i am using R, haplo.stats package. i am working in liux, redhat 5. I run this
many times, but suddenly its showing 0xf42418, cause 'memory not mapped'. i
run again again its showing the same problem. Can you help me to solve the
problem.
thank you in advance
kalai
___
Thank you Jim.
brkdn.plot() seems to be just the function I need.
Still, I wonder if there are lists of labels of the different measures
of central tendency (mct argument) and measures of dispersion (md)?
dror
-
On Wed, Nov 4, 2009 at 7:41 AM, Jim Lemon wrote:
> On
thank you for your help,it is a good way.
Steven Kang wrote:
>
> can try
>
> matrix.x <- as.matrix(x)
>
> On Mon, Nov 2, 2009 at 8:38 PM, bbslover wrote:
>
>>
>> In my disk C:/ have a a.csv file, I want to read it to R, importantly,
>> when
>> I use x=read.csv("C:/a.csv") ,the x format
On Tue, Nov 03, 2009 at 11:54:52PM -0500, R_help Help wrote:
> Hi,
>
> Assuming I have a time series on which I will perform rolling-window
> MLE. In other words, if I stand at time t, I'm using points t-L+1 to t
> for my MLE estimate of parameters at time t (here L is my rolling
> window width).
I am using gsubfn 0.5-0. When I do not specify perl = TRUE I now get
the following error on the same document:
Error in structure(.External("dotTcl", ..., PACKAGE = "tcltk"), class
= "tclObj") :
[tcl] bad index "1e+05": must be integer?[+-]integer? or end?
[+-]integer?.
Regards,
Richard
R users. Thanks in advance.
I would be glad if someone could tell me how to get a simulated data from
AR(1),
MA(1)
My email is : assaed...@yahoo.com
Thanks
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing l
This seems to be simply integer overflow in a calculation.
Changed in R-patched to use doubles.
The issue I patched for Kenneth Roy Cabrera was for perl = FALSE only.
On Tue, 3 Nov 2009, William Dunlap wrote:
Here is a more self-contained way to reproduce the problem in 2.10.0
using the prebui
On Tue, 03-Nov-2009 at 12:26PM -0800, frenchcr wrote:
|> > How do i do this?
|> >
|>
|> ok, so matrix(x, 3, 3) works.
|>
|> what if i have
|>
|> a
|> b
|> c
|> a
|> c
|> a
|> c
|>
|> and want
|>
|> a b c
|> a c
|> a c
|>
|> ??
Hint 1: A matrix can use characters or numerics
Hint 2: "
2009/11/4 NCS :
> I cannot seem to write a randomforest model in PMML - either through calling
> PMML(model) or through Rattle. It appears that it is not yet supported.
> Randomsurvivalforest is, but not randomforest. Any ideas on possible
> workarounds for this?
>
> Thanks
> ncs
It is not yet i
On 11/03/2009 10:15 PM, Dror D Lev wrote:
Hello.
I need to plot a two-way interaction (5 levels X 3) with error bars.
The x.factor will be the five-levels var and the trace.factor will be
the three level var.
I was able to find functions that draw error bars, but still couldn't
find a way to dr
Hi,
I have a data set with three variables,X Y and Time. X and Y are the
coordinates of points, i want to join these points according to the Time
sequence using arrows?
Demo Example data:
> x<-c(1:6)
> y<-c(1:6)
> time<-c(6:1)
> data<-cbind(x,y,time)
> data
x y time
[1,] 1 16
[2,] 2 2
Hello All,
I have a 40k rows long data set that is taking a lot of time to be read-in.
Is there a way to skip reading even/odd numbered rows or read-in only rows
that are multiples of, say, 10? This way I get the general trend of the data
w/o actually reading the entire thing. The option 'skip' in
Hi,
Assuming I have a time series on which I will perform rolling-window
MLE. In other words, if I stand at time t, I'm using points t-L+1 to t
for my MLE estimate of parameters at time t (here L is my rolling
window width). Next, at t+1, I'll do the same.
My question is that is there anyway to a
Thank you.
On Tue, Nov 3, 2009 at 3:08 PM, Manuel Morales
wrote:
> On Tue, 2009-11-03 at 10:20 -0600, Michael Just wrote:
>> Hello,
>>
>> Thanks for the responses. Yes, I did try to use ?bargraph.CI for the
>> colors. When I said bars, I meant the main bars on the graph not the
>> error bars. H
Note that you don't need perl = T since by default strapply uses tcl
regular expressions and they support \w. What happens if you omit the
perl = T?
