Gabor Grothendieck wrote:
There is an example at the end of the Prices and Returns section of
the zoo-quickref vignette in the zoo package.
library(zoo)
vignette("zoo-quickref")
If speed is your main concern check this recent thread that was posted
on R-sig-finance:
http://n4.nabble.com/SUMMARY
On Tue, 8 Dec 2009, Bogaso wrote:
Hi all, I would like to ask how to order a Zoo object? Consider following
code
dat <- zooreg(rnorm(5), as.yearmon(as.Date("2001-01-01")), frequency=12)
dat
Jan 2001 Feb 2001 Mar 2001 Apr 2001 May 2001
-0.8916124 -0.4516505 1.1305884 -1.4881309 0.3
On Sun, Dec 6, 2009 at 8:53 AM, Peng Cai wrote:
> Please ignore my last question. I found a way to handle that. One last
> thing:
>
> I'm defining my own y scales. In the process the bar starts from below the
> "y=0" line (or below the y-axis). Is there a way to get rid of it.
Have you tried addi
Hi all, I would like to ask how to order a Zoo object? Consider following
code
> dat <- zooreg(rnorm(5), as.yearmon(as.Date("2001-01-01")), frequency=12)
> dat
Jan 2001 Feb 2001 Mar 2001 Apr 2001 May 2001
-0.8916124 -0.4516505 1.1305884 -1.4881309 0.3703734
Here I want to order fro
Dear Sir,
Thanks for your reply
But still exists a trick . Basically I want to do Panel Tobit. I am using
the tobit function from the package (AER) on a panel data .
Suppose that Gasoline$lgaspcar is a 0 inflated data and I do
m1<- tobit (as.formula(paste("lgaspcar ~", rhs)), data=Gasoline)
then
On Dec 9, 2009, at 12:00 AM, Peng Yu wrote:
On Tue, Dec 8, 2009 at 10:37 PM, David Winsemius > wrote:
On Dec 8, 2009, at 11:28 PM, Peng Yu wrote:
I have the following code, which tests the split on a data.frame and
the split on each column (as vector) separately. The runtimes are of
10 time
Dear R-philes:
I am having an issue with exporting contingency tables with xtable().
I set up a contingency and convert it to a matrix for passing to
xtable() as shown below.
v.cont.table <- table(v_lda$class, grps,
dnn=c("predicted", "observed"))
v.cont.mat <- as.matrix(v.cont.tab
On Dec 8, 2009, at 11:37 PM, Peng Yu wrote:
I want to split a matrix, where both 'u' and 'w' are results of
possible ways. However, whenever 'n' changes, the argument passed to
mapply() has to change. Is there a way to pass elements of a list as
multiple arguments?
You need to explain what yo
On Tue, Dec 8, 2009 at 10:37 PM, David Winsemius wrote:
>
> On Dec 8, 2009, at 11:28 PM, Peng Yu wrote:
>
>> I have the following code, which tests the split on a data.frame and
>> the split on each column (as vector) separately. The runtimes are of
>> 10 time difference. When m and k increase, th
On Dec 8, 2009, at 11:50 PM, Peng Yu wrote:
For any given package (for example, data.table), is there a way to
show all the available vignette from the package (or to know whether
there is vignette)?
?vignette
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
_
For any given package (for example, data.table), is there a way to
show all the available vignette from the package (or to know whether
there is vignette)?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read t
I want to split a matrix, where both 'u' and 'w' are results of
possible ways. However, whenever 'n' changes, the argument passed to
mapply() has to change. Is there a way to pass elements of a list as
multiple arguments?
m=10
n=2
k=3
set.seed(0)
x=replicate(n,rnorm(m))
f=sample(1:k, size=m, repl
On Dec 8, 2009, at 11:28 PM, Peng Yu wrote:
I have the following code, which tests the split on a data.frame and
the split on each column (as vector) separately. The runtimes are of
10 time difference. When m and k increase, the difference become even
bigger.
I'm wondering why the performance
I have the following code, which tests the split on a data.frame and
the split on each column (as vector) separately. The runtimes are of
10 time difference. When m and k increase, the difference become even
bigger.
