Yes, here it is:
samr.obj<-samr(data,resp.type="Two class unpaired", nperms=100,
center.arrays=T)
where *data *is a matrix of microarray gene expressions with genes as rows
and tissues as columns. With putting *center.arrays=T *the *data* matrix is
normalized such as each column has median=0. I wo
Hi Ivan,
It was not clear from your original post that QA was a repeated factor.
But your problem may be reframed very much like you would do with a
McNemar chi-square: Just count the number of times both procedure give
the same result, of each kind, and different results, again of both
kinds
Hi
[EMAIL PROTECTED] napsal dne 11.06.2008 00:56:19:
> I'm tyring to fit a relatively simple nls model to some data, but keep
coming
> up against the same error (code follows):
>
> Oto=nls(Otolith ~ Linf*(1-exp(-k(AGE-to))),
> data = ages,
> start = list(Linf=1000, k=0.1, to=0.1),
>
ng,
I have a matrix (x) with binary content. Each row of the matrix holds exactly
one 1, and the rest of the row is zeros. The thing is that I need to 'collapse'
the matrix to one column where each row holds the original column index of the
1's (y). Sometimes, the matrix is quite large, so I h
DrakeGis wrote:
WinBugs doesn't run in LINUX. Neither OpenBugs.
Ah, sorry, haven't read carefully enough.
The only possible thing is to run WinBUGS under wine and use R2WinBUGS,
which supports the wine/WinBUGS combination.
Best wishes,
Uwe Ligges
DrakeGis wrote:
Hi all,
I observed that
I believe nlminb() performs *constrained* optimization, where as nlm() is for
*unconstrained* opimization
So I guess nlm() is for solving min(f[a,b]), and nlminb() min(f[a,b]) given
a+b <= c
FYI I think optim() also does constrained optimization, well I've used for
min(f[a,b]) given a <= a* and
try this:
x <- matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0),
ncol = 3, byrow = TRUE)
which(x == 1, arr.ind = TRUE)[, "col", drop = FALSE]
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kap
sorry, my previous answer was not correct; you need:
x <- matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0),
ncol = 3, byrow = TRUE)
which(t(x == 1), arr.ind = TRUE)[, "row", drop = FALSE]
Best,
Dimitris
Dimitris Rizopoulos
Biostatistical Centre
School of Public Health
Catholic Univer
you may try a matrix multiplication, which has a very high performance in R
x%*%1:ncol(x)
hth.
[EMAIL PROTECTED] schrieb:
ng,
I have a matrix (x) with binary content. Each row of the matrix holds exactly
one 1, and the rest of the row is zeros. The thing is that I need to 'collapse'
the ma
On Wed, 11 Jun 2008, Uwe Ligges wrote:
DrakeGis wrote:
WinBugs doesn't run in LINUX. Neither OpenBugs.
Ah, sorry, haven't read carefully enough.
The only possible thing is to run WinBUGS under wine and use R2WinBUGS, which
supports the wine/WinBUGS combination.
'Linux' is too vague. AFA
Hi
[EMAIL PROTECTED] napsal dne 11.06.2008 08:41:27:
> Hello to everyone,
> I am trying to compute a mean and show it to the screen and then save to
later be used
> as a boxplot.I have used the following code:
> dat<-read.table(file="C:\\Documents and Settings\\Owner\\My
Documents\\Yeast\
here
Hii
The contour plot is somewhat similar to what I want.
I am attaching the data here.
What I want to do is divide all the three variable( Recency, frequency,and
monetory ) in 5 intervals based on their values.That will make
25 rectangle(5(recency)X 5(frequency) boxes on heat plot. and then plot
"sumit gupta" <[EMAIL PROTECTED]> napsal dne 11.06.2008 07:53:54:
> Hello
> thanx for the info..
>
> I still have tthe doubt regarding the heat map.
