This reflects a problem in your locale (traditional Chinese): we cannot
reproduce it. Try running R in a different locale (e.g. append LC_ALL=en
to the target when you start R).
On Sun, 7 Sep 2008, Thomas Lo wrote:
Dear all,
I encountered a problem on starting and using the R v 2.7.2
aggregate(x$weight,list(x$fruit),max)
Group.1 x
1 apple 1.6
2 orange 1.6
Shubha Karanth | Amba Research
Ph +91 80 3980 8031 | Mob +91 94 4886 4510
Bangalore * Colombo * London * New York * San José * Singapore *
www.ambaresearch.com
-Original Message-
From: [EMAIL PROTECTED]
I know how to compute the ROC curve and the empirical AUC from the logistic
regression after fitting the model.
But here is my question, how can I compute the standard error for the AUC
estimator resulting form logistic regression? The variance should be more
complicated than AUC based on known
Hi
I have a dataframe in which some subjects appear in more than one row. I
want to extract the subject-rows which have the minimum date per subject. I
tried the following aggregate function.
attach(dataframe.xy)
aggregate(Date,list(SubjectID),min)
Unfortunately, the format of the Date-column
I set up a Facebook profile where I can post my pictures, videos and events and
I want to add you as a friend so you can see it. First, you need to join
Facebook! Once you join, you can also create your own profile.
Thanks,
Giovanna
Here's the link:
Hi R,
I have the variance-covariance matrix,
V=matrix(c(0.09238, 0.002407527, 0.002407527, 0.020739401),2,2)
I need to find a vector X=c(x1,x2), such that
1) X'VX is equal to a constant 2 (say) and
2) sum(x) should be equal to a another constant 1.5 (say).
How do we do this
On Mon, 8 Sep 2008, Prof Brian Ripley wrote:
This reflects a problem in your locale (traditional Chinese): we cannot
reproduce it. Try running R in a different locale (e.g. append LC_ALL=en to
the target when you start R).
Maybe I have found this as a bug in iconv. Please try a version of
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Antje wrote:
Hi there,
does anybody know how to return the xmlPath from a node?
For example, at several location in the xml file, I have nodes with the
same name and I'd like to process only the nodes from a certain path.
Any idea?
As with
On Mon, 8 Sep 2008, Duncan Murdoch wrote:
On 08/09/2008 6:56 AM, Prof Brian Ripley wrote:
On Mon, 8 Sep 2008, Prof Brian Ripley wrote:
This reflects a problem in your locale (traditional Chinese): we cannot
reproduce it. Try running R in a different locale (e.g. append LC_ALL=en
to the
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Uwe Ligges
Kurapati, Ravichandra (Ravichandra) wrote:
Hi
Lets say
Main(){
Shubha Vishwanath Karanth wrote:
Hi R,
I have the variance-covariance matrix,
V=matrix(c(0.09238, 0.002407527, 0.002407527, 0.020739401),2,2)
I need to find a vector X=c(x1,x2), such that
1) X'VX is equal to a constant 2 (say) and
2) sum(x) should be equal to a another
I have a problem whose solution requires non-negative least squares. That is
minimize sum(y - Xbeta)^2 subject to beta =0
Splus has the nnls.fit command. Is there an R alternative?
Terry Therneau
__
R-help@r-project.org
Hi there,
I have a very basic but still important question:
How can I free my workspace COMPLETELY?
I tried:
rm(list = ls())
rm(list=ls(all=TRUE))
remove(list = conflicts(detail=TRUE)$.GlobalEnv)
but I still have lots of conflicts() which I have never seen before.
Thanx for the help.
--
View
93354504 wrote:
clotting - data.frame(u = c(5,10,15,20,30,40,60,80,100), lot1 =
c(118,58,42,35,27,25,21,19,18), lot2 = c(69,35,26,21,18,16,13,12,12));
lot=c(clotting$lot1,clotting$lot2);
uu=c(clotting$u,clotting$u);
x=uu;
y=lot;
n=length(y);
x=cbind(rep(1,n),x);
c=2;
On 08/09/2008 6:56 AM, Prof Brian Ripley wrote:
On Mon, 8 Sep 2008, Prof Brian Ripley wrote:
This reflects a problem in your locale (traditional Chinese): we cannot
reproduce it. Try running R in a different locale (e.g. append LC_ALL=en to
the target when you start R).
