Hi all,
is it possible to perform a linear-by-linear association test (lbl_test,
package coin) for a SINGLY ordered table ? And, if yes, how to manage the
scores ?
Thanks in advance
Gilles - France
[[alternative HTML version deleted]]
__
A communication matrix K is such that : K * VEC(A) = VEC(transpose of A).
Is there any readily available R function to find that Communication matrix?
Thanks
--
View this message in context:
http://www.nabble.com/How-to-get-%22Communication-matrix%22---tp21525220p21525220.html
Sent from the R
On Sun, 18 Jan 2009, Jörg Groß wrote:
Hi,
is there a way to increase the distance beween the plot (or plot-region) and
the main-title?
I haven't found anything via ?par().
Well then, you need to look again: you want a bigger margin, so look
for mar and mai.
Once you have increaed teh
Hi,
Im using R-2.8.1 on windows vista and have 4GB RAM. Im trying to run LDA
from the MASS package on a fairly large dataset and keep running out of
memory (Cannot allocate vector of size ...)
Ive tried freeing up as much memory as possible with gc(). I tried using the
ff package but that would
Hi
I was wondering if there was a way to run system when there
is e.g. a C program which is not totally command-line driven.
That is, I have a C program which requires manual input of
options into the shell during the run. I would like to run this program from
R and as far as I have seen system()
one way is the following:
dat - read.table(textConnection(id time y
1 1 10
1 2 12
1 3 15
1 6 18
2 1 8
2 3 9
2 4 11
2 5 12
3 1 8
3 4 16
4 1 9
4 5 13
5 1 7
5 2 9
5 6 11), header = TRUE)
closeAllConnections()
val - 4
dat. - data.frame(id = unique(dat$id), time = val)
out - merge(dat, dat., all =
On 17/01/2009 10:34 PM, Paul Johnson wrote:
On Sat, Jan 17, 2009 at 4:00 PM, Duncan Murdoch murd...@stats.uwo.ca wrote:
On 17/01/2009 4:29 PM, Ben Bolker wrote:
Duncan Murdoch murdoch at stats.uwo.ca writes:
R is open source, so this is no mystery: if you use [noae] then Sweave
won't use
Please RTFM, specifically the rw-FAQ and ?Memory-limits. The short
answer is that your OS is your problem, and using a decent 64-bit OS
will solve this (so why should the very limited resources of the R
developers be spent working on this?)
People who must use Windows may like to be aware
See ?pipe. You can pipe input to a program you run that way.
On Sun, 18 Jan 2009, Hadassa Brunschwig wrote:
Hi
I was wondering if there was a way to run system when there
is e.g. a C program which is not totally command-line driven.
That is, I have a C program which requires manual input of
Dear All,
Could someone give me some pointers (just a guide as to what functions I
need to look at would be fine)
as to how I go about this simple problem please;
The problem looks like this;
Choose x1 to x4 such that you minimize the MAXIMUM ABSOLUTE value returned
in the vector result of this
Try this. 'by' splits up the data frame into one data frame
per id and then f acts separately on each such sub-dataframe
returning a ts series with NAs for the missings. cbind'ing
those all together gives us this series with one column
per id:
tt
Time Series:
Start = 1
End = 6
Frequency = 1
I'm not sure that it comes across as a matrix problem in the sense
of involving matrix algebra, but it's also possible that I don't
understand what you are saying. Seems that an appropriate application
of:
?abs
?max
?sapply
?which.min
... ought to do the trick. If you were hoping for
Hello,
i have a xls file. I will read it in r, what library-command i use for
this??
any ideas??
Thanks
Michele
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
Hi Glenn,
perhaps I'm misunderstanding your problem, but it seems to me that the
zero vector (yielding a zero vector after the multiplication) will
happily minimize everything you want.
Apart from that, you may want to look at the Optimization Task View on CRAN.
Good luck,
Stephan
glenn
See the R Import/Export manual. Also
RSiteSearch(import excel)
gives many hits. It seems as if this question is
asked almost daily.
On Sun, Jan 18, 2009 at 9:15 AM, Michele Santacatterina
miksa...@gmail.com wrote:
Hello,
i have a xls file. I will read it in r, what library-command i use for
One idea would be to use search facilities. At a guess this has
already been asked and answered 100-200 times. I know for a fact that
it was asked and answered twice in the last week. The four search
engine links on the mailing list page are here replicated:
Hello,
I have a dataset (named x) with many (966) columns. What I would like to do
is delete any columns that do not have at least 375 non-blank observations
(i.e., the cells have some value in them besides NA).
