Ross Boylan wrote:
> In the following example, the inner evaluation pulls in the global value
> of subset (a function) rather than the one I thought I was passing in (a
> vector). Can anyone help me understand what's going on, and what I need
> to do to fix the problem?
>
> f0 <- function(formula,
It's working for other users of icc. Check what CC is set to in
etc/Makeconf, and if it is not 'icc -c99', reset it.
On Wed, 13 Feb 2008, Denham Robert wrote:
> I am having trouble compiling R-2.6.2 on suse linux x86_64 using the
> intel compiler. I read section C.2.1 Intel compilers in the R
Charilaos Skiadas wrote:
> On Feb 12, 2008, at 1:31 PM, [Ricardo Rodriguez] Your XEN ICT Team
> wrote:
>
>
>> Sundar Dorai-Raj wrote:
>>
>> Could it we advisable that cast, melt or whatever function we deal
>> with
>> throws an more informative error message when this kind of conflicts
>> oc
In the following example, the inner evaluation pulls in the global value
of subset (a function) rather than the one I thought I was passing in (a
vector). Can anyone help me understand what's going on, and what I need
to do to fix the problem?
f0 <- function(formula, data,
G'day Stanley,
On Wed, 13 Feb 2008 10:40:09 +0800
"Ng Stanley" <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Given test <- matrix(c(0,2,0,1,3,5), 3,2)
>
> > test[test>0]
> [1] 2 1 3 5
>
> These are values >0
>
> > which(test>0)
> [1] 2 4 5 6
>
> These are array indices of those values >0
>
> > which
Is this what you are after?
> test <- matrix(c(0,2,0,1,3,5), 3,2)
> (x <- which(test > 0, arr.ind=TRUE))
row col
[1,] 2 1
[2,] 1 2
[3,] 2 2
[4,] 3 2
> unique(x[, 'row'])
[1] 2 1 3
>
On Feb 12, 2008 9:40 PM, Ng Stanley <[EMAIL PROTECTED]> wrote:
> Hi,
>
> Given test <- matrix
X <- matrix(c(0,2,0,1,0,0,3,5), ncol=2)
Informative version:
isPositive <- (X > 0)
nbrOfPositives <- apply(isPositive, MARGIN=1, FUN=sum)
hasPositives <- (nbrOfPositives >= 1)
positiveRows <- which(hasPositives)
Compact version:
positiveRows <- which(rowSums(X > 0) >= 1)
If you have an extremely
[Ricardo Rodriguez] Your XEN ICT Team wrote:
> Peter McMahan wrote:
>
>> try something like this:
>> username <- "abc"
>> pass <- "123"
>>
>> source(paste("http://",username,":",pass,"@mire.environmentalchange.net/~webmaster/R/3Dsurface.r",sep="";))
>>
>>
>> I'm not sure it'll wor
I am having trouble compiling R-2.6.2 on suse linux x86_64 using the
intel compiler. I read section C.2.1 Intel compilers in the R
Installation and Administration manual, and put
CC=icc
CFLAGS="-g -O3 -wd188 -ip"
F77=ifort
FFLAGS="-g -O3"
ICC_LIBS=/opt/intel/cce/10.1.012/lib
IFC_LIBS=/opt/intel/
Hi,
Given test <- matrix(c(0,2,0,1,3,5), 3,2)
> test[test>0]
[1] 2 1 3 5
These are values >0
> which(test>0)
[1] 2 4 5 6
These are array indices of those values >0
> which(apply(test>0, 1, all))
[1] 2
This gives the row whose elements are all >0
I can't seem to get indices of rows containin
Peter McMahan wrote:
> try something like this:
> username <- "abc"
> pass <- "123"
>
> source(paste("http://",username,":",pass,"@mire.environmentalchange.net/~webmaster/R/3Dsurface.r",sep="";))
>
>
> I'm not sure it'll work but it's worth a try.
Thanks, Peter. It didn't work. I a
> To: [EMAIL PROTECTED]
> From: [EMAIL PROTECTED]
> Date: Wed, 13 Feb 2008 11:38:43 +1100
> Subject: AIC Quantile Regression
>
> Dear R,
> I currently trying to fit quantile regression models to test the
> relationship between abalone and algal cover at different quadrat sizes.
> I am using the
Please, is it possible to pass an username and a password to source() to
access code stored in restricted access web server?