Maybe I have found
Hi There!
I am trying to read a grid file (.asc) file with readGDAL (library[rgdal]).
The file has 7541 rows and 5247 columns and a size of 177 Mb. My laptop has
2 Gb of RAM and 150 Gb of total hard disk memory. Is it possible to open
this raster file in R at all?
Thanks, Shams
--
View this
Terry Therneau wrote:
I have a problem whose solution requires non-negative least squares. That is
minimize sum(y - Xbeta)^2 subject to beta =0
Splus has the nnls.fit command. Is there an R alternative?
There is a package nnls on CRAN by Kate Mullen and Ivo van Stokkum:
Hi
Lets say
Main(){
Fcaproxy(user=ravi,password=db,database=oracle,host=10.0.0.3)
}
Fcaproxy -Function(...,importlimit=){
Dbproxy(...)
}
Hi there,
does anybody know how to return the xmlPath from a node?
For example, at several location in the xml file, I have nodes with the same
name and I'd like to process only the nodes from a certain path.
Any idea?
Antje
__
clotting - data.frame(u = c(5,10,15,20,30,40,60,80,100), lot1 =
c(118,58,42,35,27,25,21,19,18), lot2 = c(69,35,26,21,18,16,13,12,12));
lot=c(clotting$lot1,clotting$lot2);
uu=c(clotting$u,clotting$u);
x=uu;
y=lot;
n=length(y);
x=cbind(rep(1,n),x);
c=2;
neg_loglikehood=function(theta){
On 08/09/2008 5:59 AM, GAF wrote:
Hi there,
I have a very basic but still important question:
How can I free my workspace COMPLETELY?
I tried:
rm(list = ls())
rm(list=ls(all=TRUE))
remove(list = conflicts(detail=TRUE)$.GlobalEnv)
but I still have lots of conflicts() which I have never seen
gallon li wrote:
I know how to compute the ROC curve and the empirical AUC from the logistic
regression after fitting the model.
But here is my question, how can I compute the standard error for the AUC
estimator resulting form logistic regression? The variance should be more
complicated than
Try changing the 'class' of the numeric result back to Date:
x - as.Date('2008-09-08')
x
[1] 2008-09-08
y - as.numeric(x)
y
[1] 14130
str(y)
num 14130
class(y) - Date
y
[1] 2008-09-08
str(y)
Class 'Date' num 14130
On Mon, Sep 8, 2008 at 6:38 AM, Erich Studerus
[EMAIL PROTECTED]
Thanks, I've already tried that. The problem is, that the original date is
not restored when I change the numeric back to date. I get a totally
different date.
Maybe it has something to do with the original date format. My data are
directly imported from a SQL-database. The date column to which I
Can you provide some actual data. It sounds like the columns you are
are POSIXct, in which case you would want to do something like this:
time - structure(time, class = c(POSIXt, POSIXct))
So it is important to know what your numeric values came from and what
their actual values were. Which of
Hi
Try the following reference:
Comparison of Three Methods for Estimating the
Standard Error of the Area under the Curve in ROC
Analysis of Quantitative Data by Hajian-Tilaki and Hanley, Academic
Radiology, Vol 9, No 11, November 2002.
Below is a simple implementation that will return both the
Hi Duncan,
thanks a lot for your explanations.
I tried the following now to understand a bit more:
data - getNodeSet(doc, //Data)
xmlName(data[[1]])
xmlName(xmlRoot(data[[1]]))
xpathApply(data[[1]], ./*, xmlName)
Is it right that using data in the xpathApply() somehow sets the current node
Hi there,
See ?lroc in the epicalc package.
HTH,
Jorge
On Mon, Sep 8, 2008 at 4:02 AM, gallon li [EMAIL PROTECTED] wrote:
I know how to compute the ROC curve and the empirical AUC from the logistic
regression after fitting the model.
But here is my question, how can I compute the
Erich,
how does the data look, when it comes from SQL?
