Please read the last line of every message to r-help.
In particular simplify this as much as possible and
construct some small artificial test data to illustrate.
Anyways, func is probably not what you want.
It has the same effect as
func - function(x, j) x[72+j] + [144+j]
On Wed, Jul 15, 2009
Hello, Gabor
Generally, I follow the r-help rules and provide working code snippets
to illustrate the problem. In this case it was more methodological
question of how to loop names in merge() function.
I solved this very simply buy renaming specific columns after I have
ran merge().
Thank you
Unlike Splus, R does not use T for TRUE.
On Tue, Jul 7, 2009 at 6:05 PM, Michaelcomtech@gmail.com wrote:
Hi all,
I've got the following error message in using e1071 svm routine...
Could anybody please help me?
Thank you!
-
model -
Isn't the initial value of the variable T equal to the constant TRUE?
So unless he's modified the value of T, shouldn't it work?
-s
On 7/7/09, Max Kuhn mxk...@gmail.com wrote:
Unlike Splus, R does not use T for TRUE.
On Tue, Jul 7, 2009 at 6:05 PM, Michaelcomtech@gmail.com
Hi,
On Jul 7, 2009, at 8:37 PM, Stavros Macrakis wrote:
Isn't the initial value of the variable T equal to the constant TRUE?
So unless he's modified the value of T, shouldn't it work?
Yes, it should.
Perhaps we should be looking at your data:
model - svm(y=factor(mytraindata[, 1]),
On Thu, Jul 2, 2009 at 4:34 AM, Rolf Turner r.tur...@auckland.ac.nz wrote:
On 2/07/2009, at 12:20 PM, Hsiu-Khuern Tang wrote:
Is this expected behavior?
z - 1:5
z[1] - 0
Error in z[1] - 0 : object z not found
The documentation seems to suggest that z will be found in the global
On 01/07/2009 8:20 PM, Hsiu-Khuern Tang wrote:
Is this expected behavior?
z - 1:5
z[1] - 0
Error in z[1] - 0 : object z not found
The documentation seems to suggest that z will be found in the global
environment and modified accordingly.
z - 0 works as documented, it's the indexing that
On 01/07/2009 9:34 PM, Rolf Turner wrote:
On 2/07/2009, at 12:20 PM, Hsiu-Khuern Tang wrote:
Is this expected behavior?
z - 1:5
z[1] - 0
Error in z[1] - 0 : object z not found
The documentation seems to suggest that z will be found in the global
environment and modified accordingly.
I
On 7/2/2009 6:44 AM, Kenn Konstabel wrote:
On Thu, Jul 2, 2009 at 4:34 AM, Rolf Turner r.tur...@auckland.ac.nz wrote:
On 2/07/2009, at 12:20 PM, Hsiu-Khuern Tang wrote:
Is this expected behavior?
z - 1:5
z[1] - 0
Error in z[1] - 0 : object z not found
The documentation seems to
Duncan Murdoch wrote:
- doesn't need to find z. It will replace it if found, or create a
new one if not. (Personally I would have limited that to the first
case, i.e. it should fail if it doesn't find z.)
Possibly. It's a holdover from S, where - assigns to the global
environment
On 7/2/2009 9:46 AM, Peter Dalgaard wrote:
Duncan Murdoch wrote:
- doesn't need to find z. It will replace it if found, or create a
new one if not. (Personally I would have limited that to the first
case, i.e. it should fail if it doesn't find z.)
Possibly. It's a holdover from S, where
Dear Duncan and Rolf,
That's funny! Thanks a lot.
Best regards,
Craig
Duncan Murdoch wrote:
On 30/06/2009 5:11 PM, Craig P. Pyrame wrote:
Dear Rolf,
What do you mean?
He was talking about the fortunes package. Install it, type
fortune(), and you'll get a fortune cookie message. Maybe
On 2/07/2009, at 12:20 PM, Hsiu-Khuern Tang wrote:
Is this expected behavior?
z - 1:5
z[1] - 0
Error in z[1] - 0 : object z not found
The documentation seems to suggest that z will be found in the global
environment and modified accordingly.
I would agree that the documentation would
From the help page of rep:
Value:
An object of the same type as 'x' (except that 'rep' will coerce
pairlists to vector lists).
So, you can do:
rep(list(character), 2)
On Tue, Jun 30, 2009 at 9:00 AM, Craig P. Pyrame crap...@gmail.com wrote:
Dear list,
I am trying to construct a
Dear Henrique,
Thanks! This does work, and I find the following solution to my
original problem elegant enough:
rep(list(character, integer, numeric, ...), c(3, 2, 2, ...))
Best regards,
Craig
Henrique Dallazuanna wrote:
From the help page of rep:
Value:
An object of the same
On Tue, Jun 30, 2009 at 1:00 PM, Craig P. Pyramecrap...@gmail.com wrote:
But this fails, as above. Why? Why can c(character, character) create a
list of two functions, but rep(character, 2) can't?
Another solution to my problem I could find (and you'll hopefully suggest an
even better one)
Dear Barry,
Thank you for the suggestion, it does work. I think the documentation
might be improved, but it's probably not a good idea to submit bug
reports just because I misunderstand what R does.
Best regards,
Craig
Barry Rowlingson wrote:
On Tue, Jun 30, 2009 at 1:00 PM, Craig P.
On 1/07/2009, at 12:34 AM, Craig P. Pyrame wrote:
... it's probably not a good idea to submit bug
reports just because I misunderstand what R does.
Gotta be a fortune!!!
cheers,
Rolf
##
Dear Rolf,
What do you mean?
Best regards,
Craig
Rolf Turner wrote:
On 1/07/2009, at 12:34 AM, Craig P. Pyrame wrote:
... it's probably not a good idea to submit bug
reports just because I misunderstand what R does.
Gotta be a fortune!!!
cheers,
Rolf
On 30/06/2009 5:11 PM, Craig P. Pyrame wrote:
Dear Rolf,
What do you mean?
He was talking about the fortunes package. Install it, type fortune(),
and you'll get a fortune cookie message. Maybe one with your name on it.
Duncan Murdoch
Best regards,
Craig
Rolf Turner wrote:
On
Rowlingson
Subject: Re: [R] Question about creating lists with functions as
elements
Dear Rolf,
What do you mean?
Best regards,
Craig
Rolf Turner wrote:
On 1/07/2009, at 12:34 AM, Craig P. Pyrame wrote:
... it's probably not a good idea to submit bug
reports just because I
Please ask Bioconductor questions on the Bioconductor mailing list
https://stat.ethz.ch/mailman/listinfo/bioconductor
Martin
mau...@alice.it writes:
Can biomaRt connect to data base http://mirecords.umn.edu; or a branch of it
... for instance the validated miRNAs list ..?
Thank you
Again, please refrain from crossposting.
mau...@alice.it wrote:
Can biomaRt connect to data base http://mirecords.umn.edu; or a branch of it
... for instance the validated miRNAs list ..?
The mirecords website isn't a Biomart, so no.
Jim
Thank you very much.
Maura
tutti i telefonini
On Jun 22, 2009, at 12:04 PM, Clifford Long wrote:
Hi R-list,
I'll apologize in advance for (1) the wordiness of my note (not sure
how to avoid it) and (2) any deficiencies on my part that lead to my
difficulties.
I have an application with several stages that is meant to simulate
and
Hi David,
I appreciate the advice. I had coerced 'list4' to as.list, but forgot
to specify list=() in the call to aggregate. I made the correction,
and now get the following:
slope.mult = simarray[,1]
adj.slope.value = simarray[,2]
adj.slope.level = simarray[,2]
qc.run.violation =
On Jun 22, 2009, at 6:16 PM, Clifford Long wrote:
Hi David,
I appreciate the advice. I had coerced 'list4' to as.list, but forgot
to specify list=() in the call to aggregate. I made the correction,
and now get the following:
slope.mult = simarray[,1]
adj.slope.value = simarray[,2]
On Jun 22, 2009, at 7:55 PM, David Winsemius wrote:
On Jun 22, 2009, at 6:16 PM, Clifford Long wrote:
Hi David,
I appreciate the advice. I had coerced 'list4' to as.list, but
forgot
to specify list=() in the call to aggregate. I made the
correction,
and now get the following:
David,
Once again, many thanks for your very useful and timely feedback, and
for your patience with my learning curve.
