On Fri, May 9, 2008 at 4:04 AM, Federico Calboli
<[EMAIL PROTECTED]> wrote:
>>
>> Note that random can be a list:
>>
>> "a one-sided formula of the form ~x1+...+xn, or a pdMat object with a
>> formula
>> (i.e. a non-NULL value for formula(object)), or a list of such formulas or
>> pdMat
>> objects.
On Fri, 9 May 2008, Michael DiPersio wrote:
I am trying to use winProgressBar, however I find that although
setWinProgressBar updates the value on the screen, getWinProgressBar does
not return this value.
E.g
pb <- winProgressBar()
setWinProgressBar(pb, 0.2)
getWinProgressBar(pb)
[1] 0
It do
I am new to R and am trying to solve the following problem:
I have a data file containing tick-by-tick, millisecond level prices for
some stocks. I have another file or two containing orders and trades,
again, with millisecond time-stamps. Both of these files are irregularly
spaced and the time
Le Fri, May 09, 2008 at 11:23:37PM -0400, Dipankar Basu a écrit :
> > ex <- data.frame(id=id,year=year,x=x)
> > ex1 <- subset(ex[which(ex$id=="A"|ex$id=="B"),])
> > tapply(ex1$x, ex1$id, mean)
> ABC
> 22.5 32.5 NA
Dear Dipankar,
The reason for this behaviour is that the class of ex$id
you need to redefine the factor on ex1$id.. cast it again as.factor to
redefine the levels.
Dipankar Basu wrote:
>
> Hi!
>
> I am using R version 2.7.0 and am working on a panel dataset read into R
> as
> a dataframe; I call it "ex". The variables in "ex" are: id year x
>
> id: a character
Because id is a factor in your data frame, and the levels (including
"C") is kept when subsetted. Here is one way to get ride of "C".
> ex1$id <- factor(ex1$id)
> tapply(ex1$x, ex1$id, mean)
AB
22.5 32.5
On Sat, May 10, 2008 at 11:23 AM, Dipankar Basu <[EMAIL PROTECTED]> wrote:
> Hi!
>
>
Hi!
I am using R version 2.7.0 and am working on a panel dataset read into R as
a dataframe; I call it "ex". The variables in "ex" are: id year x
id: a character string which identifies the unit
year: identifies the time period
x: the variable of interest (which might contain NAs).
Here is an
Thanks,Hadley.
Another question is how can I know there is a grob path called
"xaxis::ticks"? Is there any easy way to figure out?
Thanks in advance.
On Wed, May 7, 2008 at 10:06 PM, hadley wickham <[EMAIL PROTECTED]> wrote:
> On Tue, May 6, 2008 at 11:44 PM, ronggui <[EMAIL PROTECTED]> wrote:
>
> -Original Message-
> From: [EMAIL PROTECTED] on behalf of array chip
> Sent: Fri 5/9/2008 2:54 PM
> To: [EMAIL PROTECTED]
> Subject: [R] how to check linearity in Cox regression
> Hi, I am just wondering if there is a test available for testing if a linear
> fit of an independent vari
Turns out I needed to add "exportselection=0". So:
foo.lb <- tklistbox(root, height = 5,
selectmode = "extended", exportselection = 0)
Got this from:
http://tolstoy.newcastle.edu.au/R/e2/help/06/12/6955.html
(That's what I get for assuming there'd be no tcl/tk questions on
I get strange behavior when a GUI has more than one list box. Variable
selection between the two boxes appears to be mutually exclusive!
So, for the contrived example below:
library(tcltk)
>
> root = tktoplevel()
>
> foo = c("foo","bar", "baz", "quux")
>
> foo.list <- tclVar()
>
> tclObj(foo.lis
Just to follow up based on offline discussion the problem is that
column one values
were being interpreted as numeric (since they are all numbers) so that,
for example, 09122007 was interpreted as 9122007 without the leading
zero. In such a case the day was regarded as 91 which, of course, can't b
Hi Group
I have a large data set of individual pairwise values (250,000 rows) that I
need to reshape into a pairwise matrix for mantel tests of these values
versus genetic distances.
the data are currently in columns formatted like so
AnimalA, AnimalB, Score
A1, A2, S1
A1, A3, S2
A1, A4, S3
A2,
Hi all,
It seems that GEE does not work when all predictive variables add up to a
constant. For example,
Y ~ X1+X2+X3. If, for each data point, we have x1+x2+x3=1, the algorithm fails.
