Hi!
Does anyone know a R package for multivariate (not univariate!) shape
preserving interpolating splines. For example spline must preserve
monotonocity in both directions, etc...
Tine Mlač
__
R-help@r-project.org mailing list
https://stat.ethz.ch
I think what you are asking for is not a tilde. That is a raised symbol
(an accent), and not as in TeX's $\sim$ . It is character 126 in the
Adobe Symbol encoding (Adobe's name is '\similar'), so one way is
expression(X*symbol("\176")*N(mu, sigma^2))
There are others, and in most circumstanc
Here's an added caveat, with subsequently a more detailed explanation of the
output desired:
The data this will apply to includes a variety of whole numbers not limited
to 1 & 0, a number of which may appear consecutively and not separated by
zeros!
e.g. x<-c(3,2,0,1,0,2,0,0,1,0,0,0,0,4,1)
ans
Here's an added caveat, with subsequently a more detailed explanation of the
output desired:
The data this will apply to includes a variety of whole numbers not limited
to 1 & 0, a number of which may appear consecutively and not separated by
zeros!
e.g. x<-c(3,2,0,1,0,2,0,0,1,0,0,0,0,4,1)
answ
On 5/29/08, Wen Huang <[EMAIL PROTECTED]> wrote:
> Hello,
>
> I have been trying to make a graph that have error bars and text at
> specific position.
>
> I used the following code from the help file of xYplot(Hmisc) as an example
> except I add a myPanel function, which is just supposed to add l
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]==0) NA
2. if (x[i]>0) log(x[i]/(number of consecutive zeros immediately preceding
it +1))
x<-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, N
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]==0) NA
2. if (x[i]>0) log(x[i]/(number of consecutive zeros preceding it +1))
x<-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA, -
Cleber Nogueira Borges wrote:
my linear model is y=c+a*x1+b*x2 i tried to found a, c, b by the
use of:
mymodel<-lm(y~1+x1+x2) where y, x1, x2 are 3 vectors with the same
length
the result is a=NA.so i want to know where is the problem.
It's the sum of x1 and x2 equal to one?
run the command:
On May 29, 2008, at 11:54 PM, Redding, Matthew wrote:
Dear R Gurus,
I am having a little difficulty with nlm. I've searched the
archives and
found nothing that tells me why this is occuring -- though there are
some slightly similar issues.
A simple example:
lev2<-function(aaa,bbb,ccc,ddd,e
Thanks, Dieter,
but as far as I understand, 'glht' does not support objects of class
'aovlist' either. I mean, I know there is a "TukeyHSD" function out
there, but that's the problem: repeated measures ANOVA yields an
aovlist object, and TukeyHSD calls for an aov object.
And I don't know if
Dear R Gurus,
I am having a little difficulty with nlm. I've searched the archives and
found nothing that tells me why this is occuring -- though there are
some slightly similar issues.
A simple example:
lev2<-function(aaa,bbb,ccc,ddd,eee){
res<-aaa+bbb+ccc+ddd+eee
res
}
nlm(l
Hello,
I have been trying to make a graph that have error bars and text at
specific position.
I used the following code from the help file of xYplot(Hmisc) as an
example except I add a myPanel function, which is just supposed to add
letters from the alphabet at the position aligned at y =
I'm stumbling my way through manipulating data in multiply imputed datasets,
and have run into a problem translating code I used to run on my pre-imputed
dataset to multiple datasets. The imputation runs just fine, as does the
reading of the mi data sets into an imputationList. I run into trouble
On May 29, 2008, at 11:02 PM, [EMAIL PROTECTED] wrote:
Suppose I have a plot
plot(1:10, pch = "")
And I want some text to indicate a Normal distrubition. I could do
this:
text(5, 6, substitute(XN(mu, sigma^2)), adj = 0)
text(5.35, 6, "~", adj = 0)
But that's clumsy, and depending on you
Suppose I have a plot
plot(1:10, pch = "")
And I want some text to indicate a Normal distrubition. I could do
this:
text(5, 6, substitute(XN(mu, sigma^2)), adj = 0)
text(5.35, 6, "~", adj = 0)
But that's clumsy, and depending on your plotting device, might not even look
sensible. I'd prefe
On May 29, 2008, at 9:56 PM, lek2k wrote:
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
I (and certainly many others) have been using multiple points calls
for a while now with no problems at
Yes I know the problem can be solved analytically -- the point was to see how
well R could simulate the theory.
