Ng Stanley gmail.com> writes:
> I am faced with a strange problem. My picture file is empty when the
> following statements were run from R script. But the picture file shows up
> correctly, when the commands are individually run from Rgui.
>
> png(file="histogram_correlation.png", bg = "white")
Hello,
I am sure this must have been asked before, but my nabble search did
not turn anything useful. Just pointer where to look will also be
nice.
So, I have the following data:
x_test1 <- c(1:10)
y_test1<-rnorm(10)
x_test2 <- c(1:15)
y_test2<-rnorm(15)
x_test3 <- c(1:20)
y_test3<-rno
Tribo,
Use data.frame() and rbind() to combine the vectors.
x_test1 <- c(1:10)
y_test1<-rnorm(10)
x_test2 <- c(1:15)
y_test2<-rnorm(15)
x_test3 <- c(1:20)
y_test3<-rnorm(20)
dataset <- rbind(data.frame(Test = "Test 1", x = x_test1, y = y_test1),
data.frame(Test = "Test 2", x = x_test2, y
Hi R,
I wanted to know how do we access the elements of a list. In particular,
v=list(c(1,2,3,4,5),c(1,2,33,4,5),c(1,2,333,4,5),c(1,2,,4,5))
I want to access all the thirds items of the elements of the list. i.e.,
I want to access the elements, 3,33,333,. This can be done throug
Hi All
I have looked but cannot find an answer to this.
I want to build a formula dynamically (in a function), using for example
paste, and using it in a model:
fr<-"F1"
f1<-formula(paste(fr, "~ SensoryTerm"))
m1<-aov(f1, data=vdata)
So this is the equivalent of
use as.formula() rather than just formula().
Bill Venables
CSIRO Laboratories
PO Box 120, Cleveland, 4163
AUSTRALIA
Office Phone (email preferred): +61 7 3826 7251
Fax (if absolutely necessary): +61 7 3826 7304
Mobile: +61 4 8819 4402
Home Phone: +61
Dear John,
this is a new bug caused by a side effect from a new bugs.log function
by Jouni Kerman that improves many other things. I have fixed it in the
master sources on sourceforge right now and a bugfix release will appear
on CRAN within a couple of days.
Best,
Uwe Ligges
John Smith wro
I find it hard to imagine you need to do it without using sapply. If
you get over your phobia, you might like to try
sapply(v, "[", 3)
If the elements of the list are all the same length, as in your example,
you could do
data.frame(v)[3, ]
I think we need to know more about why you think you
Please disregard my previous reply. Now that I have read your question,
all becomes clear.
To do what you want to do is a bit tricky. Here is one way
f1 <- as.name("F1")
fm <- eval(bquote(aov(.(f1) ~ sensoryTerm, data = vdata)))
Them fm is the fitted model object, suitably formed.
E&OE, of co
[EMAIL PROTECTED] wrote:
> use as.formula() rather than just formula().
>
>
>
Hm?
That doesn't seem to cut it for me:
> f <- as.formula(y~x)
> lm(f)
Call:
lm(formula = f)
Coefficients:
(Intercept)x
0.06437 -0.09714
> summary(lm(f))
Call:
lm(formula = f)
Residuals:
Shubha Vishwanath Karanth wrote:
> Hi R,
>
>
>
> I wanted to know how do we access the elements of a list. In particular,
>
>
>
> v=list(c(1,2,3,4,5),c(1,2,33,4,5),c(1,2,333,4,5),c(1,2,,4,5))
>
>
>
> I want to access all the thirds items of the elements of the list. i.e.,
> I want to acce
Hi,
foo<-x(...)
{ # Need to remove bad characters from string arguments "..."
# here.
# Pass on the modified string arguments.
bar(...)
}
I need to modify string arguments passed in to function foo
as "..." and then pass the modified arguments on to function
bar. Is there a way to acce
I'm no guru, but can you not use do.call here?
i.e. does
foo<-function(...) #Not x(...)
{
arglist<-list(...)
#remove bad characters from arglist here.
# then
do.call("bar",arglist)
}
do what you want?
>>> "David Keegan" <[EMAIL PROTECTED]> 01/02/2008 11:00:15 >>>
Hi,
foo<-x(
Hello,
I am sure this must have been asked before, but my nabble search did
not turn anything useful. Just pointer where to look will also be
nice.