And why not extract the data with SQL directly, so you don't
have this issue in the first place?
el
on 9/8/08 3:15 PM Erich Studerus said the following:
Thanks, I've already tried that. The problem is, that the original date is
not
Here's how the dates look like after the sql-query:
Oav$Date[1:3]
[1] 1991-11-22 00:45:00 CET 1991-12-13 00:01:00 CET 1992-02-06 00:45:00
CET
class(oav$Date[1:3])
[1] POSIXt POSIXct
x-as.numeric(oav$Date[1:3])
x
[1] 690767100 692578860 697333500
class(x)-Date
x
[1] 3226-01-31 8186-07-07
thanx for the quick reply!
so this is what I get when I type in conflicts()
[1] part buzz distance huid
in_whistle interruptionsmatch
[8] overlap part part_2 pause_dur
pause_dur2
Those are not dates! They are date-times.
aggregate is overkill for a single column. Something simple like
DT - seq(Sys.time(), by=4 hours, len=24)
grp - rbinom(24, 1, p=0.5)
res - tapply(DT, grp, min)
class(res) - class(DT)
res
would suffice.
On Mon, 8 Sep 2008, Erich Studerus wrote:
On 08/09/2008 7:45 AM, GAF wrote:
thanx for the quick reply!
so this is what I get when I type in conflicts()
[1] part buzz distance huid
in_whistle interruptionsmatch
[8] overlap part part_2
Hi,
Appologies for the simple nature of this question, I am unable to find
the answer in manuals (EG and introduciton to R).
I have written a function in a text editor and saved it with an .R
extension. It is saved in my working directory. How can I run it, do I
need to use source? If so,
You can 'source' it in to define it and then call it as usual. No
particular directory since you have control of that with the 'source'
function. File extension is by convention and anyone will work since
source is just going to read in an text file and execute it.
On Mon, Sep 8, 2008 at 10:46
On Monday, 08 September 2008, 15:46 (UTC+0100), Williams, Robin wrote:
I have written a function in a text editor and saved it with an .R
extension. It is saved in my working directory. How can I run it, do I
need to use source?
Just to add to Jim's reply: if you want to have this
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Antje wrote:
Hi Duncan,
thanks a lot for your explanations.
I tried the following now to understand a bit more:
data - getNodeSet(doc, //Data)
xmlName(data[[1]])
xmlName(xmlRoot(data[[1]]))
xpathApply(data[[1]], ./*, xmlName)
Is it
Duncan Temple Lang schrieb:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Antje wrote:
Hi Duncan,
thanks a lot for your explanations.
I tried the following now to understand a bit more:
data - getNodeSet(doc, //Data)
xmlName(data[[1]])
xmlName(xmlRoot(data[[1]]))
xpathApply(data[[1]],
Hi all,
I have some wind data (U and V components) and I would like to compute
Vorticity and Divergence of these fields. Is there any R function that
can easily do that?
Thanks in advance for any help
Igor Oliveira
CSAG, Dept. Environmental Geographical Science,
University of Cape Town,
Hello all,
I have a very large file (280k lines) containing three comma separated
variables. The first variable is a 0 or 1 depicting a pass or fail. The
other two are X and Y coordinates. Is there a good way I can represent this
data in a chart/plot form other than using a 3d histogram? If I
I apologize, I forgot to type the title.
On Mon, Sep 8, 2008 at 11:39 AM, Jason Thibodeau [EMAIL PROTECTED]wrote:
Hello all,
I have a very large file (280k lines) containing three comma separated
variables. The first variable is a 0 or 1 depicting a pass or fail. The
other two are X and Y
Bob Flagg wrote:
Dear All,
I'm reading Frank Harrell's wonderful Regression
Modeling Strategies book and ran into a problem
following the example in Chapter 8. I'm working
on
platform: Ubuntu 8.04 (i486-pc-linux-gnu)
R version: 2.7.2 (2008-08-25)
and my command sequence was:
As requested in the last line to every message to r-help a cutdown
version of the data should be posted with the question.
