
list2 - list(c('z','y','x','w','v','u','b'))
list2
mapply(function(x, y) any(x %in% y), list1, list2)
# example 2
list2 - list(c('z','y','x','w','v','u','t'))
list2
mapply(function(x, y) any(x %in% y), list1, list2)
HTH,
Jorge
On Sun, Jun 13, 2010 at 2:17 PM, GL wrote:
Silly question, but, can I test to see if any value of list a is contained
in
list b without doing a loop? A loop is easy enough, but wanted to see if
there was a cleaner way. By way of example:
List 1: a, b, c, d, e, f, g
List 2: z, y, x, w, v, u, b
Return true, since both lists contain b
List 1: a, b, c, d, e, f, g
List 2: z, y, x, w, v, u, t
Return false, since the lists have no mutual values
--
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Re: [R] Are any values in one list contained within a second list
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of GL Sent: Sunday, June 13, 2010 11:18 AM To: r-help@r-project.org Subject: [R] Are any values in one list contained within a second list Silly question, but, can I test to see if any value of list a is contained in list b without doing a loop? It is almost the same as your English description of the problem: any(a %in% b) A loop is easy enough, but wanted to see if there was a cleaner way. By way of example: List 1: a, b, c, d, e, f, g List 2: z, y, x, w, v, u, b It would be nicer if you used R syntax to create the datasets in your example. E.g., I'm guessing you mean a1 - c(a,b,c,d,e,f,g) but since you didn't quote the strings and you called it a list you could mean a2 - as.list(expression(a, b, c, d, e, f, g)) or, equivalently, a3 - list(quote(a), quote(b), quote(c), quote(d), quote(e), quote(f), quote(g)) Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com Return true, since both lists contain b List 1: a, b, c, d, e, f, g List 2: z, y, x, w, v, u, t Return false, since the lists have no mutual values -- View this message in context: http://r.789695.n4.nabble.com/Are-any-values-in-one-list-conta ined-within-a-second-list-tp2253637p2253637.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] losing line of mtext when saving to png format
I have a simple graph (oode below) which looks fine on the screen but when I save it in png format the title (actually the last mtext line) is cut off. I am pretty sure that I am doing something very stupid but other than playing around with the png height and width commands which don't seem to help I have no idea of what to look for. Any suggestions or cures would be appreciated. Thanks #===# mydata - data.frame(kel = c(13,253,263,273,283,292,303,313,323,333,573,853), cent= c( -280, -40, -30, -20, -10, 1, 10, 20, 30, 40, 280, 560)) (intervals - diff(mydata[,1])) opar - par(mar= (c(5, 4, 6, 2))) # png(C:/Rjunk/brendengraph.png, width=600, height=500) plot(1:12,mydata$kel, main= , xlab= , ylab= ,type=l, col=red, xaxt=n, yaxt=n,cex=.75) axis(1, at=1:12,labels=as.character(mydata$kel), cex.axis=.75) axis(3, at=1:12, labels=as.character(mydata$cent), cex.axis=.75) mtext(side = 1, line=2, text=Degrees Kelvin,cex=.75 ) mtext(side = 3, line=2, text=Degrees Centegrade,cex=.75 ) mtext(side=3, line=4, text=Room Temperature, font=2, cex=.85) # dev.off() par(opar) #= __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Are any values in one list contained within a second list
I think the simplest way is to translate the English
directly :-)
list1 = c('a','b','c','d','e','f','g')
list2 = c('z','y','x','w','v','u','b')
any(list2 %in% list1)
[1] TRUE
list2 = c('z','y','x','w','v','u','t')
any(list2 %in% list1)
[1] FALSE
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Sun, 13 Jun 2010, GL wrote:
Silly question, but, can I test to see if any value of list a is contained in
list b without doing a loop? A loop is easy enough, but wanted to see if
there was a cleaner way. By way of example:
List 1: a, b, c, d, e, f, g
List 2: z, y, x, w, v, u, b
Return true, since both lists contain b
List 1: a, b, c, d, e, f, g
List 2: z, y, x, w, v, u, t
Return false, since the lists have no mutual values
--
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Re: [R] HOW to install RSQLite database
select a cran mirror Packages Select Cran Mirror
In R use the following command
install.packages(RSQLite)
--- On Sun, 6/13/10, vijaysheegi vijay.she...@gmail.com wrote:
From: vijaysheegi vijay.she...@gmail.com
Subject: Re: [R] HOW to install RSQLite database
To: r-help@r-project.org
Received: Sunday, June 13, 2010, 2:38 AM
Yes i am asking how to install RSQLite packagein
windows.Please help on this
regards
On 6/11/10, david.jessop [via R]
ml-node+2251498-1601505055-288...@n4.nabble.comml-node%2b2251498-1601505055-288...@n4.nabble.com
wrote:
Are you asking how to install the RSQLite package or
how to create a
SQLite database? The two are somewhat distinct
questions. RSQLite is
just a package of functions for R to be able to access
data in an SQLite
database. There isn't a separate SQLite program - just
a library that is
compiled into RSQLite.
Regards
David
-Original Message-
From: [hidden
email]http://user/SendEmail.jtp?type=nodenode=2251498i=0[mailto:[hidden
email] http://user/SendEmail.jtp?type=nodenode=2251498i=1]
On Behalf Of vijaysheegi
Sent: 10 June 2010 16:22
To: [hidden email] http://user/SendEmail.jtp?type=nodenode=2251498i=2
Subject: [R] HOW to install RSQLite database
Please let me know where i have to type below thing
to RSQLite database
get installed.Please let me know the solution.Thanks
in advance
RSQLite -- Embedding the SQLite engine in R
(The RSQLite package includes a recent copy of the
SQLite distribution
from http://www.sqlite.org http://www.sqlite.org/?by-user=t.)
Installation
There are 3 alternatives for installation:
1. Simple installation:
R CMD INSTALL
RSQLite-.tar.gz
the installation automatically detects
whether SQLite is
available in any of your system
directories; if it's not
available, it installs the SQLite engine
and the R-SQLite
interface under the package directory
$R_PACKAGE_DIR/sqlite.