So, I have the following data:
x_test1 <- c(1:10)
y_test1<-rnorm(10)
x_test2 <- c(1:15)
y_test2<-rnorm(15)
x_test3 <- c(1:20)
y_test3<-rnorm(20)
Th
?save
?load
Gabor
ps. although i'm not sure what an Rdata-project means, so maybe you
need something else
On Fri, Feb 01, 2008 at 08:24:32AM +0200, Atte Tenkanen wrote:
> Dear R-users,
>
> How do you save a big table or matrix as an independent object and attach it
> to your Rdata-project when
Actually, you don't need apply. If there are no NA's then it is
very easy:
m[] <- y[ col(m) ]
If you want to keep the NA's then it is a bit more tricky:
m[] <- 0*m + y[ col(m) ]
G.
On Thu, Jan 31, 2008 at 07:03:51PM -0800, dxc13 wrote:
>
> useR's,
>
> Consider:
> y <- c(20, 25, 30)
> > m <
Hi,
I used the following statements to generate unsuccessfully a 5 by 5 multiple
densityplots on a single page. If I use plot, the whole thing works.
> data <- matrix(rnorm(25), 5, 5)
> op <- par(mfrow = c(5, 5))
> for (x in 1:5) {densityplot(data[,x])}
> par(op)
Thanks
Stanley
[[altern
Hi there,
I am an environmental studies masters student trying to get my thesis out the
door. I am also newbie at trees in general, but I like what I see in the
literature about the random forest algorithm. I think I get the general gist
of things, but even after reading stuff I’m unclear abo
Dear R-users,
How do you save a big table or matrix as an independent object and attach it to
your Rdata-project when needed?
Atte Tenkanen
University of Turku, Finland
Department of Musicology
+023335278
__
R-help@r-project.org mailing list
https://s
useR's,
Consider:
y <- c(20, 25, 30)
> m <- matrix(c(0.0,1,NA,0.5,1.25,0.75, 0.5, NA,
> NA),byrow=TRUE,nrow=3,ncol=3)
> m
[,1] [,2] [,3]
[1,] 0.0 1.00 NA
[2,] 0.5 1.25 0.75
[3,] 0.5 NA NA
For each numeric value, I want to replace them with their corresponding
y-value. The result s
Hi, Thierry,
That was exactly what I was looking for. Thanks.
Now I have a data frame with the series data in my workspace, and a
plot on the graphics device with color lines and respective color
patches on the legend. My next question is about ggplot. Is it
possible to make the legend show not o
On 01/02/2008, Ng Stanley <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> I used the following statements to generate unsuccessfully a 5 by 5
> multiple
> densityplots on a single page. If I use plot, the whole thing works.
>
> > data <- matrix(rnorm(25), 5, 5)
> > op <- par(mfrow = c(5, 5))
> > for (x in 1
Thank you very much. That problem has been niggling me for some time.
I slotted in your code and it worked. I just need to spend a bit of time
understanding it ...
Thanks again.
John Seers
---
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Sent: 01 Februa
Thanks for the help.
Regards
John Seers
---
-Original Message-
From: Peter Dalgaard [mailto:[EMAIL PROTECTED]
Sent: 01 February 2008 10:04
To: [EMAIL PROTECTED]
Cc: john seers (IFR); [EMAIL PROTECTED]
Subject: Re: [R] Building a formula using paste
[EMAIL PROTECTED] wrote:
> use
Hello Folks,
I'm after some help regarding mixed models.
Basically I have sampled a number of different animals at 10 independent sites
and am trying to create a mixed model to account for the variation between
sites my current model looks like this:
mm<-glmm.admb(Hep~Sex+Mass, random=~Mouse, g
> just the patch. But again the ggplot manual didn't have an example for
> that. Also, what if I wanted to plot with lines+symbol, do I have to
> use multiple layers? Is it (easily) achievable?
Not yet, but I hope to have that working (automatically) in the next
version. I'm working on the legend
Sorry - I thought about plot(density())
should read more carefull...