I'm wondering why the performance on data.frame is so bad. Is it a bug
in R? Can i
Jim, could you provide a code snippit to illustrate what you mean?
Hadley, good point, I did not know that.
Mark
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
Indiana University School of Medicine
15032 Hunter Court, Westfield, IN 46074
(317) 490-5129 Work, & Mobile & VoiceMai
On Dec 8, 2009, at 11:07 PM, rusers.sh wrote:
Hi,
In the following function, i hope to save my simulated data into the
"result" dataset, but why the final "result" dataset seems not to be
generated.
#Function
simdata<-function (nsim) {
# Instead why not:
cbind(x=runif(nsim), y=runif(nsim
Hi,
In the following function, i hope to save my simulated data into the
"result" dataset, but why the final "result" dataset seems not to be
generated.
#Function
simdata<-function (nsim) {
result<-matrix(NA,nrow=nsim,ncol=2)
colnames(result)<-c("x","y")
for (i in 1:nsim) {
set.seed(i)
resul
Also instead of 'splitting' the data frame, I split the indices and then use
those to access the information in the original dataframe.
On Tue, Dec 8, 2009 at 9:54 PM, Mark Kimpel wrote:
> Hadley, Just as you were apparently writing I had the same thought and did
> exactly what you suggested, co
Hadley, Just as you were apparently writing I had the same thought and did
exactly what you suggested, converting all columns except the one that I
want split to character. Executed almost instantaneously without problem.
Thanks! Mark
Mark W. Kimpel MD ** Neuroinformatics ** Dept. of Psychiatry
I
Hi Mark,
Why are you using factors? I think for this case you might find
characters are faster and more space efficient.
Alternatively, you can have a look at the plyr package which uses some
tricks to keep memory usage down.
Hadley
On Tue, Dec 8, 2009 at 9:46 PM, Mark Kimpel wrote:
> Charles
Charles, I suspect your are correct regarding copying of the attributes.
First off, selectSubAct.df is my "real" data, which turns out to be of the
same dim() as myDataFrame below, but each column is make up of strings, not
simple letters, and there are many levels in each column, which I did not
p
In the same function as given below: Can anyone suggest how can I bold font
the "site names" ??? Thanks!
library(lattice)
barchart(yield ~ variety | site, data = barley,
groups = year, layout = c(6,1), aspect=.7,
ylab = "Barley Yield (bushels/acre)",
scales = lis
@ David and Phil: Thanks for your suggestions.
@ Xin: Are you also working with barchart()?
-Gary
On Tue, Dec 8, 2009 at 5:23 PM, Phil Spector wrote:
> Gary -
> If you create an ordered factor, barchart will plot the
> sites in the order you specify. For example, try
>
> barley$site = ordere
On Dec 8, 2009, at 8:46 PM, Lynn Wang wrote:
Hi all,
I have two sets:
dig<-c("DAVID ADAMS","PIERS AKERMAN","SHERYLE BAGWELL","JULIAN
BAJKOWSKI","CANDIDA BAKER")
import<-c("by DAVID ADAMS","piersAKERMAN","SHERYLE BagWEL","JULIAN
BAJKOWSKI with ","Cand BAKER","smith green")
I want to
On Tue, 8 Dec 2009, Mark Kimpel wrote:
I'm having trouble using split on a very large data-set with ~1400 levels of
the factor to be split. Unfortunately, I can't reproduce it with the simple
self-contained example below. As you can see, splitting the artificial
dataframe of size ~13MB results i
Is anyone using RGoogleDocs? If so have you used it in the last few weeks
and is it working as it used to. Look at the problem I have run into.
Farrel Buchinsky
On Sat, Nov 28, 2009 at 14:25, Farrel Buchinsky wrote:
> Thank you for the interest in my problem.
>
>
> I have been using the same
It would be easier to answer if a portion of the input were shown in
the question using dput(rf) or dput(head(rf)) but lets assume it looks
like this and that our objective is to display a table of Name vs.
DNAME counts where DNAME consists of manufactured short names -- is
that right?
DF <- data
Hi,
I think you want
by(TestData[ , "RATIO"], LEAID, median)
-Ista
On Tue, Dec 8, 2009 at 8:36 PM, L.A. wrote:
>
> I'm just learning and this is probably very simple, but I'm stuck.
> I'm trying to understand the by().
> This works.
> by(TestData, LEAID, summary)
>
> But, This doesn't.
>
> by
Hi all,
I have two sets:
dig<-c("DAVID ADAMS","PIERS AKERMAN","SHERYLE BAGWELL","JULIAN
BAJKOWSKI","CANDIDA BAKER")
import<-c("by DAVID ADAMS","piersAKERMAN","SHERYLE BagWEL","JULIAN BAJKOWSKI
with ","Cand BAKER","smith green")
I want to get the following result from "import" after comp
I'm just learning and this is probably very simple, but I'm stuck.
I'm trying to understand the by().
This works.
by(TestData, LEAID, summary)
But, This doesn't.
by(TestData, LEAID, median(RATIO))
ERROR: could not find function "FUN"
HELP!
Thanks,
LA
--
View this message in context:
http
I really need help altering a function I have written. The function
currently performs a specific task on a dataframe. I now wish to be able to
use it with a splitting function by either passing it a pre-split dataframe
or by being able to designate the splitting factor.
Can anyone advise?
Her
Thanks David, it worked!
On Tue, Dec 8, 2009 at 5:03 PM, David Winsemius wrote:
>
> On Dec 8, 2009, at 4:42 PM, Gary Miller wrote:
>
> Hi R Users,
>>
>> I'm trying to re-order the "site names" ("Waseca", "Morris", ...). I'm
>> using
>> following code:
>>
>> libarry(lattice)
>>
>
> # slip this cod
You can use the "ifelse" function for vectorized conditionals like you have
here. See ?ifelse.
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Arthur Burke [art.bu...@educationnorthwest.org]
Sent: Tuesday, December 08, 2009
I am trying to use the value of an ID variable in an if statement and
not getting the results I expected.
# ID values for two school districts
> with(rf, tapply(DistrictID, DistrictName, min) )
Aberdeen School Dist. # 58 Buhl Joint School District
59
Here are a couple of solutions. The first uses by and the second sqldf:
> Lines <- " rt dur tid mood roi x
+ 55 5523 200 4 subj 9 5
+ 56 5523 52 4 subj 7 31
+ 57 5523 209 4 subj 4 9
+ 58 5523 188 4 subj 4 7
+ 70 4016 264 5 indic 9 51
+ 71 4016 195 5 indic 4 14"
>
Or perhaps just:
plot(xrange, yrange, type ="s")
--
On Dec 8, 2009, at 6:00 PM, Peter Alspach wrote:
Tena koe Oliver
?seq
should help if I understand your question correctly
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.o
Tena koe Oliver
?seq
should help if I understand your question correctly
Peter Alspach
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Wells Oliver
> Sent: Wednesday, 9 December 2009 11:19 a.m.
> To: r-help@r-project.
All the solutions presented work, but, gosh,
why not fix the source file in the first place?
One way: Open the text file in Excel or a clone thereof, and tell the
importer dialog box to use both tabs and commas as delimiters. Then
save the result, which will have your 'third column' nicely sp
I've sent last message only to Titus. Sorry :)
Below my proposition:
Instead of aggregate, try summaryBy from doBy package. It's much
faster. And this package made my life easier :)
try:
summaryBy(dur+x~index, data=d, FUN=c(sum, mean)),
but index should be in data.frame as I remember.
I haven't
On Fri, Dec 4, 2009 at 6:12 PM, Gavin Simpson wrote:
> On Fri, 2009-12-04 at 15:18 -0600, Peng Yu wrote:
>> On Fri, Dec 4, 2009 at 3:06 PM, Peng Yu wrote:
>> > On Fri, Dec 4, 2009 at 2:35 PM, Greg Snow wrote:
>> >> The invert argument seems a likely candidate, you could also do perl=TRUE
>> >>
On 12/09/2009 05:42 AM, Cable, Samuel B Civ USAF AFMC AFRL/RVBXI wrote:
Am doing some vector plots with the arrows() function. Works well. But
what I need to do is supply an arrow for scaling for the reader. I need
to plot an arrow of some known magnitude somewhere on the page
(preferably outs
On Fri, Dec 4, 2009 at 11:17 PM, Greg Snow wrote:
> I am sure that you mentioned before that your are using 2.7.1, and possibly
> even why, but with the number of posts to this list each day and the number
> of different posters, I cannot keep track of what version everyone is using
> (well, I
I have xrange which is a range of values from 1 to a max of 162.
