do.call(rbind,lapply(split(as.character(z[,2]),z[,1]),
function(x) c(x, rep(NA, max(table(z[,1]))-length(x)
Jacques VESLOT
CEMAGREF - UR Hydrobiologie
Route de Cézanne - CS 40061
13182 AIX-EN-PROVENCE Cedex 5, France
Tél + 0033 04 42 66 99 76
email [EMAIL PROTECTED]
>
Hello All,
I have had considerable bad luck with attempting the following with for
loops. Here is the problem:
# Suppose we have a data.frame with the following data, which can be
considered a type of edgelist (for those with networks backgrounds):
#
# V1 V2
# 1 A
#
Max wrote:
> Hi Everyone,
>
> I've got a question about data representation. I have some psychometric
> data with 5 scores for 15 different groups. I've been asked to show
> some kind of mean plots. The data below is the mean and SD for a given
> group, unfortunately my employer doesn't want me
Won't it be simpler to do:
for (i in 1:12){
data <- my.fun(my.list[i]))
save(data,file = paste("data",i,".RData", sep="")) }
--- Marie Pierre Sylvestre
<[EMAIL PROTECTED]> wrote:
> Dear R users,
>
> I am analysing a very large data set and I need to
> perform several data
> manipulations. T
Deepayan Sarkar wrote:
> On 12/19/07, Dylan Beaudette <[EMAIL PROTECTED]> wrote:
>
>> On Wednesday 19 December 2007, Max wrote:
>>
>>> Hi Everyone,
>>>
>>> I've got a question about data representation. I have some psychometric
>>> data with 5 scores for 15 different groups. I've been asked
Dear all,
I'm trying to estimate the parameters of a special case of a poisson
model, where the specified equation has an integral and several fixed
parameters.
I think that the MLE command in STATS4 package could be a good choice,
but it's a little complicated. I've got some problems with the off
> 50 %% 12
[1] 2
> 50 %/% 12
[1] 4
> ?Arithmetic
--- livia <[EMAIL PROTECTED]> wrote:
>
> Hello everyone,
>
> I have got a question about a simple calculation. If
> I would like to
> calculate 50/12 and return the result as 4 and the
> remainer 2. Is there a
> function of doing this?
>
> Man
On 12/19/07, Dylan Beaudette <[EMAIL PROTECTED]> wrote:
> On Wednesday 19 December 2007, Max wrote:
> > Hi Everyone,
> >
> > I've got a question about data representation. I have some psychometric
> > data with 5 scores for 15 different groups. I've been asked to show
> > some kind of mean plots. T
R-users
E-mail: r-help@r-project.org
I found the answer myself.
'.Fortran("baklo",' in lo.wam() and .Fortran("bakfit",in
s.wam() may carry out backfitting. But I cannot
create an R code which gives the same results as those of
"bakfit". If someone knows the detail of "bakfit" algorithm,
pleas
On Wednesday 19 December 2007, Max wrote:
> Hi Everyone,
>
> I've got a question about data representation. I have some psychometric
> data with 5 scores for 15 different groups. I've been asked to show
> some kind of mean plots. The data below is the mean and SD for a given
> group, unfortunately
Hi Everyone,
I've got a question about data representation. I have some psychometric
data with 5 scores for 15 different groups. I've been asked to show
some kind of mean plots. The data below is the mean and SD for a given
group, unfortunately my employer doesn't want me posting full datasets.
