Using Sweave on Windows (MikTeX 2.5 and R 2.4.1), I've ran into a bit of a
problem with the tex -> DVI process.
First, I've corrected the common \usepackage error by using
\usepackage{"C:/Program Files/R/R-2.4.1/share/texmf/Sweave"} to get around
the spacing problem. That is, LaTeX is finding the
Actually I am not sure what you want exactly, but is it
df1 <-data.frame(b=c(1,2,3,4,5,5,6,7,8,9,10))
df2 <-data.frame(x=c(1,2,3,4,5), y=c(2,5,4,6,5), z=c(10, 8, 7, 9, 3))
df1 <- cbind(df1,
"colnames<-"(sapply(with(df2,(x+y)/z),
function(a,b) a/b,b=df1
Convert them to factors in which you specify the order:
> mytable<-factor(c(rep("Disagree",37),rep("Agree",64)),
> levels=c("Disagree","Agree"))
> table(mytable)
mytable
DisagreeAgree
37 64
On 5/5/07, AP05 <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> I'm sure this is simple but I do
Hi all,
I'm sure this is simple but I don't get it.
I have a table
mytable<-c(rep("Disagree",37),rep("Agree",64))
table(mytable)
this gives me
AgreeDisagree
6437
but I didn't ask for it to be in alphabetic order.
How can I get it in original order?
DisagreeAgree
37
Hi,
I am new in R. Please help me in the following case.
I have data in hand:
http://www.nabble.com/file/8225/Data.txt Data.txt
There are some categorical (binary and nominal) and continuous variables.
How can i get a generic RXC contingency table from this table? My main
objective is to fine
Hello,
I'm trying to compare the allozyme data from two taxa. I have several
columns of data (19 loci) for each species. I want to do a Mann-Whitney
U-test or the wilcox.test (two sample Wilcoxon). When I try to run my code
(the first two columns are 1:name of the species, 2:name of individual) I
Is this what you want?
> x <- seq(as.POSIXct("2006-05-05 12:00"), by="17 min", length=20)
> x
[1] "2006-05-05 12:00:00 GMT" "2006-05-05 12:17:00 GMT" "2006-05-05
12:34:00 GMT"
[4] "2006-05-05 12:51:00 GMT" "2006-05-05 13:08:00 GMT" "2006-05-05
13:25:00 GMT"
[7] "2006-05-05 13:42:00 GMT" "2006-0
Can you proovide the example you are talking about. Here is what I got:
> ifelse(NA,1,2)
[1] NA
On 5/5/07, Jennifer Dillon <[EMAIL PROTECTED]> wrote:
> Hi Everyone,
>
> I think I found a problem with the ifelse function: If the condition
> argument is NA, it treats it as true. Anyone agree or
Jennifer Dillon wrote:
> Hi Everyone,
>
> I think I found a problem with the ifelse function: If the condition
> argument is NA, it treats it as true. Anyone agree or disagree with this?
I disagree.
> ifelse(c(5,4,NA) == 5, 1, 0)
[1] 1 0 NA
> Jen
--
Chuck Cleland, Ph.D.
NDRI, Inc.
71 We
Hi Everyone,
I think I found a problem with the ifelse function: If the condition
argument is NA, it treats it as true. Anyone agree or disagree with this?
Jen
--
Jennifer Dillon
Doctoral Student
Harvard Biostatistics
Room 414B, Building 1
[[alternative HTML version deleted]]
__
On Sat, 5 May 2007, Prof. Jeffrey Cardille wrote:
Hello,
Is there an accepted way to convey, for regression trees, something
akin to R-squared?
Why not use R-squared itself for your purposes?
Just get the fitted values from however you do the fit, and compute
R-squared from the basic formul
Prof. Jeffrey Cardille wrote:
> Hello,
>
> Is there an accepted way to convey, for regression trees, something
> akin to R-squared?
>
> I'm developing regression trees for a continuous y variable and I'd
> like to say how well they are doing. In particular, I'm analyzing the
> results of a
All-- Apologies if I have inadvertently posted this message twice; I
just joined the list today, after trying to post once.
Thanks-
Jeff
# r-help message is below #
Hello,
Is there an accepted way to convey, for regression trees, something
akin to R-squared?
I'm developing regression tree
On 5/4/07, ivo welch <[EMAIL PROTECTED]> wrote:
hi doug: yikes. could I have done better? Oh dear. I tried to make
my example clearer half-way through, but made it worse. I meant
set.seed(1);
fe = as.factor( as.integer( runif(100)*10 ) ); y=rnorm(100); x=rnorm(100);
print(summary(lm( y ~
Hello,
Is there an accepted way to convey, for regression trees, something
akin to R-squared?
I'm developing regression trees for a continuous y variable and I'd
like to say how well they are doing. In particular, I'm analyzing the
results of a simulation model having highly non-linear beha
I have data in Unixtime that I wish to plot several ways. I have
learned to convert to a vector of time strings, and for the sake of
plotting times of day vs. heights of tide (to examine seasonality
component) I can extract HH:mm pretty swiftly and readily from
Unixtime using emacs calc. However,
On Sat, 2007-05-05 at 11:35 -0500, Marc Schwartz wrote:
> On Sat, 2007-05-05 at 09:43 -0400, steve wrote:
> > Suppose I have a table constructed from structable or simply just an
> > object of class table. How can I convert it to a latex object? I looked
> > in RSiteSearch, but only found info ab
On Sat, 2007-05-05 at 09:43 -0400, steve wrote:
> Suppose I have a table constructed from structable or simply just an
> object of class table. How can I convert it to a latex object? I looked
> in RSiteSearch, but only found info about matrices or data frames.
