Thanks to all for the answers! I solved my problem now by sufficient iteration!
Have a nice day!
Luba
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Hi Steve,
Thanks for the pointer! After checking the source code myself, it's indeed the
case. And of course, the training error is always better than the CV error.
And this can be also checked if you vary the fold number in "cross", the
"fitted" result is not changing.
I wonder if the de
as tu lu l'article suggere dans l'aide:
M. Kendall and A. Stuart (1983) The Advanced Theory of Statistics, Vol.3,
Griffin, 410414.
> Date: Mon, 6 Jul 2009 13:22:17 -0700
> From: m.gha...@yahoo.fr
> To: r-help@r-project.org
> Subject: [R] Decompose function : calculation of each component
>
Check out the Optimization Task View:
http://cran.r-project.org/web/views/Optimization.html
On Wed, Jul 8, 2009 at 12:00 AM, leecholho wrote:
>
>
> thanks...
>
> i do homework.. my professor said , to use slover function in excel.
> but i used R , so i want to find same function.
>
> question is
thanks...
i do homework.. my professor said , to use slover function in excel.
but i used R , so i want to find same function.
question is " Max(x,y) x*E(R.a) + y*E(R.b)" , and x+y=1,
E(R.a) and E(R.b) is average of the special data..
special data is this
---
hi..
i used R a few days..
who can call me? same function solver in excel.
_
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htt
Debabrata Midya wrote:
>
>
> Dear R users,
>
> Thanks in advance.
>
>
>
> I am using R 2.9.1 on Windows XP.
>
> In the plot, the density value displays over 1. How can I explain it?
>
> library(urca)
> plot(density(Raotbl1[, 1]))
>
>
>
It's a probability density, not a probabilit
As mentioned by somebody before, there is no problem for the normal case - use
mvrnorm function from MASS package with any mu and make Sigma be any diagonal
matrix (with strictly positive diagonal). Note that even though all the
correlations are 0, the SAMPLE correlations won't be 0. If you wan
Dear R users,
Thanks in advance.
I am using R 2.9.1 on Windows XP.
In the plot, the density value displays over 1. How can I explain it?
library(urca)
plot(density(Raotbl1[, 1]))
Once again, thank you very much for the time you have given.
I am looking forward for your reply.
Regar
I guess this is going to be hard to debug w/o knowing more about your
data.
How big is your dataset?
* How many observations?
* How many predictors/features per observation?
How many classes are in the data?
Can you make a small reproducible example?
Does your now cleaned data work w/ e1071
I got the data working, but now I got another problem with KSVM:
line search fails -2.793708 -0.5831701 1.870406e-05 -5.728611e-06
-5.059796e-08 -3.761822e-08 -7.308871e-13Error in
prob.model(object)[[p]]$A :
$ operator is invalid for atomic vectors
On Tue, Jul 7, 2009 at 6:45 PM, Steve
Lian
On Tue, Jul 7, 2009 at 10:07 PM, Mark Kimpel wrote:
> Switching to Sun definitely did not help, still no build with rJava,
> below is the output of R CMD javareconf.
You didn't switch. The paths /usr/lib/gcj-4.3.-90 etc. - any path
containing 'gcj' - indicates you're using GNU's Java.
It seems t
Switching to Sun definitely did not help, still no build with rJava,
below is the output of R CMD javareconf.
mkimpel-m90 /home/mkimpel/bin# ./R CMD javareconf
*** JAVA_HOME is not a valid path, ignoring
Java interpreter : /usr/bin/java
Java version : 1.6.0_14
Java home path : /usr/lib/jvm/
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]), x=myt
Hi,
On Jul 7, 2009, at 6:44 PM, Michael wrote:
What's wrong? Very sad about this...
model <- ksvm(x=mytraindata[, -1], y=factor(mytraindata[, 1]),
prob.model=T)
Error in .local(x, ...) : x and y don't match.
Same problem you're having with the svm in e1071: I bet your data is
wonky.
s
This is just a guess: looks like you have GNU's Java version in your
path (aka "gcj").
Perhaps rJava relies on Sun's Java version.
If so, install Sun's Java first. apt-get install sun-java6-jdk might do it.
