Hi there,
This is now the code
%
library(grid)
vp - viewport(
x = unit(0, npc),
y = unit(0, npc),
just = c(left, bottom),
xscale = c(-1, 1) ,
yscale = c(-1, 1))
vp2 - viewport( # probably not needed but I had trouble placing the xaxis
x = unit(0,npc),
y = unit(0.5, npc),
just =
Ivo Shterev wrote:
I have a question about list indexing. Lets say we have a list of 3 lists,
each containing 3 different type elements:
(Details of your nice example code removed)
a=replicate(3, list(list(c(1,1,1), diag(3), c(2,2,2
str(a) # I prefer this to print(a) because
Hi, R users,
I'm using R to test the unit root for my time series data. I just compared the
ADF.test in uroot package and the adf.test in tseries package. It seems
it is difficult to define the time trend and intercept in adf.test. But it is
easy to do these in ADF.test. ADF.test also help
legen wrote:
I have a question about changing the height or scale of the y axis. When I
use following two R codes, I can get two plots. Please look at the y axes,
the number of indices (x1, x2, …) on the y axis in the first plot is
smaller than that in the second plot, and hence the
Hi Mark,
I frequently need to do that when importing data. This one-liner works:
data.frame(mapply(as, x, c(integer, character, factor),
SIMPLIFY=FALSE), stringsAsFactors=FALSE);
but it has two problems:
1) as() is an S4 method that does not always work
2) writting the vector of classes
This sort of experience is why 'The R Inferno'
came into existence.
Patrick Burns
patr...@burns-stat.com
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of The R Inferno and A Guide for the Unwilling S User)
Craig P. Pyrame wrote:
Dear Stavros,
What you discuss below is somewhat scary
Hello,
I tried to build an own R package on Windows XP but get an error which I can't
solve.
I called my package pack. It works to creat the file pack with
package.skeleton().
But when I try to compile it with Rcmd build --binary pack I obtain the
following error:
IO::Seekable::seek missing
Bob Ly robertly at vfemail.net writes:
I have the following:
Date-c(08/05/08,08/06/08,08/07/08)
Weight-c(209.4,211.8,210.0)
planned.meal-cbind(Date,Weight)
planned.meal
DateWeight
1 08/05/08, 209.4
2 08/06/08, 211.8
3 08/07/08, 210.0
This is strange. When I run your
On Thu, 25 Jun 2009, Stein, Luba (AIM SE) wrote:
Hello,
I tried to build an own R package on Windows XP but get an error which I can't
solve.
I called my package pack. It works to creat the file pack with
package.skeleton().
But when I try to compile it with Rcmd build --binary pack I
This will give you a greek character, see ?plotmath
grid.text(expression(mu*(s,t)), 0.5, unit(5, lines), vp=vp2)
The following works for me, it may be that you're using an outdated
version of R,
vp - viewport(
x = unit(0, npc),
y = unit(0, npc),
just = c(left, bottom),
xscale = c(-1, 1) ,
Hello,
I am a new R user. More precisely, I am working with JRI (on a Eclipse
Java6 project under Ubuntu9).
I have difficulties to access some R packages (For example: package
stats, object Normal, function pnorm).
I have tried many solutions to set the right path but nothing have succeed:
Version 1.2.6 of RODBC is now on CRAN. This has a number of bug fixes and many
workarounds for ODBC driver quirks--I've set up further testbeds for SQL Server
2008, Oracle and DB2.
More visibly, the documentation has been expanded in several ways, in
particular in collecting together advice
Hi Antoine,
I have got this to work on Windows, the process should be the same for
Linux.
For an eclipse plugin you will need to add the following to your
MANIFEST.MF file:
Bundle-ClassPath: whatever you normally have,
lib/JRI.jar
Bundle-NativeCode: /lib/jri.dll; osname=winxp; processor=x86
Dear Harry,
to complete the picture, for the packages installed on my machine help.search()
yielded:
help.search(Dickey)
Help files with alias or concept or title matching 'Dickey' using fuzzy
matching:
CADFtest::CADFtest Hansen's Covariate-Augmented Dickey Fuller
Dear R users,
I'm trying to randomly recreate a real dataset with missing data and I'm
not quite sure if I can use the sample command for this. I think it
might be better to do it in 2 steps and randomly replace the sampled
data with missing data...