2. If you have SQLite installed in a non-system
directory (e.g,
in $HOME/sqlite),
a) You can use
export
PKG_LIBS=-L$HOME/sqlite/lib -lsqlite
export
PKG_CPPFLAGS=-I$HOME/sqlite/include
R CMD INSTALL
RSQLite-.tar.gz
b) or you can use the --with-sqlite-dir
configuration argument
R CMD INSTALL
--configure-args=--with-sqlite-dir=$HOME/sqlite \
RSQLite-.tar.gz
3. If you don't have SQLite but you rather install the
version we
provide
into a directory different than the
RSQLite package, for instance,
$HOME/app/sqlite, use
R CMD INSTALL
--configure-args=--enable-sqlite=$HOME/app/sqlite \
RSQLite-.tar.gz
Usage
-
Note that if you use an *existing* SQLite library that
resides in a
non-system directory (e.g., other than /lib, /usr/lib,
/usr/local/lib)
you may need to include it in our LD_LIBRARY_PATH,
prior to invoking R.
For instance
export
LD_LIBRARY_PATH=$HOME/sqlite/lib:$LD_LIBRARY_PATH
R
library(help=RSQLite)
library(RSQLite)
(if you use the --enable-sqlite=DIR configuration
argument, the SQLite
library is statically linked to the RSQLite R package,
and you need not
worry about setting LD_LIBRARY_PATH.)
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Re: [R] losing line of mtext when saving to png format
Hey John, I believe this issue is that the png device is different from the onscreen one. You set the margins for the on screen with a call to par() but not for the png(). The code below works for me. mydata - data.frame(kel=c(13,253,263,273,283,292,303,313,323,333,573,853), cent=c( -280, -40, -30, -20, -10, 1, 10, 20, 30, 40, 280, 560)) png(C:/Rjunk/brendengraph.png, width=600, height=600) par(mar=(c(5, 4, 6, 2))) #this goes after the call to png() plot(1:12,mydata$kel, main= , xlab= , ylab= ,type=l, col=red, xaxt=n, yaxt=n,cex=.75) axis(1, at=1:12,labels=as.character(mydata$kel), cex.axis=.75) axis(3, at=1:12, labels=as.character(mydata$cent), cex.axis=.75) mtext(side = 1, line=2, text=Degrees Kelvin,cex=.75 ) mtext(side = 3, line=2, text=Degrees Centegrade,cex=.75 ) mtext(side = 3, line=4, text=Room Temperature, font=2, cex=.85) dev.off() # HTH, Josh On Sun, Jun 13, 2010 at 12:09 PM, John Kane jrkrid...@yahoo.ca wrote: I have a simple graph (oode below) which looks fine on the screen but when I save it in png format the title (actually the last mtext line) is cut off. I am pretty sure that I am doing something very stupid but other than playing around with the png height and width commands which don't seem to help I have no idea of what to look for. Any suggestions or cures would be appreciated. Thanks #===# mydata - data.frame(kel = c(13,253,263,273,283,292,303,313,323,333,573,853), cent= c( -280, -40, -30, -20, -10, 1, 10, 20, 30, 40, 280, 560)) (intervals - diff(mydata[,1])) opar - par(mar= (c(5, 4, 6, 2))) # png(C:/Rjunk/brendengraph.png, width=600, height=500) plot(1:12,mydata$kel, main= , xlab= , ylab= ,type=l, col=red, xaxt=n, yaxt=n,cex=.75) axis(1, at=1:12,labels=as.character(mydata$kel), cex.axis=.75) axis(3, at=1:12, labels=as.character(mydata$cent), cex.axis=.75) mtext(side = 1, line=2, text=Degrees Kelvin,cex=.75 ) mtext(side = 3, line=2, text=Degrees Centegrade,cex=.75 ) mtext(side=3, line=4, text=Room Temperature, font=2, cex=.85) # dev.off() par(opar) #= __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 Senior in Psychology University of California, Riverside http://www.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.
Re: [R] losing line of mtext when saving to png format
Oh, how embarrassing! I made the exact same mistake about 2-3 years ago. Thanks a lot. It was not even that important a graph but it was really bothering me. --- On Sun, 6/13/10, Joshua Wiley jwiley.ps...@gmail.com wrote: From: Joshua Wiley jwiley.ps...@gmail.com Subject: Re: [R] losing line of mtext when saving to png format To: John Kane jrkrid...@yahoo.ca Cc: R R-help r-h...@stat.math.ethz.ch Received: Sunday, June 13, 2010, 3:32 PM Hey John, I believe this issue is that the png device is different from the onscreen one. You set the margins for the on screen with a call to par() but not for the png(). The code below works for me. mydata - data.frame(kel=c(13,253,263,273,283,292,303,313,323,333,573,853), cent=c( -280, -40, -30, -20, -10, 1, 10, 20, 30, 40, 280, 560)) png(C:/Rjunk/brendengraph.png, width=600, height=600) par(mar=(c(5, 4, 6, 2))) #this goes after the call to png() plot(1:12,mydata$kel, main= , xlab= , ylab= ,type=l, col=red, xaxt=n, yaxt=n,cex=.75) axis(1, at=1:12,labels=as.character(mydata$kel), cex.axis=.75) axis(3, at=1:12, labels=as.character(mydata$cent), cex.axis=.75) mtext(side = 1, line=2, text=Degrees Kelvin,cex=.75 ) mtext(side = 3, line=2, text=Degrees Centegrade,cex=.75 ) mtext(side = 3, line=4, text=Room Temperature, font=2, cex=.85) dev.off() # HTH, Josh On Sun, Jun 13, 2010 at 12:09 PM, John Kane jrkrid...@yahoo.ca wrote: I have a simple graph (oode below) which looks fine on the screen but when I save it in png format the title (actually the last mtext line) is cut off. I am pretty sure that I am doing something very stupid but other than playing around with the png height and width commands which don't seem to help I have no idea of what to look for. Any suggestions or cures would be appreciated. Thanks #===# mydata - data.frame(kel = c(13,253,263,273,283,292,303,313,323,333,573,853), cent= c( -280, -40, -30, -20, -10, 1, 10, 20, 30, 40, 280, 560)) (intervals - diff(mydata[,1])) opar - par(mar= (c(5, 4, 6, 2))) # png(C:/Rjunk/brendengraph.png, width=600, height=500) plot(1:12,mydata$kel, main= , xlab= , ylab= ,type=l, col=red, xaxt=n, yaxt=n,cex=.75) axis(1, at=1:12,labels=as.character(mydata$kel), cex.axis=.75) axis(3, at=1:12, labels=as.character(mydata$cent), cex.axis=.75) mtext(side = 1, line=2, text=Degrees Kelvin,cex=.75 ) mtext(side = 3, line=2, text=Degrees Centegrade,cex=.75 ) mtext(side=3, line=4, text=Room Temperature, font=2, cex=.85) # dev.off() par(opar) #= __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 Senior in Psychology University of California, Riverside http://www.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.