- Godmar
On Tue, Jul 7, 2009 at 9:28 PM, Mark Kimpel wrote:
> Having difficulties getti
Hi Tao,
On Jul 7, 2009, at 8:33 PM, Tao Shi wrote:
Hi list,
Could someone help me to explain why the leave-one-out cross
validation results I got from svm using the internal option "cross"
are different from those I got manually? It seems using "cross" to
do cross validation, the result
Having difficulties getting rJava to install on my Debian Squeeze box.
Perused the R-help list and tried some things that have worked for
others but not for me. Below is the output of my attempted build, R
CMD javareconf -e, and sessionInfo(). Note I tried the R CMD
javareconf also as root, restart
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 wrote:
> Unlike Splus, R does not use T for TRUE.
>
> On Tue, Jul 7, 2009 at 6:05 PM, Michael wrote:
>> Hi all,
>>
>> I've got the fo
Hi Rene
Can you tell us the version of the Rstem package you installed.
Rstem_0.3-1 from
http://www.omegahat.org/Rstem/
or
install.packages("Rstem", repos = "http://www.omegahat.org/R";)
work fine for me.
I seem to recall this being a problem with an older version of Rstem.
Hi list,
Could someone help me to explain why the leave-one-out cross validation results
I got from svm using the internal option "cross" are different from those I got
manually? It seems using "cross" to do cross validation, the results are
always better. Please see the code below. I also
Hi Antje,
Are your measurements taken every minute (i.e. 30 minutes correspond to 30
consecutive values)?
How fast is your transition?
If you had 30 minures of upper temperature, then 1000 minutes of room
temperature and then 30 minutes of lower temperature - would you count this as
a cycle?
C
Unlike Splus, R does not use T for TRUE.
On Tue, Jul 7, 2009 at 6:05 PM, Michael wrote:
> Hi all,
>
> I've got the following error message in using e1071 svm routine...
>
> Could anybody please help me?
>
> Thank you!
>
> -
> model <- svm(y=factor(mytraindata[, 1]),
Your excel calculation does not correspond to your rollapply call.
rollapply(x, k, f) should have length(x) - k + 1 elements so in
this case it should have 37 - 20 + 1 = 18.
To take a simpler example,
> rollapply(zoo(1:5), 3, sum)
2 3 4
6 9 12
This has 5 - 3 + 1 = 3 elements and the result
On 07-Jul-09 21:59:41, Hans W Borchers wrote:
> Dear List:
> An e-mail mentioning the r-project.org address and sent to a friend
> at a German university was considered spam by the local spam filter.
>
> Its reasoning: the URL "r-project.org" is blacklisted at
> uribl.swinog.ch resp.
> at antispam
Does something like this work for you; it uses the reshape package:
> X<-data.frame(A=1:10, B=0, C=1, Ob1=1:10, Ob2=2:11, Ob3=3:12,
+ Ob4=4:13, Ob5=3:12, Ob6=2:11)
> Y<-data.frame(A=1:20, B=0, C=1, D=5, Ob1=1:10, Ob2=2:11, Ob3=3:12,
+ Ob4=4:13, Ob5=3:12, Ob6=2:11, Ob7=5:9)
> Z<-data.frame(A=1:30,
Many thanks to all who responded, and especially for the promptness of your
answers. The common thread in your solutions is the use of rle, a very
useful function of which I was ignorant.
--Krishna
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Assuming you are referring to ESRI shapefiles, options can be found in
the following packages: maptools, rgdal, and shapefiles
See the Spatial Task View for more info:
http://cran.r-project.org/web/views/Spatial.html
hth,
Kingsford
On Tue, Jul 7, 2009 at 12:28 PM, Sunny wrote:
> I am new with R
Hi,
I have been trying to re-order several items in a trellised barchart
display in lattice, but can't seem to figure it out.
###sample code, Stage and Colony have 2 and 3 levels respectively.
barchart(Activity ~ Percent | Stage + Colony, data = Percent.df,
horizontal = TRUE, layout = c(2,3),
What's wrong? Very sad about this...
model <- ksvm(x=mytraindata[, -1], y=factor(mytraindata[, 1]), prob.model=T)
Error in .local(x, ...) : x and y don't match.
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PLEASE do
Why not directly generate a large PNG file (which will be much better
for line art than JPG anyway)? Or EMF?