>
> Steve
>
>
> For example, her
Thanks. I did successfully install Friday afternoon using the configure script
from version 0.68, however the library would not load, so I removed the
package. I will attempt to build a new configure script using autoconf on
Monday morning. Thanks much for your feedback, I really appreciate i
>From you function, it looks like you want to return 5 values for each
element in the vector 'df1'. Is this what you want? I am not sure
what you expect the output to look like.
> df1 <-data.frame(b=c(1,2,3,4,5,5,6,7,8,9,10))
> df2 <-data.frame(x=c(1,2,3,4,5), y=c(2,5,4,6,5), z=c(10, 8, 7, 9, 3)
Dear Mailing-List,
I think this is a newbie question. However, i would like to integrate a
loop in the function below. So that the script calculates for each
variable within the dataframe df1 the connecting data in df2. Actually it
takes only the first row. I hope that's clear. My goal is to apply
Ivo,
I don't know whether you ever got a proper answer to this question.
Here is a kludgy one -- someone else can probably provide
a more elegant version of my diffid function.
What you want to do is sweep out the mean deviations from both y
and x based on the factor fe and then estimate the sim
Dear Mailing-List,
I think this is a newbie question. However, i would like to integrate a
loop in the function below. So that the script calculates for each
variable within the dataframe df1 the connecting data in df2. Actually it
takes only the first row. I hope that's clear. My goal is to apply
On 05/05/2007 9:36 AM, Greg Snow wrote:
> If we go the route of converting Perl scripts into windows executables, then
> there is the Perl Power Tools (ppt) project for perl that aims to create a
> cross platform set of common Unix tools (see
> http://sourceforge.net/projects/ppt/, or the curre
I think its actually here:
http://search.cpan.org/dist/ppt/
but would have the significant disadvantage of deepening the use of
perl whereas I think the direction should be to get rid of perl.
On 5/5/07, Greg Snow <[EMAIL PROTECTED]> wrote:
>
> If we go the route of converting Perl scripts into
I hope some of the authors of the package MCMCpack read this.
I don't know if there is a way to set the response in the model formula of
MCMClogit other than a (numeric) response vector. I think the MCMClogit in
the MCMCpack needs some development so that the response in the formula
could be set
Suppose I have a table constructed from structable or simply just an
object of class table. How can I convert it to a latex object? I looked
in RSiteSearch, but only found info about matrices or data frames.
Steve
For example, here is a table t2
> str(t2)
table [1:2, 1:2, 1:2] 6 8 594
If we go the route of converting Perl scripts into windows executables, then
there is the Perl Power Tools (ppt) project for perl that aims to create a
cross platform set of common Unix tools (see
http://sourceforge.net/projects/ppt/, or the current toolset can be downloaded
from CPAN). The f
I am glad to help. The pp program is the main tool to use to create the
executable.
-Original Message-
From: "Duncan Murdoch" <[EMAIL PROTECTED]>
To: "Greg Snow" <[EMAIL PROTECTED]>
Cc: "Gabor Grothendieck" <[EMAIL PROTECTED]>; "Doran, Harold" <[EMAIL
PROTECTED]>; "r-help@stat.math.ethz
Surely the idea of having a separate windows version of R is that
it works in a very Windows-like way and that would preclude
having conflicts with standard utilities on Windows. To me this
is one of the most annoying things about R since I do use other
Windows software and that includes software
Break your long running function into two parts. The first will be
the long running part and it can return a value that you can pass to
the second part. If the second part fails, you still have the
original value that was returned.
f1.1 <- function(){
line1 <- f2()
return(line1)
}
f1.2
On 05/05/2007 8:00 AM, Gabor Grothendieck wrote:
> I think that should be the default in order to protect the user. Protecting
> the user from this sort of annoying conflict is important for a professionally
> working product that gets along with the rest of the Windows system.
I don't, because R
I think that should be the default in order to protect the user. Protecting
the user from this sort of annoying conflict is important for a professionally
working product that gets along with the rest of the Windows system.
On 5/5/07, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 04/05/2007 9:32
On 04/05/2007 9:32 PM, Gabor Grothendieck wrote:
> It certainly would be excellent if installing perl could be eliminated.
>
> One additional thing that I really dislike about the R installation is that
> one needs "find" on one's path and that conflicts with "find" on Windows
> so other applicati
On Fri, May 04, 2007 at 04:55:36PM -0400, Weiwei Shi wrote:
> sorry for my English, staying in US over 7 years makes me think I were
> ok, now :(
>
> anyway, here is an example and why I need that:
>
> suppose I have a function like the following:
>
> f1 <- function(){
> line1 <- f2() # assume t
Using the builtin data frame CO2 this regresses uptake, which is
variable number 5, against each consecutive pair of variables:
for(i in 1:3) {
idx <- c(i, i+1, 5)
print(lm(uptake ~., CO2[idx]))
}
On 5/5/07, Victor Bennett <[EMAIL PROTECTED]> wrote:
> Does anyone know if there is a way t
On Sat, 5 May 2007, Alberto Vieira Ferreira Monteiro wrote:
> [for those that worry about these things, this _is_ a homework
> assignment. However, it's not an R homework, it's a Geography
> and History homework... and I want to use R to create a pretty
> map]
>
> Roger Bivand wrote:
> >
> >> Is
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