Sincerely,
Cliff
On Mon, Jun 22, 2009 at 7:11 PM, David Winsemiusdwinsem...@comcast.net wrote:
On Jun 22, 2009, at 7:55 PM, David Winsemius wrote:
On Jun 22, 2009, at
On 6/16/09, Marion Dumas mario...@gmail.com wrote:
Hello!
I am starting to use the lattice package. I generated an xyplot conditioned
on a factor that has three levels: hence I get three plots in three panels
spaces and one is left empty. I would like to add a plot to the empty panel
space.
plot(asdf[,1:7])
On Tue, Jun 16, 2009 at 10:58 AM, njhuang86njhuan...@yahoo.com wrote:
Hi all,
As of now, I have a 15x8 matrix (name is asdf). The first seven columns
contain numbers while the last column contains a string. The class of each
column is character. When I use the plot function
?data.matrix
Perhaps:
pairs(data.matrix(asdf[ , 1:7]) )
On Jun 16, 2009, at 10:58 AM, njhuang86 wrote:
Hi all,
As of now, I have a 15x8 matrix (name is asdf). The first seven
columns
contain numbers while the last column contains a string. The class
of each
column is character. When
Check out:
http://wiki.r-project.org/rwiki/doku.php?id=tips:data-io:ms_windows
On Fri, Jun 5, 2009 at 1:37 PM, jlfmssmjlfm...@gmail.com wrote:
I am woking on Windows and using read.xls to read Excel.xls file.
But it needs to install perl on my machine first, and then give the
following until
I've not ever tried something like this. You didn't quite answer the
question though. Do you need interactive sessions, or are users
choosing from a number of batch jobs?
If the latter, then perhaps you would be better forsaking Rcmdr (which
I have not used) and instead consider a shell-based, or
On 28-May-09 00:58:17, Erin Hodgess wrote:
Dear R People:
I would like to set up a plug-in for Rcmdr to do the following:
I would start on a Linux laptop. Then I would log into another
outside system and run a some commands.
Now, when I tried to do
system(ssh e...@xxx.edu)
password
Hi.
Do you need an interactive session at the remote machine, or are you
simply wanting to run a pre-written script?
If the latter, then you can ask ssh to execute a remote command, which
conceivably could be R CMD x
If you explain exactly why and what you are trying to do, then perhaps
My goal is for a user to sit down at a Linux laptop, get to an Rcmdr
type screen, submit jobs on a remote system and then get the results
back in R.
We will assume that the user is naive, and the only thing he/she can
do is get to the Rcmdr screen.
The Rcmdr plugin will have a submit jobs menu.
Subject: Re: [R] question about using a remote system
My goal is for a user to sit down at a Linux laptop, get to an Rcmdr
type screen, submit jobs on a remote system and then get the results
back in R.
We will assume that the user is naive, and the only thing he/she can
do is get
Thanks a lot - I am trying it with lattice now
On Sat, May 16, 2009 at 10:44 PM, Duncan Mackay mac...@northnet.com.au wrote:
Dimitri
You mentioned lattice so I presumed that you tried a barchart in lattice
if you found the class of table.a.percents will it comply with the lattice
Dimitri Liakhovitski wrote:
Hello!
I promise I looked into help files before asking. Still cannot figure
it out. I think it's because I am totally confused what packages use
lettice, which use trellis, etc.
Sections 1 and 2 below produce the data and the data to plot. My
question is about
Dimitri
You mentioned lattice so I presumed that you tried a barchart in lattice
if you found the class of table.a.percents will it comply with the lattice
requirements of the data type ?
class(table.a.percents)
[1] matrix
Using the lattice package and converting to a data.frame
x -
Erin Hodgess wrote:
Hi R People:
I have a question about setClass please. I'm working thru R
Programming for Bioinformatics.
Actually, I was wondering if there is such a thing as an updateClass,
in order to change a contains option, please?
that is, if I had
setClass(dog,
Mark,
Give a look:
rres-read.table(stdin(), head=T, sep=,)
ID,Traversed, v2
1,5,1
1,7,1
1,8,1
2,8,2
2,11,2
2,7,2
3,11,3
3,22,3
3,16,3
aggregate(rres[c(Traversed, v2)], list(rres$ID), sum)
bests
milton
On Fri, May 1, 2009 at 5:07 PM, Altaweel, Mark R. maltaw...@anl.gov wrote:
Hi,
I am
: r-help@r-project.org
Sent: Tuesday, April 28, 2009 9:42 AM
Subject: Re: [R] question about adaboost.
