Hi,
Sorry if this is an obvious one, but is there a simple way to modify the lm
function to test whether a slope coefficient is significantly different
from 1 instead of different from 0?
Thanks,
Martin
Martin Biuw
SEA MAMMAL RESEARCH UNIT
Gatty Marine Laboratory
School of Environmental and
Dedi,
There is a very useful package called stable by Philippe Lambert and Jim
Lindsey at
http://alpha.luc.ac.be/~jlindsey/rcode.html
Look down that page to the Probability functions and generalized regression
models for stable distributions identifier. And if you have not read it
already, I
No modification needed. Fit either of
lm(y-x ~ x + z)
lm(y ~ x + z + offset(x))
and the t-test in the summary will be a test of the coefficient of x being
one.
You can also use such models to do an anova against a modle with unit
coefficient.
On Thu, 5 Jun 2003, Martin Biuw wrote:
Sorry
Martin Biuw wrote:
Hi,
Sorry if this is an obvious one, but is there a simple way to modify the lm
function to test whether a slope coefficient is significantly different
from 1 instead of different from 0?
Thanks,
There might be an easier way, but the brute force method would be:
R
Dear all,
Please take my apologies if that has already been asked
- at least I couldn't find it in the archives.
When trying to specify a layout within library lattice,
i.e. using xyplot, I get an error when the prepanel
function tries to subscript the automatically generated
x.limits. This
Hello R-user
I want to compute a multiple regression but I would to include a check for
collinearity of the variables. Therefore I would like to use a ridge
regression.
I tried lm.ridge() but I don't know yet how to get p-values (single Pr() and p
of the whole model) out of this model. Can
Hi Folks,
Consider the following example (artificial, but it illustates the point):
r2-sqrt(2)
x-2-r2*r2
print(c(pi,sqrt(pi)),digits=5)
[1] 3.1416 1.7725
print(c(pi,sqrt(pi),x),digits=5)
[1] 3.1416e+00 1.7725e+00 -4.4409e-16
whereas I would prefer
[1] 3.1416 1.7725
Dear R experts,
On explanation of persp() parameters the last item is:
...: additional graphical parameters (see `par').
However, setting the `tcl' parameter has no any effect.
I guess that axes are added to persp() in somewhat freakish
way, and have nothing in common with axis() function.
On Thu, 5 Jun 2003 15:40:52 +
Wegmann (LIST) [EMAIL PROTECTED] wrote:
Hello R-user
I want to compute a multiple regression but I would to include a check for
collinearity of the variables. Therefore I would like to use a ridge
regression.
I tried lm.ridge() but I don't know yet how
Despite my assertion that I had checked all the lattice theme components, I
was in error. Having made the appropriate changes in print.trellis and
succeded in what I wanted to do, I noticed obvious trellis.par.get calls and
followed them up. In the end what I required did not take much at all.
Is there a simpler way then the solution to the one that was posted here? I'm
not very proficient with legend, and I don't understand this solution. All
I have is two or more lines on one plot that I want to put a legend on and I
can't figure out how to do it from the examples. Can you give
Hello R lovers
I have written a little cute function to count the number of missing value
per row in a matrix and return the percentage of missing value
it takes a lot of time to run with a 1000 rows matrix
I'd like to know if there is a function already implemented to count the
number of
Anna H. Pryor wrote:
Is there a simpler way then the solution to the one that was posted here? I'm
not very proficient with legend, and I don't understand this solution. All
I have is two or more lines on one plot that I want to put a legend on and I
can't figure out how to do it from the
[EMAIL PROTECTED] wrote:
Hello R lovers
I have written a little cute function to count the number of missing value
per row in a matrix and return the percentage of missing value
it takes a lot of time to run with a 1000 rows matrix
I'd like to know if there is a function already implemented to
Hello R lovers
I have written a little cute function to count the number of missing value
per row in a matrix and return the percentage of missing value
R M - matrix(rnorm(25), ncol=5)
R diag(M) - NA
R M[2,3] - NA
R apply(M, 2, function(x) sum(is.na(x)))/ncol(M)
[1] 0.2 0.2 0.4 0.2 0.2
the
Nolwenn Le Meur wrote:
Hi everybody,
just a quick question that drives me crazy:
Is it possible to join 2 logical vectors ?
