On 5/7/07, Bert Gunter [EMAIL PROTECTED] wrote:
Folks:
If I understand correctly, the following may be pertinent.
Note that the procedure:
min.nnet = nnet[k] such that error rate of nnet[k] = min[i] {error
rate(nnet(training data) from ith random start) }
does not guarantee a classifier
Hallo,
I just now a solution for da data frame. I'm not sure if this is what you want.
Just try if it helps. Here an example of my code where I added a column:
df - rbind( c(fred,mary,4),c(fred,mary,7),
c(fred,mary,9),c(barney,liz,3),
c(barney,liz,5))
df -
It seems that there is here a problem of reliability, as one never
knows whether the solution provided by R is correct or not. In the
case that I reported, it is fairly simple to see that the solution
provided by R (without any warning!) is incorrect, but, in general,
that is not so simple and one
Dear R-users,
I'd like to announce the release of the new version of package `ltm'
(i.e., ltm_0.8-0 soon available from CRAN) for Item Response Theory
analyses. This package provides a flexible framework for analyzing
dichotomous and polytomous data under IRT, including the Rasch model,
the
Dear all,
I'm struggling with weighted least squares, where something that I had
assumed to be true appears not to be the case. Take the following
data set as an example:
df - data.frame(x = runif(100, 0, 100))
df$y - df$x + 1 + rnorm(100, sd=15)
I had expected that:
summary(lm(y ~ x,
mister_bluesman wrote:
Hello.
I have a 2d plot which looks like this:
http://www.nabble.com/file/8242/1.JPG
This plot is derived from a file that holds statistics about each point on
the plot and looks like this:
abc d e
a
Hello,
I want to build empirical Returns of stockfonds, but I cant find a function
in R, which calculate this.
Alin
[[alternative HTML version deleted]]
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R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
Hallo,
I just tried, if my solution might be possible for you. Here is the result:
s=2
while(s!=0){ n=20
+ m-matrix(nrow=n,ncol=4)
+ colnames(m)=c(treatmentgrp,strata,censoringTime,survivalTime)
+ for(i in 1:20)
Hallo,
I just updated to the new version of R by installing everything new. Now
I have a problem with the help command:
help.start()
updating HTML package listing
updating HTML search index
If nothing happens, you should open
'Z:\Software\R\R-2.5.0\doc\html\index.html' yourself
The browser was
Which function implements the piecewise cubic Hermite interpolation?
I am looking for equivalent of matlab's interp1 with the method = 'pchip'
Here is the reference
http://www.mathworks.com/access/helpdesk/help/techdoc/index.html?/access/helpdesk/help/techdoc/ref/interp1.html;
--
View this
As a part of a simulation, I need to sample from a large vector repeatedly.
For some reason sample() builds up the memory usage ( 500 MB for this
example) when used inside a for loop as illustrated here:
X - 1:10
P - runif(10)
for(i in 1:500) Xsamp - sample(X,3,replace=TRUE,prob=P)
Hallo,
if you work with matrix you can use cbind to add columns.
Corinna
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von raymond chiruka
Gesendet: Montag, 7. Mai 2007 16:28
An: r
Betreff: [R] creating a new column
hie l would like to
Did you try manually opening the file mentioned in the browser
(Z:\Software\R\R-2.5.0\doc\html\index.html) ?
Does this file exist?
Was the html help installed?
Schmitt, Corinna wrote:
Hallo,
I just updated to the new version of R by installing everything new. Now
I have a problem with
The issue is that you are using a derivative based optimizer for a
problem for which it is well known that such optimizers will not
perform well. You should consider using a global optimizer. For
example, rgenoud combines a genetic search algorithm with a BFGS
optimizer and it works well for
As a part of a simulation, I need to sample from a large vector repeatedly.
For some reason sample() builds up the memory usage ( 500 MB for this
example) when used inside a for loop as illustrated here:
X - 1:10
P - runif(10)
for(i in 1:500) Xsamp - sample(X,3,replace=TRUE,prob=P)
How can I install the html help?
Corinna
.de
http://www.igb.fraunhofer.de
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von Vladimir Eremeev
Gesendet: Dienstag, 8. Mai 2007 12:22
An: r-help@stat.math.ethz.ch
Betreff: Re: [R] trouble with
You should select it in the list of components to install, when the installer
asks.
It happens after you 3rd time press Next in the installation wizard
window.
