R is case sensitive ;-) so in your list of start values you must have B=3 and
not b=3. Perhaps you have an object named b of a different length than your
data somewhere giving rise to your error message.
Also brute force wont work as Joerg point out. We have
A*1/(1+t/B)*1/sqrt(1+t/C) = D/(B+t)
dear r-helpers
i've got a table that in extracts looks like this:
V1 V2 V3V4 V5
1 01/01/1975 00:00:00 125.837 3.691 296.618
2 01/01/1975 01:00:00 124.799 3.679 281.307
3 01/01/1975 02:00:00 111.607 3.536 281.307
4 02/24/1976 11:00:00 21.602 2.555 93.893
5 02/
This is a question about SQL, or more precisely, Microsoft's peculiar
dialect of SQL. You haven't even mentioned (let alone credited) package
RODBC which you appear to be using.
In SQL queries you need to quote numeric values if you want them to be
treated as character. Why did you quote 'alt
On Tue, 2007-11-13 at 01:03 -0500, Rick Bilonick wrote:
>
> Is there some way to get ranef with postVar=TRUE to show what the
> variances are, or what the lower and upper bounds are? qqmath makes nice
> plots but I need to obtain the numerical values.
>
> Rick B.
>
I found a way:
attr(ranef(lm
Thank you,
Sigalit.
On 11/13/07, Moshe Olshansky <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> Look at names(log_v) in your notation to see what you
> really need.
> Then you could do something like:
>
> results <- list(600)
> for (ij in 1:600) {
> do what you did
> results[[ij]] <- log_v
> }
>
> So now
peter, thanks for your help with my questions regarding cor.test(). i have
another question though: does this function make any assumptions about the
underlying distribution of the two sequences? does it assume that they have a
gaussian distribution?
i ask because the data that i am working wit
Dear R-help,
I have to calculate the percent inclusion of each variable in a bootstrap
validation of a cox proportional hazards model(described in Sauerbrei and
Schumacher, Stat Med 11:1093, 1992).
First I need to get a bootstrap sample from my dataset, which I did with the
sample function. Th
Dear R-help,
I have to calculate the percent inclusion of each variable in a bootstrap
validation of a cox proportional hazards model(described in Sauerbrei and
Schumacher, Stat Med 11:1093, 1992).
First I need to get a bootstrap sample from my dataset, which I did with the
sample function. Th
On Mon, 2007-11-12 at 16:45 -0500, Doran, Harold wrote:
> No, don't reach into the bVar slot. Use the proper extractor function
> ranef() with postVar=T. There is no similar function for lme()
>
> > -Original Message-
> > From: [EMAIL PROTECTED]
> > [mailto:[EMAIL PROTECTED] On Behalf Of
Dear Sigalit,
On Nov 12, 2007 2:18 PM, sigalit mangut-leiba <[EMAIL PROTECTED]> wrote:
> Hello,
> I have a simple (?) simulation problem.
> I'm doing a simulation with logistic model and I want to reapet it 600
> times.
> The simulation looks like this:
>
> z <- 0
> x <- 0
> y <- 0
> aps <- 0
> ti
I am bewildered by the behaviour of tickmarks as demonstrated by the
following code. (What I'm trying to do is draw a single tick mark
extending
from the axis all the way down to the tick label, which is two lines
from
the axis to make sure it doesn't overlap with the ``ordinary'' tick
labe
Thank you very much for the help, Ben!