i.e.
x-4
ind - 1:19200
pin - c(0, rep(400, 48) * (1:48))
ind1-((pin[j] + 1) = ind) (ind = pin[j + 2])
ind2-((pin[j+x] + 1) = ind) (ind = pin[j+x + 2])
ind2 and ind1 give
Dear Vincent
I'd like to know if there is a function already implemented
to count the
number of occurence of a given values in a vector
Does
my.matrix-cbind(c(0,1,NA),c(NA,1,NA))
count.NA-function(the.matrix){
result-t(apply(the.matrix,1,function(x,n){
m-sum(is.na(x))
c(m,m/n)
On Thu, 5 Jun 2003 07:20:11 -0700
Anna H. Pryor [EMAIL PROTECTED] wrote:
Is there a simpler way then the solution to the one that was posted here? I'm
not very proficient with legend, and I don't understand this solution. All
I have is two or more lines on one plot that I want to put a
Hello
I have been using r to classify fish into groups using the lda formula and
predict(object,)$class. I am having problems finding an output that
simultaneously shows me classification function coefficients for predictors
and groups (such as is given in spss under classify:linear discriminant
Hi,
how can i visualize a regression plane in my 3d
scatterplot?
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Dear can y,
At 03:04 PM 6/4/2003 +0300, orkun wrote:
[previous messages deleted]
Dear Mr. Fox
thank you very much all.
Because of related to your answer. I ask you directly if you don't mind
I studied several ways after my email.
I wonder whether pgeo-predict.glm(glm.ob,type=terms)
gives
Dear list,
I would like to study the dynamics of
functions using R (instead of mathematica e.g.),
i.e. the behavior of points under iteration
of a function.
So I tried (in vain) writing a function
myfunction - function(f,n,x){...}
in order to compute f^{n}(x), f^{n}(x) being
the function f
Timur Elzhov wrote:
Dear R experts,
On explanation of persp() parameters the last item is:
...: additional graphical parameters (see `par').
However, setting the `tcl' parameter has no any effect.
I guess that axes are added to persp() in somewhat freakish
way, and have nothing in common
Dear Henric,
At 05:01 PM 6/4/2003 +0200, Henric Nilsson wrote:
I've now spent a couple of days trying to learn R and, in particular, the
gam() function, and I now have a few questions and reflections regarding
the latter. Maybe these things are implemented in some way that I'm not
yet aware
Torsten Hothorn [EMAIL PROTECTED] writes:
Hello R lovers
I have written a little cute function to count the number of missing value
per row in a matrix and return the percentage of missing value
R M - matrix(rnorm(25), ncol=5)
R diag(M) - NA
R M[2,3] - NA
R apply(M, 2, function(x)
Michele Grassi wrote:
Hi,
how can i visualize a regression plane in my 3d
scatterplot?
If you are using scatterplot3d(), see ?scatterplot3d, in particular its
last example, which is explained in the references given on the same
help page.
Uwe Ligges
On Thu, 5 Jun 2003 [EMAIL PROTECTED] wrote:
Hi Folks,
Consider the following example (artificial, but it illustates the point):
r2-sqrt(2)
x-2-r2*r2
print(c(pi,sqrt(pi)),digits=5)
[1] 3.1416 1.7725
print(c(pi,sqrt(pi),x),digits=5)
[1] 3.1416e+00 1.7725e+00
Dear list,
I would like to study the dynamics of
functions using R (instead of mathematica e.g.),
i.e. the behavior of points under iteration
of a function.
So I tried (in vain) writing a function
myfunction - function(f,n,x){...}
in order to compute f^{n}(x), f^{n}(x) being
the function f
Dear Tobias,
I would like to study the dynamics of
functions using R (instead of mathematica e.g.),
i.e. the behavior of points under iteration
of a function.