System:
Linux kernel 2.6.22-14
Ubuntu 7.10 gutsy
ESS 5.3.0 on Emacs 22.1.1
R version 2.6.0
Colleagues
I would like to use the user contributed package "clim.pact". I'm having
trouble with the dependency of "clim.pact" on "ncdf". Like others I am
finding that R CMD INSTALL does not find the netcdf
In your first case you seem to be missing a comma after the "^2"; also
missing value for 'y'. Probably should be something like:
optimize(function(x,y)
((327.727-(1-0.114^10)*y*(1-x)/x/(1-x^y))+(9517.336-327.727
*(1+(1-x)*(1+y)/x-327.727)))^2,
interval=c(0,1), y=1)
On Dec 19, 2007 5:12 PM, Xin
Alternatively
levels(df$binname)[which(df$freq >=
0.5*cumsum(df$freq)[nrow(df)])[1]]
--- Chuck Cleland <[EMAIL PROTECTED]> wrote:
> Martin Tomko wrote:
> > Dear list,
> > I have a vector (array, table row, whatever is
> best) of frequency values
> > for categories (or bins), and I need to find t
I used the command below, but R gives me the error message--syntax error.
can anyone see the mistakes I made?
optimize(function(x,y)
+ ((327.727-(1-0.114^10)*y*(1-x)/x/(1-x^y))+(9517.336-327.727
*(1+(1-x)*(1+y)/x-327.727)))^2
+ interval=c(0,1))
At the same time, I use nlm() but R gives me the
If all your entries are double precision then you are
using 8 bytes per entry, so 20,000*n entries are just
160,000*n bytes, i.e. less than 160*n Kb. If your n is
100 you get 16 Mb which is not that much (especially
if you pre-allocate it only once). So just use the
matrix and don't worry!
--- dxc
You appear to have fitted a regression tree, which does not seem to be
what your interpretation of 'pnV22' requires.
I have little idea what you actually did, but am confident that it is not
what you claim you did.
Also, note fortune("dog"):
Firstly, don't call your matrix 'matrix'. Would you
R does this sort of thing easily without any parse/eval acrobatics needed.
E.g., you can do:
> stu <- function(x) {return( 1 + (2*x*x) - (3*x) )}
> (x <- 0:3)
[1] 0 1 2 3
> stu(x)
[1] 1 0 3 10
> metafun <- function(FUN, data) FUN(data)
> metafun(stu, x)
[1] 1 0 3 10
> # if you want
On 12/19/07, John G. Bullock <[EMAIL PROTECTED]> wrote:
>
> I'm using lattice to draw a multi-panel figure: 5 rows, 4 columns. The
> y-axis for
> each panel is determined by
>
> yaxs<- list(draw=T
> , labels=c(0, '.5', '1', '1.5')
>
I'm using lattice to draw a multi-panel figure: 5 rows, 4 columns. The y-axis
for each panel is determined by
yaxs<- list(draw=T
, labels=c(0, '.5', '1', '1.5')
, at=c(0, .5, 1, 1.5)
, tck=c(.4, 0)
Hi.
I'm looking for an R equivalent to something like function pointers in C/C++.
I have a search procedure that evaluates the fitness of each point it reaches
as it moves along, and decides where to move next based on its fitness
evaluation. I want to be able to pass different fitness func
On 12/19/07, Spilak,Jacqueline [Edm] <[EMAIL PROTECTED]> wrote:
> Hi all
> I can't find what I am looking for so I am asking here. I have a
> dataset that looks something like this.
>
> Year season percent_below
> 2000 Winter 6.9179870
> 2000 Spring 1.6829436
> 2000 Summer 1.8463501
Good afternoon. I would like to post some positions on your site. Is
this possible?
Jen
Jennifer C. Turpin
Human Resources Coordinator
Wayne State University
Services in Support of the NIH Perinatology Research Branch
4 Brush South - Office 4723
3990 John R. - Detroit, MI 48201
31
I think this is what you want.
Y[Y>0] <- 1
--- "Armelini, Guillermo" <[EMAIL PROTECTED]>
wrote:
> Hello everyone!
>
> Is anybody can help me to solve this silly question
> that unfortunately I haven't found the right way to
> address it.
>
> Supose I have a matrix X[n,n] dimension
>
> I wo
Dear R Users,
I am working for the United Nations to construct a complete life table
from an abridged table.