Also please specify the version of gsubfn you are using and if its not
the latest then try it with the latest version.
On Tue, Nov 3, 2009 at 11:0
Here is a more self-contained way to reproduce the problem in 2.10.0
using the prebuilt Windows executable. Putting a trace on gsub in
the call to strapply showed that it died in the first call to gsub
when the replacement included "\\1" and the string was about 90
characters long (and include
Hi Steffen et al.
The development version of SSOAP and XMLSchema I have on my machine
does complete the processWSDL() call without errors. I have to finish
off some tests before releasing these. It may take a few days before
I have time to work on this, but hopefully soon.
Thanks for the info.
chaski01 wrote:
Hi all,
I'm trying to generate barplots from simple but long (~10-row) data
files, in which each bar will comprise two stacked 'sub-bars'. All the
upper sub-bars will have the same hue, and all the lower bars will,
likewise, have another uniform hue. However, I wish to sp
Kenneth,
Thanks for the hint. I downloaded and installed the latest patch, but
to no avail. I can reproduce the error on a single sentence, the
longest in the document. It contains 743,393 characters. It isn't a
true sentence, but since it is more than three standard deviations
longer
Hi Thomas,
sorry, I should have mentioned that I was using the survey package. But
thank you very much for your quick response. Your first solution worked
perfectly.
Best,
Thushyanthan
tlum...@u.washington.edu wrote:
On Mon, 2 Nov 2009, Thushyanthan Baskaran wrote:
Hi,
I am trying to
I write about R every weekday at the Revolutions blog:
http://blog.revolution-computing.com .
In case you missed them, here are some articles from last month of
particular interest to R users.
http://bit.ly/16wIdo offered a sneak peek of the debugger we've since
released for subscribers of REvolu
Given a data frame consisting of a pointID and 12 pairs of (lat, long)
variables, with names
latA, longA, latB, longB, ... latL, longL, I want to reshape it into a
data frame with the structure
point source lat long
1A ... ...
1B ......
I've looked at resha
Thanks for your help.
> Date: Mon, 2 Nov 2009 18:50:42 -0500
> Subject: Re: [R] avoiding loop
> From: jholt...@gmail.com
> To: bbom...@hotmail.com
> CC: mtmor...@fhcrc.org; r-help@r-project.org
>
> The first thing I would suggest is convert your dataframes to matrices
> so that you are not hav
Hi Jim,
Take a look at ?"%in%", especially its first example.
HTH,
Jorge
On Tue, Nov 3, 2009 at 5:02 PM, jimdare <> wrote:
>
> Hi,
>
> I have a dataset called 'fish'. fish$Species returns extract 1. When I
> use
> fish$Species != c("CRA","PHC"), i.e. I want all species except "CRA" and
> "PH
(1) Is there a (simple) way of getting cloud() to do *both*
type="p" and type="h"? I.e. of getting it to plot the points
as points *and* drop a perpendicular line to the underlying plane?
(2) Is there a way of telling cloud() to drop its lines to the
floor of the bounding box, rather than to t
Try this:
> require(reshape)
> # melt the data
> x <- melt(maps, id='point')
> # split out the labels
> labels <- do.call(rbind, strsplit(sub("(.*)(.)", "\\1 \\2", x$variable), ' '))
> x$source <- labels[, 2]
> x$variable <- labels[, 1]
> cast(x, point + source ~ variable)
point sourcelat
Hello,
I cannot figure out how to set the column widths (either relative or absolute)
for the heatmap function - the full manual hints that layout and lcm will
control this, but I haven't had any luck.
This is in R 2.9.2 compiled for either FC10 or SUSE11.1 (x86_64)... and am
sending the out
Hi,
I have a dataset called 'fish'. fish$Species returns extract 1. When I use
fish$Species != c("CRA","PHC"), i.e. I want all species except "CRA" and
"PHC", I get extract 2 which is blatantly wrong. Can anyone see what I'm
doing wrong?
