When running [1] "R version 2.15.1 (2012-06-22)" "x86_64-pc-mingw32",
rJava fails. I have installed both the 32-bit and 64-bit versions of
Java 7 update 9.
> library(rJava)
Error : .onLoad failed in loadNamespace() for 'rJava', details:
call: stop("No CurrentVersion entry in '", key, "'! Try re
"v"))
}
setMethod('v', signature(x='character'),
function(x,y=x,z=x){
v.print(x,y=y,z=z)
}
)
maybe that's not the right way... any ideas?
thx very much in advance!
robert
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just a note i liked to add:
so the idea is just to have the optional arguments y and z showing up in the
the v-function call using the tab-key
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Sent from
found the solution:
v <- function(x, ...) UseMethod("v")
setMethod('v', signature(x='character'), function(x, ...) v.print(x, ...))
setMethod('v', signature(x='numeric'), function(x, ...) v.numeric(x, ...))
v.print <- function(x,y=x,z=x) print(paste(x,y,z))
v.numeric <- function(x,u=3*x) print(u
<- as.matrix(expand.grid(rep(list(x), length(x
b[!sapply(1:nrow(b), function(r) any(duplicated(b[r,]))),]
}
It works, but words cannot describe its ugliness. And it gets really
slow really fast with growing x.
So, anyway. My two questions are:
1. Does R really, really, seriously lack a permut
== 1 ) return( matrix( x, nrow=1 ) )
>lst <- lapply( seq_along( x )
> , function( i ) {
> cbind( x[ i ], perm.jdn( x[ -i ] ) )
> }
> )
>do.call(rbind, lst)
> }
Nice, exactly what I was
Suppose that a data frame has been created by the user. Perhaps it has been
created using the library quantmod. Is there any command to find out what
the members of the data frame is?
Thanks
Bob
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I created the following file:
symbol,shares
XOM,1000
APA,400
CVX,200
I then read the file in R using the command:
stockList=read.table("/NotesOnR/stockList", header = T, sep=",")
I would then expect the following expression to evaluate to the simple
string APA:
stockList$symbol[
geom_smooth. The particular intercept does not mater ( I
don't care what the intercept is, though given a choice I'd prefer grand
mean centered) I would be happy with either conditional on unconditional
CI's.
Robert
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___
Please consider the following R Script:
x = c(1,2,3)
y = c(1,2,9)
cor(x,y)
These three lines will produce, as I expected, the correlation between
the variables x and y. However, R is going to assume that the
probability that x = 1 is the same as the probability that x = 2.
d P-values for Wald tests for the terms in the model (when
Terms and L are NULL), a combination of model terms (when Terms in not
NULL), or linear combinations of the model coefficients (when L is not
NULL).
noticeably absent is the sum of squares.
How do I get them?
Robe
I am trying to figure out how to use rgl package for animation. It
appears that this is done using the play3d() function. Below I have
some sample code that plots a 3D path and puts a sphere at the point
farthest from the origin (which in this case also appears to be at the
end of the path).
On 11/3/2012 6:47 AM, Duncan Murdoch wrote:
On 12-11-02 7:47 PM, Robert Baer wrote:
I am trying to figure out how to use rgl package for animation. It
appears that this is done using the play3d() function. Below I have
some sample code that plots a 3D path and puts a sphere at the point
ailing 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.
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Professor of Physiology
Kirksville College of Os
On 11/4/2012 4:32 AM, Robert Baer wrote:
Some hints:
For pdf(), height and width are in inches, not pixels. dev.off() is
necessary after drawing the image for pdf(). The name for the file
argument (file="c:/figure.xxx") is file not filename
hist(CO2[,5]) is more interesting
And
rgl function writeWebGL()
over the weekend, and I think it holds tremendous potential for sharing
visualizations of multidimensional data with a non-technical audience.
Who knew it could be so easy to move three dimensional data from R to
the web?
Rob Baer
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Robert W. Ba
avaScript additions you can
get there.
Rob
>
>
>
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reprodu
anyone have an idea for a more clever way to preform what is effectively a
simple quadratic programming problem on a discontinuous domain: {0, [minp,
maxp]}?
Thanks, Robert
Robert McGehee, CFA
Geode Capital Management, LLC
One Post Office Square, 28th Floor | Boston, MA | 02109
Direct: (617)392-83
w.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
__
Robert W Baer, Ph.D.