See http://wiki.r-project.org/rwiki/doku.php?id=tips:graphics-misc:export
[Of course, this doesn't answer the original question ... to which I
suspect the answer is "no".]
Mark
--- On Tue, 7/7/09, Duncan Murdoch wrote:
> From: Duncan Murdoch
> Subject: Re: [R] r-project.org address blacklisted by anti-spam software
> To: "Hans W Borchers"
> Cc: r-h...@stat.math.ethz.ch
> Received: Tuesday, July 7, 2009, 6:15 PM
> On 07/07/2009 5:59 PM, Hans W
> Borchers wrote:
> >
On Tue, Jul 7, 2009 at 3:15 PM, Duncan Murdoch wrote:
> On 07/07/2009 5:59 PM, Hans W Borchers wrote:
>>
>> Dear List:
>>
>> An e-mail mentioning the r-project.org address and sent to a friend at a
>> German
>> university was considered spam by the local spam filter.
>>
>> Its reasoning: the URL "r
Hello, R users.
I would like to display the font of Math Mode of MikTex 2.3, WinEdt 5.4
in R plots, e.g. in xlab, ylab or legend.
How can I do that?
Thank you in advance.
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On 07/07/2009 5:59 PM, Hans W Borchers wrote:
Dear List:
An e-mail mentioning the r-project.org address and sent to a friend at a German
university was considered spam by the local spam filter.
Its reasoning: the URL "r-project.org" is blacklisted at uribl.swinog.ch resp.
at antispam.imp.ch. I
Hi,
I think, I can answer my own posting. I found out, that the directory
structure
of the ODT-file created by "odfWeave" causes the error message.
The file structure of the unzipped odt-file looks like this:
Pictures
content_1-Boxplot.png
content.xml
current.xml
manifest.xml
meta.xml
My sincere apologies. This message was intended for the rpy2 mailing list.
Thanks.
Uwe Ligges wrote:
> Which previous message? Should we all look up the archives now? Please
> cite and stay within the same thread.
>
> Thank you,
> Uwe LIgges
>
>
>
>
>
> Chuck White wrote:
> > After
Hi all,
I've got the following error message in using e1071 svm routine...
Could anybody please help me?
Thank you!
-
model <- svm(y=factor(mytraindata[, 1]), x=mytraindata[, -1], probability=T)
Error in if (any(co)) { : missing value where TRUE/FALSE needed
In a
Dear All,
I just updated from Fedora 9 to Fedora 11, kernel version
2.6.29.5-191.fc11.i586. I'm running R 2.9.
I successfully installed package Rstem from source (it always ran fine
for me in F9). However:
> wordStem(c("This","is","a","test"))
Error in wordStem(c("This", "is", "a", "test")) :
Dear List:
An e-mail mentioning the r-project.org address and sent to a friend at a German
university was considered spam by the local spam filter.
Its reasoning: the URL "r-project.org" is blacklisted at uribl.swinog.ch resp.
at antispam.imp.ch. I checked the list
http://antispam.imp.ch/swi
On Jul 7, 2009, at 4:08 PM, Krishna Tateneni wrote:
Greetings, I have a vector of the form:
[10,8,1,3,0,8,NA,NA,NA,NA,2,1,6,NA,NA,NA,0,5,1,9...] That is, a
combination
of sequences of non-missing values and missing values, with each
sequence
possibly of a different length.
I'd like to cr
Here's one possibility:
vv <- c(10,8,1,3,0,8,NA,NA,NA,NA,2,1,6,NA,NA,NA,0,5,1,9)
> (1+cumsum(diff(is.na(c(vv[1],vv)))==1)) * !is.na(vv)
[1] 1 1 1 1 1 1 0 0 0 0 2 2 2 0 0 0 3 3 3 3
On Tue, Jul 7, 2009 at 5:08 PM, Krishna Tateneni wrote:
> Greetings, I have a vector of the form:
> [10,8,1,3,0,
Hi,
I'm running wine-1.0.1, OpenBUGS 3.0.3, R 2.9.0, and R2WinBUGS on a Redhat
Enterprise Linux machine.
Following various peoples' suggestions...