I know I can use commands like...
> source("http://mire.environmentalchange.net/~webmaster/R/3Dsurface.r";)
but I am not able to find if it is possible to pass authoriza
Marius,
several remarks which might help you to improve your code (and your style
of asking questions):
> I wish to calculate the Gini index (ineq from same package) and some
> other indices for the diameter distribution of each plot (df dgtot).
>
> dgtot:
> IDPlotDiameter(cm)
I would re
The trick is to fit the model in a form which has the two separate
intercepts and the two separate slopes as the parameters.
You do have to realise that a*b, a*b-1, a/b, a/b-1, ... all specify the
same model, they just use different parameters for the job. (Yes,
really!)
> dat
ldose sex numde
Charilaos Skiadas wrote:
>
> It tells you that a call to get was attempted looking for a variable
> of mode "function", and such a variable was not found. The problem is
> of course that the call tells you the variable is named "fun", while
> you expected it to be named "mean". But it alerts to
> sub("^I\\((.*)\\^.*$", "\\1", MyData)
[1] "Test1" "Test2" "Test1" "Test2"
"Test1.Test2"
In this case, there is a simple way of discovering which variable names
are present, though
> all.vars(parse(text = MyData))
[1] "Test1" "Test2" "Test1.Test2"
Bill Venables
C
If you want to use colClasses, then do:
read.table(, colClasses=rep('numeric', 50))
On Feb 12, 2008 5:40 PM, Weidong Gu <[EMAIL PROTECTED]> wrote:
> I have a data file with 50 columns. Among them, there are two
> coordinates, X and Y
>
> X
>
> Y
>
> 641673.78807
>
> 3607080.78438
>
> 641436.
It is just printing them out with that significance; the numbers are stored
with about 15 digits. If you want more, use 'options':
> x <- scan(textConnection("641673.78807
+
+ 3607080.78438
+
+ 641436.56207
+
+ 3607108.30543
+
+ 641165.28042
+
+ 3607136.82957
+
+ 640879.58373
+
+ 3607116.20568
+
> Speaking from personal experience, it can be quite a drag when one has set
> up and mostly-debugged a long computation only to have it stop with an
> error like "data are essentially constant" right near the end because of
> some condition for which the function author thought it better to stop w
I have a data file with 50 columns. Among them, there are two
coordinates, X and Y
X
Y
641673.78807
3607080.78438
641436.56207
3607108.30543
641165.28042
3607136.82957
640879.58373
3607116.20568
When I use read.table, it rounds X and Y to the maximal 8 decimal number
as.
641673.8
On Feb 12, 2008, at 1:31 PM, [Ricardo Rodriguez] Your XEN ICT Team
wrote:
> Sundar Dorai-Raj wrote:
>
> Could it we advisable that cast, melt or whatever function we deal
> with
> throws an more informative error message when this kind of conflicts
> occur? I am guessing this is a pretty frequ
?merge
Let the data.frame be xx
yy <- data.frame(ORF=
c("YAL026C","YAL041W","YAL048C","YAL007C","YAL012W",
"YAL016W" ))
merge(xx,yy, by="ORF")
should do it.
--- Roberto Olivares Hernandez <[EMAIL PROTECTED]> wrote:
> Hi,
> I have a data frame Y with the following
> information,
>
>
On Feb 12, 2008, at 3:31 PM, Thomas Lumley wrote:
> On Tue, 12 Feb 2008, Alistair Gee wrote:
>
>> I often want to temporarily modify the options() options, e.g.
>>
>> a <- seq(1001, 1001 + 10) # some wide object
>>
>> with.options <- function(..., expr) {
>> options0 <- options(...)
>> t
Petr PIKAL wrote:
> Hi
>
> [EMAIL PROTECTED] napsal dne 12.02.2008 09:09:23:
>
>> Hi,
>>
>> First problem:
>>> test <- matrix(c(1,1,2,1), 2,2)
>>> apply(test, 1, function(x) { t.test(x) $p.value })
>> Error in t.test.default(x) : data are essentially constant
>
> make your data not constant
>
>
On 12/02/2008 4:33 PM, Elizabeth Purdom wrote:
> Hi,
> My problem is more that the packages use the function internally, so
> they get the wrong function. Should that not be happening?