So I tried (in vain) writing a function
myfunction - function(f,n,x){...}
in order to compute f^{n}(x), f^{n}(x) being
the
Dear Paul,
At 08:41 PM 6/4/2003 +0100, Paul wrote:
Thanks for your reply.
I am using logistic regression because my response variable is categorical
- and this seems to be recommended in the literature (by Heckman, Smith
and others).
I think that Prof. Ripley's point here is that although one
Hotz, T. wrote:
What about the following function?
iterate-function(f,n,x){
if(n==0) return(x)
y-x
for(i in 1:n)y-f(y)
y
}
iterate(sqrt,3,256)
Or the following:
iterFun -
function(f,n,x,...){
if(n==0){return(x)}
for(i in 1:n){
x-f(x,...)
}
return(x)
}
If you
On Thu, 5 Jun 2003, Uwe Ligges wrote:
[EMAIL PROTECTED] wrote:
Hello R lovers
I have written a little cute function to count the number of missing value
per row in a matrix and return the percentage of missing value
it takes a lot of time to run with a 1000 rows matrix
I'd like to
Dear Thomas,
What about the following function?
iterate-function(f,n,x){
if(n==0) return(x)
y-x
for(i in 1:n)y-f(y)
y
}
iterate(sqrt,3,256)
Thank you very much, this certainly helps.
I'm still curious, though, to know how to
write the expression of my function
immediately as
Dear Group:
I am trying to create a model frame from a formula for the model as
follows:
formula - y ~ x1 + x2 + x3
X - model.frame(form=formula,data=mydata)
I have some missing values in some the variables, but I want them to be
included in my model frame and to be indicated as NA. Is there
On Thu, 5 Jun 2003, Tobias Verbeke wrote:
...
What about the following function?
iterate-function(f,n,x){
if(n==0) return(x)
y-x
for(i in 1:n)y-f(y)
y
}
iterate(sqrt,3,256)
Thank you very much, this certainly helps.
I'm still curious, though, to know how to
write
Here is a recursive function.
iterate - function(f, n, x){
if(n==0) return(x)
iterate(f, n-1, f(x))
}
iterate(function(x) 1/(1+x), 10, 1)
Best,
Ravi.
- Original Message -
From: Tobias Verbeke [EMAIL PROTECTED]
Date: Thursday, June 5, 2003 12:04 pm
Subject: Re: [R] dynamics of
On Thu, 5 Jun 2003, Tobias Verbeke wrote:
Dear Thomas,
What about the following function?
iterate-function(f,n,x){
if(n==0) return(x)
y-x
for(i in 1:n)y-f(y)
y
}
iterate(sqrt,3,256)
Thank you very much, this certainly helps.
I'm still curious, though, to know how to
Hello,
I am interested in R as an alternative for a statistical tool at our firm. I
do know RATS an SPSS but not S+. As I read that R is close to S+, I would
like to know if you could recommend me any books as an introduction to S+ or
R.
Best regards
Marc
Is there an easy way to integrate the generalized gamma distribution into
the current R survival package?
--james
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https://www.stat.math.ethz.ch/mailman/listinfo/r-help
At 11:12 2003-06-05 -0400, John Fox wrote:
2. John Fox has modified anova.glm() into anova.gam()
(http://www.socsci.mcmaster.ca/jfox/Books/Companion/nonparametric-regression.txt)
for comparison of two or more fitted models based on the difference
between residual deviances. Indiscriminate use
Regarding a previous question concerning the kmeans function I've tried the
same example and I also get a strange result (at least according to what is
said in the help of the function kmeans). Apparently, the function is
disregarding the initial cluster centers one gives it. According to the
Apart from the wealth of material on http://www.r-project.org/;, my
favorite book on R is Modern Applied Statistics with S, 4th ed., by
Venables and Ripley.
hope this helps. spencer graves
Fohr, Marc [AM] wrote:
Hello,
I am interested in R as an alternative for a statistical tool at our
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Fohr, Marc
[AM]
Sent: Thursday, June 05, 2003 11:46 AM
To: '[EMAIL PROTECTED]'
Subject: [R] (no subject)
Hello,
I am interested in R as an alternative for a statistical tool
at our firm. I
do know RATS
On Thursday 05 June 2003 06:41, Ernesto Jardim wrote:
Hi
I'm doing a xyplot and I wand to reduce the number of tick marks in the
x axis. My x axis are month and I want to reduce the 12 tick marks to 4.