As a follow up, is there a way to specify the labels,
through the way the text is written, rather than
reading the edge positions from the graph? For
example,
mytree =
"((A:51.78,(C:24.6,D:24.6):27.18):40.06,B:91.84):0.0;"
plot(read.tree(text = mytree))
ed
Package dse does do linear, multivariate time series with multivariate
exogenous inputs, using VAR, multivariate ARMA, and state-space models,
just like you are describing. You can specify the model or have it
automatically determined. Perhaps you could give a few more details
about what you ar
hi,
You should probably be looking at the functions in the following packages:
sna
network(s)
graph
dynamicgraph
mathgraph
igraph
Matrix
and a few others ;)
what you're describing sounds like, to my ear, a restricted social network
diagram; many of the problems you describe are typical of s
Hi
I am practically new to R, and need to construct connection diagrams,
I have a table of data, of nodes in vertical rows, and horizontally
the number of outgoing connections to other nodes, and the indices of
these nodes, each in a column, so some columns are used, and some are
not, based on how
On Tue, 13 Nov 2007 02:07:39 +0100,
Søren Højsgaard <[EMAIL PROTECTED]> wrote:
> Dear List, I want to turn the matrix
>> xm
> [,1] [,2] [1,] "a" "b" [2,] "d" "e"
> into a list "by rows" as: [[1]] [1] "a" "b" [[2]] [1] "d" "e"
> A (bad?) way of doing this is as
>> unlist(apply(xm,1, list), r
Try split:
> x <- matrix(letters[1:4], 2)
> x
[,1] [,2]
[1,] "a" "c"
[2,] "b" "d"
> split(x, row(x))
$`1`
[1] "a" "c"
$`2`
[1] "b" "d"
-Christos
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Søren Højsgaard
> Sent: Monday, November 1
Try:
split(xm, row(xm))
On Nov 12, 2007 8:07 PM, Søren Højsgaard <[EMAIL PROTECTED]> wrote:
> Dear List,
> I want to turn the matrix
> > xm
> [,1] [,2]
> [1,] "a" "b"
> [2,] "d" "e"
>
> into a list "by rows" as:
> [[1]]
> [1] "a" "b"
> [[2]]
> [1] "d" "e"
>
> A (bad?) way of doing this is a
Hi Luca,
One (not very elegant) way of doing this is to use
double loop as below:
Let A be your matrix (100x50), W your weights matrix
(10x10) and you want to produce matrix B (10X5)
You can do:
B <- matrix(0,nrow=10,ncol=5)
for (i in 1:10) for (j in 1:5) B[i,j] <-
sum(W*A[(10*(i-1)+1):10*i,(10*(
Dear List,
I want to turn the matrix
> xm
[,1] [,2]
[1,] "a" "b"
[2,] "d" "e"
into a list "by rows" as:
[[1]]
[1] "a" "b"
[[2]]
[1] "d" "e"
A (bad?) way of doing this is as
> unlist(apply(xm,1, list), recursive=F)
[[1]]
[1] "a" "b"
[[2]]
[1] "d" "e"
- but there must be a more elegant w
Here is one way of doing it that uses the row and column names:
> # create test data
> mat1 <- matrix(0, nrow=10, ncol=3)
> dimnames(mat1) <- list(paste('row', 1:10, sep=''), LETTERS[1:3])
> mat2 <- matrix(1:3, ncol=1, dimnames=list(c('row3', 'row7', 'row5'), "B"))
> mat2
B
row3 1
row7 2
row5
Well, let me have a crack at it ... (inline below)
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Rolf Turner
Sent: Monday, November 12, 2007 2:16 PM
To: r-help@r-project.org
Subject: [R] Tick mark puzzle.
You can do the following:
x <- 1:10
g <- rep(3:5,len=10)
df <- data.frame(g=g,x=x)
y <- aggregate(df$x,list(df$g))
z <- sapply(df$g,function(x) which(y[,1]==x))
df1 <- data.frame(df,group.mean=y[z,2])
--- Casey Klofstad <[EMAIL PROTECTED]> wrote:
> I need advice on how to create a variable that
Thanks to Gabor and Jim. I am not sure if the first
entry year = 2009 is all the problem I'm getting but
it is certainly seems like the worst of it.
My stupidity: Someone sent me the data set in Excel
and I didn't do the basic data checks on. I _KNEW_ the
data went from 2006 to 2007.
--- Gab
In your examples the first line of your data refers to the
year 2009 and Oct 1st is repeated. Is that really what
you meant?