Professor of Physiology
Kirksville College of Osteopathic Medicine
A. T. Still University of Health Sciences
Kirksville, MO 63501 US
[[alternative HTML
On 12/1/2012 10:50 AM, Andras Farkas wrote:
Dear all,
could you please give me some pointers on how I could make R screen for a value if it falls within a certain range?
I looked at the subset function, but is not doing it, perhaps because I only
have 1 value to screen?
aptreciate the inp
ct.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
__
Robert W. Baer, Ph.D.
Professor of Physiology
Kirksille College of Osteopathic Medicine
A. T. Still University of Health Sciences
Kirksville, MO 63501 USA
e.com.
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__
Robert W. Baer, Ph.D.
Profess
Hello fellow
R-users,
Â
Iâm stuck
with something i think is pretty stupid, but i canât find out where iâm
wrong,
and itâs turning me crazy!
Â
I am doing
a very simple linear regression with Northing/Easting data, then I plot the
data as well as the regression line :
Â
> plot(x=Dataset
x27;t know what exactly to install. Thank you.
Robert
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On 1/4/2014 7:42 PM, Peter Turner wrote:
Hi, I hope the following question is appropriate for the list; reflects
that I've yet to use R and have limited statistical sensibility.
I've two metal ion concentration data sets, one each for two nearby
watercourses recorded over the same period (2008
ls() doesn't do
it, neither does args() or alist().
How is that done?
Thanks,
robert
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and pro
On Sun, 09 Feb 2014 12:28:11 +
Rui Barradas wrote:
> Hello,
>
> Inside the function try
>
> dots <- list(...)
Hi guys,
thanks a lot. I knew it HAD to be something ultra-simple, like most
things in R.
Regards,
robert
__
R-
ich I
have, and like, but am not sure I should recommend).
Good luck,
robert
On Tue, Feb 25, 2014 at 8:34 PM, Lib Gray wrote:
> Hello,
>
> I am branching out to xyplot for the first time, and I want to layer
> several "complex" xyplots. I have tried using panel functio
anging the call to have byrow=TRUE does not change the
plot.
Thanks
Robert
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t
Error in MEEM(object, conLin, control$niterEM) :
Singularity in backsolve at level 0, block 1
I suspect I could solve this problem with ordering the levels of "Course"
so that P7A was the first level and thus the one that others were compared
to but I am unclear on how to do so.
Rob
I am trying to fit my data, attached, with the following model
CutOff <- 0
fit.full <- lme(fixed= zGrade ~ Rep + ISE +Yfrm7A +Ufrm7A +Female +White
+HSGPA +MATH +AP_TOTAL +Years +Course +
Course*Rep + Course*ISE +Course*Yfrm7A+Course*Ufrm7A
+Course*Female +Course*White +Course*H
o I need two or three
centered variables? and is there a quick way to get at all those variables
if my factor has many levels, e.g. 14?
Robert
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Dear all,
I have been trying to find a simple solution to my problem without success,
though i have a feeling a simple syntaxe detail coul make the job.
I am doing a polynomial linear regression with 2 independent variables such as :
lm(A ~ B + I(B^2) + I(lB^3) + C, data=Dataset))
R return me
test can be used to
test whether the heterogeity is needed.
Alternatively a generalized Akaike information criterion (GAIC) can be used
to compare models.
Robert Rigby
On Date: Sat, 26 Oct 2013 23:57:51 -0400 (EDTCollin Lynch <
coll...@cs.pitt.edu> wrote
Subject: [R] Heteroscedastici
I am trying to compare two different GAM fits.
I have something like
Course.bam20 <-bam(zGrade ~ Rep + ISE + White + Female + Years + AP_TOTAL
+ MATH + HSGPA+ EOP + factor(P7APrior, ordered = FALSE)+s(Yfrm7A,k=20),
data= Course, na.action = na.exclude,samfrac =0.1)
Course.bam4 <-bam(zGrade ~ Rep
variables.
This can be important for selecting mean model terms
and is particularly important when interest lies in the variance and/or
quantiles
of the response variable.
Robert Rigby
On 06/11/13 21:46, Collin Lynch wrote:
> Greetings, My question is more algorithmic than prectical. What I
On 12/2/2013 9:35 AM, Bert Gunter wrote:
> Not true, Rich.