Could anyone tell me why this would happen? Thank you!
Hi all,
I am trying to build a copula model using the Gumbel Copula and I have
two marginal distributions.I know the marginal parameters by using the
fitdistr() and optim().The problem is I dont know my copula parameter.
I am getting a bit confused of how shall I go about it.I read the
previous th
On 09/05/2008 5:55 PM, mtrp mtrp wrote:
Hi,
I am new to R. I am using spatstat package to generate some sample points but
don't know how to save the result to file. Could anyone please give me some
instructions? Thanks
I generate some random point data by using:
pp<-runifpoint(100)
I can plot
Hi,
I am looking for the ncf package version 1.0-4 (O. Bjornstad, 2003) and more
particularly for the function oldncf2D to estimate correlation function in four
directions. This function is not included in the new version of the ncf package
and I do not want to use all the spatial locations a
Hi,
Does anyone know of a package in R that has a function to convert
network data (e.g. an adjacency matrix or ) from 2-mode to 1-mode? I am
conducting social network analysis. I know that Pajek has this function
under Net --> Transform --> 2-mode to 1-mode --> Rows. I have searched
the doc
Hi,
I am new to R. I am using spatstat package to generate some sample points but
don't know how to save the result to file. Could anyone please give me some
instructions? Thanks
I generate some random point data by using:
pp<-runifpoint(100)
I can plot it out with plot(pp)
I suppose that pp co
Hi, I am just wondering if there is a test available for testing if a linear
fit of an independent variable in a Cox regression is enough? Thanks for any
suggestions.
John Zhang
[[elided Yahoo spam]]
Hi Mike,
I strongly suggest that you study Pinheiro and Bates (2000) to help
you make good decisions on the model specification and subsequent
steps.
In the meantime, you might find that exploring
lme(
x ~ gender*con*int*tone*cue
, random = ~ age | sub
, data = a
)
is he
Dear all!
Would anybody try to explain why could be impossible to use any of
constrained ordination methods under BiodiversityRGUI. All attempts are
finishing with empty function windows end next error massage:
Error in get(x, envir = RcmdrEnv(), mode = mode, inherits = FALSE) :
variab
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi Steffen.
~ Yes, the SSOAP WSDL processor can't handle that at present (thanks for the
example).
In this particular case, things are quite simple as it is getting stuck when
dealing with
the return type and this is just a string. So if we ju
I'm running a ridge regression of Y over 8 explanatory variables. The
selection of lambda gave me a value of 1 as the minimum value to be
added for bias. I got the coeficients for this new regression.
The thing is that I have 8 new values for explanatory variables that
are needed in order to predic
Tom,
I've never used lm.gls, but a quick look at the help page suggests
that, unlike gls, it doesn't have built-in functionality to estimate
parameters to structure the response/error covariance matrix. I
believe it assumes the covariance matrix is known (hopefully someone
will correct me if I'm
Hi Anh,
If I undestarstand:
# Your matrix
set.seed(123)
mat<-matrix(cbind(rnorm(20),rnorm(20)), ncol = 2)
# Scale
(mat-apply(mat[1:5,],2,mean))/apply(mat[1:5,],2,sd)
HTH,
Jorge
On Fri, May 9, 2008 at 3:51 PM, Anh Tran <[EMAIL PROTECTED]> wrote:
> Hi,
> Let say I have this matrix:
>
> > mat
Hi,
Let say I have this matrix:
> mat<-matrix(cbind(rnorm(20),rnorm(20)), ncol = 2)
And I want to rescale values of column [,1] and [,2] using values from row 1
to 5, such that the values of row 1:5 should be rescale to the same
amplitude (kinda like take the z-score of population from row 1:5).
On 5/9/08, Ola Caster <[EMAIL PROTECTED]> wrote:
>
> Sorry for spamming the list...