I can't tell what your problem is from your description
other than the data problems cited but there are lots of
examples of plotting with zoo in the following which may
he
Look at:
?ave
On Nov 12, 2007 5:01 PM, Casey Klofstad <[EMAIL PROTECTED]> wrote:
> I need advice on how to create a variable that is the group mean of
> another variable.
>
> For example, I have a variable called x for which each row in the data
> set has a value. I also have a nominal variable c
On 12/11/2007 2:56 PM, Joerg van den Hoff wrote:
> On Mon, Nov 12, 2007 at 11:09:21AM -0500, Duncan Murdoch wrote:
>> On 11/12/2007 9:14 AM, Joerg van den Hoff wrote:
>>> On Mon, Nov 12, 2007 at 07:36:34AM -0500, Duncan Murdoch wrote:
On 11/12/2007 6:51 AM, Joerg van den Hoff wrote:
> I in
Or
dat[,c1]
-Don
At 8:49 AM +1300 11/13/07, Rolf Turner wrote:
>On 13/11/2007, at 8:42 AM, Bernd Jagla wrote:
>
>> Hi,
>>
>>
>>
>> I would like to refer to a column in a data frame using a variable.
>> How
>> would I do this?
>>
>>
>>
>> Something like:
>>
>>
>>
>> c1<-"column1"
>>
>>
Dear useRs,
I would like to query an Access database through R based on a date attribute
but, unfortunately, I fail to do so. For example, the table test_table of the
test.mdb looks like:
ID cd market competitor dd price
1 1 2007-11-20 atl-bos delta 2007-11-20 210
2
I figured this out so I wanted to post a response to my own question. Thanks to
Brian Ripley for his hint, which got me looking in the right direction. The
plot I had seen before that I wanted was automatic output to the cor.plot
command in the mvoutlier library. A similar tolerance ellipse can
Package "dlm" has a function for maximum likelihood estimation of
parameters in general (linear Normal) state space model. The function,
dlmMLE, computes the likelihood based on singular value decomposition
and appears to be fairly robust.
No EM algorithm, though.
Giovanni
> Date: Sun, 11 No
Giusy,
There is also a package "dlm" that may be useful, but you need to
specify the model you want to use.
Giovanni
> Date: Sun, 11 Nov 2007 08:40:42 -0800 (PST)
> From: Giusy <[EMAIL PROTECTED]>
> Sender: [EMAIL PROTECTED]
> Precedence: list
>
>
> Hello to everyone!
> I have a question for
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi Konstantinos
I am not able to reproduce the error. I can use different
search queries of different length, etc.
The error is coming from libcurl and, if I have gotten things the
right way around, it basically means that the write handler callba
Josh Kalish <[EMAIL PROTECTED]> wrote:
> Has anyone used R to send out an email via an SMTP server?
If your system is Unix or Linux and you use a system call to "mail",
you might be interested to know that "uuencode" is a good way to include
attachments. Create a temp file containing your body t
I need advice on how to create a variable that is the group mean of
another variable.
For example, I have a variable called x for which each row in the data
set has a value. I also have a nominal variable called g that
indicates which of 100 different groups each row belongs to.
So, I want to cre
I guess this has a simple solution:
I have matrix 'mat1' which has row and column names, e.g.:
A B C
row10 0 0
row20 0 0
rown0 0 0
I have a another matrix 'mat2', essentially a subset of 'mat1' where the
rownames are all i
No, don't reach into the bVar slot. Use the proper extractor function
ranef() with postVar=T. There is no similar function for lme()
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Rick Bilonick
> Sent: Monday, November 12, 2007 4:40 PM
> To: R Hel
Using lmer in the lme4 package, you can compute the conditional
variance-covariance matrix of the random effects using the bVar slot:
bVar: A list of the diagonal inner blocks (upper triangles only) of the
positive-definite matrices on the diagonal of the inverse of ZtZ+Omega.