I have a yrange of values which really could be any values, but there's a
min and a max. I'd like to create N number of steps between the min and the
max so the length matches the xrange, so that I can plot them together.
Any tips?
On Tue, Dec 8, 2009 at 5:19 PM, Nikhil Kaza wrote:
> I suppose that is true, but the example data seem to suggest that it is
> sorted by rt.
I was not very clear on that. Sorry.
> d$count <- 1
> a <- with(d, aggregate(subset(d, select=c("dur", "x", "count"),
> list(rt=rt,tid=tid,mood=mood,roi=
I'm having trouble using split on a very large data-set with ~1400 levels of
the factor to be split. Unfortunately, I can't reproduce it with the simple
self-contained example below. As you can see, splitting the artificial
dataframe of size ~13MB results in a split dataframe of ~ 144MB, with an
in
On Dec 8, 2009, at 4:42 PM, Gary Miller wrote:
Hi R Users,
I'm trying to re-order the "site names" ("Waseca", "Morris", ...).
I'm using
following code:
libarry(lattice)
# slip this code (or one with your preferred ordering) in before the
plot call:
barley$site <- factor(barley$site,
Thanks a lot to everyone who replied. I used some of Stephan Kolassa's
suggestions (thanks, Stephan). Here's my final code which worked (with
comments to explain novices like me who come looking for the same answer)
x=read.table("file",sep="\t",colClasses="character")
prate<-x[,2]
lrate<-x[,3] #th
On Sun, Dec 06, 2009 at 04:34:18PM -0800, Brock Tibert wrote:
> I have successfully created and analyzed my network data. I am
> new to R, and Network Analysis too, but I want to color my
> vertex based on some of the centrality measures calculated.
>
> Can someone point me in the right directi
Depending on your sample size, you might be able to just label the nodes
by drawing a random sample from the variable names :-)
Frank
kaida ning wrote:
Hi all,
I used rpart to fit a model, where the covariates are categorical variables.
Then I plotted the tree (mytree) and used the command "t
On Dec 8, 2009, at 4:12 PM, Gaurav Moghe wrote:
> Hi David,
>
> 1) My code is as follows:
> x=read.table("file",sep="\t")
> prate<-x[,2]
> lrates<-(x[,3])
> When I do:
> print (typeof(lrates)): I get "integer"
You've already received several solutions from people more R-savvy
than I, so I wil
Hi R Users,
I'm trying to re-order the "site names" ("Waseca", "Morris", ...). I'm using
following code:
libarry(lattice)
barchart(yield ~ variety | site, data = barley,
groups = year, layout = c(6,1), aspect=.7,
ylab = "Barley Yield (bushels/acre)",
scal
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Gaurav Moghe
> Sent: Tuesday, December 08, 2009 12:56 PM
> To: r-help@r-project.org
> Subject: [R] Split comma separated list
>
> Hi all,
>
> I'm a beginner user of R. I am stuc
Here are a couple of others to try (using the lattice package):
dotplot(Factor1 ~ Value | Factor2, data=foo, groups=Factor3, auto.key=T)
dotplot(Factor1 ~ jitter(Value) | Factor2, data=foo, groups=Factor3, auto.key=T)
dotplot(Factor3 ~ Value | Factor2*Factor1, data=foo )
dotplot(Factor1:Factor3 ~
Hi David,
1) My code is as follows:
x=read.table("file",sep="\t")
prate<-x[,2]
lrates<-(x[,3])
When I do:
print (typeof(lrates)): I get "integer"
When I do:
for (line1 in lrates) {
lsp<-unlist(strsplit(line1,"\\,"))
}
I get some in
Hi Gaurav,
1) tell R when reading the data to consider the third column as
"character" via the colClasses argument to read.table()
2) foo <- as.numeric(strplit(dataset$List_of_values,","))
3) unlist(foo) or some such
HTH,
Stephan
Gaurav Moghe schrieb:
Hi all,
I'm a beginner user of R. I
Sebastian: Can you send me (off-list) your point pattern (ppp_cameroon)
so that I can experiment with it and try to figure out what's
going wrong? (I am one of the maintainers of spatstat.)