This works perfectly (yay!): wine Z:/opt/OpenBUGS/winbugs.exe
Within R, however, I get this:
(setup the example from ?bugs, then)
R> schools.s
Dear Krishna,
Here is one way. It is not very elegant, but seems to work:
# x is the vector you want to change
foo <- function(x){
R1 <- rle(!is.na(x))
R2 <- rle(is.na(x))
len <- R1$lengths[!R2$values]
x[!is.na(x)] <- rep(1:length(len), len)
x
}
# Example
x <- c(10, 8, 1, 3, 0, 8
Here is a couple of very simple data.frames:
X<-data.frame(A=1:10, B=0, C=1, Ob1=1:10, Ob2=2:11, Ob3=3:12,
Ob4=4:13, Ob5=3:12, Ob6=2:11)
Y<-data.frame(A=1:20, B=0, C=1, D=5, Ob1=1:10, Ob2=2:11, Ob3=3:12,
Ob4=4:13, Ob5=3:12, Ob6=2:11, Ob7=5:9)
Z<-data.frame(A=1:30, B=0, C=1, D=6, E=1:2, Ob1=1:10, O
sum(!is.na(x))
> Date: Tue, 07 Jul 2009 14:56:54 -0400
> From: Godmar Back
> Sender: r-help-boun...@r-project.org
> Precedence: list
> DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=gamma;
> DomainKey-Signature: a=rsa-sha1; c=nofws; d=gmail.com; s=gamma;
>
> Hi,
>
> is th
Which previous message? Should we all look up the archives now? Please
cite and stay within the same thread.
Thank you,
Uwe LIgges
Chuck White wrote:
After I posted the previous message, I repeated the process on a windows
machine with Python 2.6 and still get the same error.
I would app
I generate PDF images and then rasterise using imagemagick to large,
high quality JPG files. Then manually insert into powerpoint. Former
two can definitely be automated, I'm sure the latter insertion could
be automated with judicious use of scripting if really necessary.
2009/7/7 Thomas :
> Hi,
>
I have to admit that I have no idea what we are talking about here (yes,
I tend to forget many things these days) - and you have not cited the
original message, unfortunately (nor have you specifies R versions,
Tinn-R versions and both OS versions, but just one) ...
Best wishes,
Uwe
Knut K
Greetings, I have a vector of the form:
[10,8,1,3,0,8,NA,NA,NA,NA,2,1,6,NA,NA,NA,0,5,1,9...] That is, a combination
of sequences of non-missing values and missing values, with each sequence
possibly of a different length.
I'd like to create another vector which will help me pick out the sequences
Hi
You have encountered the fact that, while there are "[" methods for grid
unit objects, there are NOT any "[<-" methods for grid unit objects.
I guess that needs to go (back) onto my todo list.
A workaround for your simple example is ...
unit.c(testUnits[1:2], testUnit2, testUnits[4:5])
.
Hi,
Is it possible to dump a series of plots directly into a powerpoint
presentation (as is possible in Splus)?
Thank you,
Thomas
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After I posted the previous message, I repeated the process on a windows
machine with Python 2.6 and still get the same error.
I would appreciate any help.
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PLEASE do read
Here is one way:
> x <- rnorm(100)
> y <- rnorm(100)
> z <- residuals(lm(y ~ x))
> cor(x, z)
[1] 3.610290e-17
Best,
Giovanni
> Date: Tue, 07 Jul 2009 16:26:02 +0200
> From: "Stein, Luba (AIM SE)"
> Sender: r-help-boun...@r-project.org
> Accept-Language: en-US, de-DE
> Precedence: list
> Thread
subset(), like many R methods, has an argument list that
ends with '...', meaning that it will not tell you that an argument
you gave it by name= is not in the official list of arument names.
If the ... were not there then you would have gotten an error
message. E.g., the following makes a version
It is not clear to me whether you are talking about a covariance (a
theoretical quantity depending on the distribution of A and x) or an
empirical covariance, estimated from some data.
In the first case you don't need to solve anything because, as long as
A is fixed, i.e. non-random, its covaria
?ifelse
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Steve Jaffe
Sent: Tuesday, July 07, 2009 12:42 PM
To: r-help@r-project.org
Subject: [R] vectorizing a function
I'm sure I'm mi
Dear Colleagues,
I have faced with the problem that function rollaply with rolling window for
calculation of volatility doesn't give the all results of calculations.