Suppose such a cut down version is this:
DF - data.frame(Date = ISOdatetime(2008, 1, c(1, 2, 5, 3, 4), 0, 0, 0),
Subject = c(1, 1, 2, 2, 2))
# Then sort the data
Dear R-list,
maybe some of you could point me in the right direction:
Are you aware of any FREE Fortran or Java libraries/actual pieces of
code that are VERY efficient (time-wise) in running the regular linear
least-squares multiple regression?
More specifically, I have to run small regression
I have one more question. The below example is revised to better
reflect the problem that I am running into. The two columns for each
data frame in the list are named the same because they are subsets of
the same site, which has the same name in the larger data set. So
when xyplot plots the
I would test the speed before making such as assumption. Note that
lm.fit is faster than lm and if they have the same x matrix then
you can do many in one call by having y be a matrix.
On Mon, Sep 8, 2008 at 12:05 PM, Dimitri Liakhovitski [EMAIL PROTECTED] wrote:
Dear R-list,
maybe some of you
Hi Brian and Duncan,
Many thanks for your responses. Setting the 'Current format' in 'Regional
and language options' under Control Panel to English (Singapore) solved the
R usage problem for me. I will have a go at R-patched build 46507 after it
is released.
Regards,
Thomas
On Mon, Sep 8,
Dear List,
If I use the following code to generate a pdf,
pdf(filename)
image(x, y, !is.na(z), col=c(green,black))
dev.off()
Can someone extract the 0-1 data (that is, is.na(z)) from the Adobe source?
Thank you.
Xiaochun
[[alternative HTML version deleted]]
Add this after the zm- statement
colnames(zm) - sapply(z.l, colnames)
On Mon, Sep 8, 2008 at 12:09 PM, stephen sefick [EMAIL PROTECTED] wrote:
I have one more question. The below example is revised to better
reflect the problem that I am running into. The two columns for each
data frame in
On Mon, 8 Sep 2008, Li, Xiaochun wrote:
Dear List,
If I use the following code to generate a pdf,
pdf(filename)
image(x, y, !is.na(z), col=c(green,black))
dev.off()
Can someone extract the 0-1 data (that is, is.na(z)) from the Adobe source?
Yes (from the PDF file, which is not due to Adobe
Are you sure R's ways are not fast enough (there are many layers
underneath lm)? For an example of how you might do this at C/Fortran
level, see the function lqs() in MASS.
On Mon, 8 Sep 2008, Dimitri Liakhovitski wrote:
Dear R-list,
maybe some of you could point me in the right direction:
Thank you for reminding me, Gabor. I forgot to mention: So far, I have
run one test set of regressions using lm. It took R 270 sec. I need to
run 1,800,000 of those, which would imply 15.4 years of computing time
:)
I have not done the same for lm.fit because I am not sure how to get
model R
Yes, see my previous e-mail on how long R takes (270 seconds for one
of the 1,800,000 sets I need) - using system.time.
Not sure how to test the same for Fortran...
On Mon, Sep 8, 2008 at 12:51 PM, Prof Brian Ripley
[EMAIL PROTECTED] wrote:
Are you sure R's ways are not fast enough (there are
Dear R Users,
I have a relatively simple rmpi question.
My configation is:
- R version 2.7.2
- rmpi 0.5-6
- Deino MPI 1.1.0
- Windows XP SP2
After succesfully spawning slaves, I am trying to assign values to
variables in the environment of each slave and run some simple calcs. This
appears
Hello members!
This question had been posted by Thomas Steiner in May, but I couldn't
locate any followups thereafter, see:
https://stat.ethz.ch/pipermail/r-help/2008-May/161483.html
I want to raise up this question in a (hopefully) even simpler way:
Given a random variable X, it's
Disclaimer: I have **NO IDEA** of the details of what you want to do or why
-- but I am willing to bet that there are better ways of doing it than 1.8
mm multiple refressions that take 270 secs each!! (which I find difficult to
believe in itself -- are you sure you are doing things right?
I'd start with scatterplots of the two subsets (pass vs fail), but with
280k points, those are likely to be fairly uninformative masses of black
ink). However, there might be enough separation between them that you
don't need anything else.