I used the scales argument but it doesn't seem to work, althougth it
works on y axis if I
Could you try with the latest lattice (0.7-13) ? I don't see an error with it.
It's not immediately obvious to me where the problem could have come from,
but there were a bunch of similar problems fixed recently.
On Thursday 05 June 2003 08:21, Hotz, T. wrote:
Dear all,
Please take my
I will appreciate you explain me the reason for sending me this e-mail, and
the purpose of the attached file (aplication.pi).
Best regards
Leon Villan
At 11:11 a.m. 05/06/2003 +0200, you wrote:
Please see the attached file.
__
[EMAIL PROTECTED] mailing
Hello!
I am trying to compute minimal time on some data like this:
mt-tapply(mrsh$time1,list(mrsh$var1,mrsh$var2),min):
a b
145 1054800600 1054789800
340 1054804500 1054794600
349 1054820400 1054792800
55
Thanks to you all for the
clarifications.
I will work on my spacebar usage ;-)
Regards,
Tobias
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character is a mode
POSIXct is a class, but you want c(POSIXt, POSIXct)
Try
mt - structure(mt, class=c(POSIXt, POSIXct))
or
mt + ISOdatetime(1970,1,1,0,0,0)
or several other such tricks.
On Thu, 5 Jun 2003, Kosenkov Kirill wrote:
Hello!
I am trying to compute minimal time on some data
You might also consider S Programming, by Venables Ripley.
-roger
Spencer Graves wrote:
Apart from the wealth of material on http://www.r-project.org/;,
my favorite book on R is Modern Applied Statistics with S, 4th ed., by
Venables and Ripley.
hope this helps. spencer graves
Thanks for reply, but when i am trying to
mt - structure(mt, class=c(POSIXt, POSIXct))
or
mt + ISOdatetime(1970,1,1,0,0,0)
it seems that i loosing structre of matrix:
str(structure(mt, class=c(POSIXt, POSIXct)))
`POSIXct', format: chr [1:44] 2003-06-05 12:10:00...
instead of:
num [1:11,
All,
Sorry for bombarding you with GAM related questions, but...
I know a partial residual option in plot.gam() is on Simon Wood's todo
list, but since I'm in the midst of a project and not yet having acquired
sufficient R knowledge to code something usable myself I'll have to put my
trust in
On a RedHat 7.3 system with R-1.6.1, I did this
x11(width=3.5,height=4,colortype=gray)
Then plotted (with matplot) a nice looking no-color graph on the screen,
then I did this:
dev.copy2eps(file=test.eps,height=4,width=3.5)
I was surprised that the output in the eps file included the
On Thu, 5 Jun 2003, Paul E Johnson wrote:
On a RedHat 7.3 system with R-1.6.1, I did this
x11(width=3.5,height=4,colortype=gray)
Then plotted (with matplot) a nice looking no-color graph on the screen,
then I did this:
dev.copy2eps(file=test.eps,height=4,width=3.5)
I was
Dear R-Help:
I want to (a) subset a data.frame by several columns, (b) fit a model
to each subset, and (c) store a vector of results from the fit in the
columns of a data.frame. In the past, I've used for loops do do this.
Is there a way to use by?
Consider the following example:
On Thu, 5 Jun 2003, James D Thomas wrote:
Is there an easy way to integrate the generalized gamma distribution into
the current R survival package?