Cecilia Lezama wrote:
Hello,
I would like to know how to obtain the misclassification error when
performing a boosting analisis with ADABAG package?
With:
prop.table(Tesis.boostcv$confusion
0.00667 0.01333
Many, many thanks!
- Original Message -
From: Uwe Ligges lig...@statistik.tu-dortmund.de
To: Cecilia Lezama che...@netgate.com.uy
Cc: r-help@r-project.org
Sent: Tuesday, April 28, 2009 9:42 AM
Subject: Re: [R] question about adaboost.
Cecilia Lezama wrote
Helen Chen 96258011 at nccu.edu.tw writes:
I would like to run some panel regressions with R. Therefore I want to use
fixed effect model.
The focus of lme is on mixed models, but gls in nlme can handle fixed-only
problems. I doubt, however that it is exactly what you want for your case.
I
Cecilia Lezama wrote:
Hello,
I would like to know how to obtain the misclassification error when performing
a boosting analisis with ADABAG package?
With:
prop.table(Tesis.boostcv$confusion)
I obtain the confusion matrix, but not the overall missclassification error.
Well, the
This is a nontrivial problem. This comes up often on the Statalist
(-qreg- is for cross-section quantile regression):
You want to fit a plane through the origin using the L-1 norm.
This is not as easy as with L-2 norm (LS), as it is more
than a matter of dropping a constant predictor yet
I was trying to resist responding to this question since the original
questioner
had already been admonished twice last october about asking questions
on R-help about posted code that was not only not a part of R-base,
but not even a part of an R package. But the quoted comment about
Stata is
You should review your course material on interpreting general linear
models. The criterion you have chosen for significance (looking at p
values for indivdiual coefficients) is not the recommended one. Seek
out the section that discusses the proper method for using deviance
estimates for
I thought of testing the difference in deviance between the null model
and the fitted model, assuming it is distributed as chi-sq. However,
Faraway writes that if the outcome is binary, the deviance
distribution is far from chisq.
I've done a permutation test:
N-5000; # Towards the upper limit,
Surely Faraway does not suggest using the Wald statistic in preference
to the deviance?
Even if the distribution of deviance is not exactly chi-square, it
appears generally accepted that a comparison of the difference in
deviance to the chi-square statistic is better than using the ratio
David,
Faraway suggests using the Hosmer Lemeshow test in the case of a
binary response, and discusses the inadequacy of Wald statistics.
However, I'm not sure it applies here due to the limited number of
cases.
Thanks, Ehud.
On Wed, Apr 22, 2009 at 2:04 AM, David Winsemius
Grześ wrote:
I have a standard database - HouseVotes84
For example:
Class V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16
1 republicann y ny y y n n n y NA y y y ny
2 republicann y ny y y n n n nn y y y n NA
3
Gavin Simpson wrote:
Grześ wrote:
I have a standard database - HouseVotes84
For example:
Class V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16
1 republicann y ny y y n n n y NA y y y ny
2 republicann y ny y y n n n nn
Try:
z - cbind(rep(c(BIC, hist), each = 150), rep(rep(c(5, 10, 30),
each = 50),2))
z - as.data.frame(z)
z - cbind(z, runif(300))
names(z) - c(Method, sigma, Error)
z$sigma - factor(z$sigma, c(5, 10, 30))
library(lattice)
sigma - as.numeric(levels(z$sigma))
sigmaExprList - lapply(sigma,
Sorry, that should be:
sigma - as.numeric(levels(z$sigma))
sigmaExprList - lapply(sigma, function(s) bquote(sigma == .(s)))
sigmaExpr - as.expression(sigmaExprList)
bwplot(Error~Method | sigma, data = z,
horiz = F, xlab = Method,
strip = function(which.given, which.panel, var.name,
On Thu, Apr 16, 2009 at 1:45 PM, Sundar Dorai-Raj sdorai...@gmail.com wrote:
Sorry, that should be:
sigma - as.numeric(levels(z$sigma))
sigmaExprList - lapply(sigma, function(s) bquote(sigma == .(s)))
sigmaExpr - as.expression(sigmaExprList)
bwplot(Error~Method | sigma, data = z,
Strubbe Diederik diederik.strubbe at ua.ac.be writes:
Dear R community,
I have some questions regarding the analysis of a zero-inflated count
dataset and repeated measures design.