How can I do this? I have come up with the following code to _count_ the
Dear all,
I'm using igraph for some analysis about the network I have. I have a
question about the function power.law.fit.
I wonder if there is any test for checking whether the power.law.fit is
good for the input, i.e., under which situation, could we use this function
to get a reliable
Something like this should work:
num - apply(yourData, 2, function(x) sum(is.na(x)) 375)
yourData - youData[, num]
On Sun, Jan 18, 2009 at 9:55 AM, Josh B josh...@yahoo.com wrote:
Hello,
I have a dataset (named x) with many (966) columns. What I would like to do
is delete any columns
colSums(is,na(x) ) can replace your function and negative indexing can
eliminate the unwanted columns:
x[-(colSums(is.na(x)) 375)]
or equivalently:
x[(colSums(is.na(x)) = 375)]
You could (destructively) assign the result to x if you are brave.
--
David Winsemius
On Jan 18, 2009, at 9:55
Hi Michele, there are at least two good alternatives.
xlsReadWrite (read only xls)
RODBC (the one I prefer, read xls and xlsx), odbcConnectExcel (read xls) and
odbcConnectExcel2007 (read xlsx).
The libraries help files bring to you better details.
Here are some examples of file importing (the
Is this what you want:
y - scale(x)
str(x)
int [1:10] 1 2 3 4 5 6 7 8 9 10
str(y)
num [1:10, 1] -1.486 -1.156 -0.826 -0.495 -0.165 ...
- attr(*, scaled:center)= num 5.5
- attr(*, scaled:scale)= num 3.03
y
[,1]
[1,] -1.4863011
[2,] -1.1560120
[3,] -0.8257228
[4,] -0.4954337
Hi Brian,
Thanks for the answer. I guess ill move to a 64-bit linux. I did RTFM, but
didnt find it as informative as your reply, or amusing ;)
Maybe the FAQ can be modified a bit for clarity.
Thanks.
Best,
Vishwa
On Sun, Jan 18, 2009 at 3:55 AM, Prof Brian Ripley rip...@stats.ox.ac.ukwrote:
Dear Rxperts,
I have a varaibles data file that looks like this
p(1) 10
p(1) 3
p(1) 4
p(2) 20
p(2) 30
p(2) 40
p(3) 4
p(3) 1
p(1) 2
I cannot process these data with R because it does not like the parentheses.
How can I get these to look like:
p1 10
p1 3
p1 4
p2 20
p2 30
p2 40
p3 4
p3 1
p3 2
use
install.package(RODBC)
library(RODBC)
cnct - odbcConnectExcel(filename.xls)
hope this helps
On Sun, Jan 18, 2009 at 9:15 AM, Michele Santacatterina
miksa...@gmail.comwrote:
Hello,
i have a xls file. I will read it in r, what library-command i use for
this??
any ideas??
Thanks
Hello dear R Users,
I am working on a dataset of 928 Enterprises, of which are observed 12
different characters. I need to randomly sample, without repetition, 70% of
the entreprises, to create a testing set, and let the other 30% of the
enterprises be a validating set (holdout validation, I
If I have 2 data frames;
df1:dim(df1) = (1,10)
df2:dim(df2) = (2000,10)
Both with column header names, how do I multiply them together please. I.e
create a dateframe dim() = (2000,1)
Many Thanks in advance
Glenn
[[alternative HTML version deleted]]
The cc and cp bases in package `mgcv' provide periodic splines,
[e.g. gam(y~s(x,bs=cc))], but this may not be exactly the functionality you
want.
best,
Simon
On Friday 16 January 2009 08:42, cmr.p...@gmail.com wrote:
Hello group!
Is there a package that allows to fit smooth *periodic*
you should first transform them to matrices using data.matrix() and then
do a matrix multiplication using %*%.
This will create a matrix, which you can then convert to a data frame,
if you want, using as.data.frame(). For instance, check the following:
d1 - data.frame(x = rnorm(10), y =
Two other possibilities:
The 'DierckxSpline' package includes a function 'percur' for
fitting periodic splines. Unfortunately, it has a known bug that kills
R with a segmentation fault, though it not affect your application.
