Hi,
Good Day Everybody.
Can anybody help me
I am working with R ...and i am using nnet() function for regression, but
after trained model, i m getting all predictions as 0.99 or 1.,
why is so, i dont understand?
can anybody suggest something...???
Thanks Regards,
Uttam
On Wed, 5 Oct 2005 [EMAIL PROTECTED] wrote:
Hi,
Good Day Everybody.
Can anybody help me
I am working with R ...and i am using nnet() function for regression, but
after trained model, i m getting all predictions as 0.99 or 1.,
why is so, i dont understand?
As a guess, you
Hello,
I have point spatial information (information linked to points on a
map), and I'd like to make inferences on the full extent of my zone. I
know the Kriging method, but I also know that it seems to be obsolete
and new approaches have been given.
I'd appreciate any advice on readings about
Hi,
I have the same problem.
I found a solution but I think there is something of more simple and correct.
I take a group and I put 0 the values of the other group, after I use the
function glm to abtain the Fisher's discriminant function for this group.
I repeat the same for all the groups.
Dear R helpers,
I am having trouble with an array inside a loop.
I wish to accumulate the results of a function in an array within
the function, over several loops of a program outside the function.
The problem is that the array seems to re-set at every entry to the
function. Here is an
try it this way:
a - array(NA, c(3, 5))
for(i in 1:nrow(a))
a[i, ] - c(runif(3), i, i * 10)
a
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
On Wed, 5 Oct 2005, Leonardo Lami wrote:
Hi,
I have the same problem.
I found a solution but I think there is something of more simple and correct.
I take a group and I put 0 the values of the other group, after I use the
function glm to abtain the Fisher's discriminant function for this
Jonathan Williams a écrit :
maxrun=3; a=array(NA, c(3,5)); run=0
testf=function(x,y){
print(paste('Run:',run)) #check that the function knows about run
a[run,1:3]=runif(3); a[run,4]=x; a[run,5]=y #collect numbers into array a
}
a outside testf is a global variable.
a inside testf is a
Dear All,
I am trying to use the library(mathgraph) but I do not understand (or find
in any help file) the type of Input needed.
I have prepared the Adjacency Matrix for the data, which consists in a
380x26 cells with 0/1 (1 if there is a link between those vertices, 0
otherwise).
Can I use
Hi
I have a square matrix Ainv of size N-by-N where N ~ 1000
I have a rectangular matrix H of size N by n where n ~ 4.
I have a vector d of length N.
I need X = solve(t(H) %*% Ainv %*% H) %*% t(H) %*% Ainv %*% d
and
H %*% X.
It is possible to rewrite X in the recommended crossprod way:
X
Matt,
you may want to set the option all to TRUE in :
merge(species1.effort, species2.effort, by='date', all=TRUE)
in order to also have in the result matrix the lines with 'date' that is not
commun for the both.
see ?merge for details.
hope this helps,
Florence.
---
Transcriptome platform of
On Wed, 5 Oct 2005, Robin Hankin wrote:
I have a square matrix Ainv of size N-by-N where N ~ 1000
I have a rectangular matrix H of size N by n where n ~ 4.
I have a vector d of length N.
I need X = solve(t(H) %*% Ainv %*% H) %*% t(H) %*% Ainv %*% d
and
H %*% X.
It is possible to
I have some data in a CSV file:
time,pos,t,tl
15:23:44:350,M1_01,4511,1127
15:23:44:350,M1_02,4514,1128
15:23:44:350,M1_03,4503,1125
...
15:23:44:491,M2_01,4500,1125
15:23:44:491,M2_02,4496,1124
15:23:44:491,M2_03,4516,1129
...
15:23:44:710,M3_01,4504,1126
15:23:44:710,M3_02,4516,1129
Hello all!
I have a problem that calls for a better understanding, than mine, of
how lme() uses the random part of the call.
The dataset consists of eleven field trials (Trial) with three
replicates (Block) and four fertiliser treatments (Treat). Analysing for
example yield with lme() is
On Wed, 2005-10-05 at 22:19 +1000, sosman wrote:
I have some data in a CSV file:
time,pos,t,tl
15:23:44:350,M1_01,4511,1127
15:23:44:350,M1_02,4514,1128
15:23:44:350,M1_03,4503,1125
...