Save the point pattern using dput(), e.g.
dput(ppp_cameroon,"cameroon.dput")
and then attach `
Gaurav -
Here's one way:
x = textConnection('ID1 0.342 0.01,1.2,0,0.323,0.67
+ ID2 0.010 0.987,0.056,1.3,1.5,0.4
+ ID3 0.146 0.1173,0.1494,0.211,0.1257
+
+ ')
y = read.table(x,stringsAsFactors=FALSE)
res = apply(y,1,function(x)
Hi Marc,
You answered my question in depth, leaving me to go with another solution.
Thank you for the detailed answer!
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com
Two questions:
What is your code?
What do you get with:
> options()$dec
decimal_point
"."
--
David
On Dec 8, 2009, at 3:55 PM, Gaurav Moghe wrote:
Hi all,
I'm a beginner user of R. I am stuck at what I thought was a very
obvious
problem, but surprisingly, I havent found any
Hi all,
I'm a beginner user of R. I am stuck at what I thought was a very obvious
problem, but surprisingly, I havent found any solution on the forum or
online till now.
My problem is simple. I have a file which has entries like the following:
#ID Value1List_of_values
ID1
On Tue, Dec 8, 2009 at 8:52 PM, stephen's mailinglist account
wrote:
> On Tue, Dec 8, 2009 at 7:14 PM, wrote:
>> Hi Stephen,
>>
>> After running the script
>>
>> sudo apt-get update
>> sudo apt-get install r-base
>>
>> I launch R and find the it still refers to R 2.6.2.
>>
>> How do I insure tha
You can use the grconvertX and/or grconvertY functions to help find the
coordinates for your arrow (for plotting in the margin).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun..
On Dec 8, 2009, at 2:26 PM, Tal Galili wrote:
Is it possible?
I was hoping to find something like:
lwd
for the different bars in the barplot but couldn't find it.
Does it exist ?
Thanks,
Tal
barplot() calls rect() via a wrapper function internally to draw the
rectangles. rect() uses par(
On Dec 8, 2009, at 3:26 PM, Tal Galili wrote:
Is it possible?
I was hoping to find something like:
lwd
for the different bars in the barplot but couldn't find it.
Does it exist ?
?box
Thanks,
Tal
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
Hi,
I know there are older threads discussing the quadratcount function in
spatstat. Unfortunately, I could not find a solution to my problem there.
I'm analyzing a point pattern in an irregular polygonal window. Both the
window (an entire country) and the points are projected using WGS84.
W
Is it possible?
I was hoping to find something like:
lwd
for the different bars in the barplot but couldn't find it.
Does it exist ?
Thanks,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: w
I'm doing some data manipulations. I thought originally that I should
use data.frame, as the elements in a data.frame can have multiple
types but the elements in a matrix has to be the same, which all the
data have to convert to strings if there is a single column that is
string.
However, when I
Thank you for your help!
On Mon, Dec 7, 2009 at 9:53 AM, Viechtbauer Wolfgang (STAT) <
wolfgang.viechtba...@stat.unimaas.nl> wrote:
> If you just want a forest plot, then the forest() function.
>
> If you have the betas and corresponding variances, then you can create a
> forest plot with:
>
> fo
Hi all,
I used rpart to fit a model, where the covariates are categorical variables.
Then I plotted the tree (mytree) and used the command "text" to add labels
to the tree.
In the nodes of the tree, the values of the covariates are represented with
a, b or c (tree attached).