> I am attching a plot of a 70X3 data matrix.In which 3rd variable has
been
> shown as colour of different boxes.Could you help me to draw this
For precisely this particular type of question, the
following seems to be the simplest, most direct,
and most transparent solution:
rowSums(x%*%(1:ncol(x)))
# [1] 1 3 2 3 2 1
Ted.
On 11-Jun-08 09:21:35, Dimitris Rizopoulos wrote:
> sorry, my previous answer was not correct; you need:
>
> x <-
Hi,
Suppose I have the following data frame.
__BEGIN__
> library(MASS)
> data(crabs)
> crab.pca <- prcomp(crabs[,4:8],retx=TRUE)
> crab.pca$rotation
PC1PC2PC3PC4PC5
FL 0.2889810 0.3232500 -0.5071698 0.7342907 0.1248816
RW 0.1972824 0.8647159 0.414135
Dear Kimmo,
MCA is a rather old name (introduced, I think, in the 1960s by Songuist and
Morgan in the OSIRIS package) for a linear model consisting entirely of factors
and with only additive effects -- i.e., an ANOVA model will no interactions.
You can fit such a model with lm() -- e.g., lm(y ~
Jim Lemon wrote:
Dylan Beaudette wrote:
Hi,
Is anyone on the list familiar with an R implementation of Piper
Diagrams?
Example:
http://faculty.uml.edu/nelson_eby/89.315/IMAGES/Figure%209-78.jpg
I am thinking that two calls to triax.plot (plotrix) along with some
kind of affine-transformed
ChCh wrote:
> Hello,
>
> I have consciously avoided using step() for model simplification in favour
> of manually updating the model by removing non-significant terms one at a
> time. I'm using The R Book by M.J. Crawley as a guide. It comes as no
> surprise that my analysis does proceed as smooth
ChCh wrote:
Hello,
I have consciously avoided using step() for model simplification in favour
of manually updating the model by removing non-significant terms one at a
time. I'm using The R Book by M.J. Crawley as a guide. It comes as no
surprise that my analysis does proceed as smoothly as doe
Try this:
which(pca == min(pca), arr.ind = TRUE)
which(pca == max(pca), arr.ind = TRUE)
On Wed, Jun 11, 2008 at 7:34 AM, Gundala Viswanath <[EMAIL PROTECTED]>
wrote:
> Hi,
>
> Suppose I have the following data frame.
>
> __BEGIN__
>
> > library(MASS)
> > data(crabs)
> > crab.pca <- prcomp(crabs[
Hi,
I have matrix of bits and a target vector. Is there an
efficient way to search the rows of the matrix for the target?
I am interested in the first row index where target is found.
Example:
> source("lookup.R")
[,1] [,2] [,3] [,4] [,5]
[1,]10110
[2,]11
This suggests that one nice to have item would be to support
arr.ind= on which.min and which.max
On Wed, Jun 11, 2008 at 7:57 AM, Henrique Dallazuanna <[EMAIL PROTECTED]> wrote:
> Try this:
>
> which(pca == min(pca), arr.ind = TRUE)
> which(pca == max(pca), arr.ind = TRUE)
>
> On Wed, Jun 11, 2008
Hi
[EMAIL PROTECTED] napsal dne 11.06.2008 12:34:47:
> Hi,
>
> Suppose I have the following data frame.
>
> __BEGIN__
>
> > library(MASS)
> > data(crabs)
> > crab.pca <- prcomp(crabs[,4:8],retx=TRUE)
> > crab.pca$rotation
> PC1PC2PC3PC4PC5
> FL 0.288981
Dimitris Rizopoulos schrieb:
sorry, my previous answer was not correct; you need:
x <- matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0),
ncol = 3, byrow = TRUE)
which(t(x == 1), arr.ind = TRUE)[, "row", drop = FALSE]
Dear helpeRs,
I would like to use this thread for a proposal which ling
Firstly surely (CL,PC5) = -0.7416656 is the minima?