>
>
> I noticed that if you first produce a date histogram with the hist()
> function, like this:
>
> basic.histogram <- hist(my.data$date, breaks = "months", plot = FALSE)
>
> and then try to transfer the breaks fr
tom soyer wrote:
Hi,
Does anyone know if the RMSE is one of the values provided by the lm model,
or do we have to calculate it by hand from the residuals?
Thanks,
summary(my.fit)$sigma
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistic
Hi,
Does anyone know if the RMSE is one of the values provided by the lm model,
or do we have to calculate it by hand from the residuals?
Thanks,
--
Tom
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R-help@r-project.org mailing list
https://stat.e
Hi all,
I have collected response time data from 178 participants ('sub') for
each combination of 4 within-Ss factors ('con','int','tone','cue').
Additionally, I have recorded the gender of each participant, so this
forms a between-Ss factor ('gender'). Normally this would be analyzed
usi
Hi Chip,
Try this:
# For Q1
R> C1 = rbind(A,B)
or
R> C2 = do.call(rbind,list(A,B))
R> all.equal(C1,C2) # TRUE
# For Q2
R> set.seed(123)
R> vx=1:10
R> A = matrix(rnorm(500),ncol=10)
R> A2=t(sapply(A,rep,10))
R> A2[vx,]<- A[vx,]
R> A2
R> dim(A2) # 500 10
R> dim(A)# 50 10
HTH,
Jorge
I am trying to use winProgressBar, however I find that although
setWinProgressBar updates the value on the screen, getWinProgressBar does
not return this value.
E.g
> pb <- winProgressBar()
> setWinProgressBar(pb, 0.2)
> getWinProgressBar(pb)
[1] 0
I tried the same with tkProgressBar, and it is wo
__
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.
Hello Deepayan,
Please ignore my last e-mail and question. The information was
easily found in ?panel.segments.
Thanks,
John
On 5/5/08, John Poulsen <[EMAIL PROTECTED]> wrote:
>
>
> Hello,
>
> I am using xYplot to plot lines with confidence bands (see
test example
> below). I would lik
Hello R-Help,
I'm using R to do some optimization, specifically using the optim
method with method = 'SANN' (simulated annealing). I read the help
file, and noticed that this method does not include restarts/reheats,
which I think would help my optimization significantly. Does anyone
know of 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.
Greetings,
Q #1
--
How does one combine data frames by row ... no cleverness a la
merge(), just add rows.
For example, given A with 20 rows and B with 30 rows, I want C =
combine( A, B) having 50 rows.
Columns having matching names should be filled from both (all)
sources with suitabl
#try this
plot(survival)
lines(suvival)
On Fri, May 9, 2008 at 2:14 PM, Zhandong Liu <[EMAIL PROTECTED]>
wrote:
> Dear helpers,
> I am trying to plot two survival curves in the same figure.
>
> plot(survival)
> // in matlab, one just need to call "hold on"
> plot(survival2)
>
>
> I am wondering h
Dear helpers,
I am trying to plot two survival curves in the same figure.
plot(survival)
// in matlab, one just need to call "hold on"
plot(survival2)
I am wondering how to do it in R. Thank you very much!
--
Zhandong Liu
Genomics and Computational Biology
University of Pennsylvania
616 BRB
Thank you for your replies. I figured out that curve has an option of add
which adds it to a previously existing graph.
Thank you,
Amit
On Fri, May 9, 2008 at 10:07 AM, Amit Soni <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I have a histogram of an array of numbers.
> hist(v,10)
>
> How can I plot a fun
I have the following data of continuous time Markov chain.
Time -
[1] 0. 0.01219893 0.35903929 0.69378720 0.77247183 1.56008543
[7] 1.80607724 2.59023990 2.87196272 3.05311707 3.14737319 3.20758500
[13] 3.26668915 3.42428440 3.53324567 3.83668537 3.96784473 4.17196149
[19] 4.29982361 4.
I am glad to have joined such an active online community! I assume from your
R code that the deviance reported by R for "quasi..." families is not
"quasi-deviance" and that the following function summarizes the calculation
(note that I have added a parameter to k to account for the estimation of
Marc Schwartz wrote:
I might be tempted to take a more generic approach, where one can
provide an argument to the function to indicate that I want the 'top x'
maximum values and to give the user the option of returning the indices
or the values themselves.
Perhaps:
which.max2 <- function(x,
Dimitris Rizopoulos wrote:
try this:
v <- rnorm(10)
v
order(v, decreasing = TRUE)[1:2]
Wow .. that is slick! First I thought, wait .. I don't want to
reorder the elements, but this doesn't - it just returns the index
values in order. I don't really get that from reading the documentation,
it's
Try this:
plot(x, log(x),
xlab = "x", ylab = "h(x)",
main = bquote(Failure~rate~from~W(eta == .(eta), beta == .(beta)))
On Fri, May 9, 2008 at 11:31 AM, Gustave Lefou <[EMAIL PROTECTED]>
wrote:
> Hello,
>
> I have to do a few graphics of the same function and this function is
> parame
Hi,
I have a histogram of an array of numbers.
hist(v,10)
How can I plot a function, say a semi circle,
curve(sqrt(2.25-x*x), -1.5,1.5)
in the same graph?
Thank you
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
htt
Try:
xtabs(x ~ Group.2 + Group.1, data=x)
On Fri, May 9, 2008 at 1:25 PM, Oh Dong-hyun <[EMAIL PROTECTED]> wrote:
> Hi useRs!
>
> I would like to know how to make aggregated data.frame with aggregate()
> tabulated.
>
> For example, I run the following command to aggregate re with respect to
> gr
Hi there,
Try this:
R> tapply(aggr$x,aggr[,c(2,1)],function(x) x)
Group.1
Group.219921993199419951996199719981999
2000
15 0.16392 0.15467 0.15456 0.15391 0.16511 0.17368 0.17955 0.19805
0.20546
16 0.16237 0.18359 0.13811 0.13988
Dear All
I am attempting to use Sekhon's package Match to create a matched data
set. I keep getting an error regarding unused arguments whether I put Y
in or not.
Have I missed something?
myps id a propensity score (for treat) calculated from logistic
regression
mymatch<-match(Y=s
Thanks Kenn. I will add your solution to my ever growing colection of functions
;-))
Monica
Date: Fri, 9 May 2008 18:35:30 +0300From: [EMAIL PROTECTED]: [EMAIL PROTECTED]:
Re: [R] applying cor.test to a (m, n) matrixCC: [EMAIL PROTECTED]'ve used the
following function (I wrote it some time a
Hi useRs!
I would like to know how to make aggregated data.frame with
aggregate() tabulated.
For example, I run the following command to aggregate re with respect
to group1 and group2.
> (aggr <- with(final, aggregate(re, group1, group2, mean)))
Group.1 Group.2 x
1 1992
Hi All,
*I am still facing the problems in making R package on windows. *
-- Making package t1
adding build stamp to DESCRIPTION
installing NAMESPACE file and metadata
making DLL ...
making CGHseg_rewrite.d from CGHseg
It's a known scoping issue in lme -- you are doing this from a function.
Make sure your dataset is visible -- e.g. use with().
On Fri, 9 May 2008, Jorunn Slagstad wrote:
Dear R-help
I'm working on a large dataset which I have divided into 20 subsets based on
similar features. Each subset cons
I've used the following function (I wrote it some time ago so I don't
remember any more why I needed it, but I checked and it still works). I
don't think you can get rid of for loops altogether: if you look at the code
of apply, you'll see some there too.
The argument STATS specifies those compon
Hi,
I need to correct for ar(1) behavior of my residuals of my model. I noticed
that there are multiple gls models in R. I am wondering if anyone
has experience in choosing between gls models. For example, how
should one decide whether to use lm.gls in MASS, or gls in nlme for
correcting ar(1)? Do
On closer inspection of your specs, you might want to try upgrading your
Twisted install. I just checked and am pretty sure that the twistd
version is sync'd with the Twisted version. NWS requires Twisted >= 2.1.
Pat
Peter Tait wrote:
Hi Pat,
thank you for the response.
Yes I am running R o
Hi again,
I've got few very good options from the list and since they were not posted to
the list, I will provide a summary. Thank you very much to all who answered and
I hope this summary will benefit others interested in solving similar problems
like that.