I'm not entirely sure what the generalized gamma distribution is, but it
is fairly easy to add to survreg() other parametric models where
On Thu, 5 Jun 2003, Spencer Graves wrote:
Dear R-Help:
I want to (a) subset a data.frame by several columns, (b) fit a model
to each subset, and (c) store a vector of results from the fit in the
columns of a data.frame. In the past, I've used for loops do do this.
Is there a way
Hi, Thomas, et al.:
Thanks for the reply. Unfortunately, do.call strips off the subset
identifiers, which I want to use for further modeling:
do.call(rbind, byFits)
(Intercept) x
[1,] 0.333 -1.517960e-016
[2,] 0.667 3.282015e-016
The following does what I want
Since I don't have your by.df to test with I may not have it exactly
right, but something along these lines should work:
byFits - lapply(split(by.df,paste(by.df$A,by.df$B)),
FUN=function(data.) {
tmp - coef(lm(y~x,data.))
Thank you all.
I made a pretty basic error in using multinom rather than glm
family=binomial which needed rapid correction.
I have now rewritten the relevant part using glm.
After importing I convert all categorical variables into factors
londonpsm - sqlFetch(channel,
Hi Frank,
From: Frank E Harrell Jr [mailto:[EMAIL PROTECTED]
[snip]
The anova method for ols fits 'works' when you penalize the
model but there is some controversy over whether we should be
testing biased coefficients. Some believe that hypothesis
tests should be done using the
Try:
X - model.frame(form=formula, data=mydata, na.action=na.pass)
Andy
-Original Message-
From: Ravi Varadhan [mailto:[EMAIL PROTECTED]
Sent: Thursday, June 05, 2003 12:02 PM
To: [EMAIL PROTECTED]
Subject: [R] na.action in model.frame
Dear Group:
I am trying to create a
You are trying to load woa.obj, and you are on Windows which should be
looking for a .dll and the error message says woa.so. You say you keep
your .exe in the rw1070 directory.
You are rather confused! dyn.load on Windows loads a DLL file from the
current working directory, *as its help says*.
Is there an easy way to integrate the generalized gamma distribution into
the current R survival package?
This distribution is available in my gnlr3 function in my gnlm
library. It handles left-, right-, and interval censoring, with linear
and nonlinear regression on all three parameters.
Hi,
Is there any way to use R to present t test results for three groups of
experiments, each of which involves several parallel experiment series with
groups of control vs treated. I would like to present the average fold change
of the experimental parameter (concentration of enzymes) as bars
Looks like that OS wants all symbols resolved: those are all variables
exported by R.bin.
1) See if there is a ld flag to allow unresolved symbols.
2) (something of a kludge) configure with --enable-R-shared and
rebuild (from scratch), as this will link against the libR.so.
Whether that works
Hi,
When i load my workspace, i visualize my object with
the ls()function.
First problem is that the saved graph is open like a
list instead of an image!What is wrong?
Second problem is about scatterplot graph: how can i
identify a specific point?
Thanks.
Michele.
Michele Grassi wrote:
Hi,
When i load my workspace, i visualize my object with
the ls()function.
You list the available objects in your Workspace, I guess?
I don't know of the feature to visualize objects with ls().
First problem is that the saved graph is open like a
list instead of an
I know a partial residual option in plot.gam() is on Simon Wood's todo
list, but since I'm in the midst of a project and not yet having acquired
sufficient R knowledge to code something usable myself I'll have to put my
trust in you. Anybody got some code lying around for doing this? Or if
Hi R lovers!
I would like to know how you can write comments inside the code of a
function
is it latex style
% ?
or any other language style?
thanks a lot
*
Ce message et toutes les pieces jointes (ci-apres le message)
Dear helpers
I'm sorry to insist but I still think there is something wrong with the function
kmeans. For instance, let's try the same small example:
dados-matrix(c(-1,0,2,2.5,7,9,0,3,0,6,1,4),6,2)
I will choose observations 3 and 4 for initial centers and just one iteration. The
results
On Thu, 05 Jun 2003 21:50:13 -0400
Liaw, Andy [EMAIL PROTECTED] wrote:
Hi Frank,
From: Frank E Harrell Jr [mailto:[EMAIL PROTECTED]
[snip]
The anova method for ols fits 'works' when you penalize the
model but there is some controversy over whether we should be
testing biased
On Thu, 2003-06-05 at 18:26, Deepayan Sarkar wrote:
On Thursday 05 June 2003 06:41, Ernesto Jardim wrote:
Hi
I'm doing a xyplot and I wand to reduce the number of tick marks in the
x axis. My x axis are month and I want to reduce the 12 tick marks to 4.
