On 21/02/2012 03:31, Xiong Deng wrote:
Hi,
I am working on a stand-alone Linux Redhat machine, which is disconnected
to outside...But I need R to be installed on the machine
The only way available, so far as I know, is to download all the packages
first and then copy to the machine and fina
Hello,
I am using RStudio and have trouble finding out the problematic line in
the presence of a bug. Could I view the line NUMBER which contains a bug?
Is there any R editor able to do it?
Thanks,
Miao
[[alternative HTML version deleted]]
___
On 21-02-2012, at 04:41, arunkumar wrote:
> Hi
>
> my data
>
> looks like
>
> X=(3,5,6,7)B= 1 2 3 4 and C=( 1 2 3 4)
> 2 3 4 5
> 4 5 6 7
> 8 9 1 2
>
>
> I want my results to be
>
> XB + C.
Hello,
I try to handle the data using read.csv , zoo and aggregate functions.
The data contains NA values. After aggregating monthly data into quarterly
data, all data become NA. Is it because I don't properly aggregate the data
in the presence of NAs? What can I do?
Another problem is t
I am currently trying to find MLE of a function with four parameters. My codes
run well but i don't get the results. I get the following message:
BHHH maximisation
Number of iterations: 0
Return code: 100
Initial value out of range.
I don't know this is so because of the way i have written my l
Dear List members,
I really, like the feature that one can call R functions with
mathematical expressions, e.g.
curve(x^2, 0, 1)
I wonder, how I can construct in a simple way a function like
mycurve = function (expr) {...}
such that that a call
mycurve(x^2)
has the same effect as the call
Enomis wrote
>
>> beanplot(tmp1 ~ tmp2, log="", what=c(1,1,1,0), ylim=c(-1,2))
> Still, I do not know how I could be able to plot my original data. Anyone
> has got a clue?
>
The same problem occurred in my code after I changed to a newer R version.
The problem is related to the bandwidth sele
Dear R users,
I am trying to analyze multivariate abundance data using the mvabund
package, but errors occur for several functions.
I created a mvabund object named faunadat consisting of 8 dependent
variables each containing 64 positive integer values (bact, fung, plant,
omn, pred, orib, meso, s
Hello All,
I am relatively new to R. I would like to know if there is a way to alter
LIMMA defualt options such that the package instead of averaging signal
intensities of probesets selects the probesets with highest baseline
expression/highest signal intensity?
Any help would be greatly appreciat
Hi,
I am working on a stand-alone Linux Redhat machine, which is disconnected
to outside...But I need R to be installed on the machine
The only way available, so far as I know, is to download all the packages
first and then copy to the machine and finally install them locally...
But how can
Hi everyone,
I am trying to insert a row in sqlite table with my own unique id. I want
to create unique id using sqlite internal function
last_insert_rowid() which returns the next max rowid of the table which is
always unique. I tested this using sqlite and it works fine but when i run
the same
I have a follow up question to Dr Winsemius' post. You can use the AIC
criterion against all possible cut off values C to see which minimizes the
AIC and then that is the ideal cut off in trying to dichotomize a continuous
variable. What I am wondering here is, does the survivalROC package, or an
Hi
my data
looks like
X=(3,5,6,7)B= 1 2 3 4 and C=( 1 2 3 4)
2 3 4 5
4 5 6 7
8 9 1 2
I want my results to be
XB + C.
Please help me use apply in this function
-
Thanks in Advance
On Mon, Feb 20, 2012 at 09:33:28PM -, Ted Harding wrote:
[...]
> > Unfortunately, there seems to be no equivalent for products
> > (e.g. "rowProds"). But you can define one:
> >
> > rowProds <- function(X){ apply(X,1,FUN="prod") }
> >
> > rowProds(A)
> > # [1] 6 14 24 36 50
> >
> > Ev
Hello friends,
Thanks for the help.
Yes, It is works fine with Davids suggestion. I had to convert "Freqs " to
as.character().
Thanks again.
Regards,
mlsc
On Mon, Feb 20, 2012 at 5:52 PM, Richard M. Heiberger wrote:
> I do not see the restricted range that you report.
> There is probably some mas
Thanks again for your so far on proto. I have another question.