Regards,
James
EXTRACT 1
> fish$Species
[1] ALB
Michael -
I think this call to reshape gets you pretty close:
reshape(maps,direction='long',
+ varying=list(grep('lat',names(dat),value=TRUE),
+ grep('long',names(dat),value=TRUE)),
+ times=LETTERS[1:12])
point time latA longA id
1.A 1A 4
Try:
tokens <- strsplit(d,"[^[:space:]]+")
This splits each "sentence" in your vector into a vector of groups of
whitespace characters that you can then play with as you described, I think
(The results is a list of such vectors -- see strsplit()).
## example:
> x <- "xx xdfg; *&^%kk"
> st
I apologize for not being clear. d is a character vector of length
158908. Each element in the vector has been designated by sentDetect
(package: openNLP) as a sentence. Some of these are really
sentences. Others are merely groups of meaningless characters
separated by white space. str
Hi french,
Here is a suggestion:
x <- c(2, 3, 4, 2, 1, 6, 6, 4, 7)
matrix(x, ncol = 3)
# [,1] [,2] [,3]
# [1,]226
# [2,]314
# [3,]467
with x the column of data you have.
HTH,
Jorge
On Tue, Nov 3, 2009 at 2:05 PM, frenchcr <> wrote:
>
> say i have a colu
try this:
> x <- read.table(textConnection(" Age(yrs) country mu sigma
+ 1 0. Bolivia 11.42168 0.1014872
+ 2 0.0833 Bolivia 11.33625 0.1053837
+ 3 0.1667 Bolivia 11.28417 0.1070594
+ 4 0.2500 Bolivia 11.21125 0.1083872
+ 5 0. Boliv
Laila Alkhalfan wrote:
Hi
I need to find the Hessian matrix for a complicated function from a certain
kind of data but i keep getting this error
Error in f1 - f2 : non-numeric argument to binary operator
the data is given by
U<-runif(n)
Us<-sort(U)
tau1<- 2
Thushyanthan Baskaran wrote:
Hi,
I am trying to write a function to compute many cross-tabulations with
the -svytable- command. Here is a simplified example of the structure of
my code (adapted from the -svytable- help file):
data(api)
func.example<-function(variable){ dclus1<-svydesign(
On Tue, 2009-11-03 at 10:20 -0600, Michael Just wrote:
> Hello,
>
> Thanks for the responses. Yes, I did try to use ?bargraph.CI for the
> colors. When I said bars, I meant the main bars on the graph not the
> error bars. However, this "col=c("color", "color")" is what I was
> needed and while i
x <- matrix(sample(0:1, 1200, replace=T, prob=c(0.952, 0.048)), ncol=30)
table(x)
x
01
1131 69
x <- matrix(sample(0:1, 1200, replace=T, prob=c(0.952, 0.048)), ncol=30)
table(x)
x
01
1151 49
bikemike42 wrote:
Dear All,
I am trying to create an artificial binary matrix
say i have a column of data like this...
2
3
4
2
1
6
6
4
7
and i want it in three columns like this
226
314
467
...so i can make a contour plot.
How do i do this?
--
View this message in context:
http://old.nabble.com/one-long-column-of-data--%3E-three-small-columns-tp26163165p26163165.htm
Hi french,
Here is a suggestion:
x <- c(2, 3, 4, 2, 1, 6, 6, 4, 7)
matrix(x, ncol = 3)
# [,1] [,2] [,3]
# [1,]226
# [2,]314
# [3,]467
HTH,
Jorge
On Tue, Nov 3, 2009 at 2:05 PM, frenchcr <> wrote:
>
> say i have a column of data like this...
>
> 2
> 3
> 4
Hi,
I would like estimate a model for function of production's Coob-Douglas using
maximum likelihood. The model is log(Y)= beta[1]+beta[2]*log(L)+beta[3]*log(K).
I tried estimate this model using the tools nlm ( ) and optim ( ) using the
log-likelihood function below:
> mloglik <- fu
Hello List,
Does anyone have examples of custom formatting of tables in odfweave? I know
there is an example of this in the formatting.odt file that comes with the
package, but running that through odfweave gives the following error:
Error: chunk 13 (label=showTableStyles)
Error in names(x) <- v
I am currently using the package 'adapt' for multivariate integration.
However this package seems to be removed from CRAN (It is still referred to
in the help file for integrate(stats) though).
I assume it has been deprecated for a reason? Is there an alternative for
multivariate numerical integr
On Nov 3, 2009, at 2:52 PM, Peng Yu wrote:
2009/11/3 Uwe Ligges :
Peng Yu wrote:
I'm wondering if there is a textbook that summarize the methods on
adjusting p-values for multiple comparisons. Most of the
references on
p.adjust() are over 10 years old.
Being 10 years old does not mean
frenchcr wrote:
>
> say i have a column of data like this...
>
> 2
> 3
> 4
> 2
> 1
> 6
> 6
> 4
> 7
>
> and i want it in three columns like this
>
> 226
> 314
> 467
>
> ...so i can make a contour plot.
>
>
> How do i do this?
>
ok, so matrix(x, 3, 3) works.
what if i have
a
b
c
a
c
a
Hi,
The function `factanal' calls `factanal.fit.mle' to perform the likelihood
maximization. You can take a look at this function by saying:
getAnywhere("factanal.fit.mle")
If you do this, you will see that there is a call to `optim' and there you
will see that the only method used is "L-BFGS-B"
j.delashe...@ed.ac.uk wrote:
Quoting Peter Ehlers :
j.delashe...@ed.ac.uk wrote:
Quoting baptiste auguie :
Hi,
try this,
plot.new()
x=0.8
text(0.5, 0.5, bquote(rho == .(x)))
HTH,
baptiste
Aha!
That does exactly what i wanted! Thanks!
Jose
But does it do what it should? It's custo
Tena koe
?matrix
The exact syntax will depend on the class of your 'column of data'. If
it is a dataframe, for example, then try
matrix(yourData[,1], 3, 3)
HTH ...
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Be
On Tue, Nov 3, 2009 at 9:11 AM, wenjun zheng wrote:
> May be I can calculate p value by t testing approximately:
> 1-qnorm(Variance/Std.Dev.)
That would be a z test, not a t test, wouldn't it? And it would only
be meaningful if the distribution of the estimator is approximately
normal, which we
Quoting Peter Ehlers :
j.delashe...@ed.ac.uk wrote:
Quoting baptiste auguie :
Hi,
try this,
plot.new()
x=0.8
text(0.5, 0.5, bquote(rho == .(x)))
HTH,
baptiste
Aha!
That does exactly what i wanted! Thanks!
Jose
But does it do what it should? It's customary to use
"rho" for a _popul
On Tue, Nov 3, 2009 at 8:08 AM, wenjun zheng wrote:
> Thanks,Douglas,
> It really helps me a lot, but is there any other way if I want to show
> whether a random effect is significant in text file, like P value or other
> index.
> Thanks very much again.
> Wenjun.
Well there are p-values from th
Erik and Steve,
Thanks again for your help, both solutions do exactly what I need.
Cheers,
Mike
Steve Lianoglou wrote:
Hi,
On Nov 3, 2009, at 3:58 PM, bikemike42 wrote:
Dear All,
I am trying to fill in a blank vector ("a") with one value at a time,
with
the value of the number of rows in
Thank you Dr. Schwartz:
I did not read this post before,
> See this post from late 2007:
>
>http://tolstoy.newcastle.edu.au/R/e3/help/07/12/6478.html
>
> As far as I know, nothing has changed vis-a-via ODBC connectivity TO
> OpenOffice files. You can use ODBC FROM OpenOffice to connect to
Try this:
dd$i <- with(dd,interaction(z,x>0))
contrasts(dd$i,2) <- rbind( c(0,0),c(0,0),c(1,0),c(0,1) )
model.matrix( ~i, dd )[,-1]
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-bou
My question would be related more to time-dependent ROC (survivalROC package):
I would like to know how true positives (TP) and false positives (FP) are
related to the order of survival times and diagnostic markers. That is, having
the a set of TP and FP, each of them is calculated from which di
Dear All,
I am trying to fill in a blank vector ("a") with one value at a time, with
the value of the number of rows in a randomized dataset with rowSums=0.