The 'fda' package supports the use of finite Fourier
Dear all,
let's assume I have a vector of character strings:
x - c(abcdef, defabc, qwerty)
What I would like to find is the following: all elements where the word
'abc' does not appear (i.e. 3 in this case of 'x').
Since I am not really experienced with regular expressions, I started
slowly
Robbert,
On Friday 16 January 2009 14:30, Robbert Langenberg wrote:
Thanks for the swift reply,
I might have been a bit sloppy with describing my datasets and problem. I
showed the first model as an example of the type of GAM that I had been
able to use the predict function on. What I am
Just remove those elements that match:
x - c(abcdef, defabc, qwerty)
x[-grep('abc',x)]
[1] qwerty
On Sun, Jan 18, 2009 at 1:35 PM, Rau, Roland r...@demogr.mpg.de wrote:
Dear all,
let's assume I have a vector of character strings:
x - c(abcdef, defabc, qwerty)
What I would like to find
Part 1:
I want to plot the CO2 concentration vs. year, but extending the x-axis
using the xlim parameter to include the year 2006 (x axis range of values
are from -41210 to 0), and adjusting the ylim parameter to go up to 400 when
the range of y axis values are from 150 to 300. How do I do this?
Here is one way to do it:
x - matrix(1:100,10)
x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]1 11 21 31 41 51 61 71 8191
[2,]2 12 22 32 42 52 62 72 8292
[3,]3 13 23 33 43 53 63 73 8393
[4,]4 14
Rau, Roland wrote:
Dear all,
let's assume I have a vector of character strings:
x - c(abcdef, defabc, qwerty)
What I would like to find is the following: all elements where the word
'abc' does not appear (i.e. 3 in this case of 'x').
a quick shot is:
x[-grep(abc, x)]
which
Does this give you what you want:
x - read.table(textConnection(p(1) 10
+ p(1) 3
+ p(1) 4
+ p(2) 20
+ p(2) 30
+ p(2) 40
+ p(3) 4
+ p(3) 1
+ p(1) 2), as.is=TRUE)
# remove parenthesis
x$V1 - gsub([()], , x$V1)
x
V1 V2
1 p1 10
2 p1 3
3 p1 4
4 p2 20
5 p2 30
6 p2 40
7 p3 4
8 p3 1
9 p1 2
? gsub
gsub(\\(|\\), , var)
You can then read.table on a textConnection.
read.table(textConnection(gsub(\\(|\\), , var) ))
V1 V2
1 p1 10
2 p1 3
3 p1 4
4 p2 20
5 p2 30
6 p2 40
7 p3 4
8 p3 1
9 p1 2
On Jan 18, 2009, at 12:13 PM, oscar linares wrote:
Dear Rxperts,
I have a varaibles
Try this:
# indexes
setdiff(seq_along(x), grep(abc, x))
# values
setdiff(x, grep(abc, x, value = TRUE))
Another possibility is:
z - abc
x0 - c(x, z) # to handle no match case
x0[- grep(z, x0)] # values
On Sun, Jan 18, 2009 at 1:35 PM, Rau, Roland r...@demogr.mpg.de wrote:
Dear all,
Hi Saurabh
saurabh_koparkar schrieb:
Part 1:
I want to plot the CO2 concentration vs. year, but extending the x-axis
using the xlim parameter to include the year 2006 (x axis range of values
are from -41210 to 0), and adjusting the ylim parameter to go up to 400 when
the range of y axis values
Roland,
I think you were almost there with your first example. Howabout using:
x - c(abcdef, defabc, qwerty)
y - grep(pattern=abc, x=x)
z.char - x[-y]
z.index - (1:length(x))[-y]
z.char
[1] qwerty
z.index
[1] 3
Cheers,
eric
Rau, Roland wrote:
Dear all,
let's assume I have a vector
read.table(textConnection(gsub(\\(|\\), , var) )) #from prior
posting
V1 V2
1 p1 10
2 p1 3
3 p1 4
4 p2 20
5 p2 30
6 p2 40
7 p3 4
8 p3 1
9 p1 2
ridxs - sample(1:nrow(df),floor(0.7*nrow(df)) ) # the 70% sample
row IDs
df[ridxs,]
V1 V2
5 p2 30
6 p2 40
2 p1 3
7 p3 4
4 p2 20
8
Jorge Ivan Velez wrote:
Hi Wacek,
I think you wanted to say strings instead x in your last line : )
of course, thanks. the correct version is:
if(length(matching - grep(pattern, strings)))
strings[-matching]
else strings
btw., and in relation to a recent post complaining about
Thank you very much to all of you for your fast and excellent help.