What is the best way to "do stuff" based on data prior to calling a
function? I tried the code below without expr (and including commas after
data member assignments), but it errors out. I'd like to make decisions
based on inputs duri
The default nstart in ?bigkmeans is 1. Try
ans<-bigkmeans(data,k,nstart=8)
Good luck
On Mon, Feb 20, 2012 at 4:22 PM, Lishu Liu wrote:
> Hi, all,
>
> I have a really big matrix that I want to run k-means on.
> I tried:
>>data <-
> read.big.memory('mydata.csv',type='double',backingfile='mydata.bi
Early bird registration for useR! 2012 in Nashville June 12-15 ends on Feb.
29. After that, the registration fee increases by $75 for academics and
$120 for non-academics. Also, don't miss a chance to attend Bill Venables'
all day pre-conference course on June 11.
Abstract submissions are welc
On Feb 20, 2012, at 5:49 PM, adick wrote:
On Feb 20, 2012, at 2:15 PM, adick wrote:
Do you have missing entries? (Running your data fragment through your
code produces the same error.) You may want to use rep with each and
times arguments to construct a regular grid and then population the
mi
I have been working with the pls procedure and have problems getting the
procedure to work with matrix or frame data. I suspect the problem lies in
my understanding of frames, but can't find anything in the documentation
that will help.
Here is what I have done:
I read in an 1 x 8 table of
array chip yahoo.com> writes:
> Hi, I am wondering if we can make prediction on a linear mixed model
> by lmer() from lme4 package?
> Specifically I am fitting a very simple glmer() with binomial family
> distribution, and want to see if I can get the predicted probability like
> that in regul
For question 2, my personal favorite is options(error=recover) which
will let you take a look at all levels of the call stack as soon as an
error is signaled. The debug package on CRAN is popular but I've never
used it.
Michael
On Mon, Feb 20, 2012 at 7:18 PM, Shantanu MULLICK wrote:
> Hello Eve
Hello,
Try
seq1 <- 'PQRTUWXY'
seq2 <- 'AQSDSSDHRS'
seq3 <- 'EEZYJKFFBHO'
ref1 <- 'U'
ref2 <- 'S'
ref3 <- 'Y'
fun <- function(seq, chr){
f <- function(x, seq, chr){
pos <- regexpr(x, seq)
if(pos < 0)
99
else
Hello Everyone
I have a MCMC loop to calculate a time varying hierarchical Bayesian
structure.
This requires me to use around 5-6 matrix inversions in the loop.
I use cholesky and chol2inv for the matrix decomposition.
Because of the data I am working with I am required to invert a 167 by 167
m
On Feb 20, 2012, at 2:15 PM, adick wrote:
Do you have missing entries? (Running your data fragment through your
code produces the same error.) You may want to use rep with each and
times arguments to construct a regular grid and then population the
missing entries in matrix with NA's.
Y
Dear R-list,
I am wondering how to perform a bootstrap in R for the weighted time
dependent Cox model (Andersen–Gill format, with multiple observations
from each patients) to obtain the bootstrap standard error of the
treatment effect.
Below is an example dataset. Would 'censboot' be appropriate
On 21/02/12 12:54, Elaine Kuo wrote:
Hello,
This is Elaine.
I am drawing a plot with x-axis label with "km square" as the unit.
Now I want to print km square in the form of km2 and output "2" as the
uppercase.
Please kindly help suggest command to show the uppercase.
When you say "as [the] up
Worked like a charm! Thanks for your help. Gerard
On Feb 20, 2012, at 3:52 PM, Sarah Goslee wrote:
> This works for me:
>
>> plot(0:6, runif(7), xaxt="n")
>> axis(1, at=0:6, labels=c("08:00", "12:00", "16:00", "20:00", "24:00",
>> "04:00", "08:00"), cex=0.8)
>>
>
> You need the xaxt="n" in
Hello,
This is Elaine.
I am drawing a plot with x-axis label with "km square" as the unit.
Now I want to print km square in the form of km2 and output "2" as the
uppercase.
Please kindly help suggest command to show the uppercase.
Thank you.