15:23:44:491,M2_01,4500,1125
15:23:44:491,M2_02,4496,1124
15:23:44:491,M2_03,4516,1129
...
On 5 Oct 2005, at 12:15, Dimitris Rizopoulos wrote:
Hi Robin,
I've been playing with your question, but I think these two lines
are not equivalent:
N - 1000
n - 4
Ainv - matrix(rnorm(N * N), N, N)
H - matrix(rnorm(N * n), N, n)
d - rnorm(N)
quad.form - function (M, x){
jj -
Dear Ted,
I assumed that since Naiara was using scatter3d(), he wants a 3D dynamic
scatterplot. He could add points (actually, spheres) to the rgl graph
produced by scatter3d() -- the analog of plot() followed by points() for a
2D graph -- but doing so would be much more work than plotting by
sosman a écrit :
lines() has a color argument
from the online help :
?lines
lines(x, y = NULL, type = l, col = par(col),...
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
Hi,
I need further help with my GAMs. Most models I test are very
obviously non-linear. Yet, to be on the safe side, I report the
significance of the smooth (default output of mgcv's summary.gam) and
confirm it deviates significantly from linearity.
I do the latter by fitting a second
hi all
why does the following not work???
this was someone elses code and i couldnt explain why it doesn't work.
m=matrix(c(0,0),2,1)
v=matrix(c(1,0,0,1),2,2)
Y=function(X1,X2,mu=m,V=v)
{
X=matrix(c(X1,X2),2,1)
Hi all
I am a complete newbie to this list (just subscribed) and a newcomer to
R (an S user from olden times). I have been using scatter3d to create a
3d scatter plot with surface. The graphic is created within the rgl
package and I have used rgl.postscript to export it so I can generate a
an alternative is to use X2 below, which seems to be a little bit
faster:
N - 1000
n - 4
Ainv - matrix(rnorm(N * N), N, N)
H - matrix(rnorm(N * n), N, n)
d - rnorm(N)
quad.form - function (M, x){
jj - crossprod(M, x)
return(drop(crossprod(jj, jj)))
}
###
I think you probably should state more clearly the goal of your
analysis. In such situation, estimation and hypothesis testing are
quite different. The procedure that gives you the `best' estimate is
not necessarily the `best' for testing linearity of components. If your
goal is
Dear Denis,
Take a closer look at the anova table: The models provide identical fits to
the data. The differences in degrees of freedom and deviance between the two
models are essentially zero, 5.5554e-10 and 2.353e-11 respectively.
I hope this helps,
John
John
Dear Paul,
I don't have experience with rgl.postscript(), which is relatively new, but
find that the png graphs produced by rgl.snapshot() are of reasonably good
quality and preserve transparency. Perhaps the developers of the rgl package
can shed more light on the matter.
I hope this helps,
Assuming, as in your example, that you want to plot your
series against 1, 2, 3, ... first form a ts series, my.series,
and then plot it using the col= argument:
my.series - do.call(cbind, lapply(tpos, ts))
m - ncol(my.series)
ts.plot(my.series, col = 1:m)
Another
On 10/5/2005 9:31 AM, Prof. Paul R. Fisher wrote:
Hi all
I am a complete newbie to this list (just subscribed) and a newcomer to
R (an S user from olden times). I have been using scatter3d to create a
3d scatter plot with surface. The graphic is created within the rgl
package and I have
Prof. Paul R. Fisher a écrit :
... My problem is that the plotted surface is no
longer transparent in the postscript output ie. the rgl.spheres that are
behind the surface disappear in the postscript image. Can't seem to find
any info on this anywhere. Am I doing something wrong? Is there
Fair enough, Andy. I thought I was getting both predictive ability
and confirmation that the phenomenon I was studying was not linear. I
have two projects, in one prediction is the goal and I don't really
need to test linearity. In the second I needed to confirm a cycle was
taking place
Hi John,
Le 05-10-05 à 09:45, John Fox a écrit :
Dear Denis,
Take a closer look at the anova table: The models provide identical
fits to
the data. The differences in degrees of freedom and deviance
between the two
models are essentially zero, 5.5554e-10 and 2.353e-11 respectively.
I
On Wed, 5 Oct 2005, [EMAIL PROTECTED] wrote:
hi all
why does the following not work???
this was someone elses code and i couldnt explain why it doesn't work.