Below is the code I've got so far, but what I want to be as the last line is
" a[1st NA,]=nz"
such that this will run until all of my NAs a
Hi,
I am currently trying to execute the following command:
f<-factanal(factors=k$Components$nparallel,covmat=m,n.obs=2287,rotation="varimax",control=list(opt=list(method=c("BFGS"
but keep getting the error: L-BFGS-B needs finite values of 'fn'
I can't figure out what I am doing wrong here,
Dear All,
I am trying to create an artificial binary matrix such that each cell has a
probability of 0.048 of having a 1. So far the closest I've come is us by
using a random poisson distribution with a mean of 0.048, but I can't figure
out how to limit the max value to 1. Otherwise that would
On Mon, 2 Nov 2009, Thushyanthan Baskaran wrote:
Hi,
I am trying to write a function to compute many cross-tabulations with the
-svytable- command. Here is a simplified example of the structure of my code
(adapted from the -svytable- help file):
In the 'survey' package -- if you say what pa
Dear All,
I am trying to create an artificial binary matrix such that each cell has a
probability of 0.048 of having a 1. So far the closest I've come is us by
using a random poisson distribution with a mean of 0.048, but I can't figure
out how to limit the max value to 1. Otherwise that would
Kaushik,
The documentation doesn't quite tell (me, anyway) how the function behaves
when 'target' is a list (or data.frame). You'll need to dig into match.c
or experiment with match() or %in% to see what it is actually doing.
But it looks like it is matching whole columns of the data.frame
Thanks a lot Erik and Tony!
Both of your suggestions accomplish what I need.
Cheers,
Mike
Erik Iverson wrote:
Try
matrix(rbinom(100, 1, prob = 0.048), nrow = 10)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of bikemike42
Hi,
I can confirm this, just today
I tried to write a web service client.
Affected are both SSOAP-0.5-4 and SSOAP_0.4-6.
I can't access anonymous CVS atm. to check for recent fixes.
I am unable to map the error message to any of the items in
http://www.omegahat.org/SSOAP/Todo.html , is this alr
Hi,
On Nov 3, 2009, at 3:58 PM, bikemike42 wrote:
Dear All,
I am trying to fill in a blank vector ("a") with one value at a
time, with
the value of the number of rows in a randomized dataset with
rowSums=0.
Below is the code I've got so far, but what I want to be as the last
line is
"
Tena koe James
Have you tried something like !fish$Species%in%c('CRA','PHC')?
HTH
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of jimdare
> Sent: Wednesday, 4 November 2009 11:02 a.m.
> To: r-help@r-proj
Argument recycling is coming into play here with !=. I am guessing you want
! fish$Species %in% c("CRA", "PHC")
?
Best Regards,
Erik Iverson
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of jimdare
> Sent: Tuesday, November
Hi Ista,
Thanks so much :-) I hadn't clicked that mcmc sampling was used to determine
the estimates! Now it all makes sense!
Do you know if one would one usually test out several values of nsim to find a
number which is high enough that different runs get very similar results? And
do you kn
Try the patch version...
Maybe is the same problem I had with large
database when using gsub()
HTH
El mar, 03-11-2009 a las 20:31 +0100, Richard R. Liu escribió:
> I apologize for not being clear. d is a character vector of length
> 158908. Each element in the vector has been designated by s
Try
matrix(rbinom(100, 1, prob = 0.048), nrow = 10)
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of bikemike42
> Sent: Tuesday, November 03, 2009 11:49 AM
> To: r-help@r-project.org
> Subject: [R] Create Artificial Binary Ma
Peng Yu wrote:
2009/11/3 Uwe Ligges :
Peng Yu wrote:
I'm wondering if there is a textbook that summarize the methods on
adjusting p-values for multiple comparisons. Most of the references on
p.adjust() are over 10 years old.
Being 10 years old does not mean the calculus behind those methods
Hello all...
We are trying to use the analytical power of R in a Web based
environment, and are seeking the optimal ways of accomplishing the
same. We need R to accept requests from clients, perform analysis on
the data as required by the users and return the results ( formatted
text, graphs, etc)
Mike,
Is this what you're trying to do, it avoids the awkward use of a while loop,
taking advantage of the ability to vectorize R functions.
fun <- function() {
z <- matrix(rbinom(832, 1, prob = 0.048), nrow = 32)
sum(rowSums(z) == 0)
}
a <- as.matrix(replicate(1000, fun()))
> -Orig
richard@pueo-owl.ch wrote:
I'm running R 2.10.0 under Mac OS X 10.5.8; however, I don't think this
is a Mac-specific problem.