The package authors should use a NAMESPACE file to avoid exactly this
problem. You could contact the maintaine
Hi,
My problem is more that the packages use the function internally, so
they get the wrong function. Should that not be happening?
Best,
Elizabeth
jim holtman wrote:
> you can use:
>
> package::getNames()
>
> to reference the one that you want.
>
> On Feb 12, 2008 3:45 PM, Elizabeth Purdom <
you can use:
package::getNames()
to reference the one that you want.
On Feb 12, 2008 3:45 PM, Elizabeth Purdom <[EMAIL PROTECTED]> wrote:
> Hi,
> I am trying to use two contributed packages, both of which have a
> function 'getNames'. So if I load them both they obviously conflict.
> Currently
Thank you all,
All responses worked and focused on the use of combining the desired
column names (or numbers in Barry's solution) with variations of
'match(x, table)' and 'x %in% table' to generate the remaining column
names (numbers).
xxx <- data.frame(matrix(1:40, ncol=8))
names(xxx) <- letters
Hi,
I am trying to use two contributed packages, both of which have a
function 'getNames'. So if I load them both they obviously conflict.
Currently I manually detach one package and then reload the other to be
able to use one function right after another. Is there anything else I
can do?
Best,
Try this:
with.options <- function(...) {
L <- as.list(match.call())[-1]
len <- length(L)
old.options <- do.call(options, L[-len])
on.exit(options(old.options))
invisible(eval.parent(L[[len]]))
}
> with.options(width = 40, print(1:25))
[1] 1 2 3 4 5 6 7 8 9 10 11 12
On Tue, 12 Feb 2008, Alistair Gee wrote:
> I often want to temporarily modify the options() options, e.g.
>
> a <- seq(1001, 1001 + 10) # some wide object
>
> with.options <- function(..., expr) {
> options0 <- options(...)
> tryCatch(expr, finally=options(options0))
> }
>
> Then I can u
Not the most straightforward way, but I think it gets the job done:
x <- read.table(textConnection("Ageclass Scale MeanSex
1 21-40BP 40.26667 female
2 41-60BP 34.10714 female
3 61-79BP 37.3 female
4 21-40GH 30.25000 female
5 41-60GH 39.00926 fema
> I have following data set, which I want to plot the "Scale" variable on the
> x-axis and "Mean"´on the y-axis for each Ageclass and for each sex. The Mean
> value of each Ageclass for each sex would be connected by a line. Totally,
> there should be 6 lines, from which three present the Mean
I couldn't get that to work, b/c I need the expr block to be evaluated
after the call to options(). I suspect that list(...) evaluates its
arguments. Here's what I did to your example:
test2 <- function(...) {
dots <- list(...)# <=== I think expr is evaluated here.
if(sum(dots.missin
On Feb 12, 2008 1:51 PM, Peter McMahan <[EMAIL PROTECTED]> wrote:
> Thanks, that's very helpful. Unfortunately Gtk2 is difficult to get
> running on a Mac, so I've been trying the gWidgetstcktk interface.
> It sounds like the behavior you're describing is exactly what I want,
> so it may just be a
On 2/12/08, Tom Cohen <[EMAIL PROTECTED]> wrote:
> Dear list,
>
> I have following data set, which I want to plot the "Scale" variable on the
> x-axis and "Mean"´on the y-axis for each Ageclass and for each sex. The Mean
> value of each Ageclass for each sex would be connected by a line. Totall
Hi,
Is there a package that will estimate simple Markov models and hidden
Markov models for discrete time processes in R?
Thanks in advance,
David
--
===
David Kaplan, Ph.D.
Professor
Department of Educational Psychology
University of
Thanks, that's very helpful. Unfortunately Gtk2 is difficult to get
running on a Mac, so I've been trying the gWidgetstcktk interface.
It sounds like the behavior you're describing is exactly what I want,
so it may just be a difference in the TGtk2 and tcltk event loops?
In your example, can yo
Yes that will work, that's exactly what I was getting at in my second
paragraph. I wrote a function that uses this idea, except the (single)
unnamed argument can occur anywhere in the function (not necessarily
last). It will stop if there is more than one unnamed argument.
test2 <- function(..
It should be possible i think. You just supply all the arguments via
'...' and then cut off the last one. I don't see why this wouldn't work,
but maybe i'm missing something.