On 01/02/2008, Gavin Simpson <[EMAIL PROTECTED]> wrote:
>
> On Fri, 2008-02-01 at 14:00 +0200, Rainer M Krug wrote:
> > On 01/02/2008, Ng Stanley <[EMAIL PROTECTED]> wrote:
> > >
> > > Hi,
> > >
> > > I used the following statem
Have a look at scale_manual on http://had.co.nz/ggplot2/
This code will give you a key with lines instead of tiles.
ggplot(data = dataset, aes(x = x, y = y, colour = Test)) + geom_line() +
scale_colour_manual(values = c("red", "blue", "green"), guide = "line")
If you want both lines and dots.
On Friday 07 December 2007, Thomas Zumbrunn wrote:
> Since I changed my X11 settings to a Xinerama setup, calls to x11() result
> in windows with a width half the size of the one specified by the
> corresponding parameter. Furthermore, x axis labels are overlapping. Other
> devices (e.g. pdf()), ar
Derek,
the 0*m part zeros out everything in the matrix, expect for the NA's,
0*NA=NA by definition. If we add this to the y[ col(m) ] matrix, then
NA+anything=NA, but 0+anything=anything.
G.
ps. please answer to the list (as well)
On Fri, Feb 01, 2008 at 08:52:50AM -0500, Derek Cyr wrote:
>
Jason Liao <[EMAIL PROTECTED]> writes:
> I have a program which needs to compute squared Euclidean distance
> between two vectors million of times, which the Rprof shows is the
> bottleneck. I wondered if there is any faster way than my own simple
> function
>
> distance2 = function(x1, x2)
> {
>
hi,
i start working with PCA method i found VSS package it is very helpfull it
show me residual matrix with the number of components to extract now i want to
verify the normality of residual matrix and than to simulate can every one help
me please .
THANK YOU and excuse me for my b
Tribo
That function was added during the latest update. So you should update
to this version (0.5.7). You can update packages with update.packages()
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek /
Hi Carla,
Try paste("VAR: ",a,sep="")
Jorge
On 2/1/08, Carla Rebelo <[EMAIL PROTECTED]> wrote:
>
> Good morning!
>
> I do not speak English very well and so I will try to explain the best I
> can. I have this:
>
> > tabela[,1]
> [1] a a b b a c b a c c c c c
> Levels: a b c
>
> >unique(tabela[
Hi Hadley,
Thanks for ggplot. Great piece of work. Very intuitive. The legend
always seems to be the most difficult part to implement in any
plotting package. gnuplot and RLplot come to mind.
Good luck!
On Fri, Feb 1, 2008 at 10:17 PM, hadley wickham <[EMAIL PROTECTED]> wrote:
> > just the pat
Good morning!
I do not speak English very well and so I will try to explain the best I
can. I have this:
> tabela[,1]
[1] a a b b a c b a c c c c c
Levels: a b c
>unique(tabela[,1])
[1] a b c
Levels: a b c
>var<-unique(tabela[,1])[1]
> var
[1] a
Levels: a b c
But if I concatenate like th
Dear All,
Take this code:
> f <- function(x) exp(-x)*x-0.05
> g <- function(x) 0
> curve(f,0,5)
> curve(g,add=T)
Error in xy.coords(x, y) : 'x' and 'y' lengths differ
>
However, with
g <- function(x) x-x
no error is generated.
Is this a bug? I am using
> version
_
platform
Dear all,
Is it possible with two random effects in lme()?
lmefit1<- lme(Handling ~ Mass + factor(Prey)+ Mass*factor(Prey), random = ~
1 |Place+Age)
Here I use Place as random effect, but I also want to add Age as a random
effect. Since there could be an effect of Age (continous variable), but
hits=-2.6 tests=BAYES_00
X-USF-Spam-Flag: NO
On Fri, 2008-02-01 at 14:00 +0200, Rainer M Krug wrote:
> On 01/02/2008, Ng Stanley <[EMAIL PROTECTED]> wrote:
> >
> > Hi,
> >
> > I used the following statements to generate unsuccessfully a 5 by 5
> > multiple
> > densityplots on a single page. If I u
Carla Rebelo wrote:
> Good morning!
