Matthew Keller wrote:
Far from flaming you, I think you made a good point - one that I
imagine most people who use R have come across. The name R is a big
impediment to effective online searches. As a check, I entered R
software, SAS software, SPSS software, and S+ software into
google.
We still do not have reproducible code, but a 'dataframe' is not a matrix.
And I would expect a covariance matrix to have the same row and column
names: the examples do.
On Wed, 7 Feb 2007, Alistair Campbell wrote:
Thanks for that Brian,
I have worked through the examples. They work because
Hi everyone,
I am having installation problems, but this is how it all started:
I had some errors running the bioconductor package affyPLM that uses
LAPACK/Blas
Pset - fitPLM(Data)
Background correcting PM
Normalizing PM
Fitting models
Hi R,
Are there any limitations on the capacity of the data to hold for R data
frames or zoo objects? I mean to ask are there any restrictions on the
number of rows or column in the R data frames or the zoo objects?
Thank you,
Shubha
[[alternative HTML version deleted]]
Dear R-Users,
Background:
I have five multiple imputed datasets. For each datasets I have run a
regression analysis and combined the regression coefficients according to
Rubin (1987) rule.
Problem:
Now I want to use these combined regression coefficients on a different
dataset (with
We don't know what 'Linux' is here. What Linux distribution, what are
your C and Fortran compilers (in detail, e.g. from gcc --version and g77
--version)?
We need to see the tail of tests/Examples/base-Ex.Rout.fail to know what
went wrong.
If you can supply those pieces of information we can
When I run ./configure ,I got the message:
configure: WARNING: you cannot build info or html versions of the R manuals.
What's the matter?
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
Kuhn, Max wrote:
As someone who has (reluctantly) sent job postings to R Help, I think
that a SIG would be a good idea.
Max
Hi all,
My personnal experience also shows that it is difficult to find a job
where R is a key component, find R related material, or find companies
that would do
On Wed, 7 Feb 2007, xiaopeng hu wrote:
When I run ./configure ,I got the message:
configure: WARNING: you cannot build info or html versions of the R manuals.
What's the matter?
Search for info in the R Installation and Administration manual:
Simon P. Kempf wrote:
Background:
I have five multiple imputed datasets. For each datasets I have run a
regression analysis and combined the regression coefficients according to
Rubin (1987) rule.
So, now you have two numeric values: slope and offset. Right?
Simon P. Kempf wrote:
The official R Search place has been
http://search.R-project.org/
for quite a while now.
It does mention others including 'rseek' below.
BTW: It's main fault for me is that it does not include the
R-devel mailing list archives (hint hint :-)
Martin Maechler, ETH Zurich
IM == İbrahim
i get a message:
configure: WARNING: you cannot build info or html versions of the R manuals
what should i do ?
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
Conversely, unqualified(*) candidates are nearly guaranteed to find
support scarce here.
More seriously, free job boards, highly targeted like the one proposed
do seem to get enough traffic to make it worth the effort to post there.
One example serving the US market for market research is
Given two columns of type character in a dataframe of the form:
col1col2
31* 66
0 0*
102*66
71* 80
31 2*
66 31*
47 38*
How do I generate the following dataframe? Ie. col1 contains row item
with * and col2 contains row member without *
col1col2
31
Matthew Keller wrote:
Bob,
Far from flaming you, I think you made a good point - one that I
imagine most people who use R have come across. The name R is a big
impediment to effective online searches. As a check, I entered R
software, SAS software, SPSS software, and S+ software into
Thank you Martin
On 2/7/07, Martin Maechler [EMAIL PROTECTED] wrote:
The official R Search place has been
http://search.R-project.org/
for quite a while now.
It does mention others including 'rseek' below.
BTW: It's main fault for me is that it does not include the
R-devel
Hi,
in general Prof. Ripley is right that more information is needed, but
here's a hint that you might try first.