With the appropriate
Hi Dave,
Not sure what .lib.loc() is supposed to do, but you might want to look
at section 6.2 of the R Installation and Administration manual, which
covers library path issues.
Best,
Jim
Dave Jacoby wrote:
> I'm running R 2.5.0 (with the upgrade to 2.6.0 scheduled) on a and
> trying to wri
On Mon, 12 Nov 2007, Lam, Kelvin wrote:
> What is the actual usage of "mtry" in ctree(), or specifically,
> ctree_control() since it's a single tree? Thanks in advance.
As the associated man page explains: This is useful if you want to build
an ensemble (such as a random forest) based on ctree()
Hello, everyone.
I am writing a C code that uses quite a lot of matrix operations. I have
written my own functions for matrix-matrix multiplication, hadamard
(direct) and kronecker products.
However, I am sure using BLAS will greatly enhance speed. I did quite a
large search and found that I can us
Dear Group,
What is the actual usage of "mtry" in ctree(), or specifically,
ctree_control() since it's a single tree? Thanks in advance.
Regards,
Kelvin Lam, MSc.
Analyst, Programming & Biostatistics
Institute for Clinical Evaluative Sciences (ICES)
2075 Bayview Avenue, G179 Toronto
Thanks Jim and Ben.
Allen
On Nov 12, 2007 3:43 PM, Benilton Carvalho <[EMAIL PROTECTED]> wrote:
> it's defined in Biobase.
>
>
> On Nov 12, 2007, at 3:30 PM, affy snp wrote:
>
> > Thanks Jim. It is way simple. Great.
> > Is there any function like rowMedians() which
> > could take the median valu
it's defined in Biobase.
On Nov 12, 2007, at 3:30 PM, affy snp wrote:
> Thanks Jim. It is way simple. Great.
> Is there any function like rowMedians() which
> could take the median value across samples?
>
> Allen
>
> On Nov 12, 2007 3:10 PM, jim holtman <[EMAIL PROTECTED]> wrote:
>> Try something
you can use 'apply':
apply(A[, 48:243], 1, median)
On Nov 12, 2007 3:30 PM, affy snp <[EMAIL PROTECTED]> wrote:
> Thanks Jim. It is way simple. Great.
> Is there any function like rowMedians() which
> could take the median value across samples?
>
> Allen
>
> On Nov 12, 2007 3:10 PM, jim holtman <
I'm running R 2.5.0 (with the upgrade to 2.6.0 scheduled) on a and
trying to write a few applications that will be run as myself, as my
boss and as the web server, on one of many Solaris x86 servers.
I'd like to have my library sharable, and I thought I had found the way
by putting this line early
Thanks Jim. It is way simple. Great.
Is there any function like rowMedians() which
could take the median value across samples?
Allen
On Nov 12, 2007 3:10 PM, jim holtman <[EMAIL PROTECTED]> wrote:
> Try something like this:
>
> myAvg <- rowMeans(A[,48:243])
> B <- A[1:47,] / myAvg
>
>
> On Nov 12
Hello,
I have a simple (?) simulation problem.
I'm doing a simulation with logistic model and I want to reapet it 600
times.
The simulation looks like this:
z <- 0
x <- 0
y <- 0
aps <- 0
tiss <- 0
for (i in 1:500){
z[i] <- rbinom(1, 1, .6)
x[i] <- rbinom(1, 1, .95)
y[i] <- z[i]*x[i]
if (y[i]==1) a
Now your first data point is 9/26/09; is it supposed to be 9/26/06?
On Nov 12, 2007 1:47 PM, John Kane <[EMAIL PROTECTED]> wrote:
> I am completely misunderstanding how to handle dates.
> I want to plot a couple of data series against some
> dates. Simple example 1 below works fine.
> Unfortunate
So what is wrong with the plots? I used your example, and it appears
to plot the data correctly. What were you expecting?
On Nov 12, 2007 1:47 PM, John Kane <[EMAIL PROTECTED]> wrote:
> I am completely misunderstanding how to handle dates.