So something like this
x -
Hi,
I don't think the fill parameter can be a colour gradient. You'll need
to create small polygons, each with its own fill (200, say). Try this,
x= c(0, 0.5, 1)
y= c(0.5, 1, 0.5)
grid.polygon(x=x, y=y, gp=gpar(fill=grey90, col=grey90))
xx - seq(range(x)[1],range(x)[2], length=100)
yy -
Hi all,
I want to make multiple least squares estimation on two matrix (without
intercept)
imagine matrix a and matrix b, I want to regress each colum on matrix
a (dependent variable) on each column of matrix b, I mean, regress first
colum on a to first column on b. Second column on a to second
[I neglected to check some details in the previous post]
This one should work better,
library(grid)
x= c(0, 0.5, 1)
y= c(0.5, 1, 0.5)
grid.polygon(x=x, y=y, gp=gpar(fill=NA, col=grey90)) # outer shell
xx - seq(range(x)[1],range(x)[2], length=100)
dx - diff(xx) # width of clipped triangles
Dear R-community,
I'm struggling with a paper that reports only fragmented results of a
2by2by3 experimental design. However, Means and SDs for all cells are given.
Does anyone know a package/function that helps computing an ANOVA with only
Means and SDs as input (resulting in some sort of
If you want to average 20% missing values then you could try it in 1 step,
viz:
sample(c(1:2, rep(NA, 2000)),100)
Otherwise, 2 steps is preferable. Use code as below:
sample(1:2,100)-kk
kk[sample(1:100,20)]-NA
Paul
--
View this message in context:
Joanne,
[...snip...]
x - sample(1:2, 100) #without replacement
Now I want x to contain to 20% missing data (NA). Could anyone help me
how to do this?
See if this helps:
n - length(x)
x[sample(n, 0.2*n)] - NA
cheers,
-Girish
--
View this message
On 25-Jun-09 10:15:30, Sebastian Stegmann wrote:
Dear R-community,
I'm struggling with a paper that reports only fragmented results
of a 2by2by3 experimental design. However, Means and SDs for all
cells are given.
Does anyone know a package/function that helps computing an ANOVA
with only
Dear,
I am using wavelet function from library(dplR), I would like to have a
normalised spectrum as an output (power relative to white noise).
I was wondering if anyone can help me with that.
Thank you.
irina
__
R-help@r-project.org mailing list
Hi,
Could I please be removed from the email list!
Thank you,
Sophie
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
R 2.8
Windows XP
Excel 2003
I am trying to read an Excel spread sheet. I have looked at the RODBC help
pages and am having trouble setting up code that will work. My code and the
results are pasted below:
Hello,
how can I get a variable driven summary of one column of my data.frame?
Usually I would do
summary(data$columnname) to get a summary of column named
columnname of my data.frame named data.
In my case the columnname is not static but can be set dynamically.
So I save the chosen
Hi
I have the feeling, that the argument stringsAsFactors has no impact in the
function expand.grid:
a - c(PR, NC, A2, BS)
b - c(1, 0.5, 0.25, 0.125, 0.0625, 0.03125)
class(expand.grid(css, fscs, stringsAsFactors=FALSE)[[1]])
[1] factor
class(expand.grid(css, fscs, stringsAsFactors=TRUE)[[1]])
Another options is:
head(lapply(a, tail, 2), 2)
On Wed, Jun 24, 2009 at 8:42 PM, Ivo Shterev idc...@yahoo.com wrote:
Hello,
I have a question about list indexing. Lets say we have a list of 3 lists,
each containing 3 different type elements:
a=replicate(3, list(list(c(1,1,1), diag(3),
Try:
data[,variable]
On Thu, Jun 25, 2009 at 6:44 AM, Anne Skoeries h...@anne-skoeries.dewrote:
Hello,
how can I get a variable driven summary of one column of my data.frame?
