How about this:
> mat <- matrix(rep(1:4, each=4), nrow=4, byrow=TRUE)
> mat[rep(1:4, times=c(3,2,4,5)),]
Cheers,
Simon.
On Wed, 2009-06-17 at 01:54 -0400, Lei Liu wrote:
> Hi there,
>
> I have a question on manipulating a matrix. Say I have a matrix A
> with 3 rows. I want to generate a new m
Try:
b<- ifelse(is.na(a),1,2)
Grześ wrote:
>
> 2 - is.na(a) - it's superb! but I need call a function: wy[i]<-
> ifelse(((is.na(a))), call_fun1(x), call_fun2(x)
>
>
>
> Gabor Grothendieck wrote:
>>
>> Try:
>>
>> 2 - is.na(a)
>>
>>
>> On Tue, Jun 16, 2009 at 5:46 PM, Grześ wrote:
>>>
>
Hi Lei,
one way to do this is the following:
mat <- matrix(rnorm(4*6), 4, 6)
ind <- c(3, 2, 4, 5)
mat[rep(seq_along(ind), ind), ]
I hope it helps.
Best,
Dimitris
Lei Liu wrote:
Hi there,
I have a question on manipulating a matrix. Say I have a matrix A with 3
rows. I want to generate a
Hi there,
I have a question on manipulating a matrix. Say I have a matrix A
with 3 rows. I want to generate a new matrix B with 3 duplicates of
the first row of A, 2 duplicates of the second row, and 4 duplicates
of the third row. So B is a matrix with 9 rows. Or more general, I
want to gener
I would probably try a different approach than the other suggestions.
Paste all the columns other than pond_id together. You now have a
character vector. Then keep the rows in which an element of the
vector contains "dnv" but does not contain "0 dnv" [use grep()]. This
assumes there are no ext
Folks,
I need to test a model that has one predictor (a construct with three
indicators) influencing four other variables. Something like what I try to show
below.
X ---> Y1, Y2, Y3, Y4
/ | \
i1 i2 i3
Also, each variable was measured at 5 points in time. So, I'd like to
Hi,
I'm trying to get 2x2 (or other layouts) of cd_plot from the vcd package. I
have tried the usual commands like layout, par(mfrow...) etc and but cd_plot
seems to ignore them and send the plotting window back to 1x1. I have also
tried turning off the pop and newpage options in cd_plot but I st
Try:
2 - is.na(a)
On Tue, Jun 16, 2009 at 5:46 PM, Grześ wrote:
>
> Hi,
> I have a vector a=c(NA, 3, 4, 4, NA, NA, 3) and I would like to use is.na(a)
> function to get a vector like this:
> wy=(1,2,2,2,1,1,2) - you know, this vector create 1 or 2 depends on value in
> vector "a"
>
> This is my
H, I'm assuming that you are trying to find the solution of \beta
to the question:
argmin |y-X\beta|_2^2, which is a quadratic programming problem.
On Tue, Jun 16, 2009 at 6:29 PM, Hesen Peng wrote:
> Hello,
>
> Since the optimization goal is quadratic and constraint is linear,
> quadratic pr
On 17/06/2009, at 3:12 PM, dde...@sciborg.uwaterloo.ca wrote:
Hi all,
I'm stuck trying to get syntax correct for the follwing type of loop.
I would like to find the column with the largest value in a given row,
and create a new column with a categorical variable indicating which
column the high
I'm using R to do some file processing in Linux and am trying to read
in the output of find . -type f -print >
~/Music_Archives_search_problem/ls.output.find.txt
This command yields a text file with each line representing the full
path name of all files in the directory and subdirs. Unfortunately,
Hi all,
I'm stuck trying to get syntax correct for the follwing type of loop.
I would like to find the column with the largest value in a given row,
and create a new column with a categorical variable indicating which
column the highest value of "i" comes from.
too=data.frame(A=rnorm(10,1)
Hi,
I have a vector a=c(NA, 3, 4, 4, NA, NA, 3) and I would like to use is.na(a)
function to get a vector like this:
wy=(1,2,2,2,1,1,2) - you know, this vector create 1 or 2 depends on value in
vector "a"
This is my short code but something is wrong and I don't know what...
for (i in 1:7){
a
Thanks David! I had trouble understanding how to convert factors, and was
playing around with as.numeric but it had never occurred to me to use a
combination of both that and as.character.
I am getting really close to the graph I want - I've played around with the
arguments however I am still hit
On Tue, 16 Jun 2009, jose romero wrote:
Hello list:
(This is probably a stupid question).?Is there a "quick and easy" way to
confirm the gauss-markov conditions of a linear multiple regression
model?
Well, those 'conditions' are _assumptions_, and as often happens they
can be hard to veri
Where the did you get the argument ``na.nr''?
Perhaps you need new reading glasses.
Anyhow, unless there's something else you're not telling us,
apply(nafam,2,sum,na.rm=TRUE)
should work just fine.
cheers,
Rolf Turner
On 17/06/2009, at 1:56 PM, Germán Bonill
Hi all,
I can't compute the sum by columns to a matrix using apply() when I've got
missing values (either 0 or NA)...
I've got this matrix:
>nafam
Micro Flavo Helio Pseud Rhodo Bdello Chloro Syntro Verruco Prochloro
SAR11
I 1 1 1NANA 1 3 NA 1
Hello list:
(This is probably a stupid question). Is there a "quick and easy" way to
confirm the gauss-markov conditions of a linear multiple regression model?
That the mean of the residuals is 0 can easily be tested for. The normality of
the residuals as well (shapiro-wilk?). But what abou
On Monday 08 June 2009 08:50:35 pm Daofeng Li wrote:
> Dear list members,
>
> i am currently want to install Rpy2 in a linux box which has R 2.4.0
> installed
> RPy requries R 2.7.0 or above
> but i have no root previlleges
> so my question is how to install R 2.7.0 on my own directory?
> and repla
Hello,
Since the optimization goal is quadratic and constraint is linear,
quadratic programming functions will help. There are many such
packages available, e.g. quadprog.
On Tue, Jun 16, 2009 at 5:54 PM, Stu @ AGS wrote:
> After a few days of work, I think I nearly have it.
>
> Unfortunately, th
Thank you so much. Sorry for my basic questions that must read very silly for
Biologists.
As a physicist I was supposed to help Biology research with physics facets
(free energy minimization, Molecular Dynamics, and so on).
Unluckily, the student in charge of providing the test data (miRNA seque
After a few days of work, I think I nearly have it.
Unfortunately, theta is unchanged after I run this (as a script from a
file). I thought that theta would contain the fitted parameters.
The goal here is to find the least squares fit according to the function
defined as "rss" subject to the
On 17/06/2009, at 9:03 AM, stephen sefick wrote:
look at the archives - I don't remember who gave a wonderful
explanation on this topic, but it is there.
hth
Stephen Sefick
Don't know if this is what you had in mind, but Martin Maechler's
post of 28 April 2009
http://finzi.psych.upenn.edu/R
On Tue, Jun 16, 2009 at 3:29 AM, Cecilia Carmo wrote:
> Hi r-helpers!
>
> I need to save the output of summary() function that I’ve runned like this:
> z<- lmList(y~x1+x2| x3, na.action=na.omit,data1,subset=year==1999)
> w<-summary(z)
> The output (w) is something like this:
> Call:
> Model: y ~ x
Hi Jonathan,
I don't know if I got the point, but for thresholds assessments, give a look
at segmented package.
bests
milton
brazil=toronto
On Tue, Jun 16, 2009 at 5:23 PM, Jonathan Greenberg
wrote:
> Rers:
>
> I have some ecological data (stream velocity vs. % cover of submerged
> weeds) that
On Tue, Jun 16, 2009 at 7:24 AM, taz9 wrote:
>
> Hi All,
>
> I'm trying to format the y-axis in an xyplot to show numbers with a comma
> separating the thousands but I'm not able to do it using formatNum(y,
> big.mark=","). This is what I have:
>
> library(lattice)
>
> year<-c(2003,2004,2005,2006,2
Rers:
I have some ecological data (stream velocity vs. % cover of submerged
weeds) that shows strong evidence of a thresholding step-function, e.g.
below some velocity, % cover ranges from 0% to 100% (with no apparent
relationship to velocity within this range of velocities), but above a
cert
Hi Bill: I was trying to do below myself but was having problems. So I took
your solution and made another one. yours was working
a little weirdly because I don't think the person wants to keep rows where
there are 2 dnv's in a row and he/she also wanted to keep
the row if the sec
For linear, yes: ?summary.lm
For nonlinear, no, since there is no such thing (unambiguously, anyway) in
nonlinear regression.