/usr/local/lib/R/bin/exec/R: relocation error:
/usr/local/lib/R/lib/libRlapack.so: undefined symbol: s_copy
# thrown out of R
Could simply mean that the
how about:
t.d - data.frame(col1=c(31*,0,102*,71*,31,66,47),
col2=c(66,0*,66,80,2*,31*,38*),
stringsAsFactors = FALSE)
t.x - apply(t.d,1,function(x) x[order(unlist(x)==grep(\\*$,
unlist(x),value=TRUE))])
t.d2 -
xiaopeng hu [EMAIL PROTECTED] writes:
i get a message:
configure: WARNING: you cannot build info or html versions of the R manuals
what should i do ?
In principle:
* Use R-devel not R-help
* Read the Installation and Administration manual (sec.2.2)
However, the gist is that you are missing
xiaopeng hu [EMAIL PROTECTED] writes:
i get a message:
configure: WARNING: you cannot build info or html versions of the R manuals
what should i do ?
In principle:
* Use R-devel not R-help
* Read the Installation and Administration manual (sec.2.2)
However, the gist is that you are missing
Hi Vikas,
Exactly what do you want to label them with? Generally the purpose of
the plot is to avoid having explicit labels - you can just read the
numbers of the axes. If you want the exact numbers, presenting them
in a table might be more appropriate.
I'm not at my development computer at
Dear All,
I am looking for an R function or any other reference to generate a series of
correlated Binomial (not a Bernoulli) data. The bindata library can do this
for the binary not the binomial case.
Thank you,
Bernard
I have these two dataframes in which 'id' is the key field
tabella
id nome
1 1 PIEMONTE
2 2 VALLED'AOSTA
3 3
LOMBARDIA
4 4 TRENTINO
5 5VENETO
6 6FRIULI
AND
tab
id nome
1 1 PIEMONTE
2 2 VALLED'AOSTA
3 3
Hi R users,
I would like to know how to sort a matrix according a different order of
colnames (or rownames) ,e.g.,
mx = matrix(rnorm(1:20),5,4)
colnames(mx) = letters[1:4]
rownames(mx) = letters[1:5]
mx
a b c d
a 0.02362598 -0.7033460 0.8106089
Frank == Frank E Harrell [EMAIL PROTECTED]
on Tue, 06 Feb 2007 21:59:45 -0600 writes:
Frank Matthew Keller wrote:
Bob,
Far from flaming you, I think you made a good point - one
that I imagine most people who use R have come
across. The name R is a big
Chris,
You might be able to get it using format. You would have to convert the
data frame to a matrix if you want the row names to be dots too.
foo - function(x, top = 3, ...)
{
if(dim(x)[1] top + 3) stop(not enough rows)
charX - format(x, ...)
charX - charX[c(1:(top+2),
That's something for the %in% command. Try this (untested!)
tab[(tab$id %in% tabella$id) == FALSE, ]
Cheers,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Reseach Institute for Nature
and
Hi
you can use
%in%
tabella[tabella$x %in% tab$x,]
to select rows which are in both and
tabella[!(tabella$x %in% tab$x),]
to select only non matching ones
HTH
Petr
On 7 Feb 2007 at 0:00, Vittorio wrote:
Date sent: Wed, 7 Feb 2007 14:23:45 +0100 (GMT+01:00)
From:
Hello,
I would like to repost the question of Joerg:
Hello everybody,
a question that connect to the question of Frederik Karlsons about 'how
to stand. betas'
With the stand. betas i can compare the influence of the different
explaning variables. What do i with the betas of factors? I
try:
mx[,new.col.names]
HTH.
On 2/7/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Hi R users,
I would like to know how to sort a matrix according a different order of
colnames (or rownames) ,e.g.,
mx = matrix(rnorm(1:20),5,4)
colnames(mx) = letters[1:4]
rownames(mx) = letters[1:5]
mx
mx = matrix(rnorm(1:20),5,4)
colnames(mx) = letters[1:4]
rownames(mx) = letters[1:5]
mx
a b c d
a -0.6264538 -0.8204684 1.5117812 -0.04493361
b 0.1836433 0.4874291 0.3898432 -0.01619026
c -0.8356286 0.7383247 -0.6212406 0.94383621
d 1.5952808
On 07-Feb-07 Marc Bernard wrote:
Dear All,
I am looking for an R function or any other reference to generate a
series of correlated Binomial (not a Bernoulli) data. The bindata
library can do this for the binary not the binomial case.