Is there a way to sho
On Tue, 2009-12-08 at 13:42 -0500, Cable, Samuel B Civ USAF AFMC
AFRL/RVBXI wrote:
> Am doing some vector plots with the arrows() function. Works well. But
> what I need to do is supply an arrow for scaling for the reader. I need
> to plot an arrow of some known magnitude somewhere on the page
>
Am doing some vector plots with the arrows() function. Works well. But
what I need to do is supply an arrow for scaling for the reader. I need
to plot an arrow of some known magnitude somewhere on the page
(preferably outside the bounds of the plot, so that it can be seen
clearly) with some text
On Tue, Dec 8, 2009 at 6:24 PM, stephen's mailinglist account
wrote:
>
>
> On Tue, Dec 8, 2009 at 2:25 PM, Iain Gallagher
> wrote:
>>
>> Hi Steve
>>
>> Have you tried:
>>
>> apt-cache search gfortran
>>
>> in a terminal window.
>>
>> Then
>>
>> sudo apt-get install theRelevantPackage
>>
>> I thi
Ok, that seems to work. Thanks for the help.
Steve
Steve Friedman Ph. D.
Spatial Statistical Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034
steve_fried...@nps.gov
Office (305) 224 - 4282
Fax (305) 224 - 4147
On Tue, Dec 8, 2009 at 2:25 PM, Iain Gallagher <
iaingallag...@btopenworld.com> wrote:
> Hi Steve
>
> Have you tried:
>
> apt-cache search gfortran
>
> in a terminal window.
>
> Then
>
> sudo apt-get install theRelevantPackage
>
> I think you also need the Universe repos enabled.
>
> HTH
>
> Iain
I believe the prediction is done some some sort of grid, then
interpolated to fill in the rest. This is, however, purely for
computational reason, and not for any threoretical reasons. The formal
definition of local polynomials is to do a weighted fit of polynomial at
each point.
Andy
> -O
"worrying about df (ml vs reml) is just a silly obsession of statisticians (of
which I'm one)"
I too have often wondered about the importance of such tertiary issues. My
half-baked understanding is that the main "practical" difference between ML vs
REML is with regards to ease of computing th
On Dec 8, 2009, at 12:33 PM, David Grabiner wrote:
Hi list,
This was asked a couple of years ago but I can't find a resolution.
Is
there any way to get the coefficients from one of the local
polynomial fits
in locfit. I realize that locfit only constructs polynomials at a
handful
of in
On 8 December 2009 at 12:24, Gabor Grothendieck wrote:
| I am not sure if its still the case but one of the problems with the
| Planet R feed was that it had material in it not related to R or
| statistics or any technical subject at all so if you harvest it be
| sure to exclude such sources.
Thi
Hi list,
This was asked a couple of years ago but I can't find a resolution. Is
there any way to get the coefficients from one of the local polynomial fits
in locfit. I realize that locfit only constructs polynomials at a handful
of intelligently selected points and uses interpolation to predict
On Dec 8, 2009, at 12:17 PM, David Winsemius wrote:
On Dec 8, 2009, at 11:49 AM, bnorth wrote:
I am having a problem with the censor tick marks with plot and
prodlim
I am using the example code from ?prodlim but I have added
mark.time=T to
the plot command to get tick marks at censor t
I am not sure if its still the case but one of the problems with the
Planet R feed was that it had material in it not related to R or
statistics or any technical subject at all so if you harvest it be
sure to exclude such sources.
On Tue, Dec 8, 2009 at 12:12 PM, Elijah Wright wrote:
> Hi Tal!
>
Statistical Programmer - USDA Center for Veterinary Biologics
The USDA Center for Veterinary Biologics is seeking a statistical
programmer with good R skills. The position is a two-year term position in
the CVB Statistics Section with the possibility of becoming permanent.
Applicants must be Unit
On Dec 8, 2009, at 11:49 AM, bnorth wrote:
I am having a problem with the censor tick marks with plot and prodlim
I am using the example code from ?prodlim but I have added
mark.time=T to
the plot command to get tick marks at censor times.
The problem is the tick marks occur at exactly the
Hi Tal!
First let me say that I deeply appreciate the work that you're putting into
this. You're doing good things for our community, and that's great!
I put the planet-R stuff together rather hastily a few years ago, as a way
of seeing whether it was of enough use for the community for
it to be
I am having a problem with the censor tick marks with plot and prodlim
I am using the example code from ?prodlim but I have added mark.time=T to
the plot command to get tick marks at censor times.
The problem is the tick marks occur at exactly the same point in each of the
arms.