I think something like this should work, though I'm not sure
library(MASS)
data(crabs)
crab.pca <- prcomp(crabs[,4:8],retx=TRUE)
x<- crab.pca$rotation
c.len = length(x[,1])
r.len = length(x[1,])
maxy = which(x == min(x))
x.co <- ceiling(max
Thanks so much both.
- Gundala
On 6/11/08, Henrique Dallazuanna <[EMAIL PROTECTED]> wrote:
> Try this:
>
> which(pca == min(pca), arr.ind = TRUE)
> which(pca == max(pca), arr.ind = TRUE)
>
>
> On Wed, Jun 11, 2008 at 7:34 AM, Gundala Viswanath <[EMAIL PROTECTED]>
> wrote:
> >
> > Hi,
> >
> > Supp
Google for Rtips.
Also Rwiki
On Wed, Jun 11, 2008 at 6:35 AM, Dietrich Trenkler
<[EMAIL PROTECTED]> wrote:
> Dimitris Rizopoulos schrieb:
>>
>> sorry, my previous answer was not correct; you need:
>>
>> x <- matrix(c(1,0,0, 0,0,1, 0,1,0, 0,0,1, 0,1,0, 1,0,0),
>> ncol = 3, byrow = TRUE)
>> which
This should work for you:
> create_bin_string <- function(len)
+ {
+ sample(0:1, len, replace=T)
+ }
>
> ROWS = 10
> COLS = 5
> set.seed(2)
> pop = matrix(create_bin_string(ROWS*COLS), ROWS, COLS, byrow=T)
>
>
>
> target=c(1, 1, 0, 1, 1)
>
> # my population
> print(pop)
[,1] [,2] [,3] [,4]
try this:
match.pat <- function (mat, target, nomatch = -1) {
f1 <- do.call("paste", c(as.data.frame(mat), sep = "\r"))
f2 <- paste(target, collapse = "\r")
ind <- f1 %in% f2
if (any(ind)) which(ind)[1] else nomatch
}
##
set.seed(1234)
mat <- matrix(sample(0:1, 50, TRU
Hi all,
Currently I run R script with arguments the following ways
$ R --vanilla < myscript.R ARGUMENT1
And in my script it is encoded as:
__BEGIN__
args<-commandArgs()
do_sth(args[3])
My question is that is there a way to set a default
argument inside the R script?
In Perl analogically wou
Try this:
which(apply(t(m) == target, 2, all))
On Wed, Jun 11, 2008 at 8:58 AM, Esmail Bonakdarian <[EMAIL PROTECTED]>
wrote:
> Hi,
>
> I have matrix of bits and a target vector. Is there an
> efficient way to search the rows of the matrix for the target?
> I am interested in the first row index
Henrique Dallazuanna wrote:
Try this:
which(apply(t(m) == target, 2, all))
Wow! .. talk about concise! Neat! Thanks.
This will return all matches correct? So if I only wanted
the first I'd simply subscript [1] into it.
Do you think the fact that it searches the whole matrix instead
of stoppi
On Wed, 11 Jun 2008 21:01:54 +0800,
"Gundala Viswanath" <[EMAIL PROTECTED]> wrote:
> Hi all, Currently I run R script with arguments the following ways
> $ R --vanilla < myscript.R ARGUMENT1
I don't think that call could possibly recognize ARGUMENT1 in
myscript.R, according to the docs.
> And
Dimitris Rizopoulos wrote:
try this:
match.pat <- function (mat, target, nomatch = -1) {
f1 <- do.call("paste", c(as.data.frame(mat), sep = "\r"))
f2 <- paste(target, collapse = "\r")
ind <- f1 %in% f2
if (any(ind)) which(ind)[1] else nomatch
}
Thanks! More R for me to sink my teet
# determine which data matches
matches <- t(pop) == target # 't' due to matching in column order
# colSums equal to COLS will indicate matches
which(colSums(matches) == COLS)
Neat! .. somewhat similar to the solution I came up with in the
meantime, only yours works :-)
Thanks Jim.