Thank you,
Bernard
How do you want
Dear Simon,
In my opinion, standardized coefficients only offer the illusion of
comparison for quantitative explanatory variables, since there's no deep
reason that the standard deviation of one variable has the same meaning as
the standard deviation of another. Indeed, if the variables are in
Opening R by right clicking and choosing run as administrator' worked. Was
able to run install packages without a problem. I have not tested the other
methods suggested. thank you.
Dan O'Shea
-- Original message --
From: Duncan Murdoch [EMAIL PROTECTED]
On
Good afternoon, my name is Gorka Merino and i am a scientist working in the
Marine Science Institune in Barcelone.
I'm interested in the application of Detrended Fluctuation Analysis (DFA)
with the R packages.
I've tried to obtain some information related to DFA from the 'Help'
options but
How can I superimpose some text labels on ggplot? I could get
weighted quantiles using wtd.quantiles function in Hmisc package. But
I can't plot these as labels on the boxplot.
My code is as follows.
list(c(1:3),c(1:3),c(1:3))-t
library(Hmisc)
for (i in 1:3)
{
Matthew Keller wrote:
I do wonder if anything can/should be done about this. I generally
search using the term CRAN but of course, that omits lots of stuff
relevant to R.
Change the name in the next major version to 'Rplus'?
Barry
__
Dear experts,
when I call the step function for a coxph model with n covariates and a
dicotomous variable included as strata, the first term removed by step is
always the strata variable. This is not what I want and then I do a manual step
updating the model minus the least significant
Hallo,
I have a table of names and values:
joe 0.45
mike 0.34
jim 0.25
I would like to fill-in a table of all pairs of names (which I aleady have)
joe.mike NA NA
joe.jim NA NA
mike.jim NA NA
with the values from the first table in the order of the pairs. The outcome
looks like
joe.mike 0.45
Hi,
Let me first congratulate you for having written the reshape
package. It is very nice and I use it all the time. I wish the
documentation was a bit easier. It took me quite some time to find my
way through it!! But once I got the hang of how it worked, I just
loved it.
With ggplot, this is
On Tue, 6 Feb 2007, Muenchen, Robert A (Bob) wrote:
That sounds like a good idea. The name R makes it especially hard to
find job postings, resumes or do any other type of search. Googling
resume+sas or job opening+sas is quick and fairly effective (less a
few airline jobs). Doing that with R
There was a nice paper in The American Statistician by Johan Bring (1994.
How to standardize regression coefficients. The American Statistician
48(3):209-213) pointing out that comparing ratios of t-test statistic
values (for null hypothesis that parameter = 0) is equivalent to comparing
Hi All,
does anyone ever import old SPSS files in a sl3 format?
read.spss('file.sl3') does not seem to work... it's not recognised as
a supported SPSS format at all.
Best,
Fede
--
Federico C. F. Calboli
Department of Epidemiology and Public Health
Imperial College, St. Mary's Campus
Norfolk
Thank you Dan!
I dunno if I would have ever found that.
Thanks!
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 [EMAIL PROTECTED]
Shubha Vishwanath Karanth shubhak at ambaresearch.com writes:
Hi R,
Are there any limitations on the capacity of the data to hold for R data
frames or zoo objects? I mean to ask are there any restrictions on the
number of rows or column in the R data frames or the zoo objects?
Thank you,
If I read in an .xml file eg with
xeg - xmlTreeParse(system.file(exampleData, test.xml,
package=XML))
It appears to be OK however examining it with str() gives an apparent
error
str(xeg,2)
List of 2
$ doc:List of 3
..$ file: list()
.. ..- attr(*, class)= chr [1:2] XMLComment XMLNode
i have been trying to load the package ncdf using the command
library(ncdf).
below is my hardware information.
platform i686-pc-linux-gnu
arch i686
os linux-gnu
system i686, linux-gnu
status
Hello everyone,
I am having issues with the R copula package. Just trying to run the
example I found at this site gives me an error.
When I implement this code:
mycop - tCopula(param=0.5, dim=8, dispstr = ex, df =5)
myfit - fitCopula(x, mycop, c(0.6,10),
On 2/7/07, Vikas Rawal [EMAIL PROTECTED] wrote:
How can I superimpose some text labels on ggplot? I could get
weighted quantiles using wtd.quantiles function in Hmisc package. But
I can't plot these as labels on the boxplot.
My code is as follows.
list(c(1:3),c(1:3),c(1:3))-t
Hi,
I'm completely new to R, I am all at sea with the interface and the
confusing help files, so would appreciate some help to do a simple task.
Need to present the mean and variance of 100 different samples of poisson
distributions (N=1000, with fixed lambda) in a file in two columnns, and
Do you mean that you have 100 samples, each of size 1000. If this is so,
you can perhaps do:
N = 1000
n = 100
x = matrix(rpois(N*n, 3.1), ncol=100) # Generate the appropriate no. of
Poisson samples and rearrange into 100 columns of 1000
output = cbind(means=apply(x,2,mean),
That may indicate lack of fit. If the data is generated from the
t-copula, this worked for me:
set.seed(123)
mycop - tCopula(param=0.5, dim=8, dispstr = ex, df =5)
x - rcopula(mycop, 1000)
myfit - fitCopula(x, mycop, c(0.6,10),
optim.control=list(trace=1),method=Nelder-Mead)
myfit
On 2/7/07, aat
Hello,
I have a dataset with 90 columns and 300 rows. I am plotting a
scatterplot,
splom(~DD[1:20],data=dd,cex=1,pch=.)
on a 30 display. When i stretch the display to fill the screen the
20x20 boxes, resize, but don't fill the screen, i.e during the
initial draw there
Martin Maechler wrote:
Frank == Frank E Harrell [EMAIL PROTECTED]
on Tue, 06 Feb 2007 21:59:45 -0600 writes:
Frank Matthew Keller wrote:
Bob,
Far from flaming you, I think you made a good point - one
that I imagine most people who use R have come
Dear altogether,
I want to blank the lower (or upper) part of a correlation matrix as it
is done by dist()
example:
( d - cor(matrix(runif(12),nrow=4)) )
If I do the following
d[lower.tri(d)] -
of course everything is changed to character - that's not what should be.
Additionally, it does
On 07-Feb-07 Thor wrote:
Hi,
I'm completely new to R, I am all at sea with the interface
and the confusing help files, so would appreciate some help
to do a simple task.
Need to present the mean and variance of 100 different samples
of poisson distributions (N=1000, with fixed lambda) in
How about this:
x
12
1 joe 0.45
2 mike 0.34
3 jim 0.25
combine - combn(3, 2)
combine
[,1] [,2] [,3]
[1,]112
[2,]233
ans - cbind(x[combine[1,],2], x[combine[2,], 2])
rownames(ans) - paste(x[combine[1,], 1], x[combine[2,], 1], sep='.')
ans
[,1]
You can try
as.dist(d)
Christos Hatzis, Ph.D.
Nuvera Biosciences, Inc.
400 West Cummings Park
Suite 5350
Woburn, MA 01801
Tel: 781-938-3830
www.nuverabio.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Leo Gürtler
Sent: Wednesday, February
On 07-Feb-07 Ted Harding wrote:
On 07-Feb-07 Thor wrote:
[...]
So far I have figured out:
N - 1000
x - rpois(N, 3.1) ,
Comment: The Poisson distribution has only one parameter, lambda,
so it should be rpois(N, lambda), e.g. rpois(N, 3). You will get
an error with your second parameter
And if you want to know how it is done, take a look at
stats:::print.dist
-Christos
-Original Message-
From: Christos Hatzis [mailto:[EMAIL PROTECTED]
Sent: Wednesday, February 07, 2007 2:16 PM
To: 'Leo Gürtler'; 'r-help@stat.math.ethz.ch'
Subject: RE: [R] blank upper or lower
I tried to fit data with the following function:
fit-nls(y~ Is*(1-exp(-l*x))+Iph,start=list(Is=-2e-5,l=2.3,Iph=-0.3
),control=list(maxiter=500,minFactor=1/1,tol=10e-05),trace=TRUE)
But I get only a singular Gradient warning...
the data can by found attached(there are two sampels of data col
Dear R users,
I have a specific question about isoMDS. Imagine the following (fake)
distance table:
hamburg bremen berlin munich cologne
hamburg 0911982677 424
bremen 911 0293547 513
berlin 982293 0785 875
munich 677
Hi all.
I have what I'm guessing is a fairly easy question.
I want to plot groupedData objects, but there are a large number of subjects in
my data. When I use the simple command
plot(MyData)
where MyData is a groupedData object, there are simply too many individual
plots to see anything. I
Ashley == Ashley Ford [EMAIL PROTECTED]
on Wed, 07 Feb 2007 17:18:56 + writes:
Ashley If I read in an .xml file eg with
xeg - xmlTreeParse(system.file(exampleData, test.xml,
package=XML))
Ashley It appears to be OK however examining
Hello
I have a dataframe that looks like this
MSA CITY HIVEST YEAR YR CAT
1 0200 Albuquerque 0.50 1996 1996 5
2 0520 Atlanta13.00 1997 1997 5
3 0720 Baltimore 29.10 1994 1994 1
4
?aggregate says:
... the result is reformatted into a data frame containing the variables in
by and x. The ones arising from by contain the unique combinations of
grouping values used for determining the subsets, and the ones arising from
x the corresponding summary statistics for the subset of
Dear Listers,
I have a regression problem (x-y) with biological data, where x influences
y in two ways, (1) y increases with x and (2) the variation around the mean
(residuals) decreases with increasing x, i.e. y becomes more 'predictable'
as x increases.
The relationship is saturating, y~a
If you haven't already you might want to take a look at:
http://www.econ.uiuc.edu/~roger/research/rq/QReco.pdf
which is written by and for ecologists.
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox:
This is probably a simple problem but I don't see a
solution.
I have a data.frame with a number of columns where I
would like 0 - NA
thus I have df1[,144:157] - NA if df1[, 144: 157] ==0
and df1[, 190:198] - NA if df1[, 190:198] ==0
but I cannot figure out a way do this.
cata - c(
Robert: Just to provide a little follow up on Roger Koenker's response,
there indeed has been some interest and publications using quantile
regression for estimating species richness-productivity patterns in
ecology. And Roger's quantreg package available for R is a great
statistical tool
John -
Your initial problem uses 0, but the example uses 1 for the value that
gets an NA. My solution uses 1 to fit with your example. There may be
a better way, but try something like
data1[3:5] - data.frame(lapply(data1[3:5], function(x) ifelse(x==1, NA,
x)))
The data1[3:5] is just a
On 2/7/07, This Wiederkehr [EMAIL PROTECTED] wrote:
I tried to fit data with the following function:
fit-nls(y~ Is*(1-exp(-l*x))+Iph,start=list(Is=-2e-5,l=2.3,Iph=-0.3
),control=list(maxiter=500,minFactor=1/1,tol=10e-05),trace=TRUE)
But I get only a singular Gradient warning...
Did you
Works beautifully. I modified it a bit to handle the
discontinous ranges to:
a - c(3:4, 8)
data1[a] - data.frame(lapply(data1[a], function(x)
ifelse(x==1,
NA,
x)))
There may be a prettier way to handle the disconituity
but this works so it looks like I'm in good shape.
I had looked at
hello,
i have a couple of .R files distributed about my file system. i commonly
source() these from other files, but i have to include the full file path. this
is not always convenient if you move files around. is there a way of setting
the search path for source()?
thanks a lot!
cheers,
I have a specific question about isoMDS. Imagine the following (fake)
distance table:
hamburg bremen berlin munich cologne
hamburg 0911982677 424
bremen 911 0293547 513
berlin 982293 0785 875
munich 677
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