I want to use the code of Hydman Filter by Rob J Hydman but an error
sentence always appears and it simply doesn't run--
source("C:/R/Jamie/HymanFilter.R")
Error in
I figured out the problem: functions
gzfile/unz/file/... create a connection to a file but
do not *open* the connection unless the 'open'
argument is specified.
... a little R gotcha for people who use other
programming languages and expect similar
concepts/behavior.
> I get unexpected behavio
As Tony assumed: a data frame.
Joh
John Kane wrote:
> What was spectrum orginally?
>
>
> --- Johannes Graumann <[EMAIL PROTECTED]>
> wrote:
>
>> Why is class(spectrum[["Ion"]]) after this "factor"?
>>
>> spectrum <- cbind(spectrum,Ion=rep("",
>> nrow(spectrum)),Deviation.AMU=rep(0.0,
>> nrow
Hi all
I can't find what I am looking for so I am asking here. I have a
dataset that looks something like this.
Year season percent_below
2000 Winter 6.9179870
2000 Spring 1.6829436
2000 Summer 1.8463501
2000 Autumn 3.8184993
2001 Winter 2.8832806
2001 Sprin
Dear R Users,
I am working for the United Nations to construct a complete life table
from an abridged table.
I want to use the code of Hydman Filter by Rob J Hydman but an error
sentence always appears and it simply doesn't run--
source("C:/R/Jamie/HymanFilter.R")
Error in
I get unexpected behavior from "readLines()" and
"scan()" depending on how the file is opened with
"gzfile" or "unz". More specifically:
> file <- gzfile("file.gz")
> readLines(file,1)
[1] "a\tb\tc"
> readLines(file,1)
[1] "a\tb\tc"
> close(file)
It seems that the stream is rewound between calls
On Wed, Dec 19, 2007 at 07:51:47PM +0200, Adrian Dusa wrote:
> Hi James,
>
> On Wednesday 19 December 2007, David James wrote:
> > Hi,
> >
> > I'm sorry I'm also coming late to this discussion, but like Dirk, I
> > fail to understand what's wrong with using dbConnect() the way
> > the documentati
Hi James,
On Wednesday 19 December 2007, David James wrote:
> Hi,
>
> I'm sorry I'm also coming late to this discussion, but like Dirk, I
> fail to understand what's wrong with using dbConnect() the way
> the documentation, (see ?MySQL) suggests.
>
> RMySQL was developed in a fully distributed cl
On Wednesday 19 December 2007, Dirk Eddelbuettel wrote:
> [...]
>
> Err, I am late to this but
>
> dbConnect(drv, user="mysqluser", password="mysqlpassword",
> dbname="anydatabase", host="xxx.xxx.xxx")
>
> works fine for me (eg on Ubuntu with a remote MySQL on some other
> box). What's the proble
Hello everyone!
Is anybody can help me to solve this silly question that unfortunately I
haven't found the right way to address it.
Supose I have a matrix X[n,n] dimension
I would like to calculate the product of the vectors
Y=X[1,n]*X[n,1]
Then I would like to run the following operatio
Hi,
I'm sorry I'm also coming late to this discussion, but like Dirk, I
fail to understand what's wrong with using dbConnect() the way
the documentation, (see ?MySQL) suggests.
RMySQL was developed in a fully distributed client/server
environment, and it uses the MySQL-provided client API. The
Hi,
I'm looking for a R-package that does feature selection for PLS using a
genetic optimization algorithm.
I couldn't find one on CRAN and I wonder whether there is a free one. I
would be very appreciative for any help.
Regards
Rolf
__
R-help@r-pro
I think if you use Gabor's suggested addtion of
add.missing to the original cast command you get what
you want.
cast(dfm, ... ~ Var3, add.missing=TRUE)
--- Bert Jacobs <[EMAIL PROTECTED]> wrote:
>
> Thx Hadley,
> It works, but I need some finetuning.
>
> If I use the following expression:
> Ne
On 12/19/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> I'm analysing a survey, and creating a barchart of the different responses
> for each question. The questions are grouped according to a number of
> categories, so I'm using lattice to create a plot with each question in a
>
try this:
apply(sapply(basic_map, function(x)tapply(df$Value, df$Book, sum)[x]), 2, sum)
On 19/12/2007, Kondamani, Arjun (GMI - NY Corporate Bonds)
<[EMAIL PROTECTED]> wrote:
> Suppose I have:
>
> BookValue
> A 10
> B 11
> C 9
> D 8
> A 12
> C 4
> D 5
> Suppose I have:
>
> Book Value
> A 10
> B 11
> C 9
> D 8
> A 12
> C 4
> D 5
> B 7
>
> I want to summarize above not by Book but by groupings of Books as in
> (below)
>
> I have a list ... basic_map <- list(c("A",B"),c("C,D"))
> Big_names <- c("A1", "A2")
> Names(basi
Hi,
I have a problem with library (rpart) (and/or library(tree)).
I use a data.frame with variables
"pnV22" (observation: 1, 0 or yes, no)
"JTemp" (mean temperature)
"SNied" (summer rain)
I used function "rpart" to build a model:
library(rpart)
attach(data.frame)
res
On Wed, Dec 19, 2007 at 04:34:26PM +0200, Adrian Dusa wrote:
> On Saturday 15 December 2007, Gabor Grothendieck wrote:
> > Use ssh forwarding to forward local port 3307 to remote port 3306
> > specifying the remote account and password. Then if you use local port
> > 3306 you can access your local
I went to the article on np in R news 7/2 (October 2007). What's the
general technique to get the source code associated with the article
as a .R file that I can play with?
--
Ajay Shah http://www.mayin.org/ajayshah
[EMAIL PROTECTED]
Suppose I have:
BookValue
A 10
B 11
C 9
D 8
A 12
C 4
D 5
B 7
I want to summarize above not by Book but by groupings of Books as in
(below)
I have a list ... basic_map <- list(c("A",B"),c("
David Barron-3 wrote:
>
> You can calculate the AIC as follows:
>
> (fm1 <- lmer(Reaction ~ Days + (Days|Subject), sleepstudy))
> aic1 <- AIC(logLik(fm1))
>
>
Is AIC() [extractAIC()] "valid" for models with random effects? I noticed
that the help page for extractAIC() does not list models wi
What was spectrum orginally?
--- Johannes Graumann <[EMAIL PROTECTED]>
wrote:
> Why is class(spectrum[["Ion"]]) after this "factor"?
>
> spectrum <- cbind(spectrum,Ion=rep("",
> nrow(spectrum)),Deviation.AMU=rep(0.0,
> nrow(spectrum)))
>
> slowly going crazy ...
>
> Joh
>
> _
I am not sure I really understand what you want but
will this work?
tt<-c("03-Nov-1997","09-Oct-1991","27-Aug-1992","01-Jul-1994","19-Jan-1990",
"12-Nov-1993","08-Oct-1993","10-Nov-1982","08-Dec-1986","23-Dec-1987","02-Aug-1995",
"20-Oct-1998","29-Apr-1991","16-Mar-1994","20-May-1991","28-Dec-1987
--- Jim Lemon <[EMAIL PROTECTED]> wrote:
> John Kane wrote:
> > I think you're going to find that barchart with
> that
> > many values in a bar is going to be pretty well
> > uninterpretable.
> >
> > Jim Lemon gives the desired barchart but it is
> very
> > difficult to read.
> >
> > Steali
On Saturday 15 December 2007, Gabor Grothendieck wrote:
> Use ssh forwarding to forward local port 3307 to remote port 3306
> specifying the remote account and password. Then if you use local port
> 3306 you can access your local version of MySQL and if you
> use port 3307 you can access the remot
Dear all,
I am trying to wrap a *nix shell script around R for a particular
purpose, for which I need to get R to execute predetermined commands
but retain interactivity and allow user input during their execution.
A straight redirection of standard input is therefore not appropriate,
and
Dear Jose,
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> project.org] On Behalf Of Jose Quesada
> Sent: December-19-07 7:54 AM
> To: [EMAIL PROTECTED]
> Subject: [R] Correlation when one variable has zero variance
> (polychoric?)
>
> Hi,
>
> I'm running this
Hi Robin,
Before someone gives a better solution, you can try this :
x1<-array(1:10,c(2,5))
x2<-array(1:9,c(3,3))
ArrayAdd<-function(array1,array2){
x<-array(0,c(max(nrow(array1),nrow(array2)),max(ncol(array1),ncol(array2
x[1:nrow(array1),1:ncol(array1)]<-x[1:nrow(array1),1:ncol(array1
On Wednesday 19 December 2007 14:12:16 rašėte:
> sapply(levels(DATA$know1), function(x)subset(DATA, (know1==x &
> know2==x)), simplify=F)
Hey, thanks, that seems to work!
--
Donatas Glodenis
http://dg.lapas.info
__
R-help@r-project.org mailing list
ht
This should give you what you want:
x <- scan(textConnection("0. 2.3972 4.3500
-4.1972 0.6361
1.0806 5.9056 -1.8722 2.1333 -1.1806
3.6167 0.8778 8.3389 3.8417 1.
-3.7611 -11.6778 -2.0306 6.9
Here is another way of doing it:
> x <- c('A','B','A','C','C','B')
> ave(rep(1, length(x)), x, FUN=cumsum)
[1] 1 1 2 1 2 2
>
On Dec 19, 2007 5:36 AM, Eric Lecoutre <[EMAIL PROTECTED]> wrote:
> Dear R-help,
>
> I am trying to have a generic way to assess the replicates in a character
> vector.
>
Colin Robertson wrote:
> Dear List,
>
>
>
> I would like to calculate the Goodman-Kruskal gamma for the predicted
> classes obtained from an ordinal regression model using lrm in the Design
> package. I couldn't find a way to get gamma for predicted values in Design
> so have found previous pos
Hi,
I'm running this for a simulation study, so many combinations of parameter
produce many predictions that I need to correlate with data.
The problem
I'm using rating data with 3 to 5 categories (e.g., too low, correct, too
high). The underlying continuous scales should be
Hey,
I make a regression for Gamma distribution with log link, in R and in SAS.
In R, the dispersion is estimated by
\phi=Deviance/(#_of_observations),
In SAS, there are two options:
\phi=Deviance/(#_of_observations-#_of_params) or
\phi=Pearson/(#_of_observations-#_of_params).
I understand that
Here is an example of plotting the number of observations per month.
You can change this to any period you want.
# create some test data
x.d <- as.Date("2000-1-1") + runif(1000, 1, 500)
# create buckets of one month (or whatever period you want)
x.cut <- cut(x.d, breaks=seq(as.Date('2000-1-1'), ma
I have a sample of observations:
> yy
[1] 0. 2.3972 4.3500 -4.1972 0.6361
[6] 1.0806 5.9056 -1.8722 2.1333 -1.1806
[11] 3.6167 0.8778 8.3389 3.8417 1.
[16] -3.7611 -11.6778 -2.0306
Dear R-help,
I am trying to have a generic way to assess the replicates in a character
vector.
Say that I have the following vector:
x <- c('A','B','A','C','C','B')
I would like to obtain:
replicates <- c(1,1,2,1,2,2)
each number beeing the time we see the corresponding value in x.
Any clever
Thank you very much!
This does indeed what I am looking for and really have R-ish look and feel.
I just have turned that into a little handy function 'replicates'
Best wishes,
Eric
2007/12/19, Henrique Dallazuanna <[EMAIL PROTECTED]>:
>
> Try this:
>
> replicate <- vector("numeric", len=lengt
Perhaps you can try subset the data:
sapply(levels(DATA$know1), function(x)subset(DATA, (know1==x &
know2==x)), simplify=F)
On 19/12/2007, Donatas G. <[EMAIL PROTECTED]> wrote:
> Hi, I have a data frame DATA, which (simplified of course) looks like this:
>
> know1 = c("Y","N","N","Y","N","N","Y",
Hi,
I would like some extra information on the 'cgh' package in R. I noticed
that there isn't much activity regarding this package on the R and BioC
mailing list (I googled it).
I started using this package and I have few questions:
1/ As I have a custom tiling like array @8um features reso
Hi, I have a data frame DATA, which (simplified of course) looks like this:
know1 = c("Y","N","N","Y","N","N","Y","Y","N")
par1=c(1,4,5,3,3,2,3,3,5)
know2 = c("Y","Y","N","Y","N","N","N","Y","Y")
par2=c(3,4,4,3,5,2,4,3,2)
DATA=data.frame(know1,par1,know2,par2)
it represents answers in a questionn
Martin Tomko wrote:
> Thank you, Chuck,
> would you mind commenting a bit on the code, it is not all clear... HOw
> would you go to retrieve only the numeric value (not the category name)?
> I am just starting with R, and the functionality of replicate and levels
> is not quite clear. I tried the
Hi all,
I have observed that when using the randomForest package to do regression, the
predicted values of the dependent variable given by a trained forest are not
centred and have the wrong slope when plotted against the true values.
This means that the R^2 value obtained by squaring the Pea
Try this:
replicate <- vector("numeric", len=length(x))
replicate[order(x)] <- unlist(sapply(rle(sort(x))$lengths, seq_len))
On 19/12/2007, Eric Lecoutre <[EMAIL PROTECTED]> wrote:
> Dear R-help,
>
> I am trying to have a generic way to assess the replicates in a character
> vector.
> Say that I
Try this:
"%add%" <- function(x1, x2)
{
dim1 <- dim(x1)
dim2 <- dim(x2)
seq1 <- list();for(i in 1:length(dim1)) seq1[[i]]=seq(dim1[i])
filter1 <- paste(seq1, collapse=",")
cmd1 <- paste("out[", filter1, "]", sep="")
seq2 <- list();for(i in 1:length(dim2)) seq2[[i]]=seq(
Hello R Users,
I am interested in using R to generate quantitative structure-activity
relationships (QSARs) for small molecules given a set of molecular
descriptors and biological data tab-delimited or excel file. In which fist
value colum of each row is biological value and rest all are its
d
livia wrote:
> Hello everyone,
>
> I have got a question about a simple calculation. If I would like to
> calculate 50/12 and return the result as 4 and the remainer 2. Is there a
> function of doing this?
>
> Many thanks.
?"%%" to see how to get the remainder. You might put the "result" and
re
Hello everyone,
I have got a question about a simple calculation. If I would like to
calculate 50/12 and return the result as 4 and the remainer 2. Is there a
function of doing this?
Many thanks.
--
View this message in context:
http://www.nabble.com/Calculate-remainer-tp14414906p14414906.html
[EMAIL PROTECTED] wrote:
> I am plotting fishing vessel positions and want these points to be
> relative in size to the catch at that point. Is this possible? I am just
> begining to use R and my search of the help section didnt help in this
> area. Heres what Im using so far
>
> xyplot(data$lat
Thank you, Chuck,
would you mind commenting a bit on the code, it is not all clear... HOw
would you go to retrieve only the numeric value (not the category name)?
I am just starting with R, and the functionality of replicate and levels
is not quite clear. I tried the documentation, but am not any
On Wed, 2007-12-19 at 14:59 +0800, gallon li wrote:
> I have the following list of observations of calendar time:
>
> [1] 03-Nov-1997 09-Oct-1991 27-Aug-1992 01-Jul-1994 19-Jan-1990 12-Nov-1993
> [7] 08-Oct-1993 10-Nov-1982 08-Dec-1986 23-Dec-1987 02-Aug-1995 20-Oct-1998
> [13] 29-Apr-1991 16-Ma
Someone asked about R on the EEE PC. That was the second thing
I installed when I received mine (the first being the full KDE desktop).