Esmail
Dear list,
I just experienced some strange behavior of the densityplots provided by the
lattice package. Depending on the values for "from" and "to" that are passed on
to the function for kernel density estimates, the densityplot excludes the
jittered stripplot that is routinely included in la
On Wed, Jun 11, 2008 at 6:42 AM, Frank E Harrell Jr
<[EMAIL PROTECTED]> wrote:
> ChCh wrote:
>>
>> Hello,
>>
>> I have consciously avoided using step() for model simplification in favour
>> of manually updating the model by removing non-significant terms one at a
>> time. I'm using The R Book by M
How can i calculate the discrete L2 distance between the empirical copula and
the estimated one.
--
View this message in context:
http://www.nabble.com/Goodness-of-fit-tests-for-copula-tp17779318p17779318.html
Sent from the R help mailing list archive at Nabble.com.
stefan.petersson wrote:
>
>
> ng,
>
> I have a matrix (x) with binary content. Each row of the matrix holds
> exactly one 1, and the rest of the row is zeros. The thing is that I need
> to 'collapse' the matrix to one column where each row holds the original
> column index of the 1's (y). Som
Look at the y axis ... so try something like
densityplot(test, from=0, to=1, ylim=c(-0.1, 1.5))
The jittered plot is plotted centred on y=0.
On Wed, 11 Jun 2008, Malte Brockmann wrote:
Dear list,
I just experienced some strange behavior of the densityplots provided by the lattice package.
Dear all!!
I estimated the following OLS model with R 2.5.6:
Output R 2.5.6
Call:
lm(formula = UN ~ log(x) + time2, data = dati)
Residuals:
Min 1Q Median 3QMax
-5.649 -2.753 -1.015 1.225 16.199
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercep
Worked liked a charm. Thanks for your help.
SR
Steven H. Ranney
Graduate Research Assistant (Ph.D)
USGS Montana Cooperative Fishery Research Unit
Montana State University
PO Box 173460
Bozeman, MT 59717-3460
phone: (406) 994-6643
fax: (406) 994-7479
-Original Message-
From: Micha
And to follow FH and HW
What level of significance are you using? .05 is excessively liberal.
Are you adjusting your p-values for the number of possible models? Do
you realize the p-values for dropping a term, being selected as the
maximum of a set of p-values, do not follow their usual distributi
Can anybody help me with this problem? ** ONLY WINDOWS R - PROBLEM DOESN'T
OCCUR ON LINUX **
I want to print a long character to screen:
> getOption("width")
[1] 60
> z=(1:20)/10#z is a vector of length between 20 and 30 (depending on
user options) containing lengths in mm (i.e. each element
Hi,
I have a question about applying a function recursively through a
list. Suppose I have a list where the different elements have
different levels of recursion:
> test.list<-list("I"=list("A"=c("a", "b", "c"), "B"=c("d", "e", "f"),
> "C"=c("g", "h", "i")),
+ "II"=list("A"=lis
On Wed, 11 Jun 2008, roberto laforgia wrote:
Dear all!!
I estimated the following OLS model with R 2.5.6:
Output R 2.5.6
You need to read and follow the posting guide AND provide a minimal,
reproducible example.
What is R 2.5.6 ??
What do sessionInfo() or version report?
Where can I get
You could try passing your character string to the strwrap function first, then
use cat on the result.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTE
See ?rapply
On Wed, 11 Jun 2008, Georg Otto wrote:
Hi,
I have a question about applying a function recursively through a
list. Suppose I have a list where the different elements have
different levels of recursion:
test.list<-list("I"=list("A"=c("a", "b", "c"), "B"=c("d", "e", "f"), "C"=c("
R Users,
I'm new to R and was wondering how I can show more decimal places when I run
commands. If I'm simply running a correlation(ES,YM) how would I increase
the number of decimal places R shows? When I run this it shows me .9734044.