Re: [R] Get a list of installed commands
Try this: sapply(installed.packages()[,1], function(x)try(ls(asNamespace(x On Sun, Jun 13, 2010 at 9:30 AM, Data Monkey coco.datamon...@gmail.comwrote: Thanks Joris. Very helpful. I had thought of that, just curious to see if it was possible to get a fresh list in R. After reading your email I think perhaps my wording was a bit loose. I meant commands in the pre-installed packages. So basically, out of the box what commands will R recognize. I had found the full reference manual for R online and parsed the table of contents into a nice XML file I could use as a possible solution. However, your attached file looks a bit more complete (haven't really given much thought to the reason yet). Thanks again for sending it. I'm using Coda on OS X, which is really designed for Web dev, but I like it so much I've added support for Stata and now want R. It's got Terminal bult-in, so I can invoke R through there. Anyway, my ultimate goal is to not only get syntax highlighting, but also autocompletion. If possible in the prediction, also get the method (where applicable) signature. Coda allows all this. I use this a lot when coding in PHP in Coda and C# in Visual Studio - I think Microsoft call it Intellisense or something. Cheers for the help. On 13/06/2010, at 9:47 PM, Joris Meys wrote: Hi, Take a look at any of the R-editors, like Tinn-R, Emacs-ESS, Eclipse with StatET,... They contain lists you can use. Also the listings package of LaTeX contains a wordlist for R. Getting all installed commands out of R is not doable with a single command as far as I know. R works completely different than Stata; R is a fullblown programming language, not a statistical program. Try to find a list of all installed Perl commands for example... Attached is the recognition file of Tinn-R. Reserved 1 are special keywords, Reserved 2 is a list of the most commonly used commands in the pre-installed packages, and Reserved 3 is a list of common parameters of those functions. It's submitted with Tinn-R under the GNU license, so keep that in mind when using it. But instead of re-inventing the wheel and constructing your own editor, you could take a look at one of those mentioned above. On Windows, I recommend Tinn-R for daily scripting, and Eclipse for developing packages and the likes. Both offer the advantage of direct communication with the R console. Cheers Joris On Sun, Jun 13, 2010 at 9:31 AM, Data Monkey coco.datamon...@gmail.com wrote: I'm pretty new to R, but have experience with other languages, both OO and scripting. I'm trying to add support for R to my text editor of choice and to do this I need a list of installed commands I can markup with XML. I'd then simply feed in the marked up list into my text editor's library and I'm off. I've done this in Stata before using the following command: getcmds using ~/Desktop/StataCommands.txt Does anyone know of a way to do this in R? Any pointers much appreciated. Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joris Meys Statistical consultant Ghent University Faculty of Bioscience Engineering Department of Applied mathematics, biometrics and process control tel : +32 9 264 59 87 joris.m...@ugent.be --- Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php Tinn-R_recognized words.r __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Mid-P value for a chi-squared test
On Jun 13, 2010, at 2:19 PM, (Ted Harding) wrote:
On 13-Jun-10 17:12:45, David Winsemius wrote:
On Jun 1, 2010, at 4:17 AM, Wilson, Andrew wrote:
Can anyone tell me how to calculate a mid-p value for a chi-squared
test in R?
I cannot see that this has been answered. It has a date from 12 days
ago but I cannot see a reply in the archives.
So, what is a mid-p value and which chi-square test are you
asking
about? A simple data setup in R code with expected output would
speed
this discussion along.
David Winsemius, MD
The mid-p value is a device for improving the accuracy of a
continuous
approximation to a distribution which in reality is discrete.
Intuitively, the idea is to treat the discrete probabilities of the
discrete distribution as if they were proportions in a histogram.
Then imagine fitting a continuous curve (e.g. a chi-squared density)
to the histogram. The fit (agreement between the proportion in one
histogram bar, and the probability below that portion of the curve
which lies in the same range) will be better if the curve goes through
the midpoint of the top of the bar.
This leads to the formal definition:
mid-P = Prob(X Xobs) + 0.5*Prob(X = Xobs)
Looking at Agresti and Gottard's piece, cited by one of the R
functions, midPci {PropCIs} from a match to your suggested search
strategy below, they say the lower mid-p CI would be defined as
solving equation:
Pr_0L (X x) + 1/2 x Pr_0L (X = x) = a/2.
Is that mathematically equivalent ( perhaps the NP dual to a p-value
version) to what you offered? And is the upper CI then defined as
solution to :
Pr_0U (X x) + 1/2 x Pr_0U (X = x) = 1-a/2,,, ?
A number of R functions use this idea. Check out what you get by
going to http://finzi.psych.upenn.edu/nmz.html and entering mid-p
into the search box, and see whether any of them match (or come
close to) your particular case.
In the case of the chi-squared test, the idea is related to (but
not the same as) the Yates correction for continuity. chisq.test()
has an option correct=TRUE to force this, but only for 2x2 tables.
Ted.