Yasir Kaheil re-wrote my original
on 05/09/2008 08:07 AM Esmail Bonakdarian wrote:
Hello,
which.max() only returns one index value, the one for the
maximum value. If I want the two index values for the two
largest values, is this a decent solution, or is there a
nicer/better R'ish way?
max2 <-function(v)
{
m=which.max(v)
Hello.
I get a bit confused by the output from the predict function when used
on an object from coxph in combination with p-spline, e.g.
fit <- coxph(Surv(time1, time2, status)~pspline(x), Data)
predict(fit, newdata=data.frame(x=1:2))
It seems like the output is somewhat independent of the x-va
Hi, Martin and Kate:
KATE: Do you really want the noncentral t? It has mean zero but
strange tails created by a denominator following a noncentral
chi-square. The answer Martin gave is for a scaled but otherwise
standard t, which sounds like what you want, since you said the "sample
mean =
Hello,
I have to do a few graphics of the same function and this function is
parametrized by two arguments.
What I would like is to be able to change the value of these two arguments
without changing the plot command. So as to copy paste.
I tried the following :
x=1:100
eta=10
beta=5
plot(x,h(x
Hello Deepayan,
Thanks for the below solution to my graphing problem - just what I
was looking for.
One quick additional question, where do I change graphical
settings (lwd, pch, etc...) for the added points?
Thanks,
John
On 5/5/08, John Poulsen <[EMAIL PROTECTED]> wrote:
>
>
> Hello,
>
Charilaos Skiadas ha scritto:
On May 9, 2008, at 5:39 AM, Dieter Menne wrote:
If I understand the OP's question properly, the first value is to be a
multiple of 50, the second a multiple of 5, and the third a multiple
of 1. This can be done with this slight variation on the above theme:
a <-
Le ven. 09 mai à 03:44, Dimitris Rizopoulos a écrit :
try this:
A <- matrix(rnorm(10*4), 10, 4)
B <- matrix(rnorm(3*4), 3, 4)
C <- matrix(rnorm(5*4), 5, 4)
nrA <- nrow(A); nrB <- nrow(B); nrC <- nrow(C)
ind <- as.matrix(expand.grid(1:nrA, 1:nrB, 1:nrC))
D <- rowSums(A[ind[, 1], ] * B[ind[, 2],
Unfortunately, your data is *not* numeric. That is what the first
error message, " 'x' must be numeric", is telling you, and you should
believe it. It might look numeric, but it isn't, which is why Ingmar
mentioned you might have factors instead of numbers.
Your challenge is to discover why. T
Dear R-help
I'm working on a large dataset which I have divided into 20 subsets based on
similar features. Each subset consists of observations from different locations
and I wish to use the location as a random effect.
For each group I want to select regressors by a stepwise procedure and inclu
Hi Pat,
thank you for the response.
Yes I am running R on the same machine as the NWS server. I was doing this
on my Windows 2003 server with out any problem.
My Linux server has 4 cores and I would like R to take advantage of them.
Thank you for your help.
Cheers
Peter
Original Message Fol
ONKELINX, Thierry inbo.be> writes:
> This solutions works with R 2.7.0 under windows
>
> library(MASS)
> library(nlme)
> PredRes <- function(cal, val){
> cal <<- cal
> lmemod <- lme(distance ~ age * Sex, random = ~1|Subject, data = cal,
> method="ML")
> themod <- stepAIC(lmemod, dir=
You can take minpack.lm_1.1-0 (source code and MS Windows build,
respectively) from here:
http://www.nat.vu.nl/~kate/minpack.lm_1.1-0.tar.gz
http://www.nat.vu.nl/~kate/minpack.lm_1.1-0.zip
The bug that occurs when nprint = 0 is fixed. Also fixed is another
problem suggested your example: when th
Jorunn,
This solutions works with R 2.7.0 under windows
library(MASS)
library(nlme)
PredRes <- function(cal, val){
cal <<- cal
lmemod <- lme(distance ~ age * Sex, random = ~1|Subject, data = cal,
method="ML")
themod <- stepAIC(lmemod, dir="both", trace = FALSE)
prs <- predict(them
try this:
v <- rnorm(10)
v
order(v, decreasing = TRUE)[1:2]
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
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: http:/
Dear R-help
I'm working on a large dataset which I have divided into 20 subsets
based on similar features. Each subset consists of observations from
different locations and I wish to use the location as a random effect.