This seems wrong
A contrarian point of view:
If you have so little data (relative to the number of parameters to be
estimated, especially NONLINEAR parameters like covariance estimates)that
the ml vs reml bias could be large, then there's so little information
anyway that such bias is the least of your problems (i
Here is the correct version. The old version is the redirect only version of
the script.
### BEGIN SCRIPT
#!/usr/bin/perl
##
# --start, -s = The date you would like to start generating regressors
#--end, -e = When to stop generating holiday regressros
# --scope, -c = D, W for Daily or We
On Tue, Dec 8, 2009 at 4:50 PM, Gray Calhoun wrote:
> I think there might be a problem with this approach if roi, tid, rt,
> and mood are the same for nonconsecutive rows.
True, but I can use the index of my reshape solution. Aggregate was
the crucial ingredient. Thanks both!
For the record, th
# BEGIN CODE ##
#!/usr/bin/perl
##
#
# --start, -s = The date you would like to start generating regressors
#--end, -e = When to stop generating holiday regressros
# --scope, -c = D, W for Daily or Weekly respectively (e.g. Does this week
have a particular holiday)
# --file, -f = Umm
I think there might be a problem with this approach if roi, tid, rt,
and mood are the same for nonconsecutive rows.
--Gray
On Tue, Dec 8, 2009 at 9:29 AM, Nikhil Kaza wrote:
> How about creating an index using multiple columns.
>
> a <- with(d, aggregate(dur, list(rt=rt,tid=tid,mood=mood,roi=roi
I have a set of parameter estimates for a multivariable Cox model predicting
survival duration and a data-frame of new measurements for the variables in
the model, as well as the actual survival duration.
Is there a function to estimate the error the model has on predicting
survival on this new se
How about creating an index using multiple columns.
a <- with(d, aggregate(dur, list(rt=rt,tid=tid,mood=mood,roi=roi),
sum))
b <- with(d, aggregate(x, list(rt=rt,tid=tid,mood=mood,roi=roi),
mean))
c <- merge(a, b, by=c("rt","tid","mood", "roi"))
I suppose one could save some time by not r
jlu...@ria.buffalo.edu wrote:
> You need to give your criteria for "preferable". For normal-linear
> models, REML estimates of variances are unbiased, whereas ML estimates are
> downwardly biased.
I suspect that you can't actually say anything general about the
direction of the bias, except f
Your question is well taken. I did not give any criteria because I realized
there might be different answers based upon different criteria. Certainly one
fundamental criteria would be that the estimates are BLUE, but this is not the
only criteria one might be used.
John
-Original Message---
Hi
A quick question. Standard errors reported by gee/yags differs from the ones in
geeglm (geepack).
require(gee)
require(geepack)
require(yags)
mm <- gee(breaks ~ tension, id=wool, data=warpbreaks,
corstr="exchangeable")
mm2 <- geeglm(breaks ~ tension, id=wool, data=warpbreaks,
I have a set of parameter estimates for a multivariable Cox model predicting
survival duration and a data-frame of new measurements for the variables in
the model, as well as the actual survival duration.
Is there a function to estimate the error the model has on predicting
survival on this new se
Dear Sayan,
there is a vcovHC method for panel models doing the White-Arellano covariance
matrix, which is robust vs. heteroskedasticity *and* serial correlation,
although in a different way from that of vcovHAC. You can supply it to coeftest
as well, just as you did. The point is in estimating
I think I forgot to send the original to the mailing list, so I'm
forwarding it (see below). Sorry about that (and sorry if I did
remember and this is a duplicate). After a few more minutes of
thought, I realized that you should probably make sure that rt, tid,
and mood are also the same in conse
Sam K yahoo.co.uk> writes:
>
> Hi all,
>
> Is there function on R for calculating Modula generators? For example for
> primes above 100, e.g 157, i want
> to know which number generates the group under multiplication mod 157. i.e i
> want to find an element whose
> order is 156. The problem I
Dear List,
I need to print out each of 'k' levels of a factor 'n' times each, where
'n' is the number of elements belonging to each factor.
I know that this can normally be done using the gl() command,
but in my case, each level 'k' has an unequal number of elements.
Example with code is as b
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