How can I extend this further?
In addition I was running hist
See
?rapply
for your example
rapply( test.list, rev, how='replace' )
HTH,
Chuck
On Wed, 11 Jun 2008, Georg Otto wrote:
Hi,
I have a question about applying a function recursively through a
list. Suppose I have a list where the different elements have
different levels of
on 06/11/2008 10:51 AM Georg Otto wrote:
Hi,
I have a question about applying a function recursively through a
list. Suppose I have a list where the different elements have
different levels of recursion:
test.list<-list("I"=list("A"=c("a", "b", "c"), "B"=c("d", "e", "f"), "C"=c("g", "h",
"i"
I wrote a package which includes a number of genome sequencing project
statistics on the web like http://www.ncbi.nlm.nih.gov/genomes/lproks.cgi. I
included some generic functions to summarize, plot, and update the tables
with the most recent version
data(lproks)
update(lproks)
[1] "lproks
options(digits=...)
On Wed, Jun 11, 2008 at 12:17 PM, Neil Gupta <[EMAIL PROTECTED]> wrote:
> R Users,
>
> I'm new to R and was wondering how I can show more decimal places when I run
> commands. If I'm simply running a correlation(ES,YM) how would I increase
> the number of decimal places R shows
On Jun 11, 2008, at 12:17 PM, Neil Gupta wrote:
R Users,
I'm new to R and was wondering how I can show more decimal places
when I run
commands. If I'm simply running a correlation(ES,YM) how would I
increase
the number of decimal places R shows? When I run this it shows me .
9734044.
How
Hi All,
I have the data in the following format:
idktsaaplahto pidg
5266199119943078A
526619921005199210302968A
19930208199302093074A
2002032920020402F322
20020402
#I am having trouble figuring out this one. I have read the help and I am
at a loss. what am I missing
x <- structure(list(X = structure(c(6L, 5L, 9L, 2L, 10L, 8L, 7L, 3L,
13L, 12L, 11L, 4L, 1L, 1L, 1L), .Label = c("", "April", "August",
"December", "February", "January", "July", "June", "March"
on 06/11/2008 12:44 PM stephen sefick wrote:
#I am having trouble figuring out this one. I have read the help and I am
at a loss. what am I missing
x <- structure(list(X = structure(c(6L, 5L, 9L, 2L, 10L, 8L, 7L, 3L,
13L, 12L, 11L, 4L, 1L, 1L, 1L), .Label = c("", "April", "August",
"December",
I had to to get the data as a matrix and fix a couple of other things- If
you would like for me to post my solution then I can. I appoligize my
fingers went faster than my brain.
Stephen
On Wed, Jun 11, 2008 at 1:55 PM, Marc Schwartz <[EMAIL PROTECTED]>
wrote:
> on 06/11/2008 12:44 PM stephen
Greetings, all
I am having difficulty getting the fitdistr() function to return without
an error on my data. Specifically, what I'm trying to do is get a
parameter estimation for fracture intensity data in a well / borehole.
Lower bound is 0 (no fractures in the selected data interval), and upper
On 6/11/2008 1:29 PM, A Ezhil wrote:
Hi All,
I have the data in the following format:
idktsaaplahto pidg
5266199119943078A
526619921005199210302968A
19930208199302093074A
2002032920
Hi,
I have a table of numbers without any repeats. Is there a function that will
search the entire table for a specified value and return the indices if it is
found? Also, some rows may have columns without values.
Thank you so much,
-Nina
__
R-help@r-
Hi, there:
Does anyone know how to specify the ranges in the axises when I make
scatter plots using pairs()? In the general plot function, I can use
xlim and ylim option. But how can I do this if I use pairs()?
Thanks.
