consider something along the lines of:
>scriptdir <- '/path/to/scripts/"
>source(paste(scriptdir,'subscript.r',sep=''))
alternatively, you could try:
>workdir <- getwd()
>scriptdir <- '/path/to/scripts/"
>setwd(scriptdir)
>source('subscript.r')
>setwd(workdir)
cheers,
tim
On 10/26/07, Br
Check out:
http://tolstoy.newcastle.edu.au/R/e2/help/06/10/2242.html
On 10/25/07, Andrew Smith <[EMAIL PROTECTED]> wrote:
> Here's a simple example of the type of function I'm trying to write,
> where the first argument is a list of functions:
>
> myfun <- function(funlist, vec){
> tmp <- lap
On 25/10/2007 7:11 PM, Rolf Turner wrote:
>
> OTOH something called temp is unlikely to be something of which you are
> passionately fond anyhow.
Unless you put your thousand year temperature records into it.
Duncan Murdoch
__
R-help@r-project.org mai
On 25/10/2007 5:38 PM, Andrew Smith wrote:
> Here's a simple example of the type of function I'm trying to write,
> where the first argument is a list of functions:
>
> myfun <- function(funlist, vec){
> tmp <- lapply(funlist, function(x)do.call(x, args = list(vec)))
> names(tmp) <- name
Hi all. A question for knowledgeable folks using R on an Intel Mac running
OS X 10.4.10
For ease of maintenance, I have broken a large R script into a main script
which ³oversees² things by calling other scripts, using ³source². Let¹s
call the secondary scripts ³sub-scripts.²
I¹d like for the s
Hello, r-help.
Great opportunity to develop your career in our company within Funds
Management.
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We are looking for motivated people in A
matplot(mat, cbind(a,b,c))
On 10/25/07, Rafael Barros de Rezende <[EMAIL PROTECTED]> wrote:
>
> How to plot multiple variables on the same graph
>
> Dear R users,
>
> I want to plot the following variables (a, b, c) on the same graph. The
> x-axis must be the variable mat and the graph mus
You might want to consider another representation, but it would depend
on how you want to use it. Here is a 'list' that records for each row
the position of the boys; does this start to give you the type of data
that you want? These are the numeric values of where the boys occur.
> x.m
b1 b
Is this what you need?
> plot(mat,a,type="l",col="red",ylim=c(0,1))
> lines(mat,b,col="green")
> lines(mat,c,col="blue")
--- Rafael Barros de Rezende
<[EMAIL PROTECTED]> wrote:
>
>How to plot multiple variables on the same graph
>
>Dear R users,
>
>I want to plot the following var
Hi All,
I have data on the sequence of births for families with completed
fertility cycle (in a data frame); the relevant variables are called b1,
b2, b3, b4, b5, b6 and record the birth of the first, second, ..., sixth
child. So,
b1=1 if the first birth is male,
b1=2 if the first birth is femal
How to plot multiple variables on the same graph
Dear R users,
I want to plot the following variables (a, b, c) on the same graph. The
x-axis must be the variable mat and the graph must have the type = "l". How
can I do that??
> a
[1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
> b
It is a holdover from my past when one of the languages that I used to
use had that convention for 'local' variables in a block of code. I
was used to the convention of being able to define a variable that was
only known in the enclosing block. By convention I try to keep those
objects active for
On 26/10/2007, at 11:44 AM, Tim Calkins wrote:
> Hi everyone -
>
> This came up within the last day -- Jim's response to Deepankar is
> pasted below.
(but snipped out of this response).
> There are probably lots of reasons, but what is the advantage to using
> .temp over, say, temp?
>
Hi everyone -
This came up within the last day -- Jim's response to Deepankar is pasted below.
There are probably lots of reasons, but what is the advantage to using
.temp over, say, temp?
I often find myself writing temporary objects -- should I use the .
preface? What would be the advantages
I have a statistics question that eventually would be implemented in R but
my question is what to do statistically.
Suppose I have a model Y_t = Beta*X_t-1 and Y_t and X_t are both univariate.