Elaine
[[alternative HTML version deleted]]
This works for me:
> plot(0:6, runif(7), xaxt="n")
> axis(1, at=0:6, labels=c("08:00", "12:00", "16:00", "20:00", "24:00",
> "04:00", "08:00"), cex=0.8)
>
You need the xaxt="n" in the plot statement, and the correct form is at=0:6
Sarah
On Mon, Feb 20, 2012 at 6:39 PM, Gerard Smits wrote:
> H
Hi All,
I'm trying to label my plot axis with times (HH:MM) that correspond to a
numeric index (values 0:6) for my time variable. I'd like to plot "08:00",
"12:00", and so on, instead of 0 through 6.
I have used the following line of code:
axis(1, 0:6, labels=c("08:00", "12:00", "16:00", "2
Hi,
On Mon, Feb 20, 2012 at 6:15 PM, nitin kumar wrote:
> Dear All,
>
> I am new to calling R in C.
>
> I want to call sample R function in C. i.e. I want to do a sample
> equivalent to
>
> n <- c(1:10)
> p <- seq(0,10,length.out= 10)
>
> sample(n,size = 1, prob = p, replace = FALSE)
>
> how can
Hi, all,
I have a really big matrix that I want to run k-means on.
I tried:
>data <-
read.big.memory('mydata.csv',type='double',backingfile='mydata.bin',descriptorfile='mydata.desc')
I'm using doMC to register multicore.
>library(doMC)
>registerDoMC(cores=8)
>ans<-bigkmeans(data,k)
In system moni
Dear All,
I am new to calling R in C.
I want to call sample R function in C. i.e. I want to do a sample
equivalent to
n <- c(1:10)
p <- seq(0,10,length.out= 10)
sample(n,size = 1, prob = p, replace = FALSE)
how can I call this function directly in C?
any help would be great,
thanks,
nitin
Hi, I am wondering if we can make prediction on a linear mixed model by lmer()
from lme4 package? Specifically I am fitting a very simple glmer() with
binomial family distribution, and want to see if I can get the predicted
probability like that in regular logistic regression?
fit<-glmer(y~x+(
Hi John,
What do you mean by 'compare'? y = exp(5x) could equally be said: y =
exp(x)^5 so no, the slopes of those two lines are not the same or are
only equal when you take the natural logarithm of y (log_{e}(y) =
log_{e}(5x).
They specify _completely_ different relationships between x and y; s
Yes, with the old good cat() and results=tex, you can do anything. It
is just so unnatural. Why must a simple task like setting the size of
a plot involve with so much coding work?
% complete knitr code
<>=
opts_knit$set(eval.opts = c('fig.height', 'fig.width'))
my.height = 6; my.width = 7
@
<>=
A nomogram is a graphical device for displaying a model. Please try again
with your question. If you also wrote me privately earlier today please
choose one mode of communication.
Thanks
Frank
lijundfgd wrote
>
> I have got two kinds of nomogram for the same data set analyzed by Cox
> regressio
Hello,
Is there any formula or way to compare slopes of different functions?
If we fit 2 functions in our data, and we have 2 slope parameters, how can we
compare these slopes? Plotting y=5x and y=exp(5x) in which slope is equal to 5
in both of them.. doesn't seem that it makes sense to compa
I have three character strings represented below as seq1, seq2, and seq3. Each
string has a reference character different from the other. Thus, for seq1, the
reference character is U, seq2, S (3rd S from left where A is leftmost
character) and for seq3 Y.
seq1 = PQRTUWXYseq2 = AQSDSSDHRSseq3 = E
This isn't being plotted with any special time series methods (because
it's not a time-based object here) so ?abline will work.
Michael
On Mon, Feb 20, 2012 at 12:14 PM, anaraster wrote:
> How can I had a trend line to my plot?
>
> My data looks like this:
>
> Date=seq(as.Date("1910/1/1"), as.Da
[See at end]
On 20-Feb-2012 Ted Harding wrote:
> On 20-Feb-2012 Graziano Mirata wrote:
>> Hi all,
>> I am trying to multiply each column of a matrix such to have
>> a unique resulting vector with length equal to the number of
>> rows of the original matrix. In short I would like to do what
>> prod
On Feb 20, 2012, at 9:15 AM, BXC (Bendix Carstensen) wrote:
> Sometimes you want to compute the physical size of a plot based on data.
> In R itself this is no problem.
>
> But is there a way to compute the values of height and width in S-weave, say:
>
> <>=
>
> where xx and yy are computed and
On Feb 20, 2012, at 2:15 PM, adick wrote:
I want to use persp to graph my data and it keeps giving me the error
increasing 'x' and 'y' values expected, even though my data is in
increasing
order with respect to x and y.
Do you have missing entries? (Running your data fragment through your
Does something like the code below give you want you want?
It requires the ggplot2 package so you will likely have to install it.
John Kane
Kingston ON Canada
# sample data converted using dput
xx <- structure(list(X = c("Exp1", "Exp1", "Exp
I want to use persp to graph my data and it keeps giving me the error
increasing 'x' and 'y' values expected, even though my data is in increasing
order with respect to x and y.
Here is the code I'm currently using:
bob= scan ("SBA3dataTaxonB.txt",what="char")
labels = bob[1:3]
bob=bob[-c(1,2,3)
The http://mxcl.github.com/homebrew/ Homebrew package manager is an exelent
way to manage packages like R on osx. I highly recommend it to any
developer. I am hoping there are others in the R community that are
interested in getting this working too. Does anyone else run a Homebrew
install of
On Mon, Feb 20, 2012 at 06:17:23PM +0100, li...@mwoywod.de wrote:
> Hi! I'm deploying R behind a web-app on a linux-server and I don't want
> to grant the users shell access through the system() function for
> security reasons.
>
> Is there any safe way to deny a user access to the function?
Hi
On Feb 20, 2012, at 1:59 PM, Mirjam Appel wrote:
Hello John,
Thanks for your fast answer. I will try to be clearer and more
detailed this time. At the moment I am importing a dataframe like
below as a '.csv file'. I want to generate a boxplot for M and F
values grouped by X whereby boxpl
Hello John,
Thanks for your fast answer. I will try to be clearer and more detailed this
time. At the moment I am importing a dataframe like below as a '.csv file'. I
want to generate a boxplot for M and F values grouped by X whereby boxplots for
M and F should be above or very close to each ot
How can I had a trend line to my plot?
My data looks like this:
Date=seq(as.Date("1910/1/1"), as.Date("1920/1/1"), "day")
Values=runif(length(Date), min=-5, max=5)
dataset=data.frame(Values,as.Date(Date))
I just want to add a linear trend line to this
plot(dataset,col=rgb(1,0,0,1/8),cex=0.5,pch
I have got two kinds of nomogram for the same data set analyzed by Cox
regression. My question is how to compare the concordance index between
these two nomograms with R?
--
View this message in context:
http://r.789695.n4.nabble.com/how-to-How-to-compare-the-concordance-index-between-two-nomogra
NICE DDE
It solves my problem !
Awesome stuff
--
View this message in context:
http://r.789695.n4.nabble.com/Sorting-strings-tp4403696p4404424.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https:
Hi! I'm deploying R behind a web-app on a linux-server and I don't want
to grant the users shell access through the system() function for
security reasons.
Is there any safe way to deny a user access to the function?
I tried
a) alter the function in the R-Sources before compiling them. Doesn't
Dear all,
I need to run a quantile regression but considering a different objective
function to be minimized: instead of finding the parameters that minimize
rho(t) = u*(t - I(u<0)), I need to find the parameters beta that minimize a sum
of two rho functions rho(t1) = u1*(t1 - I(u1<0)) and
On 20-Feb-2012 Graziano Mirata wrote:
> Hi all,
> I am trying to multiply each column of a matrix such to have
> a unique resulting vector with length equal to the number of
> rows of the original matrix. In short I would like to do what
> prod(.) function in Matlab does, i.e.
>
> A <-matrix(c(1:1
I guess that is not possible with Sweave, but it is possible in the
knitr package (an alternative to Sweave). You can set
opts_knit$set(eval.opts = c('fig.height', 'fig.width'))
so that these two options will be evaluated as R expressions (e.g.
fig.height=x means it takes value from a variable x)
On Feb 20, 2012, at 11:41 AM, Simon Tragust wrote:
Many thanks for the suggestion.
I tried it already,
"it" means .. what exactly?
but as I've never wrote a function, I had no luck.
"no luck" means ... what?
If it's not asked too much, maybe you or somebody else could help me
getting
Simon,
Thanks alot for the further clarification. As i said some where in my
lengthy explanation - i don't what of the myriad steps were needed, only
that they were performed and i now have an up-to-date rJava.
Next time (actually coming up soon) i'll certainly be following the
couple simple
Hi,
That is not what I get. After running your code I get
levels(df.m$Period)
[1] "1991-00" "1901-10" "1981-90" "2001-06" "1911-20" "1881-90" "1971-80"
[8] "1921-30" "1891-00" "1961-70" "1871-80" "1851-60" "1951-60" "1861-70"
[15] "1841-50" "1941-50" "1831-40" "1931-40" "1820-30"
this is my v
On 20-Feb-2012 Petr Savicky wrote:
> On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
>> I did, but this does not give the answer to my question...
>> Anybody knows how to tweack the behaviour of sort or how to do ?
>
> Hi.
> Try this
>
> Sys.setlocale("LC_COLLATE", "C")
>
> This c
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
At a rough guess you may want to have a look at the mfrow in ?par but without
some sample data and a bit more information about what you need it is diff
On 2012-2-20 23:15, Rui Barradas wrote:
Could it be OS related?
Yes, it seems. I tried it on my local windows xp and redhat linux
server, and got different results. Hope it will be fixed in the future
versions. Maybe we should keep alert to check whether the results are
consistent when trans
Many thanks for the suggestion.
I tried it already, but as I've never wrote a function, I had no luck.
If it's not asked too much, maybe you or somebody else could help me
getting the code for the function right.
Regards
Simon
Am 20/02/2012 14:52, schrieb Terry Therneau:
Summary of the que
No, the authors are correct: the individuals (i.e. the 17 individuals) you have
need to be independent (i.e. no correlation between them, let alone any
individual running through your temporal experiment more than once, as
indicated in the citation), while the *observations* are of course depend
On Feb 20, 2012, at 10:26 AM, Thorsten Pöllinger wrote:
Dears,
I am a new R user and I am trying to analyze my data sets, R gives
me a default when I type in the regression formula as following:
fit1=gamlss(tot_remun_revenue$tot_remun.y~tot_remun_revenue
$revenue.x,family=NO)
Most regre
Sorry, just made a mistake. This is the result from windows xp.
> sort(c("X.","X0B"))
[1] "X." "X0B"
> sort(c("X.Z","X0B.Z"))
[1] "X.Z" "X0B.Z"
> R.version
_
platform i386-pc-mingw32
arch i386
os mingw32
system i386, mingw32
status
major
On Mon, Feb 20, 2012 at 04:56:21PM +0100, Petr Savicky wrote:
> On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
> > I did, but this does not give the answer to my question...
> > Anybody knows how to tweack the behaviour of sort or how to do ?
>
> Hi.
>
> Try this
>
> Sys.setlocale
It seems OS-dependent. I got different results when trying it on windows
xp and Redhat linux.
> R.version
_
platform x86_64-unknown-linux-gnu
arch x86_64
os linux-gnu
system x86_64, linux-gnu
status
major 2
minor 9.1
year
On 20-02-2012, at 16:11,
wrote:
> Folks,
>
> I'm trying to get stats from a matrix for each transition from one state to
> another.
>
> I have a matrix x as below.
>
> structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0,
> 0, 2, 2, 0.21, -0.57, -0.59, 0.16, -1.62, 0.18, -0.81, -
Try
apply(A, 1, prod)
I hope it helps.
Best,
Dimitris
On 2/20/2012 4:21 PM, Graziano Mirata wrote:
Hi all,
I am trying to multiply each column of a matrix such to have a unique resulting
vector with length equal to the number of rows of the original matrix. In short
I would like to do wha
Dears,
I am a new R user and I am trying to analyze my data sets, R gives me a default
when I type in the regression formula as following:
fit1=gamlss(tot_remun_revenue$tot_remun.y~tot_remun_revenue$revenue.x,family=NO)
Fehler in model.frame.default(formula = tot_remun_revenue$tot_rem
Ok I have :
R) str(R.Version())
List of 13
$ platform : chr "x86_64-unknown-linux-gnu"
$ arch : chr "x86_64"
$ os: chr "linux-gnu"
$ system: chr "x86_64, linux-gnu"
$ status: chr ""
$ major : chr "2"
$ minor : chr "12.2"
$ year
Hi all,
I am trying to multiply each column of a matrix such to have a unique resulting
vector with length equal to the number of rows of the original matrix. In short
I would like to do what prod(.) function in Matlab does, i.e.
A <-matrix(c(1:10),5,2)
V = A[,1]*A[,2]
Thank you
Graziano
Hello,
I am trying to do a meta-analysis where each study has two arms, similar to the
BCG data set. However, follow-up duration was different for each study arm, so
I would like to fit a model that uses the length of follow-up in each arm as a
moderator. Is this possible?
For example, if this
Hello,
statquant2 wrote
>
> Ok so it changed from 2.12.2 to 2.14.1 ??
> Can somebody tell me how to modify my sort or whatever to get the save
> resilt that I would get in 2.14.1 ?
>
> Cheers
>
I don't know about 2.12.2 but for 2.12.0 I get:
> R.version
_
On Mon, Feb 20, 2012 at 05:55:30AM -0800, statquant2 wrote:
> I did, but this does not give the answer to my question...
> Anybody knows how to tweack the behaviour of sort or how to do ?
Hi.
Try this
Sys.setlocale("LC_COLLATE", "C")
This comes from ?locale and reads there
Sys.setloca
On Mon, Feb 20, 2012 at 03:15:53PM +0200, jing tang wrote:
> Hi,
>
> I need some neat ways of determing a subset of binary strings. For example,
> x=c(0,0,1), y=c(0,1,1), z=c(0,1,0). So x is a subset of y and z is also a
> subset of y, but x is not a subset of z.
>
> I tried to search R functions
Thank you Chuck,
Here is the head of my data set (tjornres):
Fish.1 Fish.2 MORPHO DIET
1 1 20.03768 0.1559250
2 1 30.05609 0.7897060
3 1 40.03934 0.4638010
4 1 50.03363 0.120048
On Feb 20, 2012, at 10:11 AM, wrote:
Folks,
I'm trying to get stats from a matrix for each transition from one
state to another.
I have a matrix x as below.
structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0,
0, 2, 2, 0.21, -0.57, -0.59, 0.16, -1.62, 0.18, -0.81, -0.19,
-0.76,
Hi
>
> This is copy & paste from my session:
>
> > xyz<-as.vector(c(ls(),as.matrix(lapply(ls(),class
> > dim(xyz)<-c(length(xyz)/2,2)
> >
> > allobj<-function(){
> + xyz<-as.vector(c(ls(),as.matrix(lapply(ls(),class;
> + dim(xyz)<-c(length(xyz)/2,2);
> + return(xyz)
> + }
> > xyz
>
Sometimes you want to compute the physical size of a plot based on data.
In R itself this is no problem.
But is there a way to compute the values of height and width in S-weave, say:
<>=
where xx and yy are computed and not physically written in the document?
Bendix
Short answer, environments -- ls() looks (by default) in its current
environment, which is not the same as the global environment when
being called inside a function.
This would (I think) give the same answer but I haven't checked it. :
> allobj<-function(){
+ xyz<-as.vector(c(ls(.GlobalEnv),as.m
Sorry, just checked it and you need to add ".GlobalEnv" to both ls() calls.
Michael
On Mon, Feb 20, 2012 at 10:17 AM, R. Michael Weylandt
wrote:
> Short answer, environments -- ls() looks (by default) in its current
> environment, which is not the same as the global environment when
> being cal
On Feb 20, 2012, at 10:07 AM, Ajay Askoolum wrote:
This is copy & paste from my session:
xyz<-as.vector(c(ls(),as.matrix(lapply(ls(),class
dim(xyz)<-c(length(xyz)/2,2)
allobj<-function(){
+ xyz<-as.vector(c(ls(),as.matrix(lapply(ls(),class;
+ dim(xyz)<-c(length(xyz)/2,2);
+ return(x
Folks,
I'm trying to get stats from a matrix for each transition from one state to
another.
I have a matrix x as below.
structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0,
0, 2, 2, 0.21, -0.57, -0.59, 0.16, -1.62, 0.18, -0.81, -0.19,
-0.76, 0.74, -1.51, 2.79, 0.41, 1.63, -0.86, -0.