If not, then a pair of hexbin plots (from the
Anny,
You can also do the following
plot(0:10, 0:10, pch=16, type=h)
Cheers../Murli
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Peter Alspach
Sent: Sunday, September 07, 2008 5:10 PM
To: Anny Huang; r-help@r-project.org
Subject: Re: [R] how to draw
Try:
sum(lm.fit(x, y)$residuals^2)
On Mon, Sep 8, 2008 at 12:52 PM, Dimitri Liakhovitski [EMAIL PROTECTED] wrote:
Thank you for reminding me, Gabor. I forgot to mention: So far, I have
run one test set of regressions using lm. It took R 270 sec. I need to
run 1,800,000 of those, which would
I am using the function 'spectrum'. It returns two arrays that are interesting
to me. One would be the wieght or density of a given frequency with the
irequency given in another array. I would like to take the top 'n' weights
which would be the top 'n' frequencies contributing to the signal.
Dear Jason,
If I understood, you are looking for a way of represent a response surface.
I know that there are other very interesting (and most indicated)
solutions, but give a look at the akima package
Best wishes,
miltinho astronauta
brazil
===
x-runif(100)
y-rnorm(100)
Thank you everyone for your responses. I'll answer several questions.
1. Disclaimer: I have **NO IDEA** of the details of what you want
to do or why
-- but I am willing to bet that there are better ways of doing it than 1.8
mm multiple refressions that take 270 secs each!! (which I find
On Mon, Sep 8, 2008 at 1:47 PM, Dimitri Liakhovitski [EMAIL PROTECTED] wrote:
Thank you everyone for your responses. I'll answer several questions.
1. Disclaimer: I have **NO IDEA** of the details of what you want
to do or why
-- but I am willing to bet that there are better ways of doing
Please forgive me if this has been asked before but I could not readily find an
answer. First, from the example I was able to determine that graphics device
commands such as plot can be redirected using 'png' and 'bmp' type commands. I
tried this and it works as I understand it. The question is
?order
This will give you the indices that you can use.
On Mon, Sep 8, 2008 at 1:38 PM, [EMAIL PROTECTED] wrote:
I am using the function 'spectrum'. It returns two arrays that are
interesting to me. One would be the wieght or density of a given frequency
with the irequency given in another
Hi all,
I'm looking into opening an url on a server which requires
authentication.
After failing to find some kind of connection structure to fill in I
turned to explicitly stating the credentials in the url itself (e.g.
http://username:[EMAIL PROTECTED]).
Sadly this didn't do the trick either
Hello,
I need to perform a mixed-model (with nesting) MANCOVA, using Type III sums of
squares. I know how to perform each of these types of tests individually, but I
am not sure if performing a mixed-model MANCOVA is possible. Please let me know.
Erika
Erika
Hi,
There are more articles, papers, contents about statistical tools ???
(Perfomance, packages, Usability, difficulty in handling the software.
something)
Thank You
Ricardo
--
View this message in context:
Hi Tolga --
[EMAIL PROTECTED] writes:
Dear R Users,
I have a relatively simple rmpi question.
My configation is:
- R version 2.7.2
- rmpi 0.5-6
- Deino MPI 1.1.0
- Windows XP SP2
After succesfully spawning slaves, I am trying to assign values to
variables in the environment of each
Dear List,
If I use the following code to generate a pdf,
pdf(filename)
image(x, y, !is.na(z), col=c(green,black))
dev.off()
Can someone extract the 0-1 data (that is, is.na(z)) from the Adobe source?
Thank you.
Xiaochun
[[alternative HTML version deleted]]
Hello,
I need to perform a mixed-model (with nesting) MANCOVA, using Type III sums
of squares. I know how to perform each of these types of tests individually,
but I am not sure if performing a mixed-model MANCOVA is possible. Please let
me know.
Erika
On Monday, 08 September 2008, 10:59 (UTC-0700),
[EMAIL PROTECTED] wrote:
If I issue another 'plot' command will it also be redirected
to the 'png' device?
As long as that png device is the currently active device, yes.
Does dev.off essentially reset the graphics
device to the
Dear Martin,
Many thanks. This is very useful, much appreciated.