Usually I would do
summary(data$columnname) to get a summary of column named columnname of
my data.frame named
One thing you might consider when working with large dataframes is that
instead of partitioning the dataframe into smaller ones, create a list of
indices and use that to access the subset. Works especially well when using
'lapply' to cromp through many segments of a data frame:
y
suid
I also not find this in the source code, but you can rewrite the function
with this:
expand.grid -
function (..., KEEP.OUT.ATTRS = TRUE, stringsAsFactors = FALSE)
{
nargs - length(args - list(...))
if (!nargs)
return(as.data.frame(list()))
if (nargs == 1L is.list(a1 -
That's quite nice. Three comments:
- colClasses() in R.utils is similar, except for the particular
codes and classes supported, to expandClasses() here.
- not sure if this is important but if as() were the last
possibility tried rather than the first then in most
cases (in fact all cases
Dear John,
Try this:
require(RODBC)
spreadsheet - Sheet1 # please change this to your needs
channel -
odbcConnectExcel(i:\\all\\sorkinjohn\\stats\\silvermannatalie\\NEMOcombined06-24-09.xls)
mydata - sqlFetch(channel, spreadsheet)
odbcClose(channel)
# attach(mydata)
mydata
HTH,
Jorge
On Thu,
Dear List,
I'm searching for a way (package, function or something) providing
the http PUT, GET POST ... methods in R. httpRequest and RCurl
seems to have a lack of the PUT method.
Regards
Thomas
|| Thomas Bock c/o Physikalisch-Technische Bundesanstalt
|| Abbestr. 2-12, D-10587 Berlin,
Try the function sqlFetch to import the data in the spreadsheet.
jo-odbcConnectExcel(i:\\all\\sorkinjohn\\stats\\silvermannatalie\\NEMOcomb
ined06-24-09.xls,readOnly = TRUE)
mo-sqlFetch(jo,'Your Sheet Name or Number',colnames=F,rownames=F)
mo
Hope it helps.
legen wrote:
Hallo, All,
I have a question about changing the height or scale of the y axis. When I
use following two R codes, I can get two plots. Please look at the y axes,
the number of indices (x1, x2, …) on the y axis in the first plot is smaller
than that in the second plot, and hence the
Very good points :-)
- colClasses() in R.utils is similar, except for the particular codes and
classes supported, to expandClasses() here.
In fact I saw colClasses() once and got the idea from it, but when I needed the
functionallity I did not remember where had I seen it and rewrote it.
Dear Oscar,
Try this:
# Some data
set.seed(123)
a - matrix(rnorm(100*200), ncol = 200)
b - matrix(rnorm(100*200), ncol = 200)
# Auxiliar function to extract the coefficient
# after fitting models without intercept
mycoef - function(x, y) coefficients( lm(y ~ x - 1) )
# Results
res -
On Thu, Jun 25, 2009 at 1:38 PM, Henrique Dallazuanna www...@gmail.comwrote:
I also not find this in the source code, but you can rewrite the function
with this:
That's true - but this should be fixed in the package itself. I have just
converted the factor to characters.
Cheers
Rainer
...
Occasionally, (about 1 in every 100 simulations) I get the following warning:
Error in coxph(Surv(start, end, censorind) ~ binary + uniform :
X matrix deemed to be singular; variable 2
It is not uncommon for the X matrix in a Cox model to be close enough to
singular that the
On 6/25/2009 5:44 AM, Anne Skoeries wrote:
Hello,
how can I get a variable driven summary of one column of my data.frame?
Usually I would do
summary(data$columnname) to get a summary of column named columnname
of my data.frame named data.