Y_t can be negative or positive but generally ranges
Between -.0010 and + .0010. I can estimate Beta
Here's a simple example of the type of function I'm trying to write,
where the first argument is a list of functions:
myfun <- function(funlist, vec){
tmp <- lapply(funlist, function(x)do.call(x, args = list(vec)))
names(tmp) <- names(funlist)
tmp
}
> myfun(list("Summation" = sum,
Rolf Turner wrote:
> On 26/10/2007, at 10:14 AM, m p wrote:
>
>
>> Hello,
>> I'd like to check if my data can be well approximated with a function
>> (1+x/L) exp(-x/L)
>> and calculate the best value for L. Is there some package in R that
>> would simplify that task?
>> Thanks,
>> Mark
>>
No, that's not my homework. Does that seem so easy?
Mark
Rolf Turner <[EMAIL PROTECTED]> wrote:
On 26/10/2007, at 10:14 AM, m p wrote:
> Hello,
> I'd like to check if my data can be well approximated with a function
> (1+x/L) exp(-x/L)
> and calculate the best value for L. Is there some package
On 26/10/2007, at 10:14 AM, m p wrote:
> Hello,
> I'd like to check if my data can be well approximated with a function
> (1+x/L) exp(-x/L)
> and calculate the best value for L. Is there some package in R that
> would simplify that task?
> Thanks,
> Mark
Is this a homework question?
On 10/25/07, Omar Baqueiro <[EMAIL PROTECTED]> wrote:
> Hi again,
> After playing a bit further, I have got the dotplot to put different
> symbols using pch = c(1,2,3,4,5), however, the symbols do not appear
> in the label, where I still get only circles of different colors!!!
The easiest way i
Hello,
I'd like to check if my data can be well approximated with a function
(1+x/L) exp(-x/L)
and calculate the best value for L. Is there some package in R that would
simplify that task?
Thanks,
Mark
__
[[alternative HTML version de
While I can't help much in the way of assessing the correlation (at
least in a numerical sense), I have provided some code below to
visualize the data bringing in an additional variable for the preseason
ranking of the team according to the AP poll as it appears here:
http://sports.espn.go.com/ncf/
I recently called plot(x,y) where x was an array of POSIXct timestamps,
and was pleasantly surprised that it produced a nice plot right out of
the box:
z <- as.POSIXct(c("2006-10-26 08:00:00 EDT","2007-10-25 12:00:00 EDT"))
x <- seq(z[1],z[2],len=100)
y <- 1:100
plot(x,y,type="l")
The X axis ha
Doug and the football fans out there,
I'm no football expert myself. But here is what my colleague said after reading
the posting.
"I can't help you with the equation, but I can say that the polls are very poor
predictors of performance. The reason they do such a bad job is that pollsters
ra
Perhaps:
x <- barplot(data, ylim=c(0, max(data)+1))
text(x,data+.2, data)
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O
On 25/10/2007, Bernd Jagla <[EMAIL PROTECTED]> wrote:
>
> Again me.
>
>
>
> I want to plot the numbers on the bars of a barplot.
>
> This can be
Check out the tapply function.
?tapply
Julian
Bernd Jagla wrote:
> Hi,
> I am new to R and couldn't find any information on how to handle my table
> data that I just read in the way I want to use it..
>
> I read in a table from a file:
> x <- read.delim("filenam", header=TRUE)
>
> one column (
Again me.
I want to plot the numbers on the bars of a barplot.
This can be done using hist function when setting the label argument true
(i.e.
data <- c(1,2,3,4)
hist(data, labels=T)
When I try this using barplot I get an error:
> barplot(summary(data), labels=T)
Error in
I have reached the correlation section in a course that I teach and I
hit upon the idea of using data from the weekly Bowl Championship
Series (BCS) rankings to illustrate different techniques for assessing
correlation.
For those not familiar with college football in the United States
(where "foot
This is the command I want to execute with function contrast in
contrast package:
> contrast(MyObject, list(Trust="T"), list(Trust="U"));
As the two arguments with 'list' are defined as a string of
characters, contr:
> str(contr)
chr "list(Trust=\"T\"),list(Trust=\"U\")"
I would like to
GREAT!!!
Thanks so much !!!
-B
|-Original Message-
|From: John Kane [mailto:[EMAIL PROTECTED]
|Sent: Thursday, October 25, 2007 1:45 PM
|To: Bernd Jagla; 'Henrique Dallazuanna'
|Cc: r-help@r-project.org
|Subject: Re: [R] data frame usage
|
|I think I also misunderstood. It sounds like yo
I think I also misunderstood. It sounds like you want
to subset the data set.
Try
aa <- subset(x x$label == "xxx")
--- Bernd Jagla <[EMAIL PROTECTED]> wrote:
> No, I just want to the values for one class nothing
> applied to them.
> So for example tapply(x$val, x$label, hist) gives me
> all kin
No, I just want to the values for one class nothing applied to them.
So for example tapply(x$val, x$label, hist) gives me all kind of stats that
can be used for plotting the histogram. But I want to plot the actual
histogram for one class.
I guess it would be easiest to take your construct and jus
Yes, or
tapply(x$val, x$label, summary)[[1]]
for the first class.
On 25/10/2007, Bernd Jagla <[EMAIL PROTECTED]> wrote:
>
> Wow, that easy...
> And how can get only the values for a specific class?
>
> Like tapply(x$val, x$label, ?echo?)$class1
>
> What should echo be?
>
> Thanks,
> B
>
> |
It is not clear what the problem is. Please give us a
small reproducable example of the problem as suggested
in the posting guide.
--- Cameron Guenther <[EMAIL PROTECTED]>
wrote:
> Hello All,
> I have plotted a barplot and placed a line plot over
> top using
> par(new=TRUE).
> In the first plot
Wow, that easy...
And how can get only the values for a specific class?
Like tapply(x$val, x$label, ?echo?)$class1
What should echo be?
Thanks,
B
|-Original Message-
|From: Henrique Dallazuanna [mailto:[EMAIL PROTECTED]
|Sent: Thursday, October 25, 2007 1:15 PM
|To: Bernd Jagla
|Cc: r-h
I don't understand the criteria that gives you just
the three ids.
As I read the criteria if we put the data into a
data.frame you have
subset(mydata ,year==year.hatch & year.1st.reprod ==
year.hatch+1)
but this gives more than the three ids. What am I
missing?
--- [EMAIL PROTECTED] wrote:
Hi,
data.frame(GeneName=unique(df$GeneName),Value=sapply(unique(df$GeneName),
function(x)max(df$Value[df$GeneName==x])))
On 25/10/2007, Paul Christoph Schröder <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> maybe someone can help me with this:
>
> I have a matrix of genes and values:
> GeneName Value
?aggregate
b
On Oct 25, 2007, at 1:21 PM, Paul Christoph Schröder wrote:
> Hi,
>
> maybe someone can help me with this:
>
> I have a matrix of genes and values:
> GeneName Value
> Abc1 10
> Abc2 11
> Bbc1-5
> Bbc31 2
> Ccd
Hi,
tapply(x$val, x$label, summary)
On 25/10/2007, Bernd Jagla <[EMAIL PROTECTED]> wrote:
>
> Hi,
> I am new to R and couldn't find any information on how to handle my table
> data that I just read in the way I want to use it..
>
> I read in a table from a file:
> x <- read.delim("filenam", hea
Hi,
maybe someone can help me with this:
I have a matrix of genes and values:
GeneName Value
Abc1 10
Abc2 11
Bbc1-5
Bbc31 2
Ccd 5
Ccd -2
Ccd 7
Dda 5
Dd
Hi,
I am new to R and couldn't find any information on how to handle my table
data that I just read in the way I want to use it..
I read in a table from a file:
x <- read.delim("filenam", header=TRUE)
one column (x$label) hold the class labels. Another holds some values
(x$val).
I want to calcul
Is this what you want
Have a look at ?par and ?axis (note the las=1 is just
to rotate the letters)
---
aa <- 1:5
plot(aa, yaxt='n')
axis(2, at=1:5, c("a","b","c","d","e"), las =1)
aa <- 1:5
plot(aa, yaxt='n')
axis(2, at=1:5, c("a","b","c","d","e"))
---
On Thu, 25 Oct 2007, Keith Jones wrote:
> I would like to tell R what increments to put the tick marks on an
> axes, e.g. xlim=seq(-5,5,1).
>
> I know that will not work. xlim will only except the beginning and
> end of the range for the axes.
Consult the rest of ?par, especially 'xaxp' and 'lab
Hi,
I would like to tell R what increments to put the tick marks on an
axes, e.g. xlim=seq(-5,5,1).
I know that will not work. xlim will only except the beginning and
end of the range for the axes.