In any case, your assistance is greatly appreciated and it worked well.
Thankyou. R does a good job of simulating the experiment!
--
View this message in context:
http://www.nabble.c
Hi
I'm trying to scatter plot 3 data sets on the one graph.
I plot the first data set by calling:
plot
and the next two data sets by calling
points
points
and coloring them different colors to distinguish between the data sets
I've also set the xlim and ylim values appropiately
The problem is
Yes, here are 3 functions that can do it
grid::grid.rect
RGraphics::grid.imageFun
lattice::panel.levelplot
On Thu, May 29, 2008 at 10:48 PM, Wittner, Ben, Ph.D.
<[EMAIL PROTECTED]> wrote:
> Is there a way to display an image (such as is done with the function image())
> in a grid package viewpo
Hi Jenny,
A simple solution is to add your line to the function, re-load/source
the modified function to the console (e.g. by copy and paste). Then the
new function with the same name as the original one will be called next
time. If you don't want to use modified function any longer, just use
All:
I am new to R and would like your help in identifying the appropriate
process to follow in order to modify the output from an existing
package. I've had difficulty finding an answer online, perhaps because
I am using incorrect terminology.
A package that I am using (mmlcr) invokes an
I have been attempting to do some work using hclust, and have run
into a (possibly subtle) problem.
The background is that I constructed a dissimilarity matrix ``d1''
(it involved something called the ``Jaccard similarity coefficient'';
I won't go
into the details unless requested). I then
On 5/29/2008 7:48 PM, Ted Harding wrote:
Hi Folks,
I need help with a query about R on Windows, specifically
about graphics devices.
I'm advising someone remotely (so it's all by email) who
is running R on Windows, while I am not (Linux only).
Things have reached the stage where saving graphics
I need to estimate maximum tree crown radius and am looking for a package to
prepare stochastic frontier models in R. I have not found any package
references on Nabble R help, google, or R help. Any tips on a package for
this?
With regards,
Aaron Trowbridge
Researcher
BV Research Centre
Smith
Hi Mike,
if you can decompose the bimodal distribution into (eg two) known
forms, then you could try a stepwise approach, eg:
If uniform < 0.5 then double it and use it to draw from the inverse
cdf of A,
else, double (uniform - 0.5) and use it to draw from the inverse cdf of B.
You can change
Hi Folks,
I need help with a query about R on Windows, specifically
about graphics devices.
I'm advising someone remotely (so it's all by email) who
is running R on Windows, while I am not (Linux only).
Things have reached the stage where saving graphics plots
as Windows metafiles is looming.
I'
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]>0) log(x[i]/(number of consecutive zeros immediately preceding
it +1))
2. if (x[i]==0) NA
x<-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA
my linear model is y=c+a*x1+b*x2 i tried to found a, c, b by the use
of:
mymodel<-lm(y~1+x1+x2) where y, x1, x2 are 3 vectors with the same
length
the result is a=NA.so i want to know where is the problem.
It's the sum of x1 and x2 equal to one?
run the command: round( sum( c(x1,x2) ),12)==1
I'm trying to build on Jim's approach to change the parameters in the
function, with new rules:
1. if (x[i]==0) NA
2. if (x[i]>0) log(x[i]/(number of consecutive zeros preceding it +1))
x<-c(1,0,1,0,0,1,0,0,0,1,0,0,0,0,1)
# i.e. output desired = c(0, NA, -0.69, NA, NA, -1.098, NA, NA, NA, -1.38,
Hi,
Cleber Nogueira Borges wrote:
hanen wrote:
hi,
my linear model is y=c+a*x1+b*x2 i tried to found a, c, b by the use of:
mymodel<-lm(y~1+x1+x2) where y, x1, x2 are 3 vectors with the same length
the result is a=NA.so i want to know where is the problem.
It's the sum of x1 and x2 equal t
Your problem can be easily solved analytically
yielding that E(A/B=b) = (110 - b*b)/(21 - 2*b) - no
programming needed!
If you insist on writing a program, you could do
something like:
x <- sample(1:10,2000,replace=TRUE)
M <- matrix(x,nrow=2) #each column of M represents
two balls
A <- apply(M,2
This is not really addressing your problem, but I thought you might want to
know that the rtracklayer package in Bioconductor already supports parsing
BED files, as well as GFF and WIG. It's main purpose is to load the tracks
into genome browsers, like UCSC.
Michael
On Wed, May 28, 2008 at 1:11 A
__
R-help@r-project.org 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.
Carlos López wrote:
I´m trying to find datasets that will give me residuals, after applying
the lm function, with no normality, non linearity, and heteroscedacity
so I can try to exemplify
those cases in the linear regression model. Can you give any advice on
what datasets would be appropiate?
On 30/05/2008, at 9:57 AM, Jorge Ivan Velez wrote:
Dear Hanen,
You don't need "1" in your R code. Try this:
# Model
mymodel<-lm(y~x1+x2)
# Coefficients
summary(mymodel)
See also ?lm.
That's not the problem. The ``1'' is redundant but does no harm.
There is presumably something ``wrong''
hanen wrote:
hi,
my linear model is y=c+a*x1+b*x2 i tried to found a, c, b by the use of:
mymodel<-lm(y~1+x1+x2) where y, x1, x2 are 3 vectors with the same length
the result is a=NA.so i want to know where is the problem.
It's the sum of x1 and x2 equal to one?
run the command: round( sum(
Hello R Users,
I am doing a Latin Hypercube type simulation. I have found the
improvedLHS function and have used it to generate a bunch of properly
distributed uniform probabilities. Now I am using functions like qlnorm
to transform that into the appropriately lognormal or triangularly
distri
Dear Hanen,
You don't need "1" in your R code. Try this:
# Model
mymodel<-lm(y~x1+x2)
# Coefficients
summary(mymodel)
See also ?lm.
Thanks,
Jorge
On Thu, May 29, 2008 at 5:15 PM, hanen <[EMAIL PROTECTED]> wrote:
>
> hi,
>
>
> my linear model is y=c+a*x1+b*x2 i tried to found a, c, b by t
hi,
my linear model is y=c+a*x1+b*x2 i tried to found a, c, b by the use of:
mymodel<-lm(y~1+x1+x2) where y, x1, x2 are 3 vectors with the same length
the result is a=NA.so i want to know where is the problem.
--
View this message in context:
http://www.nabble.com/lm%28%29-function-tp1754607
Did you copy-paste that error message?
I get:
> optim(par=c(1,1), f3)
Error in f1(theta, theta1) : object "theta" not found
>
You seem to be confusing your formal and actual parameters.
When you invoke f3 (within optim()), theta is not defined within f3().
Ray
On Fri, 30 May 2008, threshold
Dear Gus,
If the nominal variables are exogenous then you need not use polychoric and
polyserial correlations. (Indeed, if they are polytomous and unordered, then
it would be inappropriate to do so.) Simply use dummy exogenous variables,
as you would in a dummy regression.
You could use a one-si
Neil Gupta wrote:
Hello R-Users,
I am new to R and trying my best however I need help with this simple task.
I have a dataset, YM1207.
X.Symbol Date Time Exchange TickType
ReferenceNumber Price Size
12491 3:YMZ7.EC 12/03/2007 08:32:50 ECB
85985770
Hi Ashish,
I am rather more concerned about whether what you outlined is legitimate
(your question 1 below). If you are looking at children, higher AGE will
be associated with higher TIV, so both variables would essentially
measure the same thing (see Miller & Chapman, Misunderstanding Analysi
Hi,
I think Ray has answered this question in the previous e-mail.