I have run the rolling window for calculation in Excel and obtained that the
number of outputs for Excel is 36 and for R is 18. The
For this case it is quite simple, see ?ifelse
> z <- ifelse( x > 0, y, -y )
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun.
how about sum(!is.na(x)) ?
On Tue, Jul 7, 2009 at 2:56 PM, Godmar Back wrote:
> Hi,
>
> is there a simpler way to count the number of elements in a vector
> that are not NA than this:
>
> countN <- function (v) {
> return (Reduce(function (x, y) x + y, ifelse(is.na(v), 0, 1)))
> }
>
> ?
>
> -
Nevermind, indeed it is obvious: Vectorize !
Steve Jaffe wrote:
>
> I'm sure I'm missing something obvious but I'm not seeing how to simply
> "vectorize" a function of two or more variables.
>
--
View this message in context:
http://www.nabble.com/vectorizing-a-function-tp24380064p24380136
I'm sure I'm missing something obvious but I'm not seeing how to simply
"vectorize" a function of two or more variables.
Say I have
f <- function(x,y) if (x>0) y else -y
Now I have vectors x and y of equal length and I'd like to apply f
element-wise. I.e. conceptually
z <- f(x,y) where x, y, z
Dear njhuang86,
Here is one way:
x <- c('a', 't', 'c', 'y', 'g')
y <- c('a', 'a', 'g', 's')
table(factor(y, levels = x))
# a t c y g
# 2 0 0 0 1
HTH,
Jorge
On Tue, Jul 7, 2009 at 3:28 PM, njhuang86 wrote:
>
> Hi all,
> Suppose I have x = c('a', 't', 'c', 'y', 'g')
> and also y = c('a', 'a',
On Tue, Jul 7, 2009 at 12:17 PM, Henrique Dallazuanna wrote:
> Try this:
>
> MyResults.GroupA <- subset(MyResults, PosType == 1)
>
>
Darn those small screen fonts. I never noticed that! Every example I'm
looking at jsut looks like a single '=' until you pointed it out!
Thanks to everyone who res
Thank you for the many replies! This is really a very friendly and
helpful community!
On Tue, Jul 7, 2009 at 3:06 PM, Henrique Dallazuanna wrote:
> another option should be:
>
> length(na.omit(v))
>
I think the above is what I was looking for since, presumably, it uses
the very same test as other
On 07-Jul-09 19:06:59, Mark Knecht wrote:
> Hi,
>I am apparently not understanding some nuance about either the use
> of subset or more likely my ability to test for a numerical match
> using '='. Which is it? Thanks in advance.
It looks as though you have tripped over the distinction between
Hi all,
Suppose I have x = c('a', 't', 'c', 'y', 'g')
and also y = c('a', 'a', 'g', 's')
If I do something like x%in%y, I obtain a vector like this: [TRUE, FALSE,
FALSE, FALSE, TRUE] which I can easily turn into this: [1, 0, 0, 0, 1].
I was wondering is there anyway for me to get a vector back in
On Tuesday, July 07, 2009 3:20 PM, Godmar Back wrote:
> ...That would be wrong, wouldn't it, if the
> other replies are correct
Yes. It was wrong. This isn't:
countN <- function ( v ) {
length ( v ) - sum ( is.na ( v ) )
}
But there are really tons of ways to do it. Even your
On 7/7/2009 3:06 PM, Mark Knecht wrote:
Hi,
I am apparently not understanding some nuance about either the use
of subset or more likely my ability to test for a numerical match
using '='. Which is it? Thanks in advance.
I've read a data file, reshaped it and then created MyResults by
keepi
Hi Mark,
X = 1
assings the number 1 to X whereas
X == 1
test if X is equal to 1. I suspect you want to do this :-) Here is an
example:
# R code
X = 1
X
# [1] 1
X == 1
# [1] TRUE
HTH,
Jorge
On Tue, Jul 7, 2009 at 3:06 PM, Mark Knecht wrote:
> Hi,
> I am apparently not understanding s
Try this:
MyResults.GroupA <- subset(MyResults, PosType == 1)
On Tue, Jul 7, 2009 at 4:06 PM, Mark Knecht wrote:
> Hi,
> I am apparently not understanding some nuance about either the use
> of subset or more likely my ability to test for a numerical match
> using '='. Which is it? Thanks in
How about
countN <- function ( v ) {
sum ( !is.na ( v ) ) - sum ( is.na ( v ) )
}
--
David
-
David Huffer, Ph.D. Senior Statistician
CSOSA/Washington, DC david.huf...@csosa.gov
--
Hi,
I am apparently not understanding some nuance about either the use
of subset or more likely my ability to test for a numerical match
using '='. Which is it? Thanks in advance.