Gabor
On Tue, Feb 12, 2008 at 12:58:25PM -0600, Erik Iverson wrote:
> Alistair -
>
> I don't believe this is possible. T
Alistair -
I don't believe this is possible. The only way formal arguments (like
expr) can be matched after a '...' is with *exact* name matching. Why
do you want to avoid explicitly naming the expr argument?
If you always want the expr argument last, you might be able to just use
... as the
That should have been:
do.call("rbind", by(iris[-5], iris$Species, colMeans))
On Feb 12, 2008 1:46 PM, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> Please provide reproducible code. Try this:
>
> do.call("rbind", by(iris, iris$Species, colMeans))
>
>
> On Feb 12, 2008 1:34 PM, Allen S. Rout <
Hi Adrian,
even better would be this:
1.) Install the latest Rgraphviz devel version
2.) Use the new API for graph rendering:
library(Rgraphviz)
example(randomEGraph)
x <- layoutGraph(g1)
graph.par(list(edges=list(lwd=2))) # if you want to set lwd=2 for the
whole session
# or if you just want to c
Please provide reproducible code. Try this:
do.call("rbind", by(iris, iris$Species, colMeans))
On Feb 12, 2008 1:34 PM, Allen S. Rout <[EMAIL PROTECTED]> wrote:
> "Gabor Grothendieck" <[EMAIL PROTECTED]> writes:
>
> > The point is don't use cbind -- use data.frame.
>
> Is there a conventional wa
Hello,
I'm trying to make a graphical interface for an R function
I've written. A common use for the function is to call it with
specific parameters, and then watch the output as it evolves.
There's not necessarily a logical stopping point, so I usually
use ctrl-C when I'm done to stop it.
I've mad
Mode <- function(var)rownames(table(var))[which.max(table(var))]
as.data.frame(sapply(c("mean", "median", "Mode"),
function(fun)tapply(x$mgl, x$RM, fun, na.rm=T)))
On 12/02/2008, stephen sefick <[EMAIL PROTECTED]> wrote:
> below is my data frame. I would like to compute summary statistics
> for
Thompson, David (MNR) wrote:
> Hello,
>
> I would like to reorder columns in a data frame by their names as
> demonstrated below:
>
> Take this data frame:
> > xxx <- data.frame(matrix(1:40, ncol=8))
> > names(xxx) <- letters[1:8]
> > xxx
> a b c d e f g h
>
"Gabor Grothendieck" <[EMAIL PROTECTED]> writes:
> The point is don't use cbind -- use data.frame.
Is there a conventional way to use data.frame instead of cbind when
processing a 'by'?
My code is littered with:
foo <- data.frame(cbind(by( /* yadda */ )))
and I've had a problem similar to t
Sundar Dorai-Raj wrote:
>
> Do you have an object called 'mean' that's masking the base::mean
> function? I can replicate your error using the following:
>
> HTH,
It did! I had a mean object in the current workspace. Once it was
deleted, the argument works without a glitch.
Sorry for being late
Paul Artes a écrit :
> DeaRs,
> i'm looking for some references on a statement as follows:
> "Humans are good at spotting trends and patterns in data, but they are also
> good at spotting those patterns where none really exist". This is not
> verbatim but there must be some scholarly work on this.
I often want to temporarily modify the options() options, e.g.
a <- seq(1001, 1001 + 10) # some wide object
with.options <- function(..., expr) {
options0 <- options(...)
tryCatch(expr, finally=options(options0))
}
Then I can use:
with.options(width=160, expr = print(a))
But I'd li
try this:
> x <- read.table(textConnection(" a b c d e f g h
+1 1 6 11 16 21 26 31 36
+2 2 7 12 17 22 27 32 37
+3 3 8 13 18 23 28 33 38
+4 4 9 14 19 24 29 34 39
+5 5 10 15 20 25 30 35 40"), header=TRUE)
> # initial columns
> init.cols <- c('b',
one way is the following:
xxx <- data.frame(matrix(1:40, ncol=8))
names(xxx) <- letters[1:8]
ind <- c('b', 'd', 'h')
nams <- names(xxx)
xxx[c(ind, nams[!nams %in% ind])]
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Cathol
Dear list,
I have following data set, which I want to plot the "Scale" variable on the
x-axis and "Mean"´on the y-axis for each Ageclass and for each sex. The Mean
value of each Ageclass for each sex would be connected by a line. Totally,
there should be 6 lines, from which three present th
>>> "Stas Kolenikov" <[EMAIL PROTECTED]> 11/02/2008 18:54 >>>
> I would think the FDA regulations could go as far as
>specific SAS syntax, or at least to specify SAS PROCs to be used.