Regards,
Tolga
Martin Morgan [EMAIL PROTECTED]
08/09/2008 19:08
To
[EMAIL PROTECTED]
cc
r-help@r-project.org
Subject
Re: [R] RMPI Question
Hi Tolga --
[EMAIL PROTECTED] writes:
Dear R Users,
I have a relatively
Is there a mod (like C '%' operator) operator in 'R'? I tried to
help.search(mod) and there were too many hits for the query to be useful.
Kevin
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
?%%
Gabor
On Mon, Sep 08, 2008 at 11:58:45AM -0700, [EMAIL PROTECTED] wrote:
Is there a mod (like C '%' operator) operator in 'R'? I tried to
help.search(mod) and there were too many hits for the query to be useful.
Kevin
__
Although I along with the other believe there probably is an efficient R
solution, the answer to your direct question can perhaps be found at
http://www.fortran.com/. The free GNU G95 fortran compiler is at
http://www.g95.org/
Joe
-Original Message-
From: [EMAIL PROTECTED]
On Mon, 8 Sep 2008, Damien wrote:
Hi all,
I'm looking into opening an url on a server which requires
authentication.
After failing to find some kind of connection structure to fill in I
turned to explicitly stating the credentials in the url itself (e.g.
http://username:[EMAIL PROTECTED]).
Csardi Gabor wrote:
?%%
Yup. Notice, by the way, that modulus [sic] is ambiguous:
Mod(1+1i)
[1] 1.414214
46 %% 7
[1] 4
Gabor
On Mon, Sep 08, 2008 at 11:58:45AM -0700, [EMAIL PROTECTED] wrote:
Is there a mod (like C '%' operator) operator in 'R'? I tried to
help.search(mod) and
Dear Mark,
See argument labels in ?pie.
x=c(11, 15, 16, 29, 31)
pie(x,labels=x,col=c('orange','green','blue','red','purple'))
HTH,
Jorge
On Mon, Sep 8, 2008 at 3:24 PM, polishookm [EMAIL PROTECTED]wrote:
With a pie chart
pie(c(11, 15, 16, 29, 31))
how can I generate labels for the
Thank you. I definitely did not want Mod.
Kevin
Peter Dalgaard [EMAIL PROTECTED] wrote:
Csardi Gabor wrote:
?%%
Yup. Notice, by the way, that modulus [sic] is ambiguous:
Mod(1+1i)
[1] 1.414214
46 %% 7
[1] 4
Gabor
On Mon, Sep 08, 2008 at 11:58:45AM -0700,
Thanks Jorge, that's perfect.
Jorge Ivan Velez wrote:
Dear Mark,
See argument labels in ?pie.
x=c(11, 15, 16, 29, 31)
pie(x,labels=x,col=c('orange','green','blue','red','purple'))
HTH,
Jorge
On Mon, Sep 8, 2008 at 3:24 PM, polishookm
[EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
On Monday, 08 September 2008, 15:24 (UTC-0400), polishookm wrote:
With a pie chart
pie(c(11, 15, 16, 29, 31))
how can I generate labels for the chart, such as
orange: 11
green: 15
blue: 16
red: 29
purple: 31
pie(c(11, 15, 16, 29, 31),labels=c(what,ever,you,want,duh))
?pie
would
Dear R Users,
I am attempting to use the snow package for clustering. Is there a way to
identfy, in the environment of each node, a rank for that node and also,
the total size of the cluster ?
By way of analogy, I am looking for the functions in snow equivalent to
mpi.comm.rank() and
Once the series has been decomposed into the seasonal, trend, and remainder
components with stl is there a 'predict' like function that I can use this data
to forecast the next 'n' values? I didn't see one in the stl documentation. So
if such a function does not exist I was wondering if I could
Damien schrieb:
I'm looking into opening an url on a server which requires
authentication.
Under a Windows Operating System you could try to start R with the
--internet2 option. This worked in my case.
Best regards
Rene
__
R-help@r-project.org
Hello,
how is it possible to plot a time series of monthly data over several years
such that the x-axis shows the first letter of the month and displays a grid
line at every year? I am new to R and had no real success until now. I have:
library(utils)
oneMonth = 1.0 / 12.0
tsData =
Thanks a lot, everybody!