I don't recall any issues with the installation. I used the standard debian
package (the native linux on the EEE is Xandros, a derivation of debian).
The caveat be
On 19/12/07 8:39 PM, eugen pircalabelu wrote:
> 2. How can i find the median of a variable in survey package?
> a<-c(1:10)
> b<-sample(1:20, 10, replace=T)
> b1<-sample(0:1, 10, replace=T)
> c<-data.frame(a,b, b1)
> library(survey)
> design<-svydesign(id=~1, data=c)
> svymean(~b, design)
>
> svyme
Hi,
I was just wondering if prop.trend.test() is equivalent to the
Cochran-Armitage Trend test?
Thank you!
[[alternative HTML version deleted]]
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PLEASE do read th
I just posted a vim-script at
http://www.vim.org/scripts/script.php?script_id=2104
that allows sending selected lines of R code in a vim-buffer to the R-
gui. There are alternatives for Windows and Linux (http://www.vim.org/
scripts/script.php?script_id=1048) but i never found anything that
works o
Mark Leeds schrieb:
> That's because the first factor is being used as the "baseline". This is
> explained in any intro to GLMs text.
>
>
>
Thank you Mark, I didn't found that hint.
Knut
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On Tue, 2007-12-18 at 16:27 -0600, Naiara Pinto wrote:
> Dear all,
>
> I would like to use a tree regression method to analyze my dataset. I
> am interested in the fact that random forests creates in-bag and
> out-of-bag datasets, but I also need an estimate of support for each
> split. That seems
Martin Tomko wrote:
> Dear list,
> I have a vector (array, table row, whatever is best) of frequency values
> for categories (or bins), and I need to find the median category.
> Trivial to do by hand, but I was wondering if there is a means to do it
> in R in an elegant way.
>
> The obvious me
John Kane wrote:
> I think you're going to find that barchart with that
> many values in a bar is going to be pretty well
> uninterpretable.
>
> Jim Lemon gives the desired barchart but it is very
> difficult to read.
>
> Stealing his code to create the same matrix I'd
> suggest may be lookin
Hi
suppose I have two arrays x1,x2 of dimensions a1,b1,c1 and
a2,b2,c2 respectively.
I want x = x1 "+" x2 with dimensions c(max(a1,a2), max(b1,b2),max
(c1,c2))
with
x[a,b,c] = x1[a1,b1,c1] + x2[a2,b2,c2] ifa <=min(a1,a2) , b<=min
(b1,b2), c<=min(c1,c2)
and the other bits either x1 o
Dear List,
I would like to calculate the Goodman-Kruskal gamma for the predicted
classes obtained from an ordinal regression model using lrm in the Design
package. I couldn't find a way to get gamma for predicted values in Design
so have found previous positings suggesting to use :
Rcorr.ce
Dear all,
I'm analysing a survey, and creating a barchart of the different responses
for each question. The questions are grouped according to a number of
categories, so I'm using lattice to create a plot with each question in a
category on it. The problem is that the response set for differe
On Dec 18, 2007 9:39 PM, David Winsemius <[EMAIL PROTECTED]> wrote:
> "Armin Goralczyk" <[EMAIL PROTECTED]> wrote in
> news:[EMAIL PROTECTED]:
>
> > It's not the spaces, the problem is the tag (sorry that I didn't
> > specify this), or maybe the string []. I am working on a Mac OS X 10.4
> > with R
Dear list,
I have a vector (array, table row, whatever is best) of frequency values
for categories (or bins), and I need to find the median category.
Trivial to do by hand, but I was wondering if there is a means to do it
in R in an elegant way.
The obvious medioan(vector) returns the median f
Yep, but I figured that out quite fast ;0)
Thanks for giving me a hand ... you want believe a many times I skimmed the
cbind help without actually seeing this ... well, it was 0:30 ...
Thanks again, Joh
Tony Plate wrote:
> Whoops, it looks like there's a typo in ?cbind (R version 2.6.0 Patched
>
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