Because optim can only use one single parameter thus you can not have
the parameters: theta, theta1 and x at the same time.
such as:
f1<-function(theta)
{theta[1]+theta[2]}
f2<-function(theta)
{f1(theta)*3}
f3<-function(thet
I´m trying to find datasets that will give me residuals, after applying
the lm function, with no normality, non linearity, and heteroscedacity
so I can try to exemplify
those cases in the linear regression model. Can you give any advice on
what datasets would be appropiate? I can´t use the ones
I totally agree both of you. This is a super place to mature the R.
I learn a lot from this R heaven!
Chunhao
Quoting Esmail Bonakdarian <[EMAIL PROTECTED]>:
Tubin wrote:
In the past few weeks I have had to give myself a crash course in
R, in order
to accomplish some necessary tasks for my
Try:
x <- c(1,2,1,1,6,7,-1,-1,5,-1)
plot(x, col = ifelse(x == -1, "red", "black"), pch = 16)
On Thu, May 29, 2008 at 1:23 PM, uv <[EMAIL PROTECTED]> wrote:
>
> Hi. I am plotting graphs for values ranging between -1 and 10, for example:
> (1,2,1,1,6,7,-1,-1,5,-1)
> I am trying to plot the graphs
Try using stop:
if (nAssetPositions != nAssetPrices) {
stop("Different number of assets! \n\n")
On Thu, May 29, 2008 at 3:23 PM, Bill Cunliffe <[EMAIL PROTECTED]>
wrote:
> For example, based on a certain condition, I may want to exit my code
> early:
>
>
>
>#
You might try to use "on.exit" or "stop"?
# on.exit
if (nAssetPositions != nAssetPrices) {
on.exit(cat("Different number of assets! "\n"))
}
# stop
if (nAssetPositions != nAssetPrices) {
stop("Different number of assets!")
}
You could find these in "S prog
Hi, thanks for your replay the previous post of mine. Let me ask you one more
question similar to the previous one, based on a naive example:
f1<-function(theta, theta1)
{theta[1]+theta[2]+theta1[1]}
f2<-function(theta, x, theta1)
{f1(theta, theta1)*exp(x)*theta1[2]}
function to be optimized wi
It looks like you want to stop the function execution on detecting an
error condition, in which case the appropriate function to call is
stop(), as in
if (nAssetPositions != nAssetPrices)
stop("Different number of assets!")
An alternative, if you don't want to write the error messages for each
Bill Cunliffe wrote:
For example, based on a certain condition, I may want to exit my code early:
# Are there the same number of assets in "prices" and
"positions"?
if (nAssetPositions != nAssetPrices) {
cat("Different number of assets! \n\n"
stop('Different number of assets! \n\n')
X
Bill Cunliffe 写道:
For example, based on a certain condition, I may want to exit my code early:
# Are there the same number of assets in "prices" and
"positions"?
if (nAssetPositions != nAssetPrices) {
Tubin wrote:
In the past few weeks I have had to give myself a crash course in R, in order
to accomplish some necessary tasks for my job. During that time, I've found
this forum to be helpful time and time again - usually I find the answer to
my problem by searching the archives; once or twice I
See
?return
HTH,
Chuck
On Thu, 29 May 2008, Bill Cunliffe wrote:
For example, based on a certain condition, I may want to exit my code early:
# Are there the same number of assets in "prices" and
"positions"?
if (nAssetPositions != nAssetPrices) {
Hi
Bill Cunliffe wrote:
For example, based on a certain condition, I may want to exit my code early:
# Are there the same number of assets in "prices" and
"positions"?
if (nAssetPositions != nAssetPrices) {
cat("Different number of assets! \
For example, based on a certain condition, I may want to exit my code early:
# Are there the same number of assets in "prices" and
"positions"?
if (nAssetPositions != nAssetPrices) {
cat("Different number of assets! \n\n")
thanks.
one more thing w/r this particular data set:
tri.mesh doesn't work in R but triangulation (S+SpatialStats) does work
interp.old (diff ncp) doesn't work in R but it does in Splus for the
same ncp values???
so,R or SPlus? who's right?