The point about alphabetical ordering explains why the author likely
explicitly set the levels for the factor, though.
As to why ordered factors, we may never know, but one possible
explanation is that at some point he was going to use
in, control$niterEM) :
Singularity in backsolve at level 0, block 1
but if I take out the /ISE I get no error, simmilarly if I take out the
COHORT/.
I want to test for the effects of the different cohorts within the ISE
subset and across ISE & NONE
I can send the data (the whole is too large
:
Singularity in backsolve at level 0, block 1
but if I take out the /ISE I get no error, simmilarly if I take out the
COHORT/.
I want to test for the effects of the different cohorts within the ISE
subset and across ISE & NONE.
Robert
__
R-hel
You don't really provide enough information like
R.Version()
but my guess is that you are running 64-bit R either directly or through
R Studio but that you have only 32-bit Java installed. I am doing fine
on Windows with Java 7 update 45 but had some 64-bit run issues with
only Java 7 updat
choose the effective degrees of freedom for smoothing are
available, see Rigby and Stasinopoulos (2004, 2005, 2006)
and Stasinopoulos and Rigby (2007).
Robert Rigby
Il 08/12/2013 16.45, Daniel Kehl ha scritto:
> Dear Community,
>
> I am struggling with a growth curve estimation probl
e gamlss manual, Stasinopoulos et al. (2008)
'Instructions on how to use the gamlss package in R, 2nd edition',
or see Chambers and Hastie (1992) 'Statistical Models in S').
Robert Rigby
Date: Tue, 10 Dec 2013 18:27:45 +
From: D?niel Kehl
To: David Winsemius
Cc: &quo
- Original Message -
From: "KENT V.T."
To:
Sent: Monday, June 28, 2010 8:28 AM
Subject: [R] Axes intercept
I have a plot where the values of the y axis go from a positive number to a
negative number and I want the x axis to intercept at zero rather than at
the bottom of the y axis,
0.1.
Am I missing something?
Thanks in advance!
Robert
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self
anks in advance,
Robert Farley
Metro
1 Gateway Plaza
Mail Stop 99-23-7
Los Angeles, CA 90012-2952
Voice: (213)922-2532
Fax:(213)922-2868
www.Metro.net
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3
..@ Dimnames:List of 2
.. ..$ : NULL
.. ..$ : NULL
..@ uplo: chr "U"
..@ factors : list()
i guess there must be some way of doing a "symmetric" assignment of a
value but I have not been able to figure it out myself from the
documentation nor searching the list no
)
lines(density(zTestData$Distance, weights=zTestData$BrdWeight ), col=3)
which works fine except the distances are all positive, but the densities don't
drop to 0 until around -2 or -3.
Is there a way for me to "force" the density plot to 0 at 0?
Thanks
Robert Farley
Met
individual measurements. Is there a
"variable bw" that could be set small at the extrema, and larger in the range
of the data?
Robert Farley
Metro
www.Metro.net
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Monday, July 19, 2010 19:31
To: Fa
McNemar is good for paired data. See
?mcnemar.test
You need to get your data into the form of a matrix (e.g., help example),
and you will need data organized by concordant and discordant pairing.