The dataset is arranged as follows :
Unit of analysis: point - these are points were bird were counted
Dear Diederik,
If you revisited the same points then it makes sense to use the data at
the point level. But then I would mke that explicit by using a nested
random effect. In the nlme/lme4 syntax: 1|Site/Point. Make shure that
each point has a unique ID.
Naming a variable count is not a very
, University of Antwerp
Universiteitsplein 1
B-2610 Antwerp, Belgium
http://webhost.ua.ac.be/deco
tel : 32 3 820 23 85
-Original Message-
From: ONKELINX, Thierry [mailto:thierry.onkel...@inbo.be]
Sent: Tue 14-4-2009 15:46
To: Strubbe Diederik; r-help@r-project.org
Subject: RE: [R] Question
check package adapt, http://cran.r-project.org/web/packages/adapt/index.html
Best,
Dimitris
salia a wrote:
I need the commond in R for double integral over the range - infinityxy infinity .
Could you please help me.
Thanks and regards
Salya
[[alternative HTML version
DF2 - read.table(textConnection(sidpidslope
+ 1.11.12
+ 1.14.13
+ 1.15.12
+ 2.15.13
+ 3.21.22
+ 3.21.73), header = TRUE)
tapply(DF2$slope, as.factor(DF2$pid), mean)
1.1 1.2 1.7 4.1
If you want zero if only one variable:
DF2
sid pid slope
1 1.1 1.1 2
2 1.1 4.1 3
3 1.1 5.1 2
4 2.1 5.1 3
5 3.2 1.2 2
6 3.2 1.7 3
tapply(DF2$slope, DF2$sid, function(x) if(length(x) == 1) 0 else mean(x))
1.1 2.1 3.2
2.33 0.00 2.50
On Fri,
David,
I struggled with this for a while. I think the problem with the dates
I have is that they are not specific dates, they are partial dates.
A workaround for that that I got from someone else in the list was:
as.Date(paste(x$Date, '1'), '%B %Y %d')
to make them specific dates (the
Another approach is to use yearmon. e.g.
library(zoo)
as.yearmon(January 1987, %B %Y)
[1] Jan 1987
Thus we can replace the w - line in my example code with:
w - zoo(as.matrix(W[-1]), as.yearmon(W[,1], %B %Y))
On Wed, Mar 11, 2009 at 12:48 PM, Oscar Bonilla oboni...@galileo.edu wrote:
David,
There are two header lines which is confusing it. Use pattern= to
start at the second one.
See the three vignettes in zoo for info on the rest.
library(gdata)
W -
read.xls(http://www2.standardandpoors.com/spf/pdf/index/CSHomePrice_History_022445.xls;,
pattern = LXXR)
library(zoo)
w -
You need to convert W$Date into a real date variable. At the moment it
is just a character variable.
str(W)
'data.frame': 265 obs. of 23 variables:
$ Date: Factor w/ 265 levels ,April 1987,..: 1 90
68 156 2 178 134 ...
$ AZ.Phoenix : Factor w/ 236 levels
David1234 wrote:
Hi everyone,
Im quite new to R an I have the following Question:
I would like to define Variables which I can add and multiply etc. and that
R can simplyfy the terms.
The variables should stand for integers.
For example I would like to have an entry in an array with
myArray[, 'z'] - myArray[, 'z'] + b
Is this what you want?
On Sat, Mar 7, 2009 at 9:52 AM, David1234 danielth...@web.de wrote:
Hi everyone,
Im quite new to R an I have the following Question:
I would like to define Variables which I can add and multiply etc. and that
R can simplyfy the
I fear that you are looking for a symbolic algebra system, and R is
not that sort of platform. If I am correct and you still want to
access a symbolic algebra system from R, then you should look at YACAS
and the interface to it, Ryacas.
--
David Winsemius
On Mar 7, 2009, at 9:52 AM,
On Wed, 4 Mar 2009, Vadlamani, Satish {FLNA} wrote:
Hi:
Sorry if this is a double post. I posted the same thing this morning and did
not see it.
I just started using R and am asking the following questions so that I can plan
for the future when I may have to analyze volume data.
1) What are
Are you trying to color the points themselves? This plots the
first two series in frame 1 (they are the same but one is plotted
as points and the other as a line) and the third series is shown
in frame 2 and for the series of points it colors them green or red.
The lines are all colored black:
Hi, Gabor
No, what I am trying to do is similar to:
abline(v=time(spread)[spread[,Indicator]==(-1)], col=yellow),
where spread is the multivariate zoo object (say, 5 timeseries).
That is, I want to color parts of the plots where indicator==(-1), but
do the coloring
without using layout() and
Do you mean you want to shade a portion of the plot?
There are two examples of that in the examples section of ?plot.zoo
and a further example using xyplot.zoo in the examples section of
?xyplot.zoo
On Tue, Mar 3, 2009 at 8:37 AM, Sergey Goriatchev serg...@gmail.com wrote:
Hi, Gabor
No, what
Gabor, yes, I want to color portions of EACH plot of the MULTIPLE plot
done with plot.zoo()
I tried to do:
plot(multivariate zoo object)
abline(v=...)
but that does not work.
I will check your suggestions of the examples.
Thank you for your help, as always!
Best,
Sergey
On Tue, Mar 3, 2009 at
Supreme!
Thanks Gabor!
On Tue, Mar 3, 2009 at 15:20, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
If you want to use abline on a multivariate plot it must be issued
within a panel function like this:
# same set up as in prior email
p - function(x, y, ...) { points(x, y); lines(x, y);
Waverley wrote:
Hi,
I have a question of the method as how to normalize the data sets
according to a set of the internal measurements.
For example, I have performed two batches of experiments contrasting
two different conditions (positive versus negative conditions): one at
a time.
1. each
Hi Deepankar
The code on the following page looks kind of cool, and also seems to
produce something of the type of graph you are after perhaps:
https://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/pkg/rgl/demo/regression.r?rev=702root=rglsortby=dateview=auto
[below is a copy of the code...]
the following should work
library(lattice)
x - seq(1,100)
y - seq(1,100)
gr - expand.grid(x,y)
gr$z - x + y + rnorm(1,0,100)
cloud(z ~ x + y, data = gr)
also, look for the package rgl which does similar but with more
possiblities.
On Feb 27, 4:28 pm, Dipankar Basu basu...@gmail.com wrote:
actually, I just realised you also want a line in the plot. I am not
super-sure how to do this.
On Feb 27, 5:20 pm, andrew andrewjohnro...@gmail.com wrote:
the following should work
library(lattice)
x - seq(1,100)
y - seq(1,100)
gr - expand.grid(x,y)
gr$z - x + y + rnorm(1,0,100)
The interface is so self-explaining... Data -- Import data -- text files...
# pls don't mind :)
library(fortunes)
fortune(15)
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086
Mobile: +86-15810805877
Homepage: http://www.yihui.name
School of
See the warn.conflicts arguments of require() and library(). People
thought about this a long time ago.
-- Bert Gunter
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Carl Witthoft
Sent: Friday, February 13, 2009 4:05 PM
To:
on 02/13/2009 06:05 PM Carl Witthoft wrote:
Recently I got introduced to two packages: {seewave} and {audio} .
Turns out they both have a tool to call a system audio tool, and in both
cases the name of the tool is play(). Naturally these two tools do
slightly different things with
See ?diff.zoo
dif - diff(data, arithmetic = FALSE) - 1
cbind(data, dif)
On Wed, Feb 11, 2009 at 6:21 AM, Sergey Goriatchev serg...@gmail.com wrote:
Hello, everyone!
Assume you have this data:
data - structure(c(66.609375, 67.09375, 66.40625, 66.734375, 67.109375,
66.875, 66.09375,
Sergey Goriatchev wrote:
Hello, everyone!
Assume you have this data:
data - structure(c(66.609375, 67.09375, 66.40625, 66.734375, 67.109375,
66.875, 66.09375, 65.921875, 66.546875, 66.140625, 66.140625,
65.65625, 65.875, 65.59375, 65.515625, 66.09375, 66.015625, 66.140625,
66.109375,
Dear Gabor,
Thank you as always. I will follow your suggestion.
But, I still want to know what I do wrong in the above code (with
embed(), apply(), and merge() functions). Why do I get that error
message and how can I rewrite the code to make it run.
Best Regards,
Sergey
On Wed, Feb 11, 2009 at
Dear Peter,
I tried with the comma already, it did not work.
rets
TY1.lev SP1.lev GC1.lev CL1.lev
2000-01-03 66.60938 1702.7 453.7 18.34
2000-01-04 67.09375 1647.7 447.8 18.29
2000-01-05 66.40625 1649.4 446.2 17.65
2000-01-06 66.73438 1639.9 446.5 17.52
2000-01-07
Sergey Goriatchev wrote:
Dear Peter,
I tried with the comma already, it did not work.
Ah, yes, sorry. The real issue is that you are trying to index twice,
both by apply(.,2,.) and by z=TY1.lev (the former loops over columns
and the latter selects one of them). So what func() sees is an
There is no zoo method for embed. embed is implicit in
rollapplly so its not normally needed in zoo and rollapply,
lag or diff would normally be used instead.
If you did want to use embed then unzoo data first but
that will lose the time index so add it back in later:
e -
Understood.
Peter, Gabor, thank you for your time and help.
Regards,
Sergey
On Wed, Feb 11, 2009 at 14:24, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
There is no zoo method for embed. embed is implicit in
rollapplly so its not normally needed in zoo and rollapply,
lag or diff would
Rolf Turner wrote:
On 9/02/2009, at 4:40 PM, bill.venab...@csiro.au wrote:
Store your 'matrix' as a data frame.
Surely it's a data frame already, since ``school'' is character or
factor,
and ``value'' is (must be?) numeric.
People have this unfortunate predilection to refer
Waverley,
you can also use p...@y.values to access the slot (see
help(performance-class) for a description of the slots).
You might also want have a look at the code for demo(ROCR) and at this
slide deck:
http://rocr.bioinf.mpi-sb.mpg.de/ROCR_Talk_Tobias_Sing.ppt
HTH,
Tobias
On Sat, Feb 7,
Tobias Sing wrote:
Waverley,
you can also use p...@y.values to access the slot (see
help(performance-class) for a description of the slots).
You might also want have a look at the code for demo(ROCR) and at this
slide deck:
http://rocr.bioinf.mpi-sb.mpg.de/ROCR_Talk_Tobias_Sing.ppt
HTH,
Store your 'matrix' as a data frame. Call it 'SchoolVals' say. Then
SchoolMeans - with(SchoolVals, tapply(value, school, mean))
should do it. If you have missing values you want to ignore:
SchoolMeans - with(SchoolVals, tapply(value, school, mean, na.rm = TRUE))
Bill Venables
On 9/02/2009, at 4:40 PM, bill.venab...@csiro.au wrote:
Store your 'matrix' as a data frame.
Surely it's a data frame already, since ``school'' is character or
factor,
and ``value'' is (must be?) numeric.
People have this unfortunate predilection to refer to data frames
Dear George,
You could use jitter(x) in place of x to randomly perturb the horizontal
coordinates. See ?jitter for details.
I hope this helps,
John
--
John Fox, Professor
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
web:
If you post the code that you used, prehapsy using the sample data in
the same package, you may get more useful replies:
commented, minimal, self-contained, reproducible code.
Gee, ... where have I seen that before?
Also could see the thread;
Dear Waverley,
Try this:
unlist(slot(perf,y.values))
See ?slot for more details.
HTH,
Jorge
On Sat, Feb 7, 2009 at 3:17 PM, Waverley waverley.paloa...@gmail.comwrote:
Hi,
I have a question about ROCR package. I got the ROC curve plotted
without any problem following the manual.
Hi Waverley,
I forgot to tell you that perf is your performance object. Here is an
example from the ROCR package:
## computing a simple ROC curve (x-axis: fpr, y-axis: tpr)
library(ROCR)
data(ROCR.simple)
pred - prediction( ROCR.simple$predictions, ROCR.simple$labels)
perf -
On Thu, Feb 5, 2009 at 6:31 AM, Uwe Ligges
lig...@statistik.tu-dortmund.dewrote:
milton ruser wrote:
Dear George,
I think it will depends on the amount of memory that each your session
will
need.
Case each session use a big amount of memory, may be your some of your
sessions will get
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