I've read a data file, reshaped it and then created MyResults by
keeping only lines where the value column is gr
another option should be:
length(na.omit(v))
On Tue, Jul 7, 2009 at 3:56 PM, Godmar Back wrote:
> Hi,
>
> is there a simpler way to count the number of elements in a vector
> that are not NA than this:
>
> countN <- function (v) {
>return (Reduce(function (x, y) x + y, ifelse(is.na(v), 0, 1
Dear Godmar,
Yes. One way would be
sum( !is.na( yourvector ) )
HTH,
Jorge
On Tue, Jul 7, 2009 at 2:56 PM, Godmar Back wrote:
> Hi,
>
> is there a simpler way to count the number of elements in a vector
> that are not NA than this:
>
> countN <- function (v) {
>return (Reduce(function (x
Hi,
is there a simpler way to count the number of elements in a vector
that are not NA than this:
countN <- function (v) {
return (Reduce(function (x, y) x + y, ifelse(is.na(v), 0, 1)))
}
?
- Godmar
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Dear Andriy:
1. The help file for "fit.NH" says the first argument should be
a "vector of data". Consider the following modification of the example
on that help page:
> rs <- matrix(rseries, 10, 202, dimnames=list(letters[1:10], 1:202))
> rs. <- fit.NH(rs)
Error: cannot allocate
Uwe Ligges schrieb:
Maybe, but the may depend on your "script", your OS, your R version,
used packages and so on.
Hence please read and follow the posting guide and provide commented,
minimal, self-contained, reproducible code.
Hi Uwe,
I was just asked the same question and hat the same
I am new with R and want do some analysis with a point vector data file. Any
help is appreciate. Sunny
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PLEASE do read the pos
Hi spime,
What is "x"? Did you have any other "X" defined in your R-session? Be aware
that R is case-sensitive.
Best,
Jorge
On Tue, Jul 7, 2009 at 1:30 PM, spime wrote:
>
> Hello,
>
> Consider this function for generalized ridge regression:
>
> gre <- function (X,y,D){
>n <- dim(X)[1]
Incomplete code leaves us able to do naught but guess;
Perhaps you are unaware that x != X
--
DW
On Jul 7, 2009, at 1:30 PM, spime wrote:
Hello,
Consider this function for generalized ridge regression:
gre <- function (X,y,D){
n <- dim(X)[1]
p <- dim(X)[2]
interce
I think that is because X in your function has (n + 1) columns and D with
only n coluimns:
matrix(1:9, ncol = 3) + matrix(1:8, ncol = 2)
On Tue, Jul 7, 2009 at 2:30 PM, spime wrote:
>
> Hello,
>
> Consider this function for generalized ridge regression:
>
> gre <- function (X,y,D){
>n <
Hello,
Consider this function for generalized ridge regression:
gre <- function (X,y,D){
n <- dim(X)[1]
p <- dim(X)[2]
intercept <- rep(1, n)
X <- cbind(intercept, X)
X2D <- crossprod(X,X)+ D
Xy <- crossprod(X,y)
bth <- qr.solve(
On Tue, Jul 7, 2009 at 1:22 PM, Godmar Back wrote:
>
>> apply(AA, c(2,1), cat, "\n")
> Error in cat(list(...), file, sep, fill, labels, append) :
> argument 1 (type 'list') cannot be handled by 'cat'
>
ps: but your idea of using 'apply' led me to this:
> dummy <- apply(AA, c(2), function (r) {
On Tue, Jul 7, 2009 at 12:20 PM, David Winsemius wrote:
> It looks like you are printing a matrix and that you want to print all rows
> of the first column before all rows of the second column. Apply should do
> it. Assume the matrix is named "AA"
>
> apply(AA, c(2,1), cat, "\n") # the \n is the
Thanks, Peter.