This means that you will need to organize your data differently than you
show us:
# pairs YES YE
Hello, I am having a problem figuring out how to model a continuous outcome
(y) given a continuous predictor (x1) and two levels of nested categorical
predictors (x3 nested in x2). The data are observational, not from a
designed experiment. There are about 15 levels of x2 and between 3 and 14
level
Hello, I am having a problem figuring out how to model a continuous outcome
(y) given a continuous predictor (x1) and two levels of nested categorical
predictors (x3 nested in x2). The data are observational, not from a
designed experiment. There are about 15 levels of x2 and between 3 and 14
level
Hello, I am having a problem figuring out how to model a continuous outcome
(y) given a continuous predictor (x1) and two levels of nested categorical
predictors (x3 nested in x2). The data are observational, not from a
designed experiment. There are about 15 levels of x2 and between 3 and 14
level
set.seed(180185)
#loop: create 10 times the variables (u1,u2,u3,u4,u5)
for (i in 1:10){
u1 <- c(runif(200,0,1))
u2 <- c(runif(200,0,1))
u3 <- c(runif(200,0,1))
u4 <- c(runif(200,0,1))
u5 <- c(runif(200,0,1))
u <- c(u1,u2,u3,u4,u5)
mu <- matrix(u, nrow=1000, ncol=1)
}
As you can see, when
I have the following formula for a linear model:
z <- lm(y~x + factor(a) + factor(b), data=NT2010)
where a (groups) and b (Sub-groups) are categorical variables (factors), x
is a continuous covariate, and y the response variable. Since b is nested
within a, the formula can also be written as:
It seems David and Phil have given you all the pieces and you are not
listening. Try:
h = sprintf('%06d', seq(07,18, by=1) ) # Note that h is text
myseries=sample(1:12, 12) #fake data to plot
tim=strptime(h,'%H%M%S')
library(zoo)
test=zoo(myseries, tim)
plot(test)
Rob
--
od<-c(10, 8, 6,4,2,1, 10.5,7.8,6.4,3.8,2.1,0.95)
cyto_conc=2650 # Highest cytokine concentration user defined
cyto_std_conc <-c(cyto_conc)
for (i in 1:5)
{
cyto_conc = cyto_conc /3
cyto_std_conc <-c(cyto_std_conc ,cyto_conc)
}
cyto_std_conc<-log2(rep(cyto_std_conc,2))
cyto<-cbind(cyto_s
I would expect that there will be interactions between variables eg. if
the
vegetation is grassland then the vegetation height variable will mediate
the
interaction, if the vegetation is arable then crop type will be more
significant.
Would it be possible to use GLM or GAM models for this type
Hi folks
I have inherited a package which was created before release 2.10 and I
need to have it working in release 2.12
There is a folder containing the Man, Data and R subfolders, and the
description
file, and there is also the corresponding zip file.
I can install the package from the zip f
your advice to useR CMD INSTALL source.library did the trick
and I now understand packages a little better.
many thanks
Bob Kinley
Uwe Ligges
28/10/2010 10:30
To
Robert Kinley
cc
r-help@r-project.org
Subject
Re: [R] updating a local package
I stand corrected :0)
Robert Kinley
Uwe Ligges
28/10/2010 11:54
To
Robert Kinley
cc
r-help@r-project.org
Subject
Re: [R] updating a local package
On 28.10.2010 12:26, Robert Kinley wrote:
> your advice to useR CMD INSTALL source.library
It's a
Hello R Users,
I'm wondering if there exists any elegant and simple way to do the
following: I have a data.frame X fills in numbers. I have a vector y with
numbers as well. Every value in X that is equal to any values in y should
be replaced by e.g. 1. Any idea?
R
Thank you vary much Ben and Erik.
It's exactly what I want. Below is my a little modified example.
set.seed(12345)
X = sample(c(40:60),40,replace=TRUE)
x = matrix(X,nc=5)
y = c(40,43,55,60)
x[x %in% y] <- -1
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ot;prediction")
Error in `[.data.frame`(newdata, , 2) : undefined columns selected
Robert Kinley ( 'baffled' of Berkshire )
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e:(Intercept)
-1.6958 0.5089 6.9764 2629.7526
attr(,"logScale")
[1] TRUE
attr(,"modelType")
[1] "4pl"
attr(,"method")
[1] "drc"
attr(,"class")
[1] "multFit"
Uwe Ligges
09/11/2010 16:2
You install JGR and its dependencies as you would any other package from the
RGUI. Make sure you have a recent java installed and functional. You may
find jselect.exe on the RForge site useful for looking at this.
When I did the install, things seemed a little flaky, but eventually by
caref
Dear R Users,
I'm looking for a package that allows to test hypothesis about a
homogeneity of odds ratio in k 2x2 tables. I know that Breslow-Day is
suitable but does anybody could me point out a package? I found diffR,
but as far as I see this package is for IRT theory.
Best,
R
65, 139, 138)
C6 <- c(47)
...
I would appreciate for any help.
Robert
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and provide commented, minimal, self-contained, reproducible code.
Thank you, but unfortunately it's not resolve my problem. After
writing in R console e.g. a2 a should get the values.
Do you know how to read file 2 into R?