This is unnecessary caution. FDA (like the MHRA in the UK, where I come
from) should not endorse a single suppl
On Feb 12, 2008 9:07 AM, Terry Therneau <[EMAIL PROTECTED]> wrote:
> David Scott asked
> "Views on Bengtsson's ideas would interest me as well."
>
> I have only one serious disagreement with their suggestions
>
>"6.3.2 In general, the use of comments should be minimized by making the
> code
Hi Stephen,
Try
tapply(DATA$mgl,DATA$RM,summary) # DATA is a data.frame
I hope this helps.
Jorge
On 2/12/08, stephen sefick <[EMAIL PROTECTED]> wrote:
>
> below is my data frame. I would like to compute summary statistics
> for mgl for each river mile (mean, median, mode). My apologies in
Hello,
I would like to reorder columns in a data frame by their names as
demonstrated below:
Take this data frame:
> xxx <- data.frame(matrix(1:40, ncol=8))
> names(xxx) <- letters[1:8]
> xxx
a b c d e f g h
1 1 6 11 16 21 26 31 36
2 2 7
Hi,
I have recently been attempting to find the LD50 from two predicted fits
(For male and females) in a Generalised linear model which models the effect
of both sex + logdose (and sex*logdose interaction) on proportion survival
(formula = y ~ ldose * sex, family = "binomial", data = dat (y is the
Here is one way of doing it: (no exactly sure if 'mode' makes sense
with your data)
> x <- read.table(textConnection("RM mgl
+ 1 215 0.9285714
+ 2 215 0.7352941
+ 3 215 1.6455696
+ 4 215 0.600
+ 5 sc 1.833
+ 6 sc 0.833
+ 7 sc 2.5438596
+ 8 sc 0.250
+ 9 202
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Samuel Braithwaite
> Sent: Tuesday, February 12, 2008 8:14 AM
> To: r-help@r-project.org
> Subject: [R] Random Effects
>
>
> Hi All:
>
> I have a panel data set with i individuals and t time periods
below is my data frame. I would like to compute summary statistics
for mgl for each river mile (mean, median, mode). My apologies in
advance- I would like to get something like the SAS print out of PROC
Univariate. I have performed an ANOVA and a tukey LSD and I would
just like the summary stat
The zoo package has a multi-way merge that works with zoo series
There is an example in this thread in which non-zoo data are converted
to zoo and merged:
https://stat.ethz.ch/pipermail/r-help/2008-January/151541.html
library(zoo)
?merge.zoo
Also zoo has 3 vignettes you could look at.
On Feb
Here is one way of doing it:
> # read the file in as lines, do the convert and then re-read
> x <- readLines(textConnection(" X1 X.789 LNM. X78 X56 X89 X56.1 X100
+ 1 2 700 AUW 78 56 8956 100
+ 2 3 400 TOC 78 56 8956 10
+ 3 4 389 RMN 78 56 8956 *89
+ 4 5
Try this:
data1 <- data.frame(Id=LETTERS[1:5], Value=sample(5))
data2 <- data.frame(Id=data1$Value, Value=c(10,20,30,40,50))
data3 <- data.frame(Id=data2$Value, Value=rnorm(5))
merge(merge(data1, data2, by.x="Value", by.y="Id"), data3,
by.x="Value.1", by.y="Id")
On 12/02/2008, joseph <[EMAIL P
Hi All:
I have a panel data set with i individuals and t time periods for which i am
required to estimate the model parameters allowing for Random Effects. How can
i go about this in R? Thanks
Sam
"There would be no heroes if there were no battles or no arduous tasks, and
there are no
Use 'xlim=c(1993,2008)' in your second plot to setup the same range.
On Feb 12, 2008 10:39 AM, Luis Ridao Cruz <[EMAIL PROTECTED]> wrote:
> R-help,
>
> I'm using the code below to plot a shaded area graph.
>
> At the same time I want to plot a second series on the y-axis (from
> par(new=T) on)
> b
Hi
merge() takes only 2 data frames. What can you do to it to make take more than
two data frames? or is there another function that does that?