On Thu, May 29, 2008 at 4:34 AM, Prof Brian Riple
Yes. You could install "mvnormtest" Package and perform the
multivariate normality test. By using mshapiro.test
I wish this is helpful!
Chunhao Tu
Quoting HongSheng Liao <[EMAIL PROTECTED]>:
My stat textbook tells me that using Shapiro-Wilk test for each variable
one by one is not equal
Patrizio Frederic wrote:
dear Harrell,
thank you for quick reply and suggestions. I still have the problem:
library(Design)
x = rnorm(100)
y = runif(100)<(exp(x)/(1+exp(x)))
y = 0*y+1*y
d = datadist(x,y)
options(datadist="d")
fit = lrm(y~x)
# works fine, but
plot(fit) #produce the error
Hi,
I would like to use the sem package to perform a path analysis (no
latent variables) with a mixture of 2 nominal exogenous, 1 continuous
exogenous, and 4 continuous endogenous variables. I seek advice as to
how to calculate the appropriate covariance matrix for use with the sem
package.
Hi Neil,
as i am not an advanced user,
i find reference cards very handy
(google: reference card R)
hth a bit,
Wim
Message: 70
Date: Wed, 28 May 2008 15:25:36 -0500
From: "Neil Gupta" <[EMAIL PROTECTED]>
Subject: [R] R reference Books
To: R-help@r-project.org
Message-ID:
<[EMAIL PROTEC
My stat textbook tells me that using Shapiro-Wilk test for each variable
one by one is not equal to a test for multivariate normality as a whole.
Does R have a function of testing for multivariate normality? Thanks.
Hongsheng (Hank) Liao, Ph.D.
Lab Manager
Center for Quantitative Fisheries Ecolo
In the past few weeks I have had to give myself a crash course in R, in order
to accomplish some necessary tasks for my job. During that time, I've found
this forum to be helpful time and time again - usually I find the answer to
my problem by searching the archives; once or twice I've posted que
Hello R-Users,
I am new to R and trying my best however I need help with this simple task.
I have a dataset, YM1207.
X.Symbol Date Time Exchange TickType
ReferenceNumber Price Size
12491 3:YMZ7.EC 12/03/2007 08:32:50 ECB
8598577013379
Amarjit Singh Sethi yahoo.co.in> writes:
>
> Dear all
> I wish to carry out sigma- and beta-convergence analysis in respect of
> panel data [wherein current value of
> one of the variables needs be regressed upon suitably transformed lagged
> values of another
> variable(s)]. I am quite new to
Rnabble, search on the CRAN site, and some others
On Thu, May 29, 2008 at 12:03 PM, Axel Etzold <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> I am fairly new to R and this list (this is my first post), so I am
> wondering whether
> there is a possibility to view posts on this list conveniently on a
> I am fairly new to R and this list (this is my first post), so I am
> wondering whether
> there is a possibility to view posts on this list conveniently on a
> website besides
> reading my email.
You can see them on Nabble, but there is a delay of a couple of hours.
http://www.nabble.com/R-hel
> I'm attempting to plot a cubic relationship between two variables
> controlling for the effects of a third variable. In this short
> example, I'm trying to use AGE to predict CORTEX while controlling for
> the effects of TIV (total intracranial volume):
>
>
> cortex =
Hi Axel,
Perhaps http://www.nabble.com/R-f13819.html can be useful.
Thanks,
Jorge
On Thu, May 29, 2008 at 12:03 PM, Axel Etzold <[EMAIL PROTECTED]> wrote:
> Dear all,
>
> I am fairly new to R and this list (this is my first post), so I am
> wondering whether
> there is a possibility to view p
Yes, starting from
http://www.r-project.org/
you can go to the mailing lists, choose the appropriate one, go to the
corresponding archives and read all of it sorted by the most common
criteria.