Since the -grep(...) solution seems to be favored by most of the answers, I
just wonder if there is really no regular expression which does the job?!?
Thanks again,
Roland
-Original Message-
From: Gabor Grothendieck
Hi
-Original Message-
From: r-help-boun...@r-project.org on behalf of saurabh_koparkar
Sent: Sun 1/18/2009 5:12 PM
To: r-help@r-project.org
Subject: [R] Formatting the axis of plot() to shown our own values.
Part 1:
I want to plot the CO2 concentration vs. year, but extending the
On 18/01/2009, at 7:55 PM, Pedro Mardones wrote:
Dear all;
I have a function written in R that returns as a list of values as
output that has associated some user defined attributes to it. How can
hide these attributes when printing the output on screen? I'm using
R-2.8.1 on WinXPit's like
Try this:
grep(^([^a]|a[^b]|ab[^c])*.{0,2}$, x, perl = TRUE)
On Sun, Jan 18, 2009 at 2:37 PM, Rau, Roland r...@demogr.mpg.de wrote:
Thank you very much to all of you for your fast and excellent help.
Since the -grep(...) solution seems to be favored by most of the answers,
I just wonder if
Gabor Grothendieck wrote:
Try this:
# values
setdiff(x, grep(abc, x, value = TRUE))
Another possibility is:
z - abc
x0 - c(x, z) # to handle no match case
x0[- grep(z, x0)] # values
on quick testing, these two and the if-based version have comparable
runtime, with a minor win for
In that case just add fixed = TRUE
On Sun, Jan 18, 2009 at 2:58 PM, Wacek Kusnierczyk
waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
Gabor Grothendieck wrote:
Try this:
# values
setdiff(x, grep(abc, x, value = TRUE))
Another possibility is:
z - abc
x0 - c(x, z) # to handle no match case
Hi,
I'm newish to R and Ubuntu, and I've getting this error when I'm trying
to install ggobi in R. Any suggestions?
DD
*R version 2.7.1 (2008-06-23)
Copyright (C) 2008 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
R is free software and comes with ABSOLUTELY NO WARRANTY.
You
Gabor Grothendieck wrote:
In that case just add fixed = TRUE
in general, if you want a complex pattern, you don't use 'fixed', and
then again you risk incorrect (well, correct for r, but not for the
problem) result in case no input string matches the pattern.
vQ
Dirk,
Thanks for the quick response. I've installed that package - once I
have it installed, how do I use rggobi inside R?
DD
Dirk Eddelbuettel wrote:
On 18 January 2009 at 15:19, Dirty D wrote:
| I'm newish to R and Ubuntu, and I've getting this error when I'm trying
| to install ggobi
Hello everybody!
I'm having this problem with the auto.arima function that i've not been
able to solve. I use this function on time series that contains NA values,
but every time that the resulting model contains drift I can't perform a
forecasting (using forecast.Arima function). The printed
Is there any R package which implements Pseudo-F-Statistics Clustering ?
Vogel and Wong derived a formula to evaluate the best clusters number through
such a method.
Silhouette metod only provides an average evaluation.
Thank you very much.
Maura
tutti i telefonini TIM!
Gabor Grothendieck wrote:
Try this:
grep(^([^a]|a[^b]|ab[^c])*.{0,2}$, x, perl = TRUE)
... and see how cumbersome it becomes for a pattern as trivial as 'abc'.
in perl, you typically don't invent such negative patterns, but rather
don't match positive patterns: instead of the match
Wacek Kusnierczyk wrote:
On Sun, Jan 18, 2009 at 2:37 PM, Rau, Roland r...@demogr.mpg.de wrote:
Thank you very much to all of you for your fast and excellent help.