Yulei
[[alternative HTML version deleted]]
___
See ?which
On Wed, Jun 11, 2008 at 4:19 PM, <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I have a table of numbers without any repeats. Is there a function that
> will
> search the entire table for a specified value and return the indices if it
> is
> found? Also, some rows may have columns without value
The data "one" is a vector of 553 observations
agglone<-agnes(one, metric = "manhattan", stand = TRUE)
plot(agglone,which.plots=2, nmax=150)
My problem is in the dendogram, I can not see the nodes because it is too
crowded. I have attached the diagram.
Any help is more than welcome. Thank you a l
Dear Fellow R-Users,
I am having some difficulties loading a Fortran subroutine into R and
wondering if anyone could lend me some insight to this problem. When I
load some simpler Fortran codes into R, it loads fine, but when I load
more complex codes I get the following error:
Error in
You can use also xlim and ylim, from the help page:
" ...: arguments to be passed to or from methods."
pairs(iris[1:4], main = "Anderson's Iris Data -- 3 species",
pch = 21, bg = c("red", "green3", "blue")[unclass(iris$Species)])
pairs(iris[1:4], main = "Anderson's Iris Data -- 3 specie
Dear Fellow R-Users,
I am having some difficulties loading a Fortran subroutine into R and
wondering if anyone could lend me some insight to this problem. When I
load some simpler Fortran codes into R, it loads fine, but when I load
more complex codes I get the following error:
Error i
On Wednesday 11 June 2008, Jim Lemon wrote:
> Jim Lemon wrote:
> > Dylan Beaudette wrote:
> >> Hi,
> >>
> >> Is anyone on the list familiar with an R implementation of Piper
> >> Diagrams?
> >>
> >> Example:
> >> http://faculty.uml.edu/nelson_eby/89.315/IMAGES/Figure%209-78.jpg
> >>
> >> I am think
Hi Erin,
Unfortunately no, its not named "chol", I know about the cholesky
decomposition in R, I was just testing out a more complex subroutine as
the subroutine I intend to use gets similar errors. Thanks for the
insight, though. Cheers.
- Derrick
Erin Hodgess wrote:
Hi Derrick!
Is you
Fox, Aaron wrote:
Greetings, all
I am having difficulty getting the fitdistr() function to return without
an error on my data. Specifically, what I'm trying to do is get a
parameter estimation for fracture intensity data in a well / borehole.
Lower bound is 0 (no fractures in the selected data i
Dear Fellow R-Users,
I am having some difficulties loading a Fortran subroutine into R and
wondering if anyone could lend me some insight to this problem. When I
load some simpler Fortran codes into R, it loads fine, but when I load
more complex codes I get the following error:
Error
first, verify the class of your data by the following code:
class(data)
the class must be numeric
if ok
the problem can be caused from the starting value. Try choose other value.
--
View this message in context:
http://www.nabble.com/-R--Copula-fitting-tp6947714p17786789.html
Sent from the R
i have created a wrapper C++ class that popen()s R. When i create a x11
window and plot, i cannot close it using the window button. Also, when i
minimize or maximize the window, the plot does not redraw. Can anybody
tell me if there is a way i can get back this window functionality or if
it is not
Are you serious, reporting on a 2004 version of R? Please do as the
posting guide requested, and update your R. (Only posting one copy is
also appreciated around here.)
Beyond that, normally foo.o is not a shared library but a compiled object.
You have not told us what you did, and probably
Thank you all! The answers are very helpful!
Best regards,
Hua
--- On Wed, 6/11/08, Prof Brian Ripley <[EMAIL PROTECTED]> wrote:
>
> > Hua Li wrote:
> >> Hi All,
> >>
> >> I have a problem of putting long titles on a
> graph:
> >>
> >> for example,
> >>
> >> x= seq(1:100)
> >> y=seq(1:100)
Summer Internship at ETH Zurich
"Dynamic Portfolio Asset Allocation"
We offer a 3-months internship starting
midth July 2008. The topic addresses
"Dynamic Portfolio Asset Allocation"
including alternative instruments and
hedge funds. The goal will be to compare
the robust mean-variance, the lo
Fox, Aaron golder.com> writes:
>
> Greetings, all
>
> I am having difficulty getting the fitdistr() function to return without
> an error on my data. Specifically, what I'm trying to do is get a
> parameter estimation for fracture intensity data in a well / borehole.