Best wishes,
uwe
Axel Etzold wrote:
Dear all,
I am fairly new to R and this list (this is my fir
On Thu, 29 May 2008, Axel Etzold wrote:
Dear all,
I am fairly new to R and this list (this is my first post), so I am wondering
whether
there is a possibility to view posts on this list conveniently on a website
besides
reading my email.
See
http://news.gmane.org/gmane.comp.lang.r
Message: 24
Date: Wed, 28 May 2008 05:53:26 -0700 (PDT)
From: Philip Twumasi-Ankrah <[EMAIL PROTECTED]>
Subject: [R] "rbinom" not using probability of success right
To: r-help@r-project.org
Message-ID: <[EMAIL PROTECTED]>
Content-Type: text/plain
I am trying to simulate a series of ones and zeros
Hi. I am plotting graphs for values ranging between -1 and 10, for example:
(1,2,1,1,6,7,-1,-1,5,-1)
I am trying to plot the graphs so that the points with value of -1 will be
in one specific color, and the rest of the points will be in one different
specific color. I would be grateful for any ide
Dear all,
I am fairly new to R and this list (this is my first post), so I am wondering
whether
there is a possibility to view posts on this list conveniently on a website
besides
reading my email.
I have in mind something like this :
http://www.ruby-forum.com/forum/4
for the Ruby programming
If you mean a package ... 'Writing R Extensions' says
Ideally, the @R{} code files should only directly assign @R{} objects
and definitely should not call functions with side effects such as
@code{require} and @code{options}.
Note that you are expecting the side-effect of 'print', which is calle
On Thu, 29 May 2008, Albert Vilella wrote:
Hi,
How can I use a substring match as a condition in a subset command?
Perhaps
subset(input, field1=="blah1" & regexpr("blah2",field3) != -1 )
??
Study in
example(gsub)
the regexpr example
and in
?gsub
the 'Value' se
Perhaps you want to read the files only once, and save those 85 or
whatever versions you want.
For Search&Replace you might want to take a look at ?gsub and friends.
Uwe Ligges
Romain wrote:
Hi there,
I would like to know if it is possible to modify a text file with a R function.
In fact I
Note that ?postscript tells you:
"Note that R does not embed the font(s) used in the PostScript output:
see ?embedFonts for a utility to help do so."
It makes really sense to read both of these help pages.
Uwe Ligges
[EMAIL PROTECTED] wrote:
Hello,
I'm trying to save a graph as a .ps fil
[EMAIL PROTECTED] wrote:
*Thanks* all those who took the time to help me (even if the
"question" was not related to - the use of - R).
Now I think I can soundly make my point w/ the referee (can I use your
replies? If so I intend to properly cite its use?!?).
In general, I think it is best not
What you have sent is almost unreadable.
Please do not send HTML messages (as the posting guide suggests)!
Wen-Ching Lin wrote:
Hi,
I am reading the source code of rpart. I have problems understand the following
code and would appreciate for any helps. In rpart.s, there is a line:
rpfit <- .
yoo wrote:
Hi,
I'm able to create a library with R CMD INSTALL cmd, etc... I'm just
You probably mean you are able to *install* a *package*, I guess.
wondering.. is it possible that when the user says library(boo), it runs
some initialization code?
Yes: See the manual writin
Hi all,
I am using RPART for my genetic study under ANOVA method. I wanted to know
if it is possible to see r-squared or the amount of the variance in the data
explained by a model (or a tree in this case from the RPART package. I am
guessing that there has to be one since I am using ANOVA to est
Dear all,
I'm attempting to plot a cubic relationship between two variables
controlling for the effects of a third variable. In this short
example, I'm trying to use AGE to predict CORTEX while controlling for
the effects of TIV (total intracranial volume):
corte
Hi,
I'm able to create a library with R CMD INSTALL cmd, etc... I'm just
wondering.. is it possible that when the user says library(boo), it runs
some initialization code?
I have a dumb R file that is:
print(2)
boo <- function(x){}
when I R CMD INSTALL the library, I'm able to see 2
I need to repeat an experiment 1000 times. Each experiment involves randomly
selecting one ball each from two separate bags. Each bag contains 10 balls,
numbered 1, 2, 3, ... , 10. So the probability of selecting any one pair of
balls is equal to all others.