Thanks,
Keith jones
__
R-help@r-project.org maili
Hi,
Is there any package that implements semi-supervised clustering through
'must-link' and 'cannot-link' constraints?
thanks!
__
[[alternative HTML version deleted]]
__
R-help@r-project.o
Hi,
Works for me:
samp$Resid <- as.numeric(do.call("c",by(samp, samp$Race,
function(x)resid(lm(stroke~Sex+Age, data=x)
On 24/10/2007, Jiong Zhang, PhD <[EMAIL PROTECTED]> wrote:
>
> Hi Folks,
>
> I want to get the regression residue which each Race. But using:
> sample$resid <- by(sample,sa
Hi again,
After playing a bit further, I have got the dotplot to put different
symbols using pch = c(1,2,3,4,5), however, the symbols do not appear
in the label, where I still get only circles of different colors!!!
Any help?
thanks!
On 10/25/07, Omar Baqueiro <[EMAIL PROTECTED]> wrote:
> Hell
Hello,
I am trying to plot a chart similar to the typical barley yield dotplot:
dotplot(variety ~ yield | site, data = barley,
groups = year, auto.key = list(space = "right"),
xlab = "Barley Yield (bushels/acre) ",
aspect=0.5, layout = c(1,5))
However, I need the plot have
On 10/25/2007 10:25 AM, Trochine, Carolina wrote:
> R helpers,
> I would like to know if it is possible that the last version of R is not
> giving the surrogate splits when you perform a Multivariate regression tree
> analysis? I installed the programm in different computers and i run the some
>
i'm trying to plot survival curves using the following code
n=20
n1=n/2
n2=n/4
a11=4;a12=4 ;a21=4 ;a22=4
t1<-array(1,c(n1))
t2<-array(2,c(n1))
treatgrp=matrix(c(t1,t2))
R helpers,
I would like to know if it is possible that the last version of R is not giving
the surrogate splits when you perform a Multivariate regression tree analysis?
I installed the programm in different computers and i run the some matrix and
it didn't gave me this information. With a previ
Dear all,
I have received some data on birds that looks sth like this:
# a unique id for each individual
id <- c(1,1,1,2,2,2,3,3,3,4,4,5,6)
# the year the bird was measured
year <- c(1995, 1996, 1997, 1995, 1996, 1997, 1996, 1997, 1998, 1996, 1997,
1997, 1998)
# the year the bird was hatched
y
On 10/25/2007 9:27 AM, Ranjan Bagchi wrote:
> Hi --
>
> I'm working with some data frames with fairly high nrows (call it 8
> columns, by 20,000 rows). Are there any indexes on these columns?
>
> When I do a df[df$foo == 42,] [which I think is idiomatic], am I doing a
> linear
> search or som
I think that if you want to apply a simple calculation
to each element something like
df[,3:7] <- df[,3:7] + 10
would do what you want.
--- DEEPANKAR BASU <[EMAIL PROTECTED]> wrote:
> Hi All,
>
> I have a data frame with a group of variables named
> b1, b2, b3, ..., b18. These variables take
On Thu, 2007-10-25 at 16:26 +1000, [EMAIL PROTECTED] wrote:
> I wondered if the real problem was bigger than your abstract version.
> OK. Here is one way to do it
>
> > myfunc <- function(x) {
> nam <- deparse(substitute(x))
> val <- mean(x)
> cat("mean(", nam, ") =", val, "\n")
>
You can pass oma= and mar= to plot.zoo. This is apparent from the
argument list in ?plot.zoo or from args(plot.zoo)
On 10/25/07, John Theal <[EMAIL PROTECTED]> wrote:
> Hi Gabor,
>
> Sorry to bother you again, I'm having trouble controlling the margins
> on a multiplot window. Using a previous e
De: Trochine, Carolina
Enviado el: mar 23/10/2007 13:52
Para: [EMAIL PROTECTED]
Asunto: Multivariate regression tree: problems with surrogate splits (complete
commands)
The secuence was not complete in the other message, sorry!
sprot_matrix=read.csv("sprot
Hi --
I'm working with some data frames with fairly high nrows (call it 8
columns, by 20,000 rows). Are there any indexes on these columns?