Choose Custom installation and select all you need.
Schmitt, Corinna wrote:
How can I install the html help?
Corinna
Hallo,
I found the solution. I installed everything new and in the setup I choose
Changing Startoptions and there I could mark html help.
It now works,
Corinna
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von Schmitt, Corinna
Gesendet:
Nathaniel,
On Mon, 7 May 2007, [EMAIL PROTECTED] wrote:
Date: Sun, 6 May 2007 12:02:31 + (GMT)
From: nathaniel Grey [EMAIL PROTECTED]
However what I really want to know is how well my nueral net is
doing at classifying my binary output variable. I am new to R and I
can't figure out how
I have a very large matrix with columns that have some of their
entries as zero
A small example if a=
[,1] [,2] [,3] [,4]
[,1] 0 2 0 0
[,2] 1 3 0 3
[,3] 2 0 3 5
[,4] 0 4 0 0
and what to get the minimum number
apply(a, 2, function(x) min(x[x!=0]) )
should do it. Might need some improvement if all numbers in a column
can be zero, try it.
Gabor
On Tue, May 08, 2007 at 09:50:43AM +0100, [EMAIL PROTECTED] wrote:
I have a very large matrix with columns that have some of their
entries as zero
A
I have just submitted irtoys_0.1.0, a package potentially useful for
those working with IRT models. It can fit the 1PL, 2PL, and 3PL models
through a simple and unified syntax, using either the R package ltm,
Brad Hanson's ICL program, or the commercially available BILOG-MG. The
purpose is
try this:
apply(a, 2, function(x) min(x[x 0]))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web:
See below.
hadley wickham wrote:
Dear all,
I'm struggling with weighted least squares, where something that I had
assumed to be true appears not to be the case. Take the following
data set as an example:
df - data.frame(x = runif(100, 0, 100))
df$y - df$x + 1 + rnorm(100, sd=15)
I
As a part of a simulation, I need to sample from a large vector repeatedly.
For some reason sample() builds up the memory usage ( 500 MB for this
example) when used inside a for loop as illustrated here:
X - 1:10
P - runif(10)
for(i in 1:500) Xsamp - sample(X,3,replace=TRUE,prob=P)
On 5/8/07, Adaikalavan Ramasamy [EMAIL PROTECTED] wrote:
See below.
hadley wickham wrote:
Dear all,
I'm struggling with weighted least squares, where something that I had
assumed to be true appears not to be the case. Take the following
data set as an example:
df - data.frame(x =
Hallo,
I added one row:
a=rbind(a,1:4)
a
[,1] [,2] [,3] [,4]
[1,]0120
[2,]2304
[3,]0030
[4,]0350
[5,]1234
And how looks like the command for the minimum of the rows? The result should
be minOfRows = 0 0 0 0
Corinna,
what is going on here? I've answered Oarabile's question, and then
you reply to me with this. I'm completely lost.
1. What is your question? Minimum of every row?
This was written in the original mail along with Oarabile's
question. Ok it wasn't rows but columns.
For rows it
Ravi Varadhan wrote:
Paul,
Your solution based on SVD does not work
Ooops. I really am getting rusty. The idea is based on eigenvalues
which, of course, are not always the same as singular values.
Paul
even for the matrix in your example
(the reason it worked for e=3, was that it is an
Hallo,
I just followed the discussion of the title: minimum from matrix.
I have a similar problem.
a=matrix(c(0,2, 0, 0, 1, 3, 0, 3, 2, 0, 3, 5, 0, 4, 0, 0),ncol=4)
a=rbind(a,1:4)
a
[,1] [,2] [,3] [,4]
[1,]0120
[2,]2304
[3,]0030
[4,]0
Check out the help for apply, particularly the MARGIN argument:
minOfRows=apply(a, 1, function(x) min(x[x!=0]) )
maxOfRows=apply(a, 1, function(x) max(x) )
Sarah
On 5/8/07, Schmitt, Corinna [EMAIL PROTECTED] wrote:
Hallo,
I just followed the discussion of the title: minimum from matrix.
I
Looks like you are reading manuals and these mailings insufficiently
carefully.
?apply says that if its second argument is 1, it gives you what you want.
Gabor Csardi has also written you this.
If you have several vectors, not a single matrix, you can use pmin:
Corinna,
what about reading my previous mail, which you suggested to forget?