I used the scales argument but it
In der Mailnachricht wurde vom Virenscanner
ein VIRUS gefunden:
Von: [EMAIL PROTECTED]
Gesendet: Fri, 6 Jun 2003 12:44:30 +0200
Betrefd: Re: Application
Daher konnte diese Mail nicht an den/die Empfaenger weitergeleitet werden.
[EMAIL PROTECTED]
| From: [EMAIL PROTECTED]
| Date: Fri, 6 Jun 2003 11:57:32 +0200
Hallo,
| Hi R lovers!
|
| I would like to know how you can write comments inside the code of a
| function
|
| is it latex style
| % ?
| or any other language style?
It's #, otherwise like tex. The same character is
Hi,
I am looking for some code for Moran's I. Has anyone previously done this?
I have been unable to find it in the search engines.
James
Computations and tests for Moran's I and Geray's c are available in
the spdep package.
--
Stéphane DRAY
On Thu, 05 Jun 2003 23:50:07 -0500, you wrote:
if(!is.loaded(C.symbol(woa))) dyn.load(woa.obj) # woa is my file name
in C
Error message coming like this:
Error in dyn.load(x, as.logical(local), as.logical(now)) :
unable to load shared library C:/Program Files/R/rw1070/woa.so:
Thanks you everybody for all your answers
|-+
| | [EMAIL PROTECTED]|
| | .uk |
| ||
| | 06/06/03 01:32 PM|
| ||
I have time series and need to draw simple and partial
correlograms with associated Q-statistics (the same as in
EViews). Can I do it in R? Thanks
library(ts) contains functions acf and pacf which come with
corresponding plot methods. See their help pages for details.
I don't know anything
Hi all;
I would like to compare robust distance measures,e.g. MVE, MCD, BACON, as measures
of leverage and to detect outliers in X-space . I could not find it in LQS or MASS
backages.
Please help
Osama Hussien
[[alternate HTML version deleted]]
In a contour plot having numeric values displayed
next to the contours, is it possible to modify these
numeric values so that they are bolded, enlarged,
and/or placed differently?
Much thanks in advance,
David Paul, Ph.D.
Battelle Memorial Institute
614.424.3176
Spencer,
Would sapply be better here?
R by.df - data.frame(A=rep(c(A1, A2), each=3),
R+ B=rep(c(B1, B2), each=3),
R+ x=1:6, y=rep(0:1, length=6))
R t(sapply(split(by.df, do.call(paste, c(by.df[, 1:2], sep = :))),
R+ function(x) coef(lm(y ~ x, data
On Fri, 06 Jun 2003 09:38:40 -0400, you wrote in message
[EMAIL PROTECTED]:
In a contour plot having numeric values displayed
next to the contours, is it possible to modify these
numeric values so that they are bolded, enlarged,
and/or placed differently?
See the ?contour.default help: vfont
Hi dear R-users
I try to reproduce the steps included in a LDA. In my textbook (Bortz)
it says, that the matrix with the eigenvectors
V
usually are not normalized to the length of 1, but in the way that the
following holds (SPSS does the same thing):
t(Vstar)%*%Derror%*%Vstar = I
where
Duncan Murdoch wrote:
On Fri, 06 Jun 2003 09:38:40 -0400, you wrote in message
[EMAIL PROTECTED]:
In a contour plot having numeric values displayed
next to the contours, is it possible to modify these
numeric values so that they are bolded, enlarged,
and/or placed differently?