2010/11/19 Joshua Wiley :
> Hi Robert,
>
> What about a named list? This will generalize to your two text files
> o
2
8,10, 11,12,15
17
21, 29, 18, 32, 145
22, 34, 19, 27, 20
24, 23, 33, 28, 25, 26
35
36, 40
38, 37
39
41
92
42
Robert
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to do it?
Thank you.
Robert
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Hi Mohamed,
It works. Thank you very much.
Best,
Robert
2010/11/20 :
> Hi Robert,
>
> Try to do this
>
> Len <- max(sapply(my.list, length))
>
> fun1 <- lapply(my.list, function(x){
> c(x, rep(0, Len))[1:Len]
> })
>
someone can help ...
Robert Kinley
Study: Pos tissue VC, Neg tissue VC, Pos/Neg tissue
correlation
Study*Group:Pos tissue VC, Neg tissue VC, Pos/Neg tissue
correlation
Residual (animal): Pos tissue VC, Neg tissue VC, Pos/Neg tissue
correlation
?RSiteSearch
--
From: "Stephen Liu"
Sent: Tuesday, November 23, 2010 9:09 AM
To:
Subject: [R] Searching dataset
Hi folks,
help.search("item_name")
displays a list of dataset containing the item_name. If I need to search
containing more than
I am having problems trying to get R to graph data input that is log-normal
on the horizontal axis like the example I have attached and is also below.
The data is log (base 10), and I am more interested in viewing the tails of
the distribution. The closest I can get with this is log on the vertica
this is a clearer (I hope) version of an earlier post -
My problem is formulating the random = argument to give estimates
of all 9 random components for this kind of setup where there are
(I think) 9 variance/covariance components ...
Study.1Study.2 ... Stud
I am having problems trying to get R to graph data input that is log-normal
on the horizontal (x) axis.
The data is log (base 10), and I am more interested in viewing the tails of
the distribution. The closest I can get with this is log on the vertical
(y) axis and linear on the horizontal axis.
On Mon, Dec 13, 2010 at 1:09 PM, robert denham wrote:
> in
> test <- data.frame(V1=c(1,2,3), V2=c(4,5,6))
>
> you are using arguments to the data.frame function in the tag=value format.
> From the help:
>
>...: these arguments are of either the form value or
On 12/12/2010 8:59 AM, Tim Clark wrote:
I need help with using graphics in Word 2007 that will later be converted
into a
pdf document. I have tried several formats and found that I get the best
quality of graphics using .wmf, .eps format, but when I convert it to
.pdf I get
a bunch of lines ac
at I could set
xlim=c(0,max(x)) but because of main increase equals 20 and the last
value 95 it doesn't solve the problem. The increase is ok. but only
line should be longer.
Thank you
Robert
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2010/12/21 Gerrit Eichner :
> par( mar = c( 3, 13, 2, 1), cex = 0.8)
>
> barplot( x, names.arg = NULL, horiz = TRUE, axes = FALSE)
>
> axis( side = 1, at = c( seq( 0, 80, by = 20), 95))
>
> axis( side = 2, at = 1:length(ety), line = -1, las = 1, tick = FALSE,
> labels = ety)
Thank you very mu
2010/12/21 Dieter Menne :
> Standard graphics has fallen a bit out of favor because of these quirks. Try
> lattice:
>
> library(lattice)
> x <- seq(20, 100, by=15)
> ety <- paste("Effect on treatment group",1:length(x))
> barchart(ety~x)
>
> Note that the ety labels must be different to make this w
Someone suggested me that I don´t have to check the normality of the
data, but
the normality of the residuals I get after the fitting of the linear
model.
I really ask you to help me to understand this point as I don´t find
enough
material online where to solve it.
Try the following:
# using
in the code you presented, or you can
extract whatever you want from the dates using as.POSIXlt and
extracting the component:
names(as.POSIXlt(aa$DateTime[1]))
Robert
On Sun, Jan 16, 2011 at 2:20 PM, rnick wrote:
>
> Hi all,
> I have run into a problem and some help would be highly ap
Dear
R-users,
I was
wondering if someone could give me some advices on the following problem.