For each group I want to select regressors by a stepwise procedure and
include
Hello,
which.max() only returns one index value, the one for the
maximum value. If I want the two index values for the two
largest values, is this a decent solution, or is there a
nicer/better R'ish way?
max2 <-function(v)
{
m=which.max(v)
v[m] = -v[m]
m2=which.max(v)
result=c(m,
Hi Stefano,
Take a look at ?read.table and its friends. Now, assuming that your file is
"data.txt" you can use, for example:
# Option 1
> df=read.table('your.ubication.here/data.txt',header=FALSE)
> df
# Option 2
Copy your data set in the clipboard and write this in the R Console:
> df=read.tab
Hi Stefano,
Why do you *need* to add the quotes to the text file? If you leave them
out, any of the standard methods of reading data (i.e.,
read.table(filename, sep="") should work. See help(read.table),
help(scan), or help(read.fwf) for more info.
Mike
Stefano Sofia wrote:
Dear R users,
I
On Fri, May 9, 2008 at 8:52 AM, Stefano Sofia
<[EMAIL PROTECTED]> wrote:
>
> Dear R users,
> I have a txt file of the form
>
> 10092007 24.62 24.31 24.90
> 11092007 19.20 23.17 22.10
> 13092007 24.71 27.33 23.10
> 14092007 27.33 27.90 24.10
> 15092007 28.22 28.55 24.30
> 16092007 28.53 29.24 27.40
Dear Mr Holman,
Thanks very much for your help.
This is what I was looking for.
Gerrit.
on 2008-05-08 20:31 jim holtman said the following:
Will this do it for you:
x <- readLines(textConnection("1
+ Pietje
+ I1 I2 Value
+ 1 1 0.11
+ 1 2 0.12
+ 2 1 0.21
+
+ 2
+ Jantje
+ I1 I2 I3 Value
+
Dear R users,
I have a txt file of the form
10092007 24.62 24.31 24.90
11092007 19.20 23.17 22.10
13092007 24.71 27.33 23.10
14092007 27.33 27.90 24.10
15092007 28.22 28.55 24.30
16092007 28.53 29.24 27.40
17092007 24.19 30.64 26.80
18092007 22.60 20.62 28.40
19092007 8.89 1.70 14.70
20092007 2
You indeed found a bug. I can reproduce it (which I should have tried to
do on other examples in the first place!). Thanks for finding it.
It will be fixed in version 1.1-0 which I will submit to CRAN soon.
On Fri, 9 May 2008, elnano wrote:
>
> Find the data (data_nls.lm_moyano.txt) here:
> ft
Find the data (data_nls.lm_moyano.txt) here:
ftp://ftp.bgc-jena.mpg.de/pub/outgoing/fmoyano
Katharine Mullen wrote:
>
> Thanks for the details - it sounds like a bug. You can either send me the
> data in an email off-list or make it available on-line somewhere, so that
> I and other people ca
Christian Ritz wrote:
> Hi Peter,
>
> I think one option for what anova could do in the nonlinear case is to
> report the analysis of variance (or deviance) table obtained when
> doing a lack-of-fit test, that is comparing the nonlinear regression
> model to an appropriate ANOVA model. This is for
Claire_6700 wrote:
>
> Hello,
>
> I need some help with the simulatedSNPs function from scrime package.
>
> I am trying to simulate some genotype of a case/control disease locus. The
> allele frequence are cases/controls
>
> Sample cases controls
> 2000 .5.10
> 1
Hi,
I have a SOAP service, provided by BioMoby
which I'd like to call via SSOAP.
My service breaks during genSOAPClientInterface()
genSOAPClientInterface(def=service, verbose = TRUE)
Operation MassBank_Simple_2
Error: Cannot resolve SOAP type in empty context
Pr
On 5/9/2008 6:43 AM, Yemi Oyeyemi wrote:
Dear everyone, I am having problem simulating multivariate data. Though I was able to
simulate the data, but finding the variance-covariance matrix of simulated data did not
give exact covariance matrix used in simulating the data. Unlike some other pack
I have a question regarding calculating marginal effects (change in
the probability due to a one unit change in x) for a probit or for
that matter any discrete choice model. Does R have a function that
will essentially do this:
dE[y|x]
--- = f(x'beta)*beta
dx
Stata has a
Thanks for the details - it sounds like a bug. You can either send me the
data in an email off-list or make it available on-line somewhere, so that
I and other people can download it.
On Fri, 9 May 2008, elnano wrote:
>
> Thank you Katharine. I am certain nprint is affecting my solution. Let me
Thank you Katharine. I am certain nprint is affecting my solution. Let me
know how I can send the data (~300Kb). The script I used it:
ST1 <- ST04
SM1 <- SM08
SR1 <- SRch2
ST <- ST1 [!is.na(SR1)]
SM <- SM1 [!is.na(SR1)]
SR <- SR1 [!is.na(SR1)]
q <- 0.90
p <- c("a"=-0.003, "b"=0.1
Note that random can be a list:
"a one-sided formula of the form ~x1+...+xn, or a pdMat object with a formula
(i.e. a non-NULL value for formula(object)), or a list of such formulas or pdMat
objects. "
If you can translate that into *informative* English I'd be grateful. I have the
Pinheiro a
Dear everyone, I am having problem simulating multivariate data. Though I was
able to simulate the data, but finding the variance-covariance matrix of
simulated data did not give exact covariance matrix used in simulating the
data. Unlike some other packages, like stata, using command "corr2data
On Fri, 9 May 2008, Berthold wrote:
I am using stepAIC for stepwise regression modeling.
Is there a way to change the entry and exit alpha levels for the
stepwise regression using stepAIC ?
No, because it does not use 'entry and exit alpha levels', rather AIC.
I do not believe you have consu
On May 9, 2008, at 5:39 AM, Dieter Menne wrote:
Dr. Ottorino-Luca Pantani unifi.it>
writes:
Imagine that for a particular cuvette (I have 112 different
cuvettes !!)
you have to mix the following volumes of solution A, B, and C
respectively.
c(1803.02, 193.51, 3.47)
Each solution is to
Hi S Ellison,
Thanks for your response,
you said anova in R works two ways;
one is single fit, another one is sequence of fits.
Single fit is used to fit single nls() model result. If I am wrong plz advice
me.
I want to use single fit. How to do that? In your example, why the
anova(y.nls1
Hi Peter,
I think one option for what anova could do in the nonlinear case is to report the analysis
of variance (or deviance) table obtained when doing a lack-of-fit test, that is comparing
the nonlinear regression model to an appropriate ANOVA model. This is for example the use
of anova in t
Federico Calboli imperial.ac.uk> writes:
>
> Hi everyone,
>
> I am confused on how to specify some nesting and interaction terma with lme().
>
> lme(y ~ selection * males, random = ~1|replica/selection/males, mydata)
Note that random can be a list:
"a one-sided formula of the form ~x1+...+xn
Dr. Ottorino-Luca Pantani unifi.it> writes:
>
> In a lab experiment I have to mix three solutions to get different
> concentrations of various molecules in a cuvette
>
> I've used R to calculate the necessary µliters for each of the level of
> the experiment and I must confess that it is more
>>> "Guru S" <[EMAIL PROTECTED]> 09/05/2008 08:18 >>>
>I fitted tree growth data with Chapman-Richards growth function using
nls.
>When I try to run the anova() function I get this:
>Error in anova.nls(fit.nls) : anova is only defined for sequences of
"nls" objects
There seem to be two problems
Berthold,
stepAIC does model selected based on AIC (or a similar criterion like
BIC). So it does not uses the alpha levels. Hence it's pointless to
specify them in stepAIC.
HTH,
Thierry
ir. Thierry Onkelinx
Instituu
Sorry for spamming the list...
I noticed that if you first produce a date histogram with the hist()
function, like this:
basic.histogram <- hist(my.data$date, breaks = "months", plot = FALSE)
and then try to transfer the breaks from that histogram to a lattice
equivalent, like this:
histogram(
I am using stepAIC for stepwise regression modeling.
Is there a way to change the entry and exit alpha levels for the
stepwise regression using stepAIC ?
Many thanks,
Berthold
Berthold Stegemann
Bakken Research Center
Maastricht
The Netherlands
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