Gabor
ps. it is apply(a, 1, min)
On Tue, May 08, 2007 at 03:21:26PM +0200, Schmitt, Corinna wrote:
Hallo,
I just followed the discussion of the title: minimum from matrix.
I have a similar problem.
Hi,
is it possible to find the index of a point of a plot (e.g. scatterplot) in
an easy way?
Eg.
x - c(1:5); y - c(1:5);
plot(x, y);
On the plot if I move my cursor on top of a point or click on it is it
possible to have its index printed or its exact value? Any clues?
Thanks.
Hi Peter,
thanks for the quick help. Yes it does what I want.
Thanks, Corinna
Von: Peter Konings [mailto:[EMAIL PROTECTED]
Gesendet: Dienstag, 8. Mai 2007 15:29
An: Schmitt, Corinna
Betreff: Re: [R] minimum of each row in a matrix
Hi Corinna,
does
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Try ?locator
--- [EMAIL PROTECTED] wrote:
Hi,
is it possible to find the index of a point of a
plot (e.g. scatterplot) in
an easy way?
Eg.
x - c(1:5); y - c(1:5);
plot(x, y);
On the plot if I move my cursor on top of a point or
click on it is it
possible to have its index printed
Hi Jessica
[EMAIL PROTECTED] wrote:
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R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal,
Sorry, you did not explain that your weights correspond to your
frequency in the original post. I assumed they were repeated
measurements with within group variation.
I was merely responding to your query why the following differed.
summary(lm(y ~ x, data=df, weights=rep(2, 100)))
Hi, Tony:
Are you familiar with the 'debug' command? I agree that more
informative error messages and 'traceback' would be nice, but I've found
the 'debug' facility quite useful. [I even sometimes prepare a shell of
a function 'fn', then say debug(fn) and fn(), and complete writing
Paul,
The problem lies neither with R nor with numercial methods. The onus is always
on the user to understand what the numerical schemes can do and what they can't
do. One should never blindly take the results given by a numerical scheme and
run with it. In your example, the optimization
Dear R gurus,
There is an interesting problem with accessing specific items in a
column of data frame that has incorrectly been given a duplicate
name, even though addressing the item by row and column number.
Although the column is correctly listed, an item addressed by row and
column number
Hi all,
I'm looking for a package that will give me comprehensive set of
descriptive statistics, including skew and kurtosis. Also, is there a
similar package that will provide multivariate descriptive statistics as
well? Thanks in advance,
David
--
Dear Hadley,
I think that the problem is that the term weights has different meanings,
which, although they are related, are not quite the same.
The weights used by lm() are (inverse-)variance weights, reflecting the
variances of the errors, with observations that have low-variance errors
On 08/05/07, John Kane [EMAIL PROTECTED] wrote:
Try ?locator
Thanks. Your tip also lead to another function: ?identify to add my two
cents.
--- [EMAIL PROTECTED] wrote:
Hi,
is it possible to find the index of a point of a
plot (e.g. scatterplot) in
an easy way?
Eg.
x -
Paul,
You have picked a function that is not smoothly differentiable and also started
at one of many 'stationary' points in a system with multiple solutions. In
practice, I think it'll get a zero gradient as the algorithm does things
numerically and you have a symmetric function. It probably
Thanks, Ravi, for your clear explanation!
Paul
On 5/8/07, RAVI VARADHAN [EMAIL PROTECTED] wrote:
Paul,
The problem lies neither with R nor with numercial methods. The onus is
always on the user to understand what the numerical schemes can do and what
they can't do. One should never
Doubling the length of the data doubles the apparent number of observations.
You would expect the standard error to reduce by sqrt(2) (which it just about
does, though I'm not clear on why its not exact here)
Weights are not as simple as they look. You have given all your data the same
weight,
It is not clear to me what you want here.
Errors are tagged by a 'call', and f(1:3) is the innermost 'call' (special
primitives do not set a context and so do not count if you consider '['
to be a function).
The message could tell you what the type was, but it does not and we have
lost the
Hello,
I have many subsets of x. I want to get the standard deviation
for each subset with the same factor levels. For FUN=mean and
FUN=median I am using ave().
Can anybody tell me the match of ave() for using FUN=SD?
At the beginning I used aggregate(), also for mean and median. But
Hi,
I have a bivariate longitudinal dataset. As an example say,
i have the data frame with column names
var1 var2 Unit time trt
(trt represents the treatment)
Now suppose I want to fit a joint model of the form for the *i* th unit
var1jk = alpha1 + beta1*timejk + gamma1* trtjk +
Dear All,
I sended my first mail as HTML by accident.