It's almost always a bad idea even in linear regression, though.
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: r-help-boun...@r-projec
look at the archives - I don't remember who gave a wonderful
explanation on this topic, but it is there.
hth
Stephen Sefick
On Tue, Jun 16, 2009 at 4:50 PM, Derek An wrote:
> Dear all,
>
> Is there a instruction that can help me obtain the coefficient of
> determination R^2 after doing linear/non
Perhaps I'm just being obtuse, but I don't see what ols() has to do with
the question that was asked.
Often with R it is easier to roll your own rather than struggling
with the
arcana of someone else's software.
Here is a function that seems to do what Ben wants:
pecb <- function(fit,tt,alp
This easy function you are looking for is tapply. Take a look at the
following example:
day=rep(1:30,each=30)
##There are thirty days
##with thirty obs each
y=rnorm(length(day),mean=2*day,sd=day)
##a dep. variable
##with mean=2*day index no.
##and sd=day
tapply(y,day,length)
##shows no. of obs f
Dear all,
Is there a instruction that can help me obtain the coefficient of
determination R^2 after doing linear/nonlinear regression using lm/nls?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/m
Hello, and thanks in advance.
I have a data.frame from which I want to count observations that occur
on each day and determine the mean and std.error of said counts.
For instance:
x<-split(my.df, my.df$julian.days)
Although I'm still in my R learning infancy I am under the impression
that x
Try TargetScan, Pictar, miRbase.
These are all useful miRNA databases. Data can be downloaded as cvs or tab
delimited files and parsed in R after that. In fact this may be possible with
the resource you have looked at (although I haven't checked).
Cheers
Iain
--- On Tue, 16/6/09, David Winsem
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Mark Na
> Sent: Tuesday, June 16, 2009 11:27 AM
> To: r-help@r-project.org
> Subject: [R] How to subset my dataframe? (a bit tricky)
>
> Hi R-helpers,
>
> I would like to subset
Backslashes in character strings need to be doubled. Your "\b" is a
backspace. "\\" is a backslash.
http://cran.r-project.org/doc/manuals/R-lang.html#Literal-constants
On 16/06/09 17:12, Stephen J. Barr wrote:
> Hello,
>
> I would like to create a matrix with one of the columns named
> $\del
Frank E Harrell Jr wrote:
Dear Group,
I have made significant improvements to our Sweave template, have made
the template self-contained (i.e., you can run it yourself and it will
find the datasets it needs), and have included the output pdf file.
This is at http://biostat.mc.vanderbilt.edu/
Hi Livia and everyone,
Did you ever get a response on this question from last year (Jan 2008)?
I am also looking for more explanatory documentation on the
ui and ci parameters for the function constrOptim().
The examples provided in the R help and the full refe
Hi R-helpers,
I would like to subset my dataframe, keeping only those rows which
satisfy the following conditions:
1) the string "dnv" is found in at least one column;
2) the value in the column previous to the one "dnv" is found in is not "0"
Here's what my data look like:
POND_ID 2009-05-
Dear Mark,
If I understood correctly, the following should work:
index <- apply(pvh, 1, function(x) any(x == "dnv") )
pvh_dnv <- pvh[index,]
pvh_dnv
HTH,
Jorge
On Tue, Jun 16, 2009 at 1:41 PM, Mark Na wrote:
> Hi R-helpers,
>
> I'm trying to use this code
>
> > pvh_dnv<-pvh[sapply(pvh=="dnv"
Sorry, it should be:
pvh[apply(pvh == "dnv", 1, any),]
On Tue, Jun 16, 2009 at 3:09 PM, Henrique Dallazuanna wrote:
> Try this:
>
> pvh[apply(pvh == "dnv", 2, any)]
>
>
> On Tue, Jun 16, 2009 at 2:41 PM, Mark Na wrote:
>
>> Hi R-helpers,
>>
>> I'm trying to use this code
>>
>> > pvh_dnv<-pvh[s
Try this:
pvh[apply(pvh == "dnv", 2, any)]
On Tue, Jun 16, 2009 at 2:41 PM, Mark Na wrote:
> Hi R-helpers,
>
> I'm trying to use this code
>
> > pvh_dnv<-pvh[sapply(pvh=="dnv"),]
>
> to make a new dataframe containing the rows from pvh that contain the
> value of "dnv" in ANY column.