Thanks
joseph
Looking for last minute shopping deals?
Thanks to all who responded, those were very helpful!
Henrique's solution (below) gets right to it & counts whatever repositories you
have selected. For all repositories the number is 2,758, so there must have
been a few duplicates in my manual count.
I'm trying make the case to users of other
Dear Eleni,
from a previous post regarding maximum number of variables in a
multiple linear regression analysis, posted last tuesday, and I think
it can be relevant also to Cox PH models:
"I can think of
no circumstance where multiple regression on "hundreds of thousands of
variables" is anythi
R-help,
I'm using the code below to plot a shaded area graph.
At the same time I want to plot a second series on the y-axis (from
par(new=T) on)
but as the two series have different x-axis range (first 1994:2007 and
second 1996:2007)
the corresponding x's do not match.
How can this be sorted ou
Paul,
The literature on the topic is extensive. You could start here:
@ARTICLE{Burns2004a,
author = {Burns, Bruce D.},
title = {Heuristics as beliefs and as behaviors: The adaptiveness
of the "hot
hand"},
journal = {Cognitive Psychology},
year = {2004},
volume = {48},
Hi Roberto,
One of these ?
> Y[ Y$ORF %in% X , ]
ORF spectra
3 YAL007C 2
4 YAL012W 8
5 YAL016W 24
11 YAL026C 2
16 YAL041W 4
18 YAL048C 1
> subset( Y, ORF %in% X )
ORF spectra
3 YAL007C 2
4 YAL012W 8
5 YAL016W 24
11 YAL026C
Hi,
I have a data frame Y with the following information,
ORF spectra
1 YAL001C 2
2 YAL005C 21
3 YAL007C 2
4 YAL012W 8
5 YAL016W 24
6 YAL019W 3
7 YAL020C 2
8 YAL021C 7
9 YAL022C 3
10 YAL023C 6
11 YAL026C 2
12 YAL029C
Hi,
I have a data frame Y with the following information,
ORF spectra
1 YAL001C 2
2 YAL005C 21
3 YAL007C 2
4 YAL012W 8
5 YAL016W 24
6 YAL019W 3
7 YAL020C 2
8 YAL021C 7
9 YAL022C 3
10 YAL023C 6
11 YAL026C 2
12 YAL029C
On Feb 12, 2008 9:12 AM, ONKELINX, Thierry <[EMAIL PROTECTED]> wrote:
> Just to inform the list that the problem is solved:
>
> This problem was due to DF$Hora being a factor. Converting it to numeric
> or time solved the problem.
This is because geoms are automatically broken into groups based on
?packageStatus gives you relevant information. Something like:
options(repos = "http://your.favorite.cran.mirror";)
pS <- packageStatus()
pS
And there's also a summary method with more detailed information:
summary(pS)
HTH,
Fernando Mayer.
Muenchen, Robert A (Bob) escreveu:
> Hi All,
>
> I se
Just to inform the list that the problem is solved:
This problem was due to DF$Hora being a factor. Converting it to numeric
or time solved the problem.
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzo
David Scott asked
"Views on Bengtsson's ideas would interest me as well."
I have only one serious disagreement with their suggestions
"6.3.2 In general, the use of comments should be minimized by making the
code
self-documenting by appropriate name choices and an explicit logical structu
as.data.frame(sapply(DATA, function(x){x[grep(patt="\\*", x)]<-NA;x}))
On 12/02/2008, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
>
> Dear all,
>
> I am working with a csv file.
> Some data of the file are not valid and they are marked with a star '*'.
> For example : *789.
>
> I have attached w
Hi David,
The problem is that I need all these regressors. I need a coefficient for
every one of them and then rank them according to that coefficient.