On Mon, Sep 8, 2008 at 3:11 PM, Lucke, Joseph F
[EMAIL PROTECTED] wrote:
Although I along with the other believe there probably is an efficient R
solution, the answer to your direct question can perhaps be found at
http://www.fortran.com/. The free GNU G95 fortran
On Mon, 08 Sep 2008 01:45:51 +0200, Peter Dalgaard wrote :
Emmanuel Charpentier wrote:
Dear list,
[ Snip ... ]
This looks reasonably sane, I think. The last loop could be d[] -
lapply(d, conv1, from, to), but I think that is cosmetic. You can't
really do much better because there
Hello,
I would like to sample from a Gumbell (minimum) distribution. I have
installed package {evd} but the Gumbell functions there appear to refer
to the maximum case. Unfortunately, setting the scale parameter
negative does not appear to work.
Is there a separate package for the Gumbell
Try this:
plot(tsData, ylab=Values, xlab=Zeit, xaxt = n)
axis(1, time(tsData), rep(substr(month.abb, 1, 1), length = length(tsData)),
cex.axis = .3, tcl = -.5)
jan - time(tsData)[cycle(tsData) == 1] # january
axis(1, jan, FALSE, tcl = -1)
abline(v = jan, lty = 2)
and also look at the
Hi Ralph,
My approach provides the same answer by asking a different question.
Effectively, my approach tests whether the difference between timepoints is
larger for X than for Y (and also gives you the main effects of whether X
has a higher mean than Y and whether scores increase or decrease
For what it's worth, copy/paste between R and Word 2008 works perfectly. It
doesn't work at all for Word 2004...so that's at least ONE improvement (the
only one I've seen) in versions of Word.
--Adam
On Sun, 7 Sep 2008, Stefan Evert wrote:
On 7 Sep 2008, at 16:57, John Kane wrote:
I think
Hi Steve,
You probably want to check out ?by or ?aggregate, maybe using
(rownames(df) %/% 60) : (colnames(df) %/% 60) as your index variable.
--Adam
On Sun, 7 Sep 2008, Steve Murray wrote:
Dear all,
I have a large dataset which I hope to reduce in size, to make it more
useable. I
Word in Office 2003 (and I assume Office 2007) can insert encapsulated
postscript with no problems on Windows, and display the graphic fine.
Does this not work with Mac? Doesn't dragging a file into Word lose the
full resolution of postscript?
Frank
Adam D. I. Kramer wrote:
For what it's
I think you want the ?lines function.
To connect a point (x,y) to the x-axis,
lines(x=c(x,x),y=c(y,0))
...draws a line from that point to the x-axis. You may also want to specify
pch=c(?,),type=b where ? is the original point type (which you don't
want to run over) and is the pch for theline
On Mon, 8 Sep 2008, [EMAIL PROTECTED] wrote:
I have a nested ANOVA, with a fixed factor tmt nested within site
(random). There are missing values in the data set.
aeucs, tmt and site have been defined as objects
I have tried:
model1=lme(aeucs~tmt,random=~1|tmt/site)
I think you want
word 2004
post script (plot created with postscript()) does not work but pdf
graphics work on the mac. I suggest not copying and pasting because,
in my expreience, the output is not as good as pdf et al.
On Mon, Sep 8, 2008 at 6:19 PM, Frank E Harrell Jr
[EMAIL PROTECTED] wrote:
Word in Office
Hi Srini,
This may be as simple as tapply(weight,fruit,max)
or t(that) if you want it as you specified.
--Adam
On Sun, 7 Sep 2008, Srinivas Iyyer wrote:
dear group,
i have a data matrix with some replicate items with different values. I want to
extract the row with max value.
for example:
Emmanuel Charpentier wrote:
On Mon, 08 Sep 2008 01:45:51 +0200, Peter Dalgaard wrote :
Emmanuel Charpentier wrote:
Dear list,
[ Snip ... ]
This looks reasonably sane, I think. The last loop could be d[] -
lapply(d, conv1, from, to), but I think that is cosmetic.
Hi Erich,
Since min() is defined for numbers and not dates, the problem is in the
min() function. min() is converting from date format to number format.
Your best bet is to make this conversion explicit...such that it is
reversable. So, convert the date into UTC, then UTC to seconds since
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