For each experiment, what I need to do
Gundala Viswanath gmail.com> writes:
>
> Hi all,
>
> After running this code (attaches is the input file):
>
> dat <- read.table("gene_prob.txt", sep = "\t")
> n <- length(dat$V1)
> print(n)
> print(dat$V1)
>
> I get this print out.
>
> ..
> [8541] LOC552889 GPR15 SLC2A11
Hello all,
I am trying make an graphic of 1ª PC projection, but I not get success
in adaption of code at ref[1].
The code make a 2ª PC projection...{by mistake, I guess...}
Anyone can help me? Thanks in advance.
Cleber
# adaption code :: ref[1]="Statistics with R", cap.09;
http://zoonek
*Thanks* all those who took the time to help me (even if the
"question" was not related to - the use of - R).
Now I think I can soundly make my point w/ the referee (can I use your
replies? If so I intend to properly cite its use?!?).
Regards, Eduardo Esteves
ps - Sorry for not explaining
Neil Gupta wrote:
Hello,
I have quite a simple problem that I believe can be solved quite easily. I
have a dataframe as such:
Symbol Date Time Exchange TickType ReferenceNumber Price Size
1 3:YMZ7.EC 12/03/2007 08:30:00 ECB83916044 133879
2 3:YMZ7.EC 12/03/200
The low R2 says the model does not explain much of the variance.
But the high significance arises from the very large number of degrees
of freedom.
This is not an 'incompatibility'; just what happens with large
dispersion, small effects and a very large number of observations.
But you clearly hav
Hello,
I have quite a simple problem that I believe can be solved quite easily. I
have a dataframe as such:
Symbol Date Time Exchange TickType ReferenceNumber Price Size
1 3:YMZ7.EC 12/03/2007 08:30:00 ECB83916044 133879
2 3:YMZ7.EC 12/03/2007 08:30:00 EC
dear Harrell,
thank you for quick reply and suggestions. I still have the problem:
library(Design)
x = rnorm(100)
y = runif(100)<(exp(x)/(1+exp(x)))
y = 0*y+1*y
d = datadist(x,y)
options(datadist="d")
fit = lrm(y~x)
# works fine, but
plot(fit) #produce the error message
Error in value.chk
Frank, I believe, is correct. Using the AIC/BIC for data-driven model
selection does NOT solve the "stepwise problem". This is because the
distribution of the sample AIC is changed from its original distribution to an
extreme-value distribution, e.g.., min (AIC1, AIC2, ..., AICn). Thus, whatev
Wittner, Ben, Ph.D.:
> Is there a way to display an image (such as is done with the function
> image()) in a grid package viewport?
Yes. Use levelplot() from the lattice package.
--
Karl Ove Hufthammer
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Ben Bolker wrote:
ualg.pt> writes:
Dear All,
I'me having (much) trouble understanding why it happened and answering
a referee's comment to part of a submitted manuscript. I've tried to
google for help but... I'm really confident that although this is a
R-Help list someone can help me
ualg.pt> writes:
>
> Dear All,
> I'me having (much) trouble understanding why it happened and answering
> a referee's comment to part of a submitted manuscript. I've tried to
> google for help but... I'm really confident that although this is a
> R-Help list someone can help me!
>
> I us
Patrizio Frederic wrote:
Dear R-helpers,
I'm having a problem in using plot.design in Design Library. Tho
following example code produce the error:
n <- 1000# define sample size
set.seed(17) # so can reproduce the results
age<- rnorm(n, 50, 10)
blood.pressure <- rnorm(n, 120, 15
Is there a way to display an image (such as is done with the function image())
in a grid package viewport?
Thanks.
-Ben
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Xiaohui Chen wrote:
Frank E Harrell Jr 写道:
Xiaohui Chen wrote:
step or stepAIC functions do the job. You can opt to use BIC by
changing the mulplication of penalty.
I think AIC and BIC are not only limited to compare two pre-defined
models, they can be used as model search criteria. You coul
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