I tried to
apply moranâs test to a small dataset and couldnât succeed, here is the
error
message:
Â
mor <- moran.test(x, res2)
Â
Avis dans
moran.test(x, res2) :
 Out-of-range p-value: r
Did you do:
update.packages(checkBuilt=TRUE)
Rob
--
From: "Joshua Wiley"
Sent: Wednesday, January 19, 2011 11:43 AM
To: "Joseph Boyer"
Cc:
Subject: Re: [R] how to get old packages to work on R 2.12.1
Dear Joe,
On Wed, Jan 19, 2011 at 7:49 AM,
I recently started using R and I have a simple question. I am running R
(v.
2.12.1) and Rcmdr (v.1-6.3) on Mac (Snow Leopard).
I am using a data set I used before for practicing ANOVA with R, so I
know
what the results should look like. I can get ANOVA table using both Rcmdr
and GUI. However,
modelfn(count ~ spray, data=test01)
Error: could not find function "modelfn"
What will be +var3 ?
I think the general advice applied to your specific situation was meant to
suggest using the construct:
InsectSprays.aov <- aov(count ~ spray, data=InsectSprays)
rather than:
InsectSprays.aov <-
The R Journal article by Paul Murrell may be of interest:
http://journal.r-project.org/2009-1/RJournal_2009-1_Murrell.pdf
The documentation for the Diagram Package may also be of interest:
http://cran.r-project.org/web/packages/diagram/diagram.pdf
http://cran.r-project.org/web/packages/di
- Original Message -
From: "Stephen Liu"
To:
Sent: Tuesday, August 17, 2010 8:25 AM
Subject: Re: [R] Missing values
- Original Message
From: Michael Bedward
To: Stephen Liu
Cc: r-help@r-project.org
Sent: Tue, August 17, 2010 3:57:05 PM
Subject: Re: [R] Missing values
As is often the case in R, just because you shouldn't do something, doesn't
mean you can't do it. Still, I'd urge you to consider the "visual honesty"
of what you propose.
If you're still insistent:
dat <- (-3:4)
dat1=dat-min(dat)
barplot(dat1,axes=FALSE)
axis(2,dat1,labels=dat)
Typically, ze
It occurred to me after my initial post that you will need a little more
"fixing" of your axis labels if you have data that is not trivial integers
as in your example. Consider the following solution for some irrational
random numbers:
dat=rnorm(20)
dat1=dat-min(dat)
barplot(dat1,axes
all
points are set to be equally spaced in time. I'm wondering if I
miss-specified any parameters or not. Any advice would be much
appreciated. Thank you in advance.
Robert
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osuda-devel
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--
Robert W. Baer, Ph.D.
Professor of Physiology
Kirksville College of Osteopathic Medicine
A. T. Still University of Health Sciences
800 W. Jefferson St.
thanks in advance for any help.
Robert
--
Robert Cassidy, PhD
Department of Psychology
Concordia University
7141 Sherbrooke W.
Montreal (QC) H4B 1R6
tel: (514) 848-2424 x2244
fax: (514) 848-4523
office: PY-119.2
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I have a erdos-renyi game with 6000 nodes and probability 0.003.
g1 = erdos.renyi.game(6000, 0.003)
How to create a Watts Strogatz game with the same probability.
g1 = watts.strogatz.game(1, 6000, ?, ?)
What should be the third and fourth parameter to this argument.
According to ?watts.stroga
each of the 3 groups. Somehow, I
can't quite see how to do it without manually sub-setting the parent data frame
and then putting it back together again.
Anyone able to get me started on a cleaner (more R-like) approach?
Thanks,
Rob
--
Robert W. B
orm(100),y=rnorm(100));
qtile <- quantile(dftest$x,probs=c(0.05,0.95));
badIdx <- which((dftest$x < qtile[1]) | (dftest$x > qtile[2]));
if (length(badIdx) > 0) {
dftest <- dftest[-idx,];
}
My question is that is there a more streamlined way to achieve this? Thank
Thanks Bill and William. Both solutions work, and both have introduced me
to whole new R functions which I'm still wrapping my head around.
What a great resource this list is.
Thanks again,
Rob
--
Robert W. Baer, Ph.D.
Professor of Physi
tained, reproducible code.
At least on Windows (not sure about other OS's
------
Robert W. Baer, Ph.D.
Professor of Physiology
Kirksville College of Osteopathic Medicine
A. T. Still University of Health Sciences
800 W. Jefferson St.
Kirksville, MO
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Cheers,
Robert
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