In my case the columnname is not static but can
Zeljko Vrba wrote:
On Wed, Jun 10, 2009 at 08:21:06AM +0200, Poizot Emmanuel wrote:
Error in fun(...) :
GDAL Error 1: libgrass_I.so: Ne peut ouvrir le fichier d'objet
partagé: Aucun fichier ou dossier de ce type (sorry for the french :) )
It would have been far more useful had
It works for me. Try a more recent version of R.
a - c(PR, NC, A2, BS)
b - c(1, 0.5, 0.25, 0.125, 0.0625, 0.03125)
class(expand.grid(a, b, stringsAsFactors=FALSE)[[1]])
[1] character
class(expand.grid(a, b, stringsAsFactors=TRUE)[[1]])
[1] factor
R.version.string
[1] R version 2.9.1 RC
Have a look at ddply from the plyr package, http://had.co.nz/plyr.
It's made for exactly this type of operation.
Hadley
On Wed, Jun 24, 2009 at 10:34 PM, Stephan Lindnerlindn...@umich.edu wrote:
Dear all,
I have a code where I subset a data frame to match entries within
levels of an factor
On Thu, Jun 25, 2009 at 2:55 PM, Gabor Grothendieck ggrothendi...@gmail.com
wrote:
It works for me. Try a more recent version of R.
Good to see that it is fixed in 2.9.1. I am using the latest release and am
going to wait till 2.9.1 is released.
Thanks,
Rainer
a - c(PR, NC, A2, BS)
Your request for a more general approach is precisely the reason that
Hadley Wickham wrote the plyr package. He describes a split-apply-
combine strategy for a variety of data structures and tools to
implement those strategies here:
http://had.co.nz/plyr/plyr-intro-090510.pdf
The argument
Hi,
I have a big dataframe with a POSIXct column and I'd like to extract a
subset contained in a given time interval, from Date 1 to Date 2.
Paulo E. Cardoso
__
R-help@r-project.org mailing list
Look at the anova.mean function in the HH package.
It does what you are asking, although limited to one-way ANOVA.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Dear R-Users,
I need to lookup values from a 2-d table using the row names and column
names as indices. I was wondering if there's a way to do this without an
explicit loop.
Example:
#x is the 2-d table that holds the values
x - matrix(rnorm(26*12),nrow=26)
rownames(x) - letters
colnames(x) -
Resending after fixing a mistake in the earlier email ... sorry for the
confusion.
**
Dear R-Users,
I need to lookup values from a 2-d table using the row names and column
names as indices. I was wondering if there's a way to do this without an
explicit loop.
Example:
#x is the 2-d table
That works!! Very nice way to do it! Thank you, Henrique!
Rama Ramakrishnan
On Thu, Jun 25, 2009 at 10:11 AM, Henrique Dallazuanna www...@gmail.comwrote:
Try this:
y$values - diag(x[y$ltrs, y$mnths])
On Thu, Jun 25, 2009 at 11:02 AM, Rama Ramakrishnan r...@alum.mit.eduwrote:
Dear R-Users,
Try this:
y$values - diag(x[y$ltrs, y$mnths])
On Thu, Jun 25, 2009 at 11:02 AM, Rama Ramakrishnan r...@alum.mit.eduwrote:
Dear R-Users,
I need to lookup values from a 2-d table using the row names and column
names as indices. I was wondering if there's a way to do this without an
explicit
Dear all,
I'm training an SVM with default settings on a matrix csr (SparseM
package). I realized that if I train
the SVM with the (hopefully) equivalent matrix (Matrix package)
representation, the returned models and predictions
sometimes differ. I expected both representations of the same
Follow-on question: is there a way to do this for higher-dimensional (i.e.
more than 2 dimensions) arrays?
On Thu, Jun 25, 2009 at 10:17 AM, Rama Ramakrishnan r...@alum.mit.eduwrote:
That works!! Very nice way to do it! Thank you, Henrique!
Rama Ramakrishnan
On Thu, Jun 25, 2009 at 10:11
On Jun 25, 2009, at 10:02 AM, Rama Ramakrishnan wrote:
Dear R-Users,
I need to lookