Le jeu. 8 mai à 17:20, Giuseppe Paleologo a écrit :
I am struggling with R code optimization, a recurrent topic on this
list.
I have three arrays, say A, B and C, all having the same number of
columns.
I need to compute an array D whose generic element is
D[i, j, k] - sum_n A[i, n]*B[j,
interesting request..I'm looking forward to the replies
All I could come up with is putting it in two lines..
pr-array(0,c(dim(x)[2],dim(x)[2]));
for (i in 1:dim(x)[2]) for (j in 1:dim(x)[2])
pr[i,j]-cor.test(x[,i],x[,j])$p.val;
y
Monica Pisica wrote:
Hi everybody,
I would like to
Hi all,
I have a data management question. I am using an panel dataset read into
R as a dataframe, call it ex. The variables in ex are: id year x
id: a character string which identifies the unit
year: identifies the time period
x: the variable of interest (which might contain NAs).
Here is
Thanks a lot Deepayan. Could you please inform me what update are you
referring to, and give me some very vague sense when it might happen (within
weeks, months, or years)?
Many thanks
Ola
2008/5/8 Deepayan Sarkar [EMAIL PROTECTED]:
On 5/8/08, Ola Caster [EMAIL PROTECTED] wrote:
Dear help
k == kate [EMAIL PROTECTED]
on Thu, 8 May 2008 10:45:04 -0500 writes:
k In my data, sample mean =-0.3 and the histogram looks like t
distribution;
k therefore, I thought non-central t distribution may be a good fit.
Anyway, I
k try t distribution to get MLE. I found some
I want to draw a subset of ex by selecting only the A and B units:
ex1 - subset(ex[which(ex$id==A|ex$id==B),])
or a bit simpler:
ex1 - subset(ex, ex$id %in% c('A','B'))
In your expresion you don't need the subset function, as you are already
using indexing to extract the desired subset.
Hello,
I am hoping you can help me with a question concerning kmeans clustering
in R. I am working with the following data-set (abbreviated):
BMW Ford Infiniti Jeep Lexus Chrysler Mercedes Saab Porsche
Volvo
[1,] 6828 4544 7
Hi,
I fitted tree growth data with Chapman-Richards growth function using nls.
summary(fit.nls)
Formula:
Parameters:
Estimate Std. Error t value Pr
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.879 on 713 degrees of freedom
Algorithm
have a look also at function rcor.test() from package ltm.
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:
Hi,
I fitted tree growth data with Chapman-Richards growth function using nls.
summary(fit.nls)
Formula: HEIGHT ~ A * (1 - exp(-B * AGE))^C
Parameters:
Estimate Std. Error t value Pr(|t|)
A 29.007627 0.270485 107.24 2e-16 ***
B 0.030813 0.001095 28.13 2e-16 ***
C 1.849405
Hi,
I fitted tree growth data with Chapman-Richards growth function using nls.
summary(fit.nls)
Formula: HEIGHT ~ A * (1 - exp(-B * AGE))^C
Parameters:
Estimate Std. Error t value Pr(|t|)
A 29.007627 0.270485 107.24 2e-16 ***
B 0.030813 0.001095 28.13 2e-16 ***
C 1.849405
Hi,
I would like to thank Gabor, Duncan, John and Vincent for patiently helping me
with tips.
I found John's method to be the simplest to implement. Actually, I should have
been able to think of that myself. But I got distracted by trying to digest too
much information all at once.
Vincent,
Hi,
I fitted tree growth data with Chapman-Richards growth function using nls.
summary(fit.nls)
Formula: HEIGHT ~ A * (1 - exp(-B * AGE))^C
Parameters:
Estimate Std. Error t value Pr(|t|)
A 29.007627 0.270485 107.24 2e-16 ***
B 0.030813 0.001095 28.13 2e-16 ***
C
On 9 May 2008, at 09:12, Jordan van Rijn wrote:
Hello,
I am hoping you can help me with a question concerning kmeans
clustering
in R. I am working with the following data-set (abbreviated):
BMW Ford Infiniti Jeep Lexus Chrysler Mercedes Saab Porsche
Volvo
[1,] 6
Guru S wrote:
Hi,
I fitted tree growth data with Chapman-Richards growth function using nls.
summary(fit.nls)
Formula: HEIGHT ~ A * (1 - exp(-B * AGE))^C
Parameters:
Estimate Std. Error t value Pr(|t|)
A 29.007627 0.270485 107.24 2e-16 ***
B 0.030813 0.001095 28.13
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], ] * C[ind[, 3], ])
dim(D) - c(nrA, nrB, nrC)
D
I hope it
Dear R-users,
I have the following problem
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 useful and
Still I did not find any suggestion. Is my problem not elaborate enough?
Megh Dal [EMAIL PROTECTED] wrote: I think I should be clear exactly what I
want :
take following example :
a = b = seq(1, 5, by=500)
v = matrix(0, nrow=length(a), ncol=length(a))
for (i in 1:length(a))
{
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
[[alternative HTML version deleted]]
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:
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
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.
The first is
Dr. Ottorino-Luca Pantani ottorino-luca.pantani at 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
Federico Calboli f.calboli at 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
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
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
On May 9, 2008, at 5:39 AM, Dieter Menne wrote:
Dr. Ottorino-Luca Pantani ottorino-luca.pantani at 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,
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
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
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
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=0.13, c=0.50,
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
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
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
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
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
1500
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
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 can
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:
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
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
+
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
17092007
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
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:
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,
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
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:
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
ONKELINX, Thierry Thierry.ONKELINX at 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,
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
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
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.
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], ] *
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 -
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,
I
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
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 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
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)
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 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
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)?
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
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
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
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
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
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
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
group1 and
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]]
__
R-help@r-project.org mailing list
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
parametrized by
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
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,
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
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
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 function,
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
#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 how to do it
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
__
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 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
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 like to
__
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.
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
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
On Fri,
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
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
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)$sigma
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of
On 5/9/08, Ola Caster [EMAIL PROTECTED] wrote:
Sorry for spamming the list...
I noticed that if you first produce a date histogram with the hist()
function, like this:
basic.histogram - hist(my.data$date, breaks = months, plot = FALSE)
and then try to transfer the breaks from that
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).
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:
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
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
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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
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) :
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
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,
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
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
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
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