David Winsemius, MD
West Hartford, CT
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] ERROR need finite 'ylim' values
Giuseppe, See comments below. On 2010-06-13 10:24, David Winsemius wrote: Giuseppe wrote: Hello: I use R with MAC I have a simple data table, numeric and text columns, named dt. The table is imported through read.csv from a csv file. Row numbers are automatically assigned, header is set to TRUE. there are 599 rows and several columns. I am trying to plot using the stripchart command: one numeric variable (say dt$fnatg) vs a text column (say dt$pat). dt$pat contains one of 3 values: pos, neg, So I issue the following command: stripchart (dt$fnatg~dt$pat) and works well. it works well also with several options and nuances: stripchart (dt$fnatg ~ dt$pat, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) Now I want my graph to exclude values for which dt$pat == I tried: stripchart (dt$fnatg ~ dt$pat, method =jitter, subset (dt, dt$pat!=),jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) there is no effect: the plot contains the same values as before the I tried first subsetting the table: patonly-(dt, dt$pat!=) which works well in creating a new table excluding the unwanted rows. I have noticed that the new table keeps the same row numbers assigned in the previous table. So row numbers now go 1 to 599 but with some intervals (for example there is no row 475 etc.). then I use: stripchart (patonly$fnatg ~ patonly$pat, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) and I get the following error: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In min(x) : no non-missing arguments to min; returning Inf 2: In max(x) : no non-missing arguments to max; returning -Inf 3: In min(x) : no non-missing arguments to min; returning Inf 4: In max(x) : no non-missing arguments to max; returning -Inf I f I try the same command but I use another text variable (for example patonly$gr) in the same table to split the plot, it now works: stripchart (patonly$fnatg ~ patonly$gr, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) My question is two fold: Why does not the subset command work within the stripchart command? Why the subsetted table cannot be used in the stripchart command, when the plotting variable is the same previously used in the subsetting process? You appear to have adopted a strategy of using positional matching. Naming your arguments will often result in more informative error messages. Looking at the help page for stripchart, it appears that there is no subset parameter to set in any of its methods and only the formula method has a data argument. It should work with: stripchart(formula1 , data=subset(dta, subset=criteria), rest of arguments preferably named ) Your other option might be to use the with() function: with( subset(patonly, pat!=), stripchart(fnatg ~ gr, ...named arguments) ) HTH. and if it doesn't, then submit a reproducible data example to work with. Actually, Giuseppe appears to have stumbled upon a bug in the stripchart() function. First, here's a fix: After your command patonly-(dt, dt$pat!=) which I assume is meant to be patonly - subset(dt, dt$pat!=) and which can be written as patonly - subset(dt, pat!=) you should issue this: patonly$pat - factor(patonly$pat) which will remove the empty level; stripchart() should work well after that (and do use the data= argument rather than dt$...). Alternatively, you could change your text variables (which I assume are factors) to character values (or re-import your data with stringsAsFactors = FALSE). Now for the bug in stripchart(): If the *first* group of the grouping variable is empty, then stripchart() has a problem determining the range of data values (x-values for horizontal charts, y-values otherwise). I can replicate your problem withe OrchardSprays dataset: # this works: stripchart(decrease ~ treatment, data = OrchardSprays, subset = treatment != B) # this doesn't stripchart(decrease ~ treatment, data = OrchardSprays, subset = treatment != A) I'll be submitting a bug report (and I think the fix is easy). -Peter Ehlers __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Count of unique factors within another factor
You can also use the sqldf package:
x
unit species
1 123ACMA
2 123LIDE
3 123LIDE
4 123SESE
5 123SESE
6 123SESE
7 345HEAR
8 345LOHI
9 345QUAG
10 345TODI
require(sqldf)
sqldf('select unit, count(distinct species) as count from x group by unit')
unit count
1 123 3
2 345 4
On Sun, Jun 13, 2010 at 12:07 PM, Birdnerd haaszool...@gmail.com wrote:
I have a data frame with two factors (sampling 'unit', 'species'). I want to
calculate the number of unique 'species' per 'unit.' I can calculate the
number of unique values for each variable separately, but can't get a count
for each ‘unit’.
data=read.csv(C:/Desktop/sr_sort_practice.csv)
attach(data)
data[1:10,]
unit species
1 123 ACMA
2 123 LIDE
3 123 LIDE
4 123 SESE
5 123 SESE
6 123 SESE
7 345 HEAR
8 345 LOHI
9 345 QUAG
10 345 TODI…..
sr.unique- lapply (data, unique)
$unit
[1] 123 345 216
$species
[1] ACMA LIDE SESE HEAR LOHI QUAG TODI UMCA ARSP LIDE
sapply (sr.unique,length)
unit species
3 10
Then, I get stuck here because this unique species count is not given for
each ‘unit’.
What I'd like to get is:
unit species
123 3
345 4
216 --
Thanks--
--
View this message in context:
http://r.789695.n4.nabble.com/Count-of-unique-factors-within-another-factor-tp2253545p2253545.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Get a list of installed commands
On Jun 13, 2010, at 3:44 PM, Henrique Dallazuanna wrote: Try this: sapply(installed.packages()[,1], function(x)try(ls(asNamespace(x You could clean that up a bit with: funlist - sapply(installed.packages()[,1], function(x)try(ls(asNamespace(x function.list - fun.list[lapply(fun.list, length) 1] # removes all entries for the package with no namespace. I was left wondering if the speed characteristics of an implementation that needed to search against 32K names might be a bit slow? length(function.list) [1] 211 # number of installed packages on my machine with namespaces sum(unlist(lapply(function.list, length))) [1] 31960 # number of function names You might want to limit your search to base R packages minus datasets ( from R-FAQ listing): http://cran.r-project.org/doc/FAQ/R-FAQ.html#Add_002don-packages-from-CRAN core.pkgs - c(base, grDevices, graphics, grid, methods, splines, stats, stats4, tcltk, tools, utils) core.fun.list - sapply(core.pkgs, function(x) try(ls(asNamespace(x sum(unlist(lapply(core.fun.list[lapply(core.fun.list, length) 1], length))) [1] 4036 -- David. On Sun, Jun 13, 2010 at 9:30 AM, Data Monkey coco.datamon...