This is copy & paste from my session:
> xyz<-as.vector(c(ls(),as.matrix(lapply(ls(),class
> dim(xyz)<-c(length(xyz)/2,2)
>
> allobj<-function(){
+ xyz<-as.vector(c(ls(),as.matrix(lapply(ls(),class;
+ dim(xyz)<-c(length(xyz)/2,2);
+ return(xyz)
+ }
> xyz
[,1] [,2]
I don't *think* it's version specific, but rather it depends on your
(still unstated) locale, as the documentation goes to great lengths to
point out. Change that and you might see different behaviors.
Michael
On Mon, Feb 20, 2012 at 8:55 AM, statquant2 wrote:
> I did, but this does not give the
I do not see the restricted range that you report.
There is probably some masking as David pointed out.
Try again in a fresh R session with --vanilla.
You probably want three additional arguments.
update(.Last.value, xlim=c(0, 100), between=list(x=1, y=1), origin = 0)
to the barchart function call
The 10% rule does not provide a unique answer. Should it apply to the
cumulative probability, its logarithm, or log-log (log hazard scale)? Many
studies are too small to achieve 10% at any time point. I think it is more
traditional (but not without bias) to stop where fewer than 10 subjects are
I did, but this does not give the answer to my question...
Anybody knows how to tweack the behaviour of sort or how to do ?
--
View this message in context:
http://r.789695.n4.nabble.com/Sorting-strings-tp4403696p4404091.html
Sent from the R help mailing list archive at Nabble.com.
_
On Feb 20, 2012, at 5:57 AM, Louise Mair wrote:
Hello,
I am trying to fit gamma, negative exponential and inverse power
functions
to a dataset, and then test whether the fit of each curve is good.
To do
this I have been advised to calculate predicted values for bins of
data (I
have group
On Feb 20, 2012, at 5:04 AM, Marion Wenty wrote:
Dear people,
I created a plot which looks like this:
Ee1<-matrix(c(88,86,74,62,41),ncol=5)
colnames(Ee1)<-
c
("Lehrer
","Lehrerinnen","Klassenkollegen","Klassenkolleginnen","Geschwister")
par(las=1)
par(mar=c(5,13,4,2))
barplot(Ee1,horiz=T,
Hi Jing,
I am not sure I got your definition of a "subset" right, but maybe this
helps.
Regards,
Enrico
x <- c(0,0,1)
y <- c(0,1,1)
z <- c(0,1,0)
## is x a 'subset' of y?
isSubset <- function(x, y) {
x <- as.logical(x)
y <- as.logical(y)
all(y[x] == TRUE)
}
isSubset(x, y)
isSub
On Feb 20, 2012, at 5:34 AM, MLSC wrote:
Hi friends,
I have following data and would like to plot this with barchart()
availble
with lattice package.
RsID Freqs Genotype
AAA 63.636 1/1
AAA 32.727 1/2
AAA 3.636 2/2
BBB 85.965 2/2
BBB 14.035 2/1
CCC 63.158
On Feb 20, 2012, at 1:45 AM, johnmark wrote:
MIchael -
Thanks for your insight. I think I see where you're going with this.
To make '==' comparisons for subsetting against an ordered factor,
I've had
to create a lookup table for all possible values I'd ever want to
compare
against (all d
Summary of the query: "update" does not work on a coxme object
I ran into this bug myself 2 days ago -- I rarely use update() so
hadn't encountered it before. The problem is that coxme breaks the
formula into fixed and random portions, and this confuses the default
method for formula.
Solution:
On 20-02-2012, at 14:15, jing tang wrote:
> Hi,
>
> I need some neat ways of determing a subset of binary strings. For example,
> x=c(0,0,1), y=c(0,1,1), z=c(0,1,0). So x is a subset of y and z is also a
> subset of y, but x is not a subset of z.
>
> I tried to search R functions and packages b
Hello,
I would like to perform triangular test for clinical trial with R.
can you help me please ?
Jan
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PLEASE do read the po
I like it better. Thanks!
Ben
On Fri, Feb 17, 2012 at 11:38 PM, Gabor Grothendieck <
ggrothendi...@gmail.com> wrote:
> On Sat, Feb 18, 2012 at 12:44 AM, Ben quant wrote:
> > The code below works as expected but:
> > Using the proto package, is this the best way to 1) make a parameter
> > persis
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