It has probably been stripped off... (see first mail below)
During that time, I was actually able to find a solution to my problem : I
wanted to plot times on a graph representing precipitation=f(time)
here is an example:
time-c(2004-10-18
Hi - I encounter two problems with
SystemFit. I have a matrix of 20 variables (38 observations each).
I am trying to fit using 2SLS in Systemfit because I want to be able
to simulate new observations with the resulting model. The first
problem I found is that it ran out of memory given that I
Thanks it worked.
Oarabile
Gabor Csardi wrote:
apply(a, 2, function(x) min(x[x!=0]) )
should do it. Might need some improvement if all numbers in a column
can be zero, try it.
Gabor
On Tue, May 08, 2007 at 09:50:43AM +0100, [EMAIL PROTECTED] wrote:
I have a very large matrix with columns
Thanks it worked
Oarabile
Dimitris Rizopoulos wrote:
try this:
apply(a, 2, function(x) min(x[x 0]))
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35,
Hi - I encounter two problems with SystemFit. I have a matrix of 20 variables
(38 observations each). I am trying to fit using 2SLS in Systemfit
because I want to be able to simulate new observations with the resulting
model. The first problem I found is that it ran out of memory given that
Paul,
I believe the model you describe below can be fitted in GENMOD and GLIMMIX
in SAS.
Alternatively, as Brian Ripley suggests, you could use MCMC. BUGS has a
nice example of a multinomial logit model in the second example manual.
While this example considers only fixed effects, it's not
thanks for the help But l ened up using
act.surv.time-pmin(m[,censoringTime],m[,survivalTime])
m-cbind(m,act.surv.time)
which seems to work
thanks
Schmitt, Corinna [EMAIL PROTECTED] wrote: Hallo,
I just tried, if my solution might be possible for you. Here is the result:
s=2
Dear all,
I actually would like to improve the label orientation on the x-axis (turn
them to 45 degrees)
I tried the par(las=2) ... but doesn't work...
Do anyone knows how to do ?
Jessica
__
R-help@stat.math.ethz.ch mailing list
Hadley,
You asked
.. what is the usual way to do a linear
regression when you have aggregated data?
Least squares generally uses inverse variance weighting. For aggregated data
fitted as mean values, you just need the variances for the _means_.
So if you have individual means x_i and sd's
This is not an R question but if anyone can help me, it's much
appreciated.
Suppose I have a series ( stationary ) y_t and a series x_t ( stationary
)and x_t has variance sigma^2_x and epsilon is normal
(0, sigma^2_epsilon )
and the two series have the relation
Dear Prof. Ripley:
1. I very much appreciate your contributions to the R project.
2. Whether with release 2.4.0 or earlier, I noticed that
'traceback()' had become less informative. This loss was more than
offset when I learned to use the 'debug' function in the 'base'
in R when carring out the log rank test is the censored variable denoted by 1
or 0 or its of no consequence.
thanks
-
always stay connected to friends.
[[alternative HTML version deleted]]
__
First, you should not be using colnames-, which is for a matrix, on a
data frame. Use names- for data frames (and as.data.frame to convert to
a data frame).
Second, whereas duplicate row names are not allowed in a data frame,
duplicate column names are but at your own risk.
Third, there is a
Mark, I suppose you make the usual assumptions, ie. E[x]=0, E[x*epsilon]=0, the
correlation is just simply,
corr(x,y) = beta * ( var(x) / var(y) )
And you could get var(y) from var(x) and var(epsilon).
HTH.
Horace
Leeds, Mark (IED) [EMAIL PROTECTED] 5/8/2007 10:25:11 AM
This is not an R
Well, I guess it makes quite a difference in survival analysis whether
you know that a person was alive/censored or experienced the event of
interest at a certain point of time/age. You could have tried it easily
for yourself by slightly modifying the example on the help page of
'survdiff'.
Hi,
summary(yourdata) is often a very good starting point for descriptive
data statistics. Or you can write your own little function which returns
what you actually like to see (the code below was written very quickly.
No care is taken for the presence of missing values or anything else).