When I do a df[df$foo == 42,] [which I think is idiomatic], am I doing a linear
search or something better? If the column contents is ordered, I'd like
De: Trochine, Carolina
Enviado el: mar 23/10/2007 13:45
Para: [EMAIL PROTECTED]
Asunto: Multivariate regression tree: problems with surrogate splits
R helpers,
I am working with the R program performing multivariate regression trees (MRT).
I have a matrix wi
Hi,
I have R 2.6.0 with updated lme4 and Matrix packages, and I am trying to
fit a nonlinear multilevel model. I get the following error message:
Error in nlmer(f ~ grModel(x, w, Tmin, Tmax, Topt, kopt, m) ~ kopt |
flat, :
gradient attribute of evaluated model must be a numeric matrix
and I
I think a better idea would be to make a jEdit plugin that embeds an R
console component (perhaps the one from JGR, but the last I checked it was
not embeddable, unfortunately). Anyway, it would be easy enough to roll your
own by extending the existing jEdit Console plugin using JRI. I would have
d
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Alacaðý araþtýrma ücreti 1 Eylül 2007 tarihinden itibaren 1615EUR = 2668CHF
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Hi,
Thanks to everyone for their advice! It was really helpfull.
Geertje
Geertje van der Heijden
PhD student
Tropical Ecology
School of Geography
University of Leeds
Leeds LS2 9JT
Tel: (+44)(0)113 3433345
Email: [EMAIL PROTECTED]
From: Vic
- Original Message -
From: "Petr PIKAL" <[EMAIL PROTECTED]>
To: "sun" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Wednesday, October 24, 2007 4:46 PM
Subject: Re: [R] convert factor dataframe into numeric matrix
> Hi
>
> [EMAIL PROTECTED] napsal dne 24.10.2007 16:24:29:
>
>>
>> --
HI,
Perhaps:
cbind(ID=DF$ID,as.data.frame(sapply(levels(VAL$type),
function(x){DF[,x]=VAL$number[VAL$type==x];DF[,x]})))
On 25/10/2007, Lauri Nikkinen <[EMAIL PROTECTED]> wrote:
>
> Hi folks,
>
> I have two dataframes like these:
>
> DF <- data.frame(ID=c("AA1234","AB3233","AC4353","AD2345","AE
when i am trying to start R-Engine from my JAVA web application it is
throwing this exception
rShowMessage "Fatal error: unable to restore saved data in .RData
if any one knows about this issue plz let me know
--
View this message in context:
http://www.nabble.com/Fatal-error-tf4690072.html#a134
Hi Gabor,
Sorry to bother you again, I'm having trouble controlling the margins
on a multiplot window. Using a previous example you posted in the
archives:
library(zoo)
# test data
z <- structure(c(21,34,33,41,39,38,37,28,33,40),
index = structure(c(8044,8051,8058,8065,8072,
On Thu, 25 Oct 2007, Nic Surawski wrote:
> Dear R users,
>
> I have been using the zicounts package (verson 1.1.4) in R (version
> 2.4.1). I have been fitting zero inflated Poisson regressions to model
> the number of trips made by a household. Whilst I can get the best fit
> parameter set fro
On 10/24/07, Paul Murrell <[EMAIL PROTECTED]> wrote:
> Hi
>
>
> Gustaf Rydevik wrote:
> > Hi all,
> >
> > I'm trying to generate a plot containing a scatterplot, with marginal
> > densityplots for x and y.
> > However, when I try to generate a vertical densityplot, I get the
> > message "warning: c
I think 'usr' can be used to specify the extent of the plotting
region. So it is useful if, say, you are creating a multiplot window
and want to scale all plots for comparison. I think that (for
multiplots) xlim has limited usefulness.
On 23-Oct-07, at 10:12 PM, Dave Hewitt wrote:
> No p
Dieter Menne menne-biomed.de> writes:
Trying again with a minimal example. When I run the following batch file under
Windows 2000, R 2.6.0, I get the error message below. Rtools are installed and
at the front of the path.
rem Begin test package; File test.cmd
rem Check the following line for li
Dear listers,
I am trying to use an old script which was working well in the previous
R version. It looks like if it no longer works in R.6.0. I have a model
of the form:
glm(nath2$Positif ~ n + yearday + x + y + I(x^2) + I(y^2) + yearday:x +
yearday:y, family = poisson, data = nath2)
and wan
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