> I want to plot a couple of data series against some
>
On 13/11/2007, at 8:27 AM, Horacio Castellini wrote:
> Hi all, I'd like fit this function:
>
> G(t)=A*1/(1+t/B)*1/sqrt(1+t/C)
>
> where A, B and C are fitting constants that one like search. I have
> got a fcs<-(t,G(t)) value table in a data frame.
>
> Any one can help me? Tahnks Horacio.
I ***
Was supposed to be:
B <- A[,1:47] / myAvg
On Nov 12, 2007 3:10 PM, jim holtman <[EMAIL PROTECTED]> wrote:
> Try something like this:
>
> myAvg <- rowMeans(A[,48:243])
> B <- A[1:47,] / myAvg
>
>
> On Nov 12, 2007 1:37 PM, affy snp <[EMAIL PROTECTED]> wrote:
> > Dear list,
> >
> > Hi! I have a tab
Try something like this:
myAvg <- rowMeans(A[,48:243])
B <- A[1:47,] / myAvg
On Nov 12, 2007 1:37 PM, affy snp <[EMAIL PROTECTED]> wrote:
> Dear list,
>
> Hi! I have a table A, 238304 rows and 243 columns (representing
> samples). First of all, I would like to pool a group of samples
> from 48th
On Mon, Nov 12, 2007 at 04:27:26PM -0300, Horacio Castellini wrote:
> Hi all, I'd like fit this function:
>
> G(t)=A*1/(1+t/B)*1/sqrt(1+t/C)
>
> where A, B and C are fitting constants that one like search. I have
> got a fcs<-(t,G(t)) value table in a data frame.
>
> Any one can help me? Tahnks
What happens if you have multiple matches in the comparison between
the content_feat and ob_feat? Why don't you just iterate through the
content_feat and use 'match' to find the corresponding match in
ob_feat? This should speed it up. Also why are you using 'as.matrix'
when the values in the 'if
On Mon, Nov 12, 2007 at 11:09:21AM -0500, Duncan Murdoch wrote:
> On 11/12/2007 9:14 AM, Joerg van den Hoff wrote:
> >On Mon, Nov 12, 2007 at 07:36:34AM -0500, Duncan Murdoch wrote:
> >>On 11/12/2007 6:51 AM, Joerg van den Hoff wrote:
> >>>I initially thought, this should better be posted to r-deve
Jim,
That works - thank you; but still doesn't explain why stringsAsFactors
doesn't seem to have an effect.
Regards,
David
-Original Message-
From: jim holtman [mailto:[EMAIL PROTECTED]
Sent: 08 November 2007 13:41
To: Jessop, David
Cc: r-help@r-project.org
Subject: Re: [R] Bug (?)
On 13/11/2007, at 8:42 AM, Bernd Jagla wrote:
> Hi,
>
>
>
> I would like to refer to a column in a data frame using a variable.
> How
> would I do this?
>
>
>
> Something like:
>
>
>
> c1<-"column1"
>
> dat${c1}
>
>
>
> where names(dat) includes 'column1'
Use dat[[c1]].
Hi,
I would like to refer to a column in a data frame using a variable. How
would I do this?
Something like:
c1<-"column1"
dat${c1}
where names(dat) includes 'column1'
Thanks,
Bernd
[[alternative HTML version deleted]]
___
Hi all, I'd like fit this function:
G(t)=A*1/(1+t/B)*1/sqrt(1+t/C)
where A, B and C are fitting constants that one like search. I have
got a fcs<-(t,G(t)) value table in a data frame.
Any one can help me? Tahnks Horacio.
__
R-help@r-project.org mailin
Hua Li wrote:
>
> Dear Simon and everyone,
>
> Thanks for response.