I think this is a case of FAQ 7.17: Why does outer() behave strangely with
my function?
-thomas
m=matrix(c(0,0),2,1)
Thanks! That's what I had come up with, but was unsure about it. I'm
checking the marginals against the published manuals now. Nice to have fresh
eyes on the problem! Thanks, also, for the helpful link.
On 10/4/05, Thomas Lumley [EMAIL PROTECTED] wrote:
On Tue, 4 Oct 2005, David L. Van Brunt,
On Wed, 5 Oct 2005, Duncan Murdoch wrote:
On 10/5/2005 9:31 AM, Prof. Paul R. Fisher wrote:
Hi all
I am a complete newbie to this list (just subscribed) and a newcomer to
R (an S user from olden times). I have been using scatter3d to create a
3d scatter plot with surface. The graphic is
On 10/5/2005 11:10 AM, Prof Brian Ripley wrote:
On Wed, 5 Oct 2005, Duncan Murdoch wrote:
On 10/5/2005 9:31 AM, Prof. Paul R. Fisher wrote:
Hi all
I am a complete newbie to this list (just subscribed) and a newcomer to
R (an S user from olden times). I have been using scatter3d to create a
I can output two histograms of variables
AXFILTERED and AZPTOP as follows:
win.metafile(filename=C:/AXFILTERED.emf,pointsize=12)
hist(AXFILTERED,breaks=40)
dev.off()
win.metafile(filename=C:/AZPTOP.emf,pointsize=12)
hist(AZPTOP,breaks=40)
dev.off()
But, I actually have a dataframe of 120
(xnew -- edit(data.frame()).
is equivalent to:
(xnew -(- edit(data.frame())).
make sense?
spencer graves
Nathan Dieckmann wrote:
Hey there,
I apologize if this is an irritatingly simple question ... I'm a
new user. I can't understand why R flips the sign of all data values
when
Dear Denis,
The chi-square test is formed in analogy to what's done for a GLM: The
difference in residual deviance for the nested models is divided by the
estimated scale parameter -- i.e., the estimated error variance for a model
with normal errors. Otherwise, as you point out, the test would be
Paul E. Green a écrit :
... Any solutions that would
save me from repeating this code 120 times? Can I
pass arguments inside quotes? Can I write a function
to do this?
as a toy example, with some of the usable functions :
(see ?dir)
fct00 = function()
{
rep0 = data;
fichier = dir(rep0,
Sehr geehrte Frau/Herr, send
Suchen Sie eine Arbeit? hard
Eine der besten Finanzgesellschaften in Osteuropa freut sich sehr
darauf,
Ihnen eine ausgezeichnete Arbeit vorzuschlagen, die wirklich ein gro?es
Einkommen erm?glicht!! Sie sollen nichts investieren oder keine Waren
kaufen, um bei uns
On Mon, 2005-10-03 at 15:28 -0500, lforzani wrote:
Hello, I am using library fda and I can not run a lot of functions because
I receive the error:
Error in bsplineS(evalarg, breaks, norder, nderiv) :
couldn't find function spline.des
do you know how I can fix that? Thnaks.
On 10/5/2005 11:33 AM, Duncan Murdoch wrote:
On 10/5/2005 11:10 AM, Prof Brian Ripley wrote:
On Wed, 5 Oct 2005, Duncan Murdoch wrote:
On 10/5/2005 9:31 AM, Prof. Paul R. Fisher wrote:
Hi all
I am a complete newbie to this list (just subscribed) and a newcomer to
R (an S user from olden
Hi, there:
I am wondering if anyone here can provide an example using pca doing
dimension reduction for a dataset.
The dataset can be n*q (n=q or n=q).
As to dimension reduction, are there other implementations for like ICA,
Isomap, Locally Linear Embedding...
Thanks,
weiwei
--
Weiwei Shi,
Thank you everyone for your help, but my introduction to GAM is
turning my brain to mush. I thought the one part of the output I
understood the best was r-sq (adj), but now even this is becoming foggy.