See the
Hello,
I have time series and need to draw simple and partial correlograms with
associated Q-statistics (the same as in EViews). Can I do it in R? Thanks
Hello Shutnik,
see ?Box.test in the ts-package for Q-statistics.
To my knowledge a plot function as in EViews (left-hand panel are ACFs
Dear all,
I have data like 3 coulmns and many rows. Each entry
is less than 10.
Example
x y z
1 5 3 2
2 3 7 8
3 8 9 5
4 5 4 6
--
---
I have to sum
I am interested in R as an alternative for a statistical tool
at our firm.
Ditto...
I have recently moved to this agency from a company where I had access to
Splus. There is also a coworker here who had used Splus at a previous
employer. We both would like some access to the S language. We
I don't completely understand what you want, but might the following help?
cumsum(1:11)
[1] 1 3 6 10 15 21 28 36 45 55 66
which(cumsum(1:11)9)
[1] 4 5 6 7 8 9 10 11
which(cumsum(1:11)9)[1]
[1] 4
hth. spencer graves
N Dey wrote:
Dear all,
I have data like 3 coulmns and many rows.
Hello,
I have a question about using the layout command within a function. I've
written function that uses layout to create a figure from 2 plots. This
works fine to create a figure. When I use par(mfrow = c(2,2)) to create multiple
plots, it seems that the layout command resets the mfrow
I'm sorry to insist but I still think there is something wrong with the function
kmeans. For instance, let's try the same small example:
dados-matrix(c(-1,0,2,2.5,7,9,0,3,0,6,1,4),6,2)
I will choose observations 3 and 4 for initial centers and just one iteration. The
results are
I have a question about using the layout command within a
function. I've
written function that uses layout to create a figure from 2
plots. This
works fine to create a figure. When I use par(mfrow =
c(2,2)) to create multiple
plots, it seems that the layout command resets the mfrow
On Fri, 6 Jun 2003 12:04:47 -0400 , you wrote in message
[EMAIL PROTECTED]:
1) I want a test suite for R. I noted in the messages (Date: Mon Feb 24
2003 - 22:18:03 EST) that Prof Ripley wrote Well, R itself has lots of
tests in its test suite (see directory tests in the sources) packages...
but
I like R more than SAS. My job is doing research on clinical trial. But I was told
that FDA only accepts the result from SAS. Is that true? TOO BAD.
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Thank you for your detailed response.
R is more responsive to bug reports.
and blithering from frustrated code converters :)
boot(vname, median, na.rm=T)
should work, ...
agreed
but myfunc - function(x, str) median(x, na.rm=str)
boot(vname, myfunc, str=T)
would not.
agreed
What I
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of wensui liu
Sent: Friday, June 06, 2003 1:16 PM
To: [EMAIL PROTECTED]
Subject: [R] sas vs. r
I like R more than SAS. My job is doing research on clinical
trial. But I was told that FDA only accepts the
Please check http://www.r-project.org/; - search - R site
search. My search just now for R vs. SAS for FDA produced 14 matches.
My bottom line from reading comments on this issue is that it is like
the comment 20 years ago that, Nobody in management ever got fired for
buying IBM. The
On Fri, 6 Jun 2003, wensui liu wrote:
I like R more than SAS. My job is doing research on clinical trial. But
I was told that FDA only accepts the result from SAS. Is that true? TOO
BAD.
No, it isn't true. The FDA does not approve or certify statistical
software. Its main regulation
wensui liu [EMAIL PROTECTED] writes:
I like R more than SAS. My job is doing research on clinical
trial. But I was told that FDA only accepts the result from SAS. Is
that true? TOO BAD.
No. But it's not that simple.
See previous postings (within the last 2 months) on validation and
FDA.
Dear R help-list reader,
I have a grouping variable C-c(1,2,3) which gives me the plot
symbols in a xyplot (lattice library) in three different colors,
according the group variable.
xyplot(A ~ B), group=factor(C))
if I now apply the panel.xyplot / panel.loess all the colors are
changing back
1 - 100 of 107 matches
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