I haven't been following your discussion, but based on your method, the
following should also work and would be the one-line equivalent:
m$act.surv.time-pmin(m[,censoringTime],m[,survivalTime])
In other words, you can add a column by just assigning a value to a new column
name. You can also
Dear r-helpers,
In an xyplot I have
xyplot(y ~ x | c, groups = g,
scales = list(
x = list(
at = v1,
labels = c('A', 'B', 'C', 'D'),
rot = 45
)
)
)
g consists of two groups.
Thanks people! Sorry to sound thick but after I download the poltrix package,
how do i install it?
Thank you!
--
View this message in context:
http://www.nabble.com/Representing-a-statistic-as-a-colour-on-a-2d-plot-tf3703885.html#a10383035
Sent from the R help mailing list archive at
The short answer is that you can't.
The longer answer is that you need to replace them
with text. Have a look at the FAQ 7.27 How can I
create rotated axis labels?
It provides a pretty good example.
--- [EMAIL PROTECTED] wrote:
Dear all,
I actually would like to improve the label
How did you download it? In Windows if you used the
Packages Install packages route it is installed.
Just type library(plotrix) to load it. No idea for
Linux or Mac.
--- mister_bluesman [EMAIL PROTECTED]
wrote:
Thanks people! Sorry to sound thick but after I
download the poltrix package,
mister_bluesman wrote:
Thanks people! Sorry to sound thick but after I download the poltrix package,
how do i install it?
Thank you!
which platform are you on?
On Windows I would suggest you use the Menu of the GUI Install packages
from local zip files in the main category 'packages'
On
Thanks for you all your help guys. Much appreciated.
--
View this message in context:
http://www.nabble.com/Representing-a-statistic-as-a-colour-on-a-2d-plot-tf3703885.html#a10383998
Sent from the R help mailing list archive at Nabble.com.
__
I wrote an ugly algorithm to set certain elements of a matrix to 1
without looping and below works and you can see what
The output is below the code.
K-6
lagnum-2
restrictmat-matrix(0,nrow=K,ncol=K*3)
restrictmat[((col(restrictmat) - row(restrictmat) = 0 )
(col(restrictmat)-row(restrictmat)) %%
I work on Windows, R version 2.4.1
I need to leave out many rows and columns when reading a matrix. The 'skip'
commands in read.table, seem to be just for skipping rows, if I understand
well, but I need to skip many columns as well. I tried something very simple
and left one row and one column
Suggestion:
You might make it easier for folks to help if you explained in clear and
simple terms what you are trying to do. Code is hard to deconstruct.
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of
Does anyone know of an R function for computing the Greenland-Robins
variance for Mantel-Haenszel relative risks?
Thanks
Frank
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
Prof. Ripley,
Maybe the fact that few R plot regions have a 1:1 aspect ratio
is not a major problem here. One has to deal with the same
issue when drawing a circle parametrically. Depending on the
window size, a (cos(t),sin(t)) plot appears as an ellipse.
To get a circle parametrically, one has
That's a good idea : I didn't realize that my matrices would look so bad
in the final email. All I want
To do is output 1's in the diagonal elements and zero's everywhere else
but the matrix is not square so by diagonals I
Really mean if
Lagnum = 1 then the elements are (1,1), (2,2),
zzz - zzz[-c(1,45,23,321), -c(23,34,45,67)]
On 5/8/07, Silvia Lomascolo [EMAIL PROTECTED] wrote:
I work on Windows, R version 2.4.1
I need to leave out many rows and columns when reading a matrix. The 'skip'
commands in read.table, seem to be just for skipping rows, if I understand
well,
Hi Mark,
Is this of any help?
resMat-function(K=6,lag=2,ncol=3*K){
X-matrix(0,K,ncol)
X[,1:(K*lag)]-diag(K)
return(X)
}
Cheers,
Anders.
On Tuesday 08 May 2007 11:21 am, Leeds, Mark (IED) wrote:
I wrote an ugly algorithm to set certain elements of a matrix to 1
without
Is this what you want?
n - 6
lagnum - 2
result - NULL
for (i in 1:lagnum) result - cbind(result, diag(n))
result
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,]100000100 0 0 0
[2,]0100000
Would this function help:
http://www.csm.ornl.gov/~frome/ES/RRMHex/MHanalysis.txt ?
Regards, -Cody
Frank E Harrell
Jr
Dear all,
I have two data frame, on with a complete list of my field survey with
frequency data of a sample species. This data frame looks like:
simulation-data.frame(cbind(my.year=c(rep(2000,8),rep(2001,12),rep(2002,12)),my.month=c(5:12,1:12,1:12)))
thanks anders : that works perfectly. I'll have to study it better to
understand but it's much appreciated.