In my original message I mentioned a gam fit that turns out to be a
linear fit. By
?princomp ?prcomp give examples
-- Bert Gunter
Genentech Non-Clinical Statistics
South San Francisco, CA
The business of the statistician is to catalyze the scientific learning
process. - George E. P. Box
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On
Something similar to this
(but I haven't tested)
mydf - data.frame(xx=rnorm(100), yy=rnorm(100), zz=rnorm(100))
for (nm in names(mydf)) {
fnm - file.path('c:',paste(nm,'.emf',sep=''))
cat('Creating histogram in file',fnm,'\n')
win.metafile( filename=fnm)
hist(mydf[[nm]])
Dear Denis,
You got me: I would have thought from ?summary.gam that this would be the
same as the adjusted R^2 for a linear model. Note, however, that the
percentage of deviance explained checks with the R^2 from the linear model,
as expected.
Maybe you should address this question to the
Thanks. I got what I needed.
for example:
USA.pca-prcomp(USArrests, scale = TRUE)
predict(USA.pca)
PC1 PC2 PC3 PC4
Alabama -0.97566045 1.12200121 -0.43980366 0.154696581
Alaska -1.93053788 1.06242692 2.01950027 -0.434175454
Arizona -1.74544285 -0.73845954 0.05423025 -0.826264240
Arkansas
Hi,
I have a matrix A and a vector b, and would like to apply a function
f(a,b) to the rows of A and the elements of b. Eg
A - matrix(1:4,2,2)
b - c(5,7)
f - function(a,b) {sum(a)*b}
myapply(f,A=A,b=b)
would give
(1+3)*5 = 20
(2+4)*7 = 42
I found mapply, but it does not work for matrices.
I'm using R in Windows XP. I created a package myself.
I've used R CMD check to check it. Everything seems OK
except the latex. I get the error message:
* checking bbHist-manual.tex ... ERROR
LaTeX errors when creating DVI version.
This typically indicates Rd problems.
I ignored it because I
At the risk of being dense or R-ically incorrect, why do it without loops
when it is natural and easy to do it with them? More to the point:
1. vectorization speeds things up
2. apply commands are basically looping, not vectorization. Their advantage
is coding transparency, not speed
Flog me if
Try this:
mapply(f, split(A, 1:nrow(A)), b)
On 10/5/05, Tamas K Papp [EMAIL PROTECTED] wrote:
Hi,
I have a matrix A and a vector b, and would like to apply a function
f(a,b) to the rows of A and the elements of b. Eg
A - matrix(1:4,2,2)
b - c(5,7)
f - function(a,b) {sum(a)*b}
Hello all
I am playing with R for to make a animated GIF.
any suggestions, improvements are welcome :-)
case somebody could help me, i thanks!
Cleber N. Borges ( klebyn )
my objective:
(steps TODO)
---
1) to save PNG files;
- i don't know the best way to make
This probably has nothing to do with your software but on my Windows
XP system I just get a static image on Internet Explorer with the
animated GIF but with Firefox and the same GIF the animation comes
out as expected.
On 10/4/05, Tuszynski, Jaroslaw W. [EMAIL PROTECTED] wrote:
Hi,
I was
Hi all
I'm doing some things with a colleague comparing different
sorts of models. My colleague has fitted a number of glms in
Genstat (which I have never used), while the glm I have
been using is only available for R.
He has a spreadsheet of fitted means from each of his models
obtained from
Well I haven't seen any replies to this, so I have had a stab at the
problem of getting the data into a data frame.
The approach I took was to break up the data into a list, and then fill in
a matrix, row by row, filling down a la spreadsheet style when necessary,
taking advantage of the
Is it possible to write functions in such a way that, rather than
having to write a=function(a), one can just write function(a) and
have the variable passed as the argument be modified?
My real interest here is being able to invoke the editor by writing
ed(filename) rather than
Check out:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/38536.html
On 10/5/05, Paul Baer [EMAIL PROTECTED] wrote:
Is it possible to write functions in such a way that, rather than
having to write a=function(a), one can just write function(a) and
have the variable passed as the argument be
Works well with both IE and Firefox on my 2 year old DELL WinXP machine.
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Gabor
Hello Everybody,
I am reffering David Meyer's Benchmarking Support Vector Machines ,
Report No.78 (Nov.2002), i am newly working with R but i am not sure how
it is handling missing values in the benchmark datasets, I would be very
thankful to you if you could let me know how to handle those
Greetings all
I promised a summary of the responses that I got to my question:
Next year I will be teaching a third year course in applied statistics
about 1/3 of which is multivariate statistics. I would be interested in
hearing experiences from those who have taught multivariate statistics
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