>
> Specifically, I have a tree with species "A, B, C" and
> can be written as
> "((A:45.15,C:45.15):46.19,B:91.34):0.0;" . If I use
> the R command
>> mytreeABC
>
> [1] "((A:45.15,C:45.15):46.19,B:91.34):0.0;"
>
>> pl
Hi
I am performing a lmer on count data with several explanatory variables both
continuous and categorical. All go in fine apart from two of the continuous
variables that have na's in them. They give errors like:
Leading minor of order 5 in downdated X'X is not positive definite
I used the as.f
I am completely misunderstanding how to handle dates.
I want to plot a couple of data series against some
dates. Simple example 1 below works fine.
Unfortunately I have multiple observations per day (no
time breakdowns) and observations across years.
(example 2 very simplistic version )
Can any
Hello,
We're installing R v.2.6 on HP Proliant 32-bit server and RedHat Enterprise
Linux v.4 using the binaries and RPM's in CRAN in /bin/linux/redhat/el4/i386.
I could not find instructions on how to use the RPM's. I assume you issue:
rpm -i R-2.6.0-3.rh4.i386.rpm
rpm -i libRmath-2.6.0-3.rh4.
Dear list,
Hi! I have a table A, 238304 rows and 243 columns (representing
samples). First of all, I would like to pool a group of samples
from 48th column to 243rd column and take the average across
them and make a single column,saying as the reference column.
Second, I want to use each column o
Hello,
I am trying to access Microsoft Live Search Using SOAP through R.
In R I am using the RCurl packages to make the calls.
I have the following situation that looks crazy and cannot figure out how to
solve it:
#SOAP Request
library(RCurl)
h = basicTextGatherer()
body='
http://schemas.xmls
I have a matrix of values... for exemple the matrix could be 100 rows
and 50 columns...
I would like to resample the matrix to obtain a new matrix -> a 10x5 matrix
So i need to calculate a new value for every cell, i would like to use a
weightened mean to do this. I explain this thing better:
In th
I am a very beginner user of R. I have yet sample weights that I would like
to apply to all my data set.
Which function can I apply? The survey package begins from the survey
design.. I have yet, in my dataset, the weights column.
thanks a lot,
Vincenzo
--
View this message in context:
http://w
Download of the fEcofin fails from:
http://cran.r-project.org/bin/windows/contrib/r-release/fEcofin_260.72.z
ip
results in IE complaining
Internet Explorer cannot display the webpage
Most likely causes:
You are not connected to the Internet.
The website is encountering problems.
There
hi,
please help me out in the following case. seems like it stuck in some
where(already 7 hrs passed). what I want is to combine 4 matrix in to one
matrix of desired length.
final_matrix<-function(ob_feat,content_feat,link_feat,link_feat_transformed){
complete_feat<-matrix(rep(-1),nrow=11402,nc
Muenchen, Robert A (Bob) wrote:
> You probably don't want to spend time figuring out the .spo format. From
> SPSS 16 on, that format is obsolete and replaced by the Unicode
> XML-based .spv file. SPSS 16 users need a separate Legacy Viewer to read
> .spo files. -Bob
Great. I've had a quick Google
Hi all!
I encountered precision problems using pdf().
So far, I found out, that pdf() sets the Mediabox
to even values in dtp points (1/72 inch).
This can be seen in the following example.
page.width <- 13.1/2.54 # in centimeters
page.height <- 19/2.54
pdf(file = "foo.pdf", width = page.width,
x is a 1XN sparse matrix of numerics. I am using the Matrix package to
represent as a sparse matrix; the representation has a numeric vector
representing the positions within the matrix. My goal is find the columns
with the n largest values, here positive correlations. Part of my strategy
is to o
Thanks for your answer.
I downloaded the recent version today (1.3-8)
B
Am 12.11.2007 um 17:11 schrieb John Fox:
> Dear Birgit,
>
> I'm not a Mac user, so can only offer limited help. I've just
> verified,
> however, that the most recent version of the Rcmdr package on CRAN
> (version 1.3-8)
Dear Simon and everyone,
Thanks for response.
Specifically, I have a tree with species "A, B, C" and
can be written as
"((A:45.15,C:45.15):46.19,B:91.34):0.0;" . If I use
the R command
> mytreeABC
[1] "((A:45.15,C:45.15):46.19,B:91.34):0.0;"
> plot(read.tree(text = mytreeABC))
I would g
Dear Birgit,
I'm not a Mac user, so can only offer limited help. I've just verified,
however, that the most recent version of the Rcmdr package on CRAN
(version 1.3-8) installs and runs fine under R 2.6.0 on my Ubuntu Linux
system.
What is the version of the Rcmdr package that you installed?
I'm
On 11/12/2007 9:14 AM, Joerg van den Hoff wrote:
> On Mon, Nov 12, 2007 at 07:36:34AM -0500, Duncan Murdoch wrote:
>> On 11/12/2007 6:51 AM, Joerg van den Hoff wrote:
>> >I initially thought, this should better be posted to r-devel
>> >but alas! no response.
>>
>> I think the reason there was no
replace the second plot() by
lines(d7, (a*d7)+b)
b
On Nov 12, 2007, at 10:42 AM, elyakhlifi mustapha wrote:
> Hi,
> I have been searched how to do to display in the same chart this
> following functions:
>
> plot(d7,d6,type="b",col="dark red")
> plot(d7,(a*d7)+b,type="l")
>
> in using par it
Hi,
I have been searched how to do to display in the same chart this following
functions:
plot(d7,d6,type="b",col="dark red")
plot(d7,(a*d7)+b,type="l")
in using par it display in 2 diffenrent charts.
_
l
On about line 547 in nls.R there is
mf$formula <- # replace by one-sided linear model formula
as.formula(paste("~", paste(varNames[varIndex], collapse = "+")),
env = environment(formula))
If this is replaced with
mf$formula <- # replace by one-sided linear model f
On Mon, 12 Nov 2007, Stefano Ghirlanda wrote:
> i am using the boot package for some bootstrap calculations in place
> of anovas. one reason is that my dependent variable is distributed
> bimodally, but i would also like to learn about bootstrapping in
> general (i have ordered books but they have
Dear Prof. Ripley (and other readers) --
Unfortunately, I already did this (using the packages menu to install
RMySQL and copying libMySQL.dll in the RMySQL/libs directory). But after
the installation, when I load the package, I have the same error message.
Regards,
- Joachim Claudet
--
<º)))
On Mon, Nov 12, 2007 at 03:25:38PM +0100, Martin Maechler wrote:
> > "DM" == Duncan Murdoch <[EMAIL PROTECTED]>
> > on Mon, 12 Nov 2007 07:36:34 -0500 writes:
>
> DM> On 11/12/2007 6:51 AM, Joerg van den Hoff wrote:
> >> I initially thought, this should better be posted to r-de
I don't understand why you are trying to install a 2.5.x binary on R
2.6.0. Perhaps if you tried not to make things so hard for yourself and
just used the Packages menu to install RMySQL this might work: it does for
me and quite a few others.
Otherwise, you do need libMySQL.dll on the path or
> "DM" == Duncan Murdoch <[EMAIL PROTECTED]>
> on Mon, 12 Nov 2007 07:36:34 -0500 writes:
DM> On 11/12/2007 6:51 AM, Joerg van den Hoff wrote:
>> I initially thought, this should better be posted to r-devel
>> but alas! no response.
DM> I think the reason there was no
On Mon, Nov 12, 2007 at 07:36:34AM -0500, Duncan Murdoch wrote:
> On 11/12/2007 6:51 AM, Joerg van den Hoff wrote:
> >I initially thought, this should better be posted to r-devel
> >but alas! no response.