> Lower bound is 0 (no frac
This should be posted on R-devel, not here.
-- Bert
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of [EMAIL PROTECTED]
Sent: Wednesday, June 11, 2008 1:30 PM
To: r-help@r-project.org
Subject: [R] Close Window Button Problems
i have created a wrapper C++
Lucke, Joseph F uth.tmc.edu> writes:
>
> And to follow FH and HW
>
> What level of significance are you using? .05 is excessively liberal.
> Are you adjusting your p-values for the number of possible models? Do
> you realize the p-values for dropping a term, being selected as the
> maximum of a
Jim, it worked perfectly. thanks a lot
On Mon, Jun 9, 2008 at 8:58 PM, jim holtman <[EMAIL PROTECTED]> wrote:
> This should do it:
>
>> x <- read.table(textConnection("subject hospitaldate_enrollment
>> hospital_beds
> + 1 hospitalA 1/3/2002300
> + 2 hospitalA
Hi
In case cohort study, we can fit proportional hazard regression model to
case-cohort data. In R, the function is cch() in Survival package
Now I am working on case cohort analysis with time dependent covariates
using cch() of "Survival" R package. I wonder if cch() provide this utility
or not?
nlminb provides unconstrained optimization and optimization subject to
box constraints (i.e. upper and/or lower constraints on individual
elements of the parameter vector). The nlm function provides
unconstrained optimization.
I created the nlminb function because I was unable to get reliable
con
Ben Bolker wrote:
Lucke, Joseph F uth.tmc.edu> writes:
And to follow FH and HW
What level of significance are you using? .05 is excessively liberal.
Are you adjusting your p-values for the number of possible models? Do
you realize the p-values for dropping a term, being selected as the
maximu
Hi
Bernhard wrote:
> Hi everyone.
>
> I want to solve the following problem. I have a data.frame and I
> create a dotplot using lattice.
>
> Then I want to use the grid-package to create a combined graphic which
> contains the dotplot as well as a textplot() (using package gplots) of
> the data
Data into R from Excel csv file
xd<-read.csv("court.dates1.txt",as.is=T, header = F)
> str(xd)
'data.frame': 5 obs. of 1 variable:
$ V1: chr "6/6" "5/27" "5/16" "5/2" ...
>xd
V1
1 6/6
2 5/27
3 5/16
4 5/2
5 4/29
cdates <- as.Date(xd, format = " %m/ %d")
Error in as.Date.default(xd, f
Using
xd <- structure(list(V1 = c("6/6", "5/27", "5/16", "5/2")), .Names =
"V1", row.names = c(NA,
-4L), class = "data.frame")
Try:
> as.Date(paste("2008", xd$V1, sep = "/"), "%Y/%m/%d")
[1] "2008-06-06" "2008-05-27" "2008-05-16" "2008-05-02"
See R News 4/1 article on dates.
On Wed, Jun 11, 20
since it's a dataframe, use xd$V1.
also, it doesn't seem like your data has a space between the / and the
numbers, so take that out of the pattern:
cdates <- as.Date(xd$V1, format = "%m/%d")
On Wed, Jun 11, 2008 at 6:22 PM, Mr Natural <[EMAIL PROTECTED]> wrote:
>
> Data into R from Excel csv fil
on 06/11/2008 05:53 PM Frank E Harrell Jr wrote:
Ben Bolker wrote:
Lucke, Joseph F uth.tmc.edu> writes:
And to follow FH and HW
What level of significance are you using? .05 is excessively liberal.