>
> I do not speak English very well and so I will try to explain the best I
> can. I have this:
>
> > tabela[,1]
> [1] a a b b a c b a c c c c c
> Levels: a b c
>
> >unique(tabela[,1])
> [1] a b c
> Levels: a b c
>
> >var<-unique(tabela[,1])[1]
>
> > var
>
Greetings:
FORM/SORM (First/Second Order Reliability Methods) are very popular among some
structural engineers but almost never discussed among statisticians. I am
quite predudiced as to the reasons for this, but would enquire among the
world's statistical thinkers for their perspectives on th
Carla Rebelo wrote:
> Good morning!
>
> I do not speak English very well and so I will try to explain the best I
> can. I have this:
>
> > tabela[,1]
> [1] a a b b a c b a c c c c c
> Levels: a b c
>
> >unique(tabela[,1])
> [1] a b c
> Levels: a b c
>
> >var<-unique(tabela[,1])[1]
>
> > v
Paul Smith wrote:
>
>> f <- function(x) exp(-x)*x-0.05
>> g <- function(x) 0
>> curve(f,0,5)
>> curve(g,add=T)
>> Error in xy.coords(x, y) : 'x' and 'y' lengths differ
>>
>
> However, with
>
> g <- function(x) x-x
>
> no error is generated.
>
> Is this a bug?
>
No; simplify it:
g1 <- functio
For some reason I get the following error:
"Error: could not find function "scale_colour_manual"
so I couldn't make the plots.
Otherwise I completely agree that color is good way to summarize in
the key several different models and operations performed on the same
data (it is always a trick to f
Paul Smith gmail.com> writes:
>
> Dear All,
>
> Take this code:
>
> > f <- function(x) exp(-x)*x-0.05
> > g <- function(x) 0
> > curve(f,0,5)
> > curve(g,add=T)
> Error in xy.coords(x, y) : 'x' and 'y' lengths differ
> >
>
> However, with
>
> g <- function(x) x-x
>
> no error is generated.
This is discussed in this thread:
http://tolstoy.newcastle.edu.au/R/e2/help/07/01/9353.html
This mostly goes over the ground discussed here plus there
is a do.call based solution there too.
On Feb 1, 2008 7:10 AM, john seers (IFR) <[EMAIL PROTECTED]> wrote:
>
>
>
> Thanks for the help.
>
> Regar
> Does anyone have any ideas how I could do a power calculation for a log
> rank test. I would like to know what the suggested sample sizes would
> be to pick a difference when the control to active are in a ratio of 80%
> to 20%.
Power for a log-rank test is the same as power for a Cox model.
On Feb 1, 2008 2:46 PM, Martin Maechler <[EMAIL PROTECTED]> wrote:
> >> Take this code:
> >>
> >> > f <- function(x) exp(-x)*x-0.05
> >> > g <- function(x) 0
> >> > curve(f,0,5)
> >> > curve(g,add=T)
> >> Error in xy.coords(x, y) : 'x' and 'y' lengths differ
> >> >
>
Dear all,
I am creating functions with the "three dot" strategy. I wold like to have
suggestion when writing multiple functions that calls multiple functions
with "...".
I will give you a couple of example:
>a=1:5
>b=6:10
>d=3:7
>example=function(x,y, z, ...){
plot(x,y, ...)
points(x,z, .
I had a problem:
I saved a matrix:
> save(Tondistmatrix1, file="/Users/atte/Skriptit/Tondistmatrix1")
then I tried to open it with R:
> Tondistmatrix1=load("/Users/atte/Skriptit/Tondistmatrix1")
> Tondistmatrix1[1:10,]
Error in Tondistmatrix1[1:10, ] : incorrect number of dimensions
> dim(Tondis
Is there a way to perform a dynamic SQL query in R, using RMySQL? Thanks
My D. Coyne
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PLEASE do read the posting
R is treating tablea[,1] as a factor. Try converting
it to character.
--- Carla Rebelo <[EMAIL PROTECTED]> wrote:
> Good morning!
>
> I do not speak English very well and so I will try
> to explain the best I
> can. I have this:
>
> > tabela[,1]
> [1] a a b b a c b a c c c c c
> Levels: a b
> "BB" == Ben Bolker <[EMAIL PROTECTED]>
> on Fri, 1 Feb 2008 14:03:15 + (UTC) writes:
BB> Paul Smith gmail.com> writes:
>>
>> Dear All,
>>
>> Take this code:
>>
>> > f <- function(x) exp(-x)*x-0.05
>> > g <- function(x) 0
>> > curve(f,0,5)
Try this:
x[x==""] <- NA
On 01/02/2008, My Coyne <[EMAIL PROTECTED]> wrote:
> I would like to replace all null ("") in a column of a matrix to NA, how
> would I do that without a do loop?