Thanks,
Eleni
On Feb 12, 2008 4:54 PM, <[EMAIL PROTECTED]> wrote:
> Hi Eleni,
>
> I am not an expert in R or statistics but in my opinion you h
Using brute force you can do something like:
my.df<-read.table(stdin(),head=T,sep=",")
X1,X.789,LNM.,X78,X56,X89,X56.1,X100
1,2,700,AUW,78,56,89,56,100
2,3,400,TOC,78,56,89,56,10
3,4,389,RMN,78,56,89,56,*89
4,5,400,LNM,78,56,*452,56,100
5,6,200,UTC,78,*40,89,56,100
6,7,100,GAT,78,56,8,56,*100
7,8
DeaRs,
i'm looking for some references on a statement as follows:
"Humans are good at spotting trends and patterns in data, but they are also
good at spotting those patterns where none really exist". This is not
verbatim but there must be some scholarly work on this. I can't remember
where I came
x <- available.packages()
length(unique(rownames(x)))
On 12/02/2008, Muenchen, Robert A (Bob) <[EMAIL PROTECTED]> wrote:
> Hi All,
>
> I searched around to find the number of R packages currently available,
> but didn't find anything, so I choose all repositories & told it to
> install. The list c
Hi Eleni,
I am not an expert in R or statistics but in my opinion you have too
many regressors compared to the number of observations and that might
be the reason why you get the error. Others might say better but as
far as I know, having only 80 observations, it is a good idea to first
filter
Dear all,
I am working with a csv file.
Some data of the file are not valid and they are marked with a star '*'.
For example : *789.
I have attached with this email a example file (test.txt) that looks like
the data I have to work with.
I see 2 possibilities ..thast I cannot manage anyway in R
1305 by my count.
Go to CRAN and click on Packages. Then copy and paste the list into your
text processor and look at the line count.
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL
Hi All,
I searched around to find the number of R packages currently available,
but didn't find anything, so I choose all repositories & told it to
install. The list contained about 2,856 (correcting roughly for those
installed). But the list includes repetitions such as 19 names that
begin with "
Richard:
Assuming that you have installed R and are not trying to find everything on
the website, you can list the code of any function by just typing the
function name and
You will necessarily have to read the R code, but that is much easier than
it may seem when you look at a couple pages of i
On 2/12/2008 8:54 AM, Kes Knave wrote:
> Dear all,
>
> How do I run a basic one-way anova in R?
Have you read the relevant sections of An Introduction to R, as the
posting guide requests? Have you tried searching for documentation
using, for example:
RSiteSearch("oneway ANOVA")
or
help.s
Hi Kes,
May be:
my.df<-data.frame(FT=rep(c("Fa","Fb"),10),Y=runif(20))
aov(Y ~ FT, data = my.df ) # fixed effects
Kind regards,
miltinho
On 2/12/08, Kes Knave <[EMAIL PROTECTED]> wrote:
>
> Dear all,
>
> How do I run a basic one-way anova in R?
>
> Regards Kes
>
>[[alternative HTML ver
look at the online help file for ?aov().
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
?aov
?anova
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
G
Here is one way of doing it:
> MyData <- c("Test1","Test2","I(Test1^2)","I(Test2^3)","I(Test1.Test2^2)")
> x <- gsub("^(.*\\(|)([^^)]*|.*).*", "\\2", MyData)
> x
[1] "Test1" "Test2" "Test1" "Test2" "Test1.Test2"
> unique(x)
[1] "Test1" "Test2" "Test1.Test2"
>
Dear all,
How do I run a basic one-way anova in R?
Regards Kes
[[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/pos
A way to do it is to use groups (in perl terminology) in connection with
regular expressions. My (limited) understanding of it is as follows: Consider
> s <-"BBBEEE"
> gsub("BBB(.*)EEE(.*)", "\\1AAA\\2\\", s)
[1] "AAA"
>
The terms in the parentheses are groups which you
unique(gsub("^.*\\((.+)\\^.*", "\\1", MyData))
?
b
On Feb 12, 2008, at 5:44 AM, Tom.O wrote:
Hi
I have this vector of strings.
MyData <-
c("Test1","Test2","I(Test1^2)","I(Test2^3)","I(Test1.Test2^2)")
where I want to extract only the text after "I(" and before "^" so
that the
string re
Can someone direct me to a resource or resources that list the formulae used
by R functions (i.e. predict.lm ) to calculate the statistic reported. I
am not a programmer and studying the r code is extremely slow going. I
have searched r-project.org and all the function help files wi
Hello R-community,
It's been a week now that I am struggling with the implementation of a cox
model in R. I have 80 cancer patients, so 80 time measurements and 80
relapse or no measurements (respective to censor, 1 if relapsed over the
examined period, 0 if not). My microarray data contain around
1 - 100 of 133 matches
Mail list logo