>
> I think the reason there was no response is that your example is too
> complicated. Yo
Dear list members --
I have problems in loading the pre-compiled library RMySQL_0.5-7 taken
here (http://stat.bell-labs.com/RS-DBI/download/index.html) or
RMySQL_0.6-0 taken here
(http://www.stats.ox.ac.uk/pub/RWin/bin/windows/contrib/2.5/) on R 2.6.0
under Windows XP. I am using the MySQL fea
Hi all,
Two questions:
1. Is there a way to evaluate models from lmer() with a poisson
distribution? I get the following error message:
library(lme4)
lmer(tot.fruit~infl.treat+def.treat+(1|initial.size),family=poisson)->model
plot(fitted(model),resid(model))
Error: 'resid' is not implemented y
You probably don't want to spend time figuring out the .spo format. From
SPSS 16 on, that format is obsolete and replaced by the Unicode
XML-based .spv file. SPSS 16 users need a separate Legacy Viewer to read
.spo files. -Bob
=
Bob Muenchen
On Fri, 9 Nov 2007, David Kaplan wrote:
> Hi all,
>
> It seems that I can get White's (HC3) test using MASS. The syntax I
> used for the particular problem is
>
> anova(scireg3, white.adjust="hc3")
I don't think this is true. My guess is that you are using Anova() from
package "car".
> where s
Michael,
I look at it.
Thanks for this first path.
Pierre-Olivier
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l
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https://s
Hello,
I am trying to remove nodes from a dendrogram - specifically nodes that have
less than 3 members. I am trying to create a function to do so, but I am
not sure how to reference the node inside of the function. Here is the
code:
rem.nodes <<- function(n) {
if(!is.leaf(n) & attr(n,
Sorry forgot subject.
B
> Von: Birgit Lemcke <[EMAIL PROTECTED]>
> Datum: 12. November 2007 13:26:55 GMT+01:00
> An: R Hilfe <[EMAIL PROTECTED]>
>
> Hello members of the mailinglist!
>
> I am running R 2.6.0 GUI 1.21 (4815) (4815) on a PPC with Mac OS X
> 10.4.
>
> I just tried to use the R c
On 11/12/2007 6:51 AM, Joerg van den Hoff wrote:
> I initially thought, this should better be posted to r-devel
> but alas! no response.
I think the reason there was no response is that your example is too
complicated. You're doing a lot of strange things (fitfunc as a result
of deriv, using a
Hello members of the mailinglist!
I am running R 2.6.0 GUI 1.21 (4815) (4815) on a PPC with Mac OS X 10.4.
I just tried to use the R commander.
I followed these instructions:
4) Start R (e.g. double-click the R icon in /Applications).
5) Start X11 (e.g. click the X11 button on the R Console or d
>
> Hello,
>
I am trying to remove nodes from a dendrogram - specifically nodes that
> have less than 3 members. I am trying to create a function to do so, but I
> am not sure how to reference the node inside of the function. Here is the
> code:
>
> rem.nodes <<- function(n) {
> if(!is
Hi,
I have been having trouble finding references that compare the classical
decomposition methods [decompose()] with the STL method of
Cleveland et al. (1990) [stl()].
Can anybody indicate some references that make comparisons between the two
methods and the advantage/disadvantages of each one i
See and *read* the help file ?glm
the object returned by glm() includes the `data' component
hence:
aa<-glm(..)
aa$data
or also eval(aa$call$data)
leffgh ha scritto:
> my function is
> glm(a~log(b)+c+d+e,family=binomial,data=f)->aa
>
>
> I want to extract the original data set of a
I can confirm this behavior on R-2.6.0 but don't have time to look into it
further at the moment.
On Mon, 12 Nov 2007, Joerg van den Hoff wrote:
>
> I initially thought, this should better be posted to r-devel
> but alas! no response. so I try it here. sory for the
> lengthy explanation bu
Try:
aa$data
On 12/11/2007, leffgh <[EMAIL PROTECTED]> wrote:
>
> my function is
> glm(a~log(b)+c+d+e,family=binomial,data=f)->aa
>
>
> I want to extract the original data set of aa. How to do it ?
>
> You may suggest the model.frame() function. In fact ,i have tried it.
> model.frame returns
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