@gmail.com wrote: Thanks Joris. Very helpful. I had thought of that, just curious to see if it was possible to get a fresh list in R. After reading your email I think perhaps my wording was a bit loose. I meant commands in the pre-installed packages. So basically, out of the box what commands will R recognize. I had found the full reference manual for R online and parsed the table of contents into a nice XML file I could use as a possible solution. However, your attached file looks a bit more complete (haven't really given much thought to the reason yet). Thanks again for sending it. I'm using Coda on OS X, which is really designed for Web dev, but I like it so much I've added support for Stata and now want R. It's got Terminal bult-in, so I can invoke R through there. Anyway, my ultimate goal is to not only get syntax highlighting, but also autocompletion. If possible in the prediction, also get the method (where applicable) signature. Coda allows all this. I use this a lot when coding in PHP in Coda and C# in Visual Studio - I think Microsoft call it Intellisense or something. Cheers for the help. On 13/06/2010, at 9:47 PM, Joris Meys wrote: Hi, Take a look at any of the R-editors, like Tinn-R, Emacs-ESS, Eclipse with StatET,... They contain lists you can use. Also the listings package of LaTeX contains a wordlist for R. Getting all installed commands out of R is not doable with a single command as far as I know. R works completely different than Stata; R is a fullblown programming language, not a statistical program. Try to find a list of all installed Perl commands for example... Attached is the recognition file of Tinn-R. Reserved 1 are special keywords, Reserved 2 is a list of the most commonly used commands in the pre-installed packages, and Reserved 3 is a list of common parameters of those functions. It's submitted with Tinn-R under the GNU license, so keep that in mind when using it. But instead of re-inventing the wheel and constructing your own editor, you could take a look at one of those mentioned above. On Windows, I recommend Tinn-R for daily scripting, and Eclipse for developing packages and the likes. Both offer the advantage of direct communication with the R console. Cheers Joris On Sun, Jun 13, 2010 at 9:31 AM, Data Monkey coco.datamon...@gmail.com wrote: I'm pretty new to R, but have experience with other languages, both OO and scripting. I'm trying to add support for R to my text editor of choice and to do this I need a list of installed commands I can markup with XML. I'd then simply feed in the marked up list into my text editor's library and I'm off. I've done this in Stata before using the following command: getcmds using ~/Desktop/StataCommands.txt Does anyone know of a way to do this in R? Any pointers much appreciated. Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joris Meys Statistical consultant Ghent University Faculty of Bioscience Engineering Department of Applied mathematics, biometrics and process control tel : +32 9 264 59 87 joris.m...@ugent.be --- Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php Tinn-R_recognized words.r __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25°
Re: [R] ERROR need finite 'ylim' values
On Jun 13, 2010, at 3:49 PM, Peter Ehlers wrote: Giuseppe, See comments below. On 2010-06-13 10:24, David Winsemius wrote: Giuseppe wrote: Hello: I use R with MAC I have a simple data table, numeric and text columns, named dt. The table is imported through read.csv from a csv file. Row numbers are automatically assigned, header is set to TRUE. there are 599 rows and several columns. I am trying to plot using the stripchart command: one numeric variable (say dt$fnatg) vs a text column (say dt$pat). dt$pat contains one of 3 values: pos, neg, So I issue the following command: stripchart (dt$fnatg~dt$pat) and works well. it works well also with several options and nuances: stripchart (dt$fnatg ~ dt$pat, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) Now I want my graph to exclude values for which dt$pat == I tried: stripchart (dt$fnatg ~ dt$pat, method =jitter, subset (dt, dt$pat!=),jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) there is no effect: the plot contains the same values as before the I tried first subsetting the table: patonly-(dt, dt$pat!=) which works well in creating a new table excluding the unwanted rows. I have noticed that the new table keeps the same row numbers assigned in the previous table. So row numbers now go 1 to 599 but with some intervals (for example there is no row 475 etc.). then I use: stripchart (patonly$fnatg ~ patonly$pat, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/ mL),xlab=Surgical Pathology) and I get the following error: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In min(x) : no non-missing arguments to min; returning Inf 2: In max(x) : no non-missing arguments to max; returning -Inf 3: In min(x) : no non-missing arguments to min; returning Inf 4: In max(x) : no non-missing arguments to max; returning -Inf I f I try the same command but I use another text variable (for example patonly$gr) in the same table to split the plot, it now works: stripchart (patonly$fnatg ~ patonly$gr, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/ mL),xlab=Surgical Pathology) My question is two fold: Why does not the subset command work within the stripchart command? Why the subsetted table cannot be used in the stripchart command, when the plotting variable is the same previously used in the subsetting process? You appear to have adopted a strategy of using positional matching. Naming your arguments will often result in more informative error messages. Looking at the help page for stripchart, it appears that there is no subset parameter to set in any of its methods and only the formula method has a data argument. It should work with: stripchart(formula1 , data=subset(dta, subset=criteria), rest of arguments preferably named ) Your other option might be to use the with() function: with( subset(patonly, pat!=), stripchart(fnatg ~ gr, ...named arguments) ) HTH. and if it doesn't, then submit a reproducible data example to work with. Actually, Giuseppe appears to have stumbled upon a bug in the stripchart() function. I thought when Guisseppe wrote: I tried: stripchart (dt$fnatg ~ dt$pat, method =jitter, subset (dt, dt$pat!=),jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) ... that the expression: subset(dt, dt$pat!=) , would get erroneously matched to dlab, the third parameter in the formula method arg list, but maybe it would get matched to the data, the second parameter. I would also worry that specifying a data object, but then in a sense contravening that specification by reference to the full name in the formula specification (dt$fnatg) might cause problems. First, here's a fix: After your command patonly-(dt, dt$pat!=) which I assume is meant to be patonly - subset(dt, dt$pat!=) and which can be written as patonly - subset(dt, pat!=) you should issue this: patonly$pat - factor(patonly$pat) which will remove the empty level; stripchart() should work well after that (and do use the data= argument rather than dt$...). Alternatively, you could change your text variables (which I assume are factors) to character values (or re-import your data with stringsAsFactors = FALSE). Now for the bug in stripchart(): If the *first* group of the grouping variable is empty, then stripchart() has a problem determining the range of data values (x-values for horizontal charts, y-values otherwise). I can replicate your problem withe OrchardSprays dataset: # this works: stripchart(decrease ~ treatment, data = OrchardSprays, subset = treatment != B) # this doesn't stripchart(decrease ~ treatment, data = OrchardSprays, subset = treatment != A) I'll be submitting a
Re: [R] Count of unique factors within another factor
On Sun, 13 Jun 2010, Birdnerd wrote: I have a data frame with two factors (sampling 'unit', 'species'). I want to calculate the number of unique 'species' per 'unit.' I can calculate the number of unique values for each variable separately, but can't get a count for each ‘unit’. If I understand you colSums( xtabs( ~ specie + unit , data ) !=0 ) HTH, Chuck data=read.csv(C:/Desktop/sr_sort_practice.csv) attach(data) data[1:10,] unit species 1 123ACMA 2 123LIDE 3 123LIDE 4 123SESE 5 123SESE 6 123SESE 7 345HEAR 8 345LOHI 9 345QUAG 10 345TODI….. sr.unique- lapply (data, unique) $unit [1] 123 345 216 $species [1] ACMA LIDE SESE HEAR LOHI QUAG TODI UMCA ARSP LIDE sapply (sr.unique,length) unit species 3 10 Then, I get stuck here because this unique species count is not given for each ‘unit’. What I'd like to get is: unit species 1233 3454 216-- Thanks-- -- View this message in context: http://r.789695.n4.nabble.com/Count-of-unique-factors-within-another-factor-tp2253545p2253545.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Charles C. Berry(858) 534-2098 Dept of Family/Preventive Medicine E mailto:cbe...@tajo.ucsd.edu UC San Diego http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ERROR need finite 'ylim' values
Peter Ehlers wrote: Giuseppe, See comments below. On 2010-06-13 10:24, David Winsemius wrote: Giuseppe wrote: Hello: I use R with MAC I have a simple data table, numeric and text columns, named dt. The table is imported through read.csv from a csv file. Row numbers are automatically assigned, header is set to TRUE. there are 599 rows and several columns. I am trying to plot using the stripchart command: one numeric variable (say dt$fnatg) vs a text column (say dt$pat). dt$pat contains one of 3 values: pos, neg, So I issue the following command: stripchart (dt$fnatg~dt$pat) and works well. it works well also with several options and nuances: stripchart (dt$fnatg ~ dt$pat, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) Now I want my graph to exclude values for which dt$pat == I tried: stripchart (dt$fnatg ~ dt$pat, method =jitter, subset (dt, dt$pat!=),jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) there is no effect: the plot contains the same values as before the I tried first subsetting the table: patonly-(dt, dt$pat!=) which works well in creating a new table excluding the unwanted rows. I have noticed that the new table keeps the same row numbers assigned in the previous table. So row numbers now go 1 to 599 but with some intervals (for example there is no row 475 etc.). then I use: stripchart (patonly$fnatg ~ patonly$pat, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) and I get the following error: Error in plot.window(...) : need finite 'ylim' values In addition: Warning messages: 1: In min(x) : no non-missing arguments to min; returning Inf 2: In max(x) : no non-missing arguments to max; returning -Inf 3: In min(x) : no non-missing arguments to min; returning Inf 4: In max(x) : no non-missing arguments to max; returning -Inf I f I try the same command but I use another text variable (for example patonly$gr) in the same table to split the plot, it now works: stripchart (patonly$fnatg ~ patonly$gr, method =jitter, jitter = 0.3, vertical =TRUE,log=y, pch=1, ylab=Thyroglobulin (ng/mL),xlab=Surgical Pathology) My question is two fold: Why does not the subset command work within the stripchart command? Why the subsetted table cannot be used in the stripchart command, when the plotting variable is the same previously used in the subsetting process? You appear to have adopted a strategy of using positional matching. Naming your arguments will often result in more informative error messages. Looking at the help page for stripchart, it appears that there is no subset parameter to set in any of its methods and only the formula method has a data argument. It should work with: stripchart(formula1 , data=subset(dta, subset=criteria), rest of arguments preferably named ) Your other option might be to use the with() function: with( subset(patonly, pat!=), stripchart(fnatg ~ gr, ...named arguments) ) HTH. and if it doesn't, then submit a reproducible data example to work with. Actually, Giuseppe appears to have stumbled upon a bug in the stripchart() function. First, here's a fix: After your command patonly-(dt, dt$pat!=) which I assume is meant to be patonly - subset(dt, dt$pat!=) and which can be written as patonly - subset(dt, pat!=) you should issue this: patonly$pat - factor(patonly$pat) which will remove the empty level; stripchart() should work well after that (and do use the data= argument rather than dt$...). Alternatively, you could change your text variables (which I assume are factors) to character values (or re-import your data with stringsAsFactors = FALSE). Now for the bug in stripchart(): If the *first* group of the grouping variable is empty, then stripchart() has a problem determining the range of data values (x-values for horizontal charts, y-values otherwise). I can replicate your problem withe OrchardSprays dataset: # this works: stripchart(decrease ~ treatment, data = OrchardSprays, subset = treatment != B) # this doesn't stripchart(decrease ~ treatment, data = OrchardSprays, subset = treatment != A) I'll be submitting a bug report (and I think the fix is easy). -Peter Ehlers __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Thank you Peter, both your solutions worked for me. Why do you insist for using data = argument? Giuseppe Giuseppe -- View this message in context: http://r.789695.n4.nabble.com/ERROR-need-finite-ylim-values-tp2253388p2253767.html Sent from the R
[R] Standard error of regression coefficient
Hi all, This should be a very simple question for you, whereas it is proving devilish for me. How do I output the STANDARD ERROR of the regression coefficient (i.e., the standard error of b) from a simple linear regression? Consider this data, taken directly from ?lm: ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) group - gl(2,10,20, labels=c(Ctl,Trt)) weight - c(ctl, trt) The problem is that neither of these lines of code seem to show me what I'm looking for: lm(weight ~ group) Call: lm(formula = weight ~ group) Coefficients: (Intercept) groupTrt 5.032 -0.371 anova(lm.D9 - lm(weight ~ group)) Analysis of Variance Table Response: weight Df Sum Sq Mean Sq F value Pr(F) group 1 0.6882 0.6882 1.4191 0.249 Residuals 18 8.7292 0.4850 So what's the secret, then? Thanks very much R users! Josh [[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] Standard error of regression coefficient
Josh B wrote: Hi all, This should be a very simple question for you, whereas it is proving devilish for me. How do I output the STANDARD ERROR of the regression coefficient (i.e., the standard error of b) from a simple linear regression? The first 'See Also' in ?lm is for ?summary.lm, which will give you what you want. So simply pass your lm object to summary. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 remove double precision round off chaff
I would like to get rid of the double precision round off chaff, so is the following the best way to handle it? 0.625-0.8+0.45-0.275 [1] -5.551115e-17 round(0.625-0.8+0.45-0.275, digits=4) [1] 0 Motivation for removing the chaff is for no other reason than to titty up the digits for display purposes and to help quicken the analysis of values that most definitely should be zero. Thanks for any insights and feedback. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Best way to remove double precision round off chaff
Jason: Many print methods have specific options to control this: ?print.default See also the digits and scipen settings in ?options for global control (these can be put in your startup file, for example). See also ?format and e.g. ?sprintf for more precise control of print format. All of this has to do with what is displayed and does not change the underlying values, of course, which round() and friends do. Cheers, Bert Gunter Genentech Nonclinical Statistics -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jason Rupert Sent: Sunday, June 13, 2010 4:24 PM To: R Project Help Subject: [R] Best way to remove double precision round off chaff I would like to get rid of the double precision round off chaff, so is the following the best way to handle it? 0.625-0.8+0.45-0.275 [1] -5.551115e-17 round(0.625-0.8+0.45-0.275, digits=4) [1] 0 Motivation for removing the chaff is for no other reason than to titty up the digits for display purposes and to help quicken the analysis of values that most definitely should be zero. Thanks for any insights and feedback. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Registration deadline, useR! 2010
The final registration deadline for the R User Conference is June 20, 2010, one week away. Later registration will not be possible on site! Conference webpage: http://www.R-project.org/useR-2010 Conference program: http://www.R-project.org/useR-2010/program.html Registration: http://www.R-project.org/useR-2010/registration/registration.html The conference is scheduled for July 21-23, 2010, and will take place at the campus of the National Institute of Standards and Technology (NIST) in Gaithersburg, Maryland, USA. Following the successful useR! 2004, useR! 2006, useR! 2007, useR! 2008, and useR! 2009, conferences, the conference is focused on: 1. R as the `lingua franca' of data analysis and statistical computing, 2. providing a platform for R users to discuss and exchange ideas on how R can be used to do statistical computations, data analysis, visualization and exciting applications in various fields, 3. giving an overview of the new features of the rapidly evolving R project. As for the predecessor conferences, the program will consist of two parts: invited lectures and user-contributed sessions. Prior to the conference, there will be tutorials on R, descriptions of which are available at http://www.R-project.org/useR-2010/tutorials INVITED LECTURES Invited speakers will include Mark Handcock, Frank Harrell Jr, Friedrich Leisch, Michael Meyer, Richard Stallman, Luke Tierney, Diethelm Wuertz. USER-CONTRIBUTED SESSIONS The sessions will be a platform to bring together R users, contributors, package maintainers and developers in the S spirit that `users are developers'. People from different fields will show us how they solve problems with R in fascinating applications. The sessions are organized by members of the program committee, including Dirk Eddelbuettel, John Fox, Virgilio Gomez-Rubio, Richard Heiberger, Torsten Hothorn, Aaron King, Jan de Leeuw, Nicholas Lewin-Koh, Andy Liaw, Uwe Ligges, Martin Maechler, Katharine Mullen, Heather Turner, Ravi Varadhan, H. D. Vinod, John Verzani, Alan Zaslavsky, Achim Zeileis. The program will cover topics such as * Applied Statistics Biostatistics * Bayesian Statistics * Bioinformatics * Chemometrics and Computational Physics * Data Mining * Econometrics Finance * Environmetrics Ecological Modeling * High Performance Computing * Machine Learning * Marketing Business Analytics * Psychometrics * Robust Statistics * Social network analysis * Spatial Statistics * Statistics in the Social and Political Sciences * Teaching * Visualization Graphics * and many more. IMPORTANT DATES * ** 2010-06-20 registration deadline **(later registration NOT possible on site) * 2010-07-20 tutorials 2010-07-21 conference start 2010-07-23 conference end We hope to meet you in Gaithersburg! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Get a list of installed commands
Enrique, That's fantastic. Thanks for that1 I got a couple of warnings but for the most part it looks like it gives what I want. Cheers. On 14/06/2010, at 5:44 AM, Henrique Dallazuanna wrote: Try this: sapply(installed.packages()[,1], function(x)try(ls(asNamespace(x On Sun, Jun 13, 2010 at 9:30 AM, Data Monkey coco.datamon...@gmail.com wrote: Thanks Joris. Very helpful. I had thought of that, just curious to see if it was possible to get a fresh list in R. After reading your email I think perhaps my wording was a bit loose. I meant commands in the pre-installed packages. So basically, out of the box what commands will R recognize. I had found the full reference manual for R online and parsed the table of contents into a nice XML file I could use as a possible solution. However, your attached file looks a bit more complete (haven't really given much thought to the reason yet). Thanks again for sending it. I'm using Coda on OS X, which is really designed for Web dev, but I like it so much I've added support for Stata and now want R. It's got Terminal bult-in, so I can invoke R through there. Anyway, my ultimate goal is to not only get syntax highlighting, but also autocompletion. If possible in the prediction, also get the method (where applicable) signature. Coda allows all this. I use this a lot when coding in PHP in Coda and C# in Visual Studio - I think Microsoft call it Intellisense or something. Cheers for the help. On 13/06/2010, at 9:47 PM, Joris Meys wrote: Hi, Take a look at any of the R-editors, like Tinn-R, Emacs-ESS, Eclipse with StatET,... They contain lists you can use. Also the listings package of LaTeX contains a wordlist for R. Getting all installed commands out of R is not doable with a single command as far as I know. R works completely different than Stata; R is a fullblown programming language, not a statistical program. Try to find a list of all installed Perl commands for example... Attached is the recognition file of Tinn-R. Reserved 1 are special keywords, Reserved 2 is a list of the most commonly used commands in the pre-installed packages, and Reserved 3 is a list of common parameters of those functions. It's submitted with Tinn-R under the GNU license, so keep that in mind when using it. But instead of re-inventing the wheel and constructing your own editor, you could take a look at one of those mentioned above. On Windows, I recommend Tinn-R for daily scripting, and Eclipse for developing packages and the likes. Both offer the advantage of direct communication with the R console. Cheers Joris On Sun, Jun 13, 2010 at 9:31 AM, Data Monkey coco.datamon...@gmail.com wrote: I'm pretty new to R, but have experience with other languages, both OO and scripting. I'm trying to add support for R to my text editor of choice and to do this I need a list of installed commands I can markup with XML. I'd then simply feed in the marked up list into my text editor's library and I'm off. I've done this in Stata before using the following command: getcmds using ~/Desktop/StataCommands.txt Does anyone know of a way to do this in R? Any pointers much appreciated. Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Joris Meys Statistical consultant Ghent University Faculty of Bioscience Engineering Department of Applied mathematics, biometrics and process control tel : +32 9 264 59 87 joris.m...@ugent.be --- Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php Tinn-R_recognized words.r __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Count of unique factors within another factor