I have a very large (158,908 possible sentences, ca. 58 MB) plain text
document d which I am
trying to tokenize: t <- strapply(d, "\\w+", perl = T). I am
encounte
Peng Yu wrote:
I'm wondering if there is a textbook that summarize the methods on
adjusting p-values for multiple comparisons. Most of the references on
p.adjust() are over 10 years old.
Being 10 years old does not mean the calculus behind those methods has
changed.
I feel it would be b
2009/11/3 Uwe Ligges :
>
>
> Peng Yu wrote:
>>
>> I'm wondering if there is a textbook that summarize the methods on
>> adjusting p-values for multiple comparisons. Most of the references on
>> p.adjust() are over 10 years old.
>
> Being 10 years old does not mean the calculus behind those methods
Thanks Andreas.
This is just the start point I was needing.
Best,
Rick
From: Andreas Hary
Sent: Tuesday, November 03, 2009 7:19 AM
To: Ricardo Gonçalves Silva
Subject: Re: [R] AR Simulation with non-normal innovations - Correct
Have a look at function arima.sim. It allows you to specify a r
?"%in%" says "x" and "table" must be vectors. You supplied
data.frames. So %in% is coercing your today.sequence to a vector using
as.character(today.sequence)
Perhaps you should paste the columns together first:
x <- do.call("paste", c(sequence, sep = "::"))
table <- do.call("paste", c(today.seq
Hi all,
I'm trying to generate barplots from simple but long (~10-row) data
files, in which each bar will comprise two stacked 'sub-bars'. All the
upper sub-bars will have the same hue, and all the lower bars will,
likewise, have another uniform hue. However, I wish to specify the
luminance
I'm running R 2.10.0 under Mac OS X 10.5.8; however, I don't think this
is a Mac-specific problem.
I have a very large (158,908 possible sentences, ca. 58 MB) plain text
document d which I am
trying to tokenize: t <- strapply(d, "\\w+", perl = T). I am
encountering the following error:
Error in
Hi folks
I have two data frames. I know that the nth (let's say the 7th) row
in the first data frame (sequence) is there in the second
(today.sequence). When I try to check that by doing 'sequence[7,]
%in% today.sequence', I get all FALSE when it should be all TRUE.
I'm certain I'm making some
Hi all,
I'm trying to generate barplots from simple but long (~10-row) data
files, in which each bar will comprise two stacked 'sub-bars'. All the
upper sub-bars will have the same hue, and all the lower bars will,
likewise, have another uniform hue. However, I wish to specify the
luminance
Dear R-helpers,
I have a data.frame (bcpe.lat.m) containing 13 countries, ages 0-50yrs per
month, and the corresponding mu&sigma (see below).
*I would like to limit the age range to include all 12 months for the
1st 5 years and only whole years for all ages thereafter for each of the
c
j.delashe...@ed.ac.uk wrote:
Quoting baptiste auguie :
Hi,
try this,
plot.new()
x=0.8
text(0.5, 0.5, bquote(rho == .(x)))
HTH,
baptiste
Aha!
That does exactly what i wanted! Thanks!
Jose
But does it do what it should? It's customary to use
"rho" for a _population_ correlation coeffi
To extract various portions of the coxph standard printout, look at
summary.coxph
help('summary.coxph')
fit <- coxph(...
sfit <- summary(fit)
Terry Therneau
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do r
By default orthogonal polynomial contrasts are used for ordered
factors. Drop the 'ordered = TRUE' and you will get treatment
contrasts.
hth,
Kingsford Jones
On Tue, Nov 3, 2009 at 8:52 AM, Jen-Chien Chang wrote:
> Hi,
>
> I am wondering if there is a way to display the full anme of the regressi
LinkedIn
Konrad Banachewicz requested to add you as a connection on LinkedIn:
--
Arnaud,
I'd like to add you to my professional network on LinkedIn.
- Konrad
Accept invitation from Konrad Banachewicz
http://www.linkedin.com/e/qlt6CtWzi7sEoE_
Hi Joel,
On Nov 3, 2009, at 11:30 AM, Joel Fürstenberg-Hägg wrote:
> > However, I get the following error:
> >
> > Error in if (x[i] < 0) { : argument is of length zero
>
> This is telling you that x[i] is a zero length object, so you're
> indexing is wrong
>
Doesn't x[i] means index i in vecto
Quoting baptiste auguie :
Hi,
try this,
plot.new()
x=0.8
text(0.5, 0.5, bquote(rho == .(x)))
HTH,
baptiste
Aha!