Since the -grep(...) solution seems to be favored by most of the answers,
I just wonder if there is really no regular
Dear Rxperts,
I have the following data:
Study Study.Name C Category TC Time QC QO SD FSD Theta
1 NONE 0P(22) 0 0.00 7.5596 0 0 8.0361e-03 0
1 NONE 6G(50) 0 0.00 1. 0 0 0.e+00 0
1 NONE 2F(02) 0 0.00
Wacek Kusnierczyk wrote:
# r code
ungrep = function(pattern, x, ...)
grep(paste(pattern, (*COMMIT)(*FAIL)|(*ACCEPT), sep=), x,
perl=TRUE, ...)
strings = c(abc, xyz)
pattern = a[a-z]
(filtered = strings[ungrep(pattern, strings)])
# xyz
this was a toy example, but if you need this
Dear Oscar,
You don't need grep() to do what you want:
Data[Data$Category == G(50), c(Time, QC)]
Time QC
2 0.00 1.
6 0.01 1.0001
7 0.02 1.0003
8 0.03 1.0004
9 0.04 1.0005
10 0.05 1.0007
11 0.06 1.0008
12 0.07 1.0009
13 0.08 1.0011
14 0.09 1.0012
15 0.10 1.0013
I hope this
Thanks! (I have to admit, though, that I expected something simple)
Thanks,
Roland
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: Sun 1/18/2009 8:54 PM
To: Rau, Roland
Cc: r-help@r-project.org
Subject: Re: [R] regex - negate a word
Try this:
Well, that's why it was only provided when you insisted. This is
not what regexp's are good at.
On Sun, Jan 18, 2009 at 4:35 PM, Rau, Roland r...@demogr.mpg.de wrote:
Thanks! (I have to admit, though, that I expected something simple)
Thanks,
Roland
-Original Message-
From:
On 19/01/2009, at 10:44 AM, Gabor Grothendieck wrote:
Well, that's why it was only provided when you insisted. This is
not what regexp's are good at.
On Sun, Jan 18, 2009 at 4:35 PM, Rau, Roland r...@demogr.mpg.de
wrote:
Thanks! (I have to admit, though, that I expected something simple)
That's an entirely different point from whether regular expressions
can do it as grep -v is just another way to do it without using a regular
expression to specify the entire job.
On Sun, Jan 18, 2009 at 5:02 PM, Rolf Turner r.tur...@auckland.ac.nz wrote:
On 19/01/2009, at 10:44 AM, Gabor
On Sun, Jan 18, 2009 at 2:22 PM, Wacek Kusnierczyk
waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
x - c(abcdef, defabc, qwerty)
...[find] all elements where the word 'abc' does not appear (i.e. 3 in this
case of 'x').
x[-grep(abc, x)]
which unfortunately fails if none of the strings in x
Note that the variation of this that I posted already handles that case.
On Sun, Jan 18, 2009 at 5:32 PM, Stavros Macrakis macra...@alum.mit.edu wrote:
On Sun, Jan 18, 2009 at 2:22 PM, Wacek Kusnierczyk
waclaw.marcin.kusnierc...@idi.ntnu.no wrote:
x - c(abcdef, defabc, qwerty)
...[find] all
Hi All, some help on this would be appreciated;
Understood combinations(7,4) returns all the possible 4 part combinations
out of 7. Is is possible to substitute the ³7² for a list of stuff you would
like to see the mix of; c(³a²,...,²g²) say ?
Thanks
Glenn
[[alternative HTML version
Dear Glenn,
Try this:
# Your example
choose(7,4)# 35
require(forward)
fwd.combn(7,4) # a 4x35 matrix
# Other possibilities
sapply(4:10,function(x) choose(x,4))
sapply(4:10,function(x) fwd.combn(x,4))
HTH,
Jorge
On Sun, Jan 18, 2009 at 6:06 PM, glenn g1enn.robe...@btinternet.com
Hi all,
I am trying to write a random walk metropolis hastings algorithm, I using
the latest debian distribution of R.
Can anyone tell me what I need to insert in the below code? I have tried
putting various combinations of curly brackets and punctuation between the
acc=1 and if statements, all
Thank you everybody, problem solved! :)
David Winsemius wrote:
read.table(textConnection(gsub(\\(|\\), , var) )) #from prior
posting
V1 V2
1 p1 10
2 p1 3
3 p1 4
4 p2 20
5 p2 30
6 p2 40
7 p3 4
8 p3 1
9 p1 2
ridxs - sample(1:nrow(df),floor(0.7*nrow(df)) ) # the
this may be slightly off-topic, as it doesn't pertain directly to the
R application, but some of the documentation.
when reading R's fullrefman.pdf (available from
http://cran.r-project.org/doc/manuals/fullrefman.pdf) in Mac OS X's
preview.app (version 4.1, on Mac OS 10.5.x), if i try to do a
Hi,
I'm planning to access R from my perl scripts.
The only noteworthy bridge seems to be
Statistics-R-0.03http://search.cpan.org/%7Ectbrown/Statistics-R/lib/Statistics/R.pm.
Would anyone like to share their experience with this Perl-R bridge?
I'd like to install it in a Mac OS X.
Suggestions on
The comma *before* acc=1 ?
I also wondered whether (further up) this should work:
s2y[i,]=s2y[i-1] # would think this to result in a dimension mismatch
This looks sketchy as well:
s2y[i,j] = s2y[b[i-1,j] + rnorm(1,mean=0, sd=s2yscale[j])
^ ^ ^ # unmatched sqr-brackets
It
I am working on a methodology for qualifying R, for GLP and GCP.
If I quailfy only the base R install, with no contributed packages, it
seems relatively simple to qualify R. However, from time to time I
will want to use a contributed package. If I use a contributed package,
does it leave
hi
are you using a 64bit system? 32 bit systems can only allocate about
3GB to a single process.
http://msdn.microsoft.com/en-gb/library/aa366778.aspx
I used to use 32bit winXP, then moved to 64bit Ubuntu8.04 to solve my
memory problems.
if you are 64bit, you should try playing around with the
I've been going back to old difficult R-list evaluation emails that I
save in order to understand evaluation better and below still confuses
me. Could someone explain why A) works and B) doesn't. A variant of
below is in the Pat's Inferno book also but I'm still not clear on what
is
Note that
rm(i)
for(j in 1:4) F(j)
raises an error due to scoping issues.
On Sun, Jan 18, 2009 at 10:02 PM, markle...@verizon.net wrote:
I've been going back to old difficult R-list evaluation emails that I save
in order to understand evaluation better and below still confuses me. Could
On 19/01/2009, at 3:14 PM, Murray Cooper wrote:
I am working on a methodology for qualifying R, for GLP and GCP.
If I quailfy only the base R install, with no contributed packages, it
seems relatively simple to qualify R. However, from time to time I
will want to use a contributed package. If
Dear All,
I am economist and working on poverty / income inequality. I need descriptive
statitics like the ratio of education expentitures between different income
quintiles where each household has a different weight. After a bit of
google search I found 'Hmisc' and 'quantreg' libraries for
Thanks to everyone who helped me when I was totally clueless. Now I'm only
partially clueless, and in writing functions, which is major progress.
This is a function that is a major improvement over which I was using MS
Excel for:
function()
{
files - list.files(pattern=ABN*) # All the
Dear Gregg,
Take a look at ?duplicates. Here is an example:
z-data.frame(
ID=c(1,1,2,1,1,1,1,2,2,3,4,3,2,2,2,2,3,4,4,5,5,6,6,7),
y=rnorm(24)
)
z[!duplicated(z$ID),]
See ?duplicated for more information.
HTH,
Jorge
On Sun, Jan 18, 2009 at 11:27 PM, greggal...@gmail.com wrote:
Hello,
On 1/18/09, Michele Santacatterina miksa...@gmail.com wrote:
i have a xls file. I will read it in r, what library-command i use for
this??
This has been discussed recently. Please search the archives for `excel'.
Liviu
--
Do you know how to read?
John,
Thanks for the tip on that document; after some searching I found a PDF copy
of it on the 'Net. It may take me a little while to work through it, but it
looks to be full of good info and worth the effort.
I'll also take a look at leaps as I get time.
Thanks,
Monte
[[alternative
Hello R-help community,
I have another question about filtering datasets.
Please consider the following toy data matrix example, called x for
simplicity. There are 20 different individuals (ID), with information about
the alleles (A,T, G, C) at six different loci (Locus1 - Locus6) for each
On 1/8/09, René J.V. Bertin rjvber...@gmail.com wrote:
Hello - and happy newyear to all of you!
I've got some data that I'm plotting with bwplot, a 3x2x3 design where
the observable decreases with the principle independent factor, but at
different rates.
I'd like to get lattice to
84 matches
Mail list logo