>
> Thanks
>
> --mc
>
>
>
>
>
>
>
>
>
>
>
>
>
>
> [[alternative HTML version deleted]]
>
> __
I would like to replace all null ("") in a column of a matrix to NA, how
would I do that without a do loop?
Thanks
--mc
[[alternative HTML version deleted]]
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Hi list,
It seems that as.numeric is rounding up my numbers that are in a character
format.
Example:
a
[1] "776554.45" "776985.31" "776076.03" "776092.01" "776151.42" "776276.97"
b <- as.numeric(a)
b
[1] 776554.4 776985.3 776076.0 776092.0 776151.4 776277.0
I've tried as.numeric(a,2) and
You can try something like:
>a <- 1:5
>b <- 6:10
>d <- 3:7
>example <- function(x,y, z, ..., plot.dots, points.dots){
plot.list <- list(x=x,y=y)
plot.list <- c(plot.list,list(...), as.list(plot.dots))
do.call(plot, plot.list)
points.list <- list(x=x,y=z)
points.list <- c(points.li
options(digits=10)
> b
or just
print(b, 10)
On 01/02/2008, Monica Pisica <[EMAIL PROTECTED]> wrote:
>
>
> Hi list,
>
> It seems that as.numeric is rounding up my numbers that are in a character
> format.
>
> Example:
>
> a
> [1] "776554.45" "776985.31" "776076.03" "776092.01" "776151.42" "7762
So this is only a "cosmetic" printing and actually my numbers do have 2
decimal numbers .
Thanks,
Monica> Date: Fri, 1 Feb 2008 16:27:00 -0200> From: [EMAIL PROTECTED]> To:
[EMAIL PROTECTED]> Subject: Re: [R] as.numeric rounds up> CC:
r-help@r-project.org> > options(digits=10)> > >
It's ugly, but you could use something like
sum(tmp[i,] == "A") > 0
on each column.
pax,
Scott
On Fri, Feb 1, 2008 at 1:58 PM, Scot W. McNary <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I have a question about how to obtain the union of several data frame
> rows. I'm trying to create a common key f
Perhaps:
data <- data.frame(key, row.names=1)
names(data) <- paste("q", 1:6, sep="")
apply(data, 2, function(x)unique(x)[unique(x) != " "])
On 01/02/2008, Scot W. McNary <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I have a question about how to obtain the union of several data frame
> rows. I'm trying
Hi,
I have a question about how to obtain the union of several data frame
rows. I'm trying to create a common key for several tests composed of
different items. Here is a small scale version of the problem. These
are keys for 4 different tests, not all mutually exclusive:
id q1 q2 q3 q4 q5
Is there any implementation in R for finding the phase shift between
two continuous signals. I would like to find the average phase shift
for tow signals over two years.
thanks
Stephen
--
Let's not spend our time and resources thinking about things that are
so little or so large that all they r
First things first: There's no "final tree" in random forests. You get
a set of trees (i.e., a forest). Secondly, a forest "cannot be
interpreted" because of the complexity, not because the splits can't
possibly make sense. You can try to interpret the trees, as long as you
understand the poten
Hi, does some one know how to run the latent class analysis in R?
like which package and command I should use?
thanks a lot,
Suyan Tian
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
hi!!
Below I have 4 columns vector of c and d which are unequal in length.These c
and d have 2 columns each where these 2 columns represent an interval values.
How am I going to get an overlapping over these interval values?? Please help
me sort this problem!! Thanks in advance..
On Fri, 1 Feb 2008, stephen sefick wrote:
> Is there any implementation in R for finding the phase shift between
> two continuous signals. I would like to find the average phase shift
> for tow signals over two years.
Following the _posting guide_, we have:
RSiteSearch("phase shift")
On Fri, 1 Feb 2008, mohamed nur anisah wrote:
> hi!!
>
> Below I have 4 columns vector of c and d which are unequal in
> length.These c and d have 2 columns each where these 2 columns
> represent an interval values. How am I going to get an overlapping over
> these interval values?? Please hel
try flexmix.
On Feb 1, 2008 3:23 PM, Suyan Tian <[EMAIL PROTECTED]> wrote:
> Hi, does some one know how to run the latent class analysis in R?
> like which package and command I should use?
>
> thanks a lot,
>
> Suyan Tian
>
> __
> R-help@r-project.org m
Very sorry about my incomplete email from 2 days ago, here is the full
version.
---
Dear all,
I have read various descriptions of employing resampling techniques, such as
the bootstrap, to estimate the uncertainties of the eigenval
On Fri, 1 Feb 2008, Markus Loecher wrote:
> Very sorry about my incomplete email from 2 days ago, here is the full
> version.
> ---
> Dear all,
> I have read various descriptions of employing resampling techniques, such as
> the b
Here is one way of doing it:
> c.i <- read.table(textConnection(" 17130612 17587118
+ 17712302 18221688
+ 21225764 21387314
+ 25012714 30748348
+ 33852816 34480192
+ 36012944 36209144
+ 36252300 36280276
+ 36737468 36971144
+ 43693832 43878548"))
>
> d.i <- read.table(textConnect
Hi all,
I know this question has been asked in the past, but I am wondering if
anyone running R on Linux has any guidance as to a text editor that works
well with R. At the present time I am running R on Windows and using
TINN-R. For a number of reasons I want to switch to Linux, but can't find
Hi,
Does anyone know if R has a Lagrange multiplier (LM) test for ARCH
effects for univariant time series?
Thanks!
--
Tom
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Wade,
I switched from R-Windows to R-Linux about 9 months ago and have had
great success with GNU-Emacs with the ESS add-on. Your distro should
have Emacs available for install, if it does not you can get the source at:
http://ftp.gnu.org/pub/gnu/emacs/
ESS can be found at:
http://ess.r-project
r-help
c(2.43, 3.22, 6.9, 3.03, 5.36, 6.9, 2.29, 6.13, 6.11, 4.25, 3.85,
5.09, 7.44, 2.86, 2.82, 3.64, 3.22, 7, 2.65, 4.5, 3.73, 5.09,
5.8, 7.87, 2.87, 2.9, 6.63, 6.8, 2.45, 7.68, 2.56, 2.54, 7.35,
4.61, 2.58, 3.27, 5.8, 3.13, 3.29, 2.56, 5.79, 2.67, 7.55, 7.13,
8.4, 6.16, 3.34, 6.49, 1.85, 7.
Hi, Tom:
The 'arch' function in the 'vars' package is supposed to be able
to do that. Unfortunately, I was unable to make it work for a
univariate series. Bernhard Pfaff, the author of 'vars', said that if I
read the code for 'arch', I could easily retrieve the necessary lines
and put
On Linux platform, nothing beats Emacs + ESS.
Shige
On Feb 2, 2008 11:06 AM, Wade Wall <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I know this question has been asked in the past, but I am wondering if
> anyone running R on Linux has any guidance as to a text editor that works
> well with R. At th
Spencer, how about something like this:
archTest=function (x, lags= 16){
#x is a vector
require(vars)
s=embed(x,lags)
y=VAR(s,p=1,type="const")
result=arch(y,multi=F)$arch.uni[[1]]
return(result)
}
can you, or maybe Bernhard, check and see whether this function gives the
correct result?
th
John Lande wrote:
> Dear all,
>
> I am creating functions with the "three dot" strategy. I wold like to have
> suggestion when writing multiple functions that calls multiple functions
> with "...".
>
> I will give you a couple of example:
>
>
>>a=1:5
>>b=6:10
>>d=3:7
>
>
>>example=function(x,
bgchen wrote:
> r-help
>
> c(2.43, 3.22, 6.9, 3.03, 5.36, 6.9, 2.29, 6.13, 6.11, 4.25, 3.85,
> 5.09, 7.44, 2.86, 2.82, 3.64, 3.22, 7, 2.65, 4.5, 3.73, 5.09,
> 5.8, 7.87, 2.87, 2.9, 6.63, 6.8, 2.45, 7.68, 2.56, 2.54, 7.35,
> 4.61, 2.58, 3.27, 5.8, 3.13, 3.29, 2.56, 5.79, 2.67, 7.55, 7.13,
> 8.4
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