That does exactly what i wanted! Thanks!
Jose
--
The University of Edinburgh is a charitable body, registered in
Scotland, with registration number SC005336.
__
Hello,
Thanks for the responses. Yes, I did try to use ?bargraph.CI for the
colors. When I said bars, I meant the main bars on the graph not the
error bars. However, this "col=c("color", "color")" is what I was
needed and while it didn't find in the ?bargraph.ci help, I suspect
its a more funda
Hi,
I am wondering if there is a way to display the full anme of the
regression coeffients/factors in the summary?
Suppose I have a bogus data set using weekday as factor which has 7 levels
such as:
mydata <- sample(364)
wk <- rep(1:7, 52)
weekday <-
factor(wk,1:7,c("Mon","Tue","Wed","Thu"
Thank you so much it did solve my purpose.
Regards
Our Thoughts have the Power to Change our Destiny.
Sunita
On Tue, Nov 3, 2009 at 9:15 PM, Gabor Grothendieck
wrote:
> Next time please provide sample input.
>
> library(chron)
>
> # input
> ch <- c("02:24:00", "04:48:00", "07:12:00", "09:36:00
Paul Heinrich Dietrich wrote:
This weekend I noticed that my R2WinBUGS connection was no longer working on
my Windows computer at work AND my Ubuntu linux computer at home. As soon
as WinBUGS opens, the message reads Index Out Of Range. I have un-installed
Yes, thanks, problem known for R-2
Hi,
On Nov 3, 2009, at 9:56 AM, Joel Fürstenberg-Hägg wrote:
Hi all,
I'm trying to write a script that changes all negative values in a
data frame column to a small positive value, based on the the
minimum value of the column.
However, I get the following error:
Error in if (x[i] < 0) {
Next time please provide sample input.
library(chron)
# input
ch <- c("02:24:00", "04:48:00", "07:12:00", "09:36:00", "12:00:00",
"14:24:00", "16:48:00", "19:12:00", "21:36:00")
# convert to times
tt <- times(ch)
# calculate mean
mean(tt)
On Tue, Nov 3, 2009 at 10:35 AM, Sunita22 wrote:
>
>
Hello
I have time data which is in hh:mm:ss format and I need to calculate
averages for this data. I tried converting the data as integer so that I
could calculate the average, but it doesn't serve the purpose. I tried using
Chron package also. But I am stuck up as to how to deal with the time da
coef(summary(fit_cox)) should give you what you wanted.
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-6609 Web: http://yihui.name
Department of Statistics, Iowa State University
3211 Snedecor Hall, Ames, IA
On Tue, Nov 3, 2009 at 12:31 AM, 孟欣 wrote:
> Hi all:
> I finished cox analysis like this:
I cannot seem to write a randomforest model in PMML - either through calling
PMML(model) or through Rattle. It appears that it is not yet supported.
Randomsurvivalforest is, but not randomforest. Any ideas on possible
workarounds for this?
Thanks
ncs
Hi,
On Nov 3, 2009, at 2:41 AM, Antje wrote:
Hi there,
currently, I've updated R on my Mac (OS X) to version 2.10. I was
wondering if I have to install all additional packages again???
In Windows, I just needed to copy the library folder of the old
installation but how does it work with Ma
May be I can calculate p value by t testing approximately:
1-qnorm(Variance/Std.Dev.)
But which function can help me to extract Variance and Std.Dev values from
the results below:
>print(fm2 <- lmer(Yield ~ 1 + (1|Stand) + (1|Variety) +
(1|Variety:Stand),Rice))
Linear mixed model fit by REML
For
Hi all,
I'm trying to write a script that changes all negative values in a data frame
column to a small positive value, based on the the minimum value of the column.
However, I get the following error:
Error in if (x[i] < 0) { : argument is of length zero
As well, I would "minimum
Hi,
try this,
plot.new()
x=0.8
text(0.5, 0.5, bquote(rho == .(x)))
HTH,
baptiste
2009/11/3 :
>
> I'm trying something that I thought would be pretty simple, but it's proving
> quite frustrating...
>
> I want to display, for instance, the correlation coefficient "rho" in a
> graph.
>
> I can d
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