Are you adjusting your p-values for the number of possible models? Do
you realize the p-values
Hello, I have the following task I'd like to accomplish:
A file contains 5 columns of data (several hundred rows), let's call
them a, b, c, d and e (ie these are their column headers)
I also have a set of definitions, e.g.,
f = a + b
g = a * 3
h = c + d
etc.
I would like to write out a new .rd
> > Good points Ben. For now I'd recommend simply that the allergic
> > reaction to insignificant statistical tests be treated with an
> > antihistamine :-)
>
>
> A vote for Frank's comment to be added to the 'fortunes' package.
>
Seconded! :-)
> :-)
>
> Regards,
>
> Marc
>
> ___
yourDF <- cbind(yourDF, f=yourDF$a+yourDF$b, g=yourDF$a * 3,
h=yourDF$c + yourDF$d)
On Wed, Jun 11, 2008 at 9:26 PM, Esmail Bonakdarian <[EMAIL PROTECTED]> wrote:
> Hello, I have the following task I'd like to accomplish:
>
> A file contains 5 columns of data (several hundred rows), let's call
> t
Esmail -
Esmail Bonakdarian wrote:
Hello, I have the following task I'd like to accomplish:
A file contains 5 columns of data (several hundred rows), let's call
them a, b, c, d and e (ie these are their column headers)
Are these 5 vectors of data stored in a data.frame? I assume so.
I als
Sometimes, for specific models, I get this error from predict.gam in library
mgcv:
Error in complete.cases(object) : negative length vectors are not allowed
Here's an example:
model.calibrate <-
gam(meansalesw ~ s(tscore,bs="cs",k=4),
data=toplot,
weights=weight,
gam.method
Hi Erik,
Erik Iverson wrote:
Esmail -
Are these 5 vectors of data stored in a data.frame? I assume so.
Yes, I do a simple load() call first to read the .rda file ...
test2 <- transform(test, d = 2*a + b, e = 3*c)
save(test2, file = "test2.Rdata")
Does this help?
Yes it does .. this is j
jim holtman wrote:
yourDF <- cbind(yourDF, f=yourDF$a+yourDF$b, g=yourDF$a * 3,
h=yourDF$c + yourDF$d)
Thanks Jim, I also learned about the transform() method from Erik
which will also work beautifully.
Esmail
__
R-help@r-project.org mailing list
ht
Hi,
With the following codes, I attempt to convert
the data.frame into a matrix.
However I notice that data.matrix function doesn't
seem to work.
__ BEGIN__
dat <- read.table("mydata", comment.char = "!" , na.strings = "null");
# Select n-genes by random sample
# n = 1
nosamp <- 1
geneid <- sequ
I would like, for possibly invalid reasons, to have the bars of the
barchart
in each panel to be the same colour, but with *different* colours
from panel
to panel. Can this be done? If so, how?
I've read the help as much as I am capable of, and done an RSiteSearch
() without
becoming any
Try this:
barchart(y ~ x | f, data=X, as.table=TRUE,
panel = function(...) panel.barchart(..., col = panel.number()))
On Wed, Jun 11, 2008 at 11:46 PM, Rolf Turner <[EMAIL PROTECTED]> wrote:
>
> I would like, for possibly invalid reasons, to have the bars of the barchart
> in each panel to be
Try
gexp.arr <- data.matrix(gexp, rownames.force = FALSE)
You are still supposed to get a matrix with one row (not a vector -
dim(gexp.arr) is c(1,8) and not NULL).
--- On Thu, 12/6/08, Gundala Viswanath <[EMAIL PROTECTED]> wrote:
> From: Gundala Viswanath <[EMAIL PROTECTED]>
> Subject: [R] Da
On 12/06/2008, at 4:08 PM, Martin Morgan wrote:
I think this works -- 'custom' variables get sent to the panel
function, where they're picked up and used for making garish plots.
barchart(y ~ x | f, data=X,
mycolors=rainbow(length(levels(X$f))),
panel=function(..., mycolors)
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