Hi: Another possibility: as.data.frame(with(data[!duplicated(data), ], table(unit)) unit Freq 1 1233 2 3454 HTH, Dennis On Sun, Jun 13, 2010 at 9:07 AM, Birdnerd haaszool...@gmail.com wrote: I have a data frame with two factors (sampling 'unit', 'species'). I want to calculate the number of unique 'species' per 'unit.' I can calculate the number of unique values for each variable separately, but can't get a count for each unit. data=read.csv(C:/Desktop/sr_sort_practice.csv) attach(data) data[1:10,] unit species 1 123ACMA 2 123LIDE 3 123LIDE 4 123SESE 5 123SESE 6 123SESE 7 345HEAR 8 345LOHI 9 345QUAG 10 345TODI .. sr.unique- lapply (data, unique) $unit [1] 123 345 216 $species [1] ACMA LIDE SESE HEAR LOHI QUAG TODI UMCA ARSP LIDE sapply (sr.unique,length) unit species 3 10 Then, I get stuck here because this unique species count is not given for each unit. What I'd like to get is: unit species 1233 3454 216-- Thanks-- -- View this message in context: http://r.789695.n4.nabble.com/Count-of-unique-factors-within-another-factor-tp2253545p2253545.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. [[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] Multiple regressions
Hi, Could you please tell me whether SVM can do multiple regression or not? Cheers, Amy _ Browse profiles for FREE! Meet local singles online. [[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] logistic regression with 50 varaibales
Hi, this is not R technical question per se. I know there are many excellent statisticians in this list, so here my questions: I have dataset with ~1800 observations and 50 independent variables, so there are about 35 samples per variable. Is it wise to build a stable multiple logistic model with 50 independent variables? Any problem with this approach? Thanks John __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Multiple regressions
Hi, On Sun, Jun 13, 2010 at 8:30 PM, Amy Hessen amy_4_5...@hotmail.com wrote: Hi, Could you please tell me whether SVM can do multiple regression or not? Yes, an SVM can. -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] font path problem in Ubuntu
I am not exactly sure whether this is the right place to ask this but I hope this might be relevant to some other users of ubuntu linux as well. I've installed R-2.11.1 in my ubuntu 10.04 LTS. When I tried to generate a simple plot, however, it crashed: plot(rnorm(100)) Error in axis(side = side, at = at, labels = labels, ...) : could not find any X11 fonts Check that the Font Path is correct. In addition: Warning messages: 1: In function (display = , width, height, pointsize, gamma, bg, : locale not supported by Xlib: some X ops will operate in C locale 2: In function (display = , width, height, pointsize, gamma, bg, : X cannot set locale modifiers sessionInfo() R version 2.11.1 (2010-05-31) i686-pc-linux-gnu locale: [1] LC_CTYPE=en_SG.utf8 LC_NUMERIC=C [3] LC_TIME=en_SG.utf8 LC_COLLATE=en_SG.utf8 [5] LC_MONETARY=C LC_MESSAGES=en_SG.utf8 [7] LC_PAPER=en_SG.utf8 LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_SG.utf8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base It seems like this is due to genuine font path problem in Ubuntu (and so has nothing to do with R itself) but I'm just asking if anyone was able to solve this problem. When I googled the font path in ubuntu, it produced some suggestions regarding xorg.conf but following them didn't solve the problem for me. Any ideas?dlgmlwj Thanks in advance, TH -- Tae-Hoon Chung, PhD [[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] using latticeExtra plotting confidence intervals
On Sun, Jun 13, 2010 at 10:10 AM, Joe P King j...@joepking.com wrote: I am wanting to plot a 95% confidence band using segplot, yet I am wanting to have groups. For example if I have males and females, and then I have them in different races, I want the racial groups in different panels. I have this minor code, completely made up but gets at what I am wanting, 4 random samples and 4 samples of confidence, I know how to get A B into one panel and CD in to another but how do I get the x axis to label them properly and have it categorized as two. I am not sure what to have to the left side of the formula. This is the example code: (1) Your code results in length(categories) [1] 2 length(mu) [1] 4 which makes the formula mu~ci.upper+ci.lower|categories meaningless. (2) You are effectively plotting xyplot(mu ~ mu | categories), with additional confidence intervals in one direction. I'm sure that's not what you want, but it's not clear what it is that you do actually want. -Deepayan library(lattice) library(latticeExtra) sample1-rnorm(100,10,2) sample2-rnorm(100,50,3) sample3-rnorm(100,20,2) sample4-rnorm(100,40,1) mu1-mean(sample1) ci.upper1-mu1+2*2 ci.lower1-mu1-2*2 mu2-mean(sample2) ci.upper2-mu2+2*3 ci.lower2-mu2-2*3 mu3-mean(sample3) ci.upper3-mu3+2*2 ci.lower3-mu3-2*2 mu4-mean(sample4) ci.upper4-mu4+2*1 ci.lower4-mu4-2*1 categories-c(A,B) mu-cbind(mu1,mu2,mu3,mu4) ci.upper-cbind(ci.upper1,ci.upper2,ci.upper3,ci.upper4) ci.lower-cbind(ci.lower1,ci.lower2,ci.lower3,ci.lower4) segplot(mu~ci.upper+ci.lower|categories, centers = mu, horizontal=FALSE) I also tried this seq1-seq(1,4,1) segplot(seq1~ci.upper+ci.lower|categories, centers = mu,horizontal=FALSE) but it also gives poor x axis, I know this is probably an elementary problem but any help would be greatly appreciated. Heres my data structure, sorry for bombarding with all the code. structure(c(9.85647167881417, 50.1856561919426, 19.8477661576365, 39.8575819498655, 13.8564716788142, 56.1856561919426, 23.8477661576365, 41.8575819498655, 5.85647167881417, 44.1856561919426, 15.8477661576365, 37.8575819498655), .Dim = c(1L, 12L), .Dimnames = list(NULL, c(mu1, mu2, mu3, mu4, ci.upper1, ci.upper2, ci.upper3, ci.upper4, ci.lower1, ci.lower2, ci.lower3, ci.lower4 ))) --- Joe King, M.A. Ph.D. Student University of Washington - Seattle Office: 404A Miller Hall 206-913-2912 mailto:j...@joepking.com j...@joepking.com --- Never throughout history has a man who lived a life of ease left a name worth remembering. --Theodore Roosevelt [[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.