You're right, I mistyped and getOption('na.action') shows na.omit.
Perhaps my question was more commentary about my perceived lack of
rationale and orthogonality in R than it should have been. Presumably,
q[[i]] is a data frame and q[[i]][,1] is a numeric vector, so cor and
cor.tes
Hi,
I am confused about how to select elements from a list.
I'm trying to select all rows of a table 'crossRsorted' such that the
mean of a related vector is > 0. The related vector is accessible as
a list element l[[i]] where i is the row index.
I thought this would work:
> crossRsorted[mean(
Hi all,
I just wanted to send a general word of thanks to the list for
making my first week using R successful (by my measures) and
reasonably pleasurable. (Not a single literal RTFM!) ;-)
I appreciate all the help I've received from folks. I have a long
way to go but I'm starting to get dat
It looks like you are printing a matrix and that you want to print all
rows of the first column before all rows of the second column. Apply
should do it. Assume the matrix is named "AA"
apply(AA, c(2,1), cat, "\n") # the \n is the line-feed character
--
DW
On Jul 7, 2009, at 10:06 AM, Go
Here are some examples that may get you started (note that there is no
guarantee that a variable follows a given distribution after it has been
adjusted to have 0 correlation with another variable):
library(MASS)
tmp <- mvrnorm(25, c(0,0), diag(2), empirical=TRUE)
zapsmall(cor(tmp))
tmp2 <- exp
Hi Albart,
This bugged me also for quite some time. After some experiments the
following syntax worked best:
library(lattice)
a = 0.11
xyplot(1:10~10:11, xlab = as.expression(bquote(R^2~" equals "~.(a
With the combination of as.expression and bquote you can mix text, math
expression and
On Tue, 7 Jul 2009, Stein, Luba (AIM SE) wrote:
Hi,
more precisely I consider a matrix with three column vectors a_i (i=1,2,3),
i.e. A=(a_1,a_2,a_3). On the other hand x should take vectors as values, i.e.
x=v_j, while j goes also from 1 till 3.
Now I just want to calculate the equation Cov(a
Be careful to be clear what you are referring to when you say
"correlation is zero".
The commands
x <- rnorm(100)
y <- rnorm(100)
will produce two vectors of given length (100) which (to within the
effectively ignorable limitations of the ransom number generator)
will have been produced indepe
?cor says that cor() can be applied to
'numeric vector, matrix or data frame'
?cor.test requires
'numeric vectors of data values'
So, what's your q?
As to na.action:
?cor.test makes no reference to na.action for the default method.
Looking at the code of cor.test.default shows that only compl
How many rows does xx have?
Let's look at your example for chunksize 1, you initially fit the first
1 observations, then the seq results in just the value 1 which means
that you do the update based on vaues 10001 through 2, if xx only has 1
rows, then this should give at lea
You could use the mvtnorm package to generate correlated vectors of random
normal deviates where the nominal correlation is 0.
library(mvtnorm)
rho <- 0.0
Cor <- array(c(1, rho, rho, 1), dim=c(2,2))
Y <- rmvnorm(1000, sigma=Cor)
plot(Y)
cor(Y)
cor.test(Y[,1],Y[,2])
Any given random set will hav
Thank you for your help!
But is it possible to produe two vectors x and y with a given length such that
there correlation is zero.
For me ist not enough just to simulate two vectors with there correlation.
Thank you,
Luba
-Urspr?ngliche Nachricht-
Von: ONKELINX, Thierry [mailto:thie
Hi,
I am trying to use R for some survey analysis, and need to compute the
significance of some correlations. I read the man pages for cor and
cor.test, but I am confused about
- whether these functions are intended to work the same way
- about how these functions handle NA values
- whether cor.t
Thank you for your help!
But is it possible to produe two vectors x and y with a given length such that
there correlation is zero.
For me ist not enough just to simulate two vectors with there correlation.
Thank you,
Luba
-Urspr?ngliche Nachricht-
Von: r-help-boun...@r-project.org
Hey Whit,
That worked! I was able to consume all the memory on the server!
Thanks!
-scz
Whit Armstrong wrote:
> Seems strange. I can go all the way up to 50GB on our machine which
> has 64GB as well. It starts swapping after that, so I killed the
> process.
>
> try this:
> ans <- list()
>
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