>
> But, i
No example. You are a regular now and examples ARE requested.
Would have thought something along the lines of:
pvh[ unlist( apply(pvh, 1, function(x) "dnv" %in%
as.character(x) ) ) , ]
Not sure if the unlist or as.character are really needed because ...
no example on which to test.
On J
See ?window.zoo
Also suggest you use dput to display sample data in posts to r-help.
On Tue, Jun 16, 2009 at 1:19 PM, wrote:
> Hi guys
>
> Does anyone know if it is possible to index a zoo series by a sequence? For
> instance, with the following irregular zoo object, I can calculate the range
>
Hi R-helpers,
I'm trying to use this code
> pvh_dnv<-pvh[sapply(pvh=="dnv"),]
to make a new dataframe containing the rows from pvh that contain the
value of "dnv" in ANY column.
But, it's not working. I get this error
Error in match.fun(FUN) : element 1 is empty;
the part of the args list o
Patrick,
Thanks for your suggestion!
The R-Inferno was especially useful!! The first chapter had me
chuckling aloud despite the fact that I work alone.
Well worth the price!
Thanks!
Stu
> -Original Message-
> From: Patrick Burns [mailto:pbu...@pburns.seanet.com]
>
Hi guys
Does anyone know if it is possible to index a zoo series by a sequence? For
instance, with the following irregular zoo object, I can calculate the
range of its time-based index:
> r <- range(index(l.zoo))
> r
[1] "2009-06-15 01:44:20.802 GMT" "2009-06-15 16:54:24.124 GMT"
If I just w
On Fri, Jun 12, 2009 at 10:54 AM, Chosid, David
(FWE) wrote:
> I'm wondering if I am dealing with a limitation in lattice. It's
> probably due to my own limitations though.
>
> I'm working with a lattice dotplot. The x-axis is set at "free". In
> one panel, there are only two data points that ha
Hello list,
I wonder if anyone might be able to help me troubleshoot an attempt at
porting some simple Python code to R.
The function below is supposed to take a matrix containing item ratings from
various users and, given a vector containing at least 1 rating and 1 missing
value, employ a 'weigh
I have not figured out why you seem so attached to the c(type)
strategy but try:
numbers <-summary(rawdata[ , c(type)] )
On Jun 16, 2009, at 12:06 PM, John Fitzgerald wrote:
Hi everyone,
I experience some problems with adressing of data.frames when I
retrieve
some information for geo
Kevin W wrote:
Frank,
Your template is very interesting. The "pretty-ifing" of the left arrow
and tilde in the input chunk has the unfortunate side-effect of making
the code non-paste-able into R.
Kevin,
I haven't needed to do that. It's best to make the .Rnw file available
to others, or
Liviu,
Thanks for your comments.
With continued study and experimentation, I have discovered the
following:
a. I need to rewrite the function to return a 1x1 as you suggested;
b. it seems that constrOptim() is the most appropriate routine to use
on a nonlinear
Hello,
On 6/16/09, Stu @ AGS wrote:
> Thanks for your response!
> No, my basic equation does not use matrices at all. It takes scalar values
> and returns a scalar.
>
Not quite. Taking the example above, if you run the following:
> with(observs , {1*x1*x2^2*x3^3})
[1] 0.000e+00 4.267e+16 8.910
Hello,
I would like to create a matrix with one of the columns named
$\delta$. I have also created columns $\beta_1$ , $\beta_2$, etc.
However, it seems like \d is an escape sequence which gets
automatically removed. (Using these names such that they work right in
xtable -> latex)
colnames(simple
Hi everyone,
I experience some problems with adressing of data.frames when I retrieve
some information for geographical position (ypos, xpos) ot of a MySQL
Database and want to perform some simple statistics. The problem is
adressing the dataframes with a construct like
rawdata[c(type)] vs. raw
Hi, your problem has nothing to do with your specific dataset. Further, it
is - let's say - suboptimal to push a 5 MB dataset through the mailing list.
The posting guides asks to provide an example with self-contained code; this
is what Milton has asked for. Such an example is not out of reach for
Hi,
yes, that's amazing
follows my .bashrc file
# .bashrc
# User specific aliases and functions
# Source global definitions
if [ -f /etc/bashrc ]; then
. /etc/bashrc
fi
alias rm='rm -i'
if [ ! -d $HOME/trash ]; then
mkdir $HOME/trash
fi
del ( ) { mv "$@" $HOME/trash/.; }
PATH=/hom
Frank,
Your template is very interesting. The "pretty-ifing" of the left arrow and
tilde in the input chunk has the unfortunate side-effect of making the code
non-paste-able into R.
I seem to recall that your reports used to include the latex source code as
an appendix. Maybe one option could b
?data.matrix
Perhaps:
pairs(data.matrix(asdf[ , 1:7]) )
On Jun 16, 2009, at 10:58 AM, njhuang86 wrote:
Hi all,
As of now, I have a 15x8 matrix (name is "asdf"). The first seven
columns
contain numbers while the last column contains a string. The class
of each
column is "character". W
The apply function is the way to iterate over rows or columns of
dataframes or matrices. An example would have made this process easier
testing and I have given up doing that job for
You might try:
apply(observs, 1, function(x) Fn(par, x) )
On Jun 16, 2009, at 11:08 AM, Stu @ AGS wrote:
Looks like a BioConductor question.
On Jun 16, 2009, at 11:05 AM, wrote:
I wonder whether R provides an interface to access miRecords data.
Particularly, I am looking for extracting humans miRNA and target
genes sequences.
All such information is stored in there in a set of structured web
I'm trying to identify the positions of all genes within a specific
chromosomal region using biomart. When using the current biomart
database I'm able to do this without issue. However, I need to use
build 36 of the mouse genome which was last included in ensembl mart
46. I selected this ma
On Tue, 16 Jun 2009, Prof Brian Ripley wrote:
On Tue, 16 Jun 2009, jim holtman wrote:
I think the only way that you are going to get it to stop on the first
mismatch is to write your own function in C if you are concerned about the
time. Matching on character vectors will be even more costly
Why not just use rowSums?
Ronggui
2009/6/16 Stu @ AGS :
> Hello!
>
>
>
> I am trying to write a function with vector and data.frame parameters that
> uses the sum() function and values from the rows of the data.frame.
>
> I need to pass this function as a parameter to optim().
>
>
>
> My starting
At 10:57 16/06/2009, Daofeng Li wrote:
Hi,
i just add /home/lidaof/R/bin and /home/lidaof/R/lib to the end of the PATH
variable
yet i type R command runs the system installed R 2.4.0...
[snip]
>>
>> On Sat, 13-Jun-2009 at 03:43PM +1000, bill.venab...@csiro.au wrote:
>>
>> |> You need to adj
plot(as.numeric(asdf[,1:7]))
On Tue, Jun 16, 2009 at 11:10 AM, stephen sefick wrote:
> plot(asdf[,1:7])
>
> On Tue, Jun 16, 2009 at 10:58 AM, njhuang86 wrote:
>>
>> Hi all,
>>
>> As of now, I have a 15x8 matrix (name is "asdf"). The first seven columns
>> contain numbers while the last column cont
plot(asdf[,1:7])
On Tue, Jun 16, 2009 at 10:58 AM, njhuang86 wrote:
>
> Hi all,
>
> As of now, I have a 15x8 matrix (name is "asdf"). The first seven columns
> contain numbers while the last column contains a string. The class of each
> column is "character". When I use the plot function to displa
On Tue, 16 Jun 2009, jim holtman wrote:
I think the only way that you are going to get it to stop on the first
mismatch is to write your own function in C if you are concerned about the
time. Matching on character vectors will be even more costly since it is
having to loop to check the equality
Hello!
I am trying to write a function with vector and data.frame parameters that
uses the sum() function and values from the rows of the data.frame.
I need to pass this function as a parameter to optim().
My starting point is:
observs <- data.frame(y, x1, x2, x3)
Fn <- function(par,
I wonder whether R provides an interface to access miRecords data.
Particularly, I am looking for extracting humans miRNA and target genes
sequences.
All such information is stored in there in a set of structured web site pages
(http://mirecords.umn.edu/miRecords)
I would greatly appreciate any
Hi all,
As of now, I have a 15x8 matrix (name is "asdf"). The first seven columns
contain numbers while the last column contains a string. The class of each
column is "character". When I use the plot function to display a scatter
plot between any of the two columns, ie. plot(asdf[, 1], asdf[, 2])
Antonio,
I just got the Cowpertwait and Metcalfe book (released today in the U.S.) and I
think it's what you're looking for. It has good, practical examples and lots of
sample code. I looked at Cryer's book, but it had too much on time series
theory for my purposes; Cowpertcalfe helpfully marks
Hello!
I am starting to use the lattice package. I generated an xyplot
conditioned on a factor that has three levels: hence I get three plots
in three panels spaces and one is left empty. I would like to add a
plot to the empty panel space. Is it possible?
Thank you
___
Many thanks to John Fox, who replied on-list,
and to David Hajage, who replied to me
off list. David suggested this quick hack of the
print.anova.mlm function, which I am sharing with his
permission
# From print.Anova.mlm
xtable.Anova.mlm <- function (x, ...) {
test <- x$test
repeated <-
Hi All,
I'm trying to format the y-axis in an xyplot to show numbers with a comma
separating the thousands but I'm not able to do it using formatNum(y,
big.mark=","). This is what I have:
library(lattice)
year<-c(2003,2004,2005,2006,2007,2008,2003,2004,2005,2006,2007,2008)
dept<-c("Politics","P
This is a sure way to get a biased variance estimate.
Instead, use a robust dispersion (scale) estimator such as Gini's mean
difference (average absolute difference between any two observations).
The median is a robust location estimator. There are others. If your
ultimate goal is a comparis
Hi Cecilia,
Could you send us a reproducible example?
cheers
milton
On Tue, Jun 16, 2009 at 4:29 AM, Cecilia Carmo wrote:
> Hi r-helpers!
>
> I need to save the output of summary() function that Ive runned like this:
> z<- lmList(y~x1+x2| x3, na.action=na.omit,data1,subset=year==1999)
> w<-s
Thanks Gabor - I'll check it out.
Actually I just realised I can also do what I am looking for in a
ridiculously simple manner (as the data I have is intra-day):
aggregate(l.zoo, hours(index(l.zoo)), mean)
Cheers
-- Rory
On Jun 16, 2009 2:46pm, Gabor Grothendieck wrote:
> See R News 4/1.
Dear R-helpers,
Very small amount of outliers can greatly affect the mean and many other
statistic of a numeric variable. So, usually we must deal with the outliers
properly in the process of data analysis. Here, I want to replace outliers
with the group median of the variable. But, I can not cons
See R News 4/1.
On Tue, Jun 16, 2009 at 9:04 AM, wrote:
> Hi all
>
> I have an irregular zoo series, where the time index looks like the
> following:
>
>> head(time(l.zoo))
>
> [1] "2009-06-15 01:44:20.802 GMT" "2009-06-15 01:44:20.812 GMT" "2009-06-15
> 01:44:20.837 GMT" "2009-06-15 01:44:20.848
Not to fix your fundamental problem, but if you don't mind using the
next-to-last versions of the packages you can install them quite easily
as debian/Ubuntu packages. They are named r-cran-hmisc and
r-cran-design. You can use sudo apt-get install ... or the adept package
manager etc.
Frank
Dear Maria,
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
> Behalf Of Maria Wolters
> Sent: June-16-09 6:31 AM
> To: r-help@r-project.org
> Subject: [R] Output of Anova (CAR package) in Sweave
>
> Dear list,
>
> I use Sweave almost exc
Thnaks to both of you, with a combined effort of the Wiki and your code I
have now managed to get teh result I was after.
When I have a better understanding of what the code is doing I shall make an
effort to contribut to the Wiki!
Cheers,
Kenny
Kenny Larsen wrote:
>
> Hi All,
>
> I have hu
Hello,
I want to generate data set from Cox PH model with gamma frailty effects.
theta(parameter for frailty distribution)=2
beta=1.5
n=300
cluster size=30
number of clusters=10
I think I should first generate u from Gamma(Theta,theta) and then using
this theta I could not decide how I should g
Not in the direction I was looking, but your function "fff" really speed
up the process, and using it along with "fff3" is fine for me.
Thanks
Petr PIKAL wrote:
> Hi
>
> r-help-boun...@r-project.org napsal dne 16.06.2009 14:31:21:
>
>
>> Hi Jim,
>>
>> What you are saying is correct. Although
[correcting a stupid error in my previous post]
testTwoStages <- function(x, y, head.stop = 100){
if(!isTRUE(all(head(x, head.stop) == head(y, head.stop
{
print(paste("quick test returned FALSE"))
return(FALSE)
} else {
full.test = isTRUE(all(tail(x, length(x) - head.stop) == tail(
-- begin included message
Thank you for the reply. I really appreciate it.
I calculated the Scoenfeld residual per event and my results are the
following:
finage race
17 -0.33942334 -2.0722187270.29024804
20 0.394600944 5.3039687
Thanks for your response!
No, my basic equation does not use matrices at all. It takes scalar values and
returns a scalar.
What I am trying to accomplish is to find the "best-fit" coefficients to the
equation as follows:
y ~ c1 * x1 * x2^c2 * x3^c3
where y, x1, x2, and x3 are observed data and c
Hi all
I have an irregular zoo series, where the time index looks like the
following:
> head(time(l.zoo))
[1] "2009-06-15 01:44:20.802 GMT" "2009-06-15 01:44:20.812 GMT" "2009-06-15
01:44:20.837 GMT" "2009-06-15 01:44:20.848 GMT" "2009-06-15 06:00:01.320
GMT"
[6] "2009-06-15 06:00:01.330 G
Hi
r-help-boun...@r-project.org napsal dne 16.06.2009 14:31:21:
> Hi Jim,
>
> What you are saying is correct. Although, my computer might not have
> same speed and I am getting the following for 10M entries:
>
>user system elapsed
> 0.559 0.038 0.607
With numbers you may speed it a
utkarshsinghal wrote:
Hi Jim,
What you are saying is correct. Although, my computer might not have
same speed and I am getting the following for 10M entries:
user system elapsed
0.559 0.038 0.607
Moreover, in the case of character vectors, it gets more than double.
In my modeling,
Dear list,
I use Sweave almost exclusively for writing papers, and I have become
quite spoiled by the excellent xtable export facilities. Has anybody
written an xtable method for the Anova function in CAR, or has anybody
used a different set of functions to import Anova results into
a table in an
?rev
On Jun 16, 2009, at 5:13 AM, Naoki Irie wrote:
Dear list
I'm having trouble with inverting color gradation.
As seen in the "rgl" example,
library(rgl)
example(rgl.surface)
I understand that I can assign colors to heights for each point.
col <- colorlut[ y-ylim[1]+1 ] # assign color
Hi
I want to use anova-pca method, but I don't know what package I use to.
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Dear All,
I am running a Ubuntu 8.04 System and trying to install the
Design-package. Hmisc is already installed, all fortran compilers and a
Tex-Package are also on board. I searched the net and the help list for
analogue threads, but didn't find any.
Whenever I run the install.packages("Des
I think the only way that you are going to get it to stop on the first
mismatch is to write your own function in C if you are concerned about the
time. Matching on character vectors will be even more costly since it is
having to loop to check the equality of each character in each element.
This is
> I was using older version of R (installed early). I install new version of R
> (R.2.9.0) but i could not find package "xlsReadWrite" to read Excel file.
As others have pointed out (thanks) you can find it here:
http://treetron.googlepages.com/.
It runs fine in 2.9.0.
> Is there any alternative
Hi Jim,
What you are saying is correct. Although, my computer might not have
same speed and I am getting the following for 10M entries:
user system elapsed
0.559 0.038 0.607
Moreover, in the case of character vectors, it gets more than double.
In my modeling, which is already highly
Hi,
In the Sweave output for summary for several types
of model objects and also for the comparison of models
with anova, I find that that the display of the call(s)
or formula does not obey the width option, even with
keep.source=TRUE set, so that a long formula will overshoot
the margins in the
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