-Original Message-
From: Anders Nielsen [mailto:[EMAIL PROTECTED]
Sent: Tuesday, May 08, 2007 5:55 PM
To: Leeds, Mark (IED)
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] Looking
[EMAIL PROTECTED] wrote:
Would this function help:
http://www.csm.ornl.gov/~frome/ES/RRMHex/MHanalysis.txt ?
Regards, -Cody
I think so. Thank you Cody. If you have time would you mind defining,
probably offline, the input data elements? I assume that one of them is
a stratification
yes, this works. It didn't think of concatenating... thanks!
jim holtman wrote:
zzz - zzz[-c(1,45,23,321), -c(23,34,45,67)]
On 5/8/07, Silvia Lomascolo [EMAIL PROTECTED] wrote:
I work on Windows, R version 2.4.1
I need to leave out many rows and columns when reading a matrix. The
Hi,
I have 2 dataset,
plot(data1)
plot(data2),
but it comes as two graphs, can I draw both on a single panel so I can
compare them?
Thanks
Pat
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R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
See ?points and ?lines for a simple solution. Eg:
plot(data1)
points(data2)
if mean this by 'panel'.
Gabor
On Tue, May 08, 2007 at 04:49:47PM -0700, Patrick Wang wrote:
Hi,
I have 2 dataset,
plot(data1)
plot(data2),
but it comes as two graphs, can I draw both on a single panel so I
You have to have a valid 'date' object on the x-axis. Try this:
simulation-data.frame(cbind(my.year=c(rep(2000,8),rep(2001,12),rep(2002,12)),my.month=c(5:12,1:12,1:12)))
http://en.wikipedia.org/wiki/Weighted_least_squares gives a formulaic
description of what you have said.
I believe the original poster has converted something like this
y x
0 1.1
0 2.2
0 2.2
0 2.2
1 3.3
- use proper spacing to make it easier to read
- start off with set.seed to make it reproducible
- omit cbind and combine all the rep's into one rep in first line
- make the date column a known date class (here Date),
set.seed(1)
sim - data.frame(
my.year = rep(2000:2002, c(8, 12, 12)),
Columns that correspond to NULLs in the colClasses arg of read.table
are ignored. ?read.table
On 5/8/07, Silvia Lomascolo [EMAIL PROTECTED] wrote:
I work on Windows, R version 2.4.1
I need to leave out many rows and columns when reading a matrix. The 'skip'
commands in read.table, seem to
R understands only numerical and Date class values for axis. So either
a) plot them using the sequence 1, ..., 32 and then explicitly label
them. Here is an example:
n - length(year.month)
plot( 1:n, freq, xaxt=n)
mtext( text=year.month, side=1, at=1:n, las=2 )
b) or create the dates
Try this:
kronecker(t(rep(1:0, c(lagnum, 3-lagnum))), diag(K))
On 5/8/07, Leeds, Mark (IED) [EMAIL PROTECTED] wrote:
That's a good idea : I didn't realize that my matrices would look so bad
in the final email. All I want
To do is output 1's in the diagonal elements and zero's everywhere else
One other idea. If your dates are continguous, as they are here,
you might want to use a ts series for this. Using the same sim
as in my prior post:
set.seed(1)
x - with(sim, ts(freq, start = c(my.year[1], my.month[1]), freq = 12))
x
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
2000
This will get you started.
Rich
tmp - data.frame(y=rnorm(20), x=rnorm(20),
c=factor(rep(letters[1:5], 4)),
g=factor(rep(1:2, each=10)))
v1 - seq(-1.5, 0, .5)
v2 - 0:2
xyplot(y ~ x | c, groups = g, data=tmp,
scales = list(x = list(
Hi all,
I am running R 2.5.0 under Windows XP Media Center Edition. Here's a
problem that's been stumping me for a few days now, and I can't find
anything useful in the archives. I am using voronoi.mosaic (tripack
package) to create proximity polygons for a study of vegetation
competition
Have you tried 'debug'?
With one of the datasets that crashes your system, run the first
voronoi.mosaic, then the torus function and then
debug(voronoi.mosaic)
and run through the second call of voronoi.mosaic. You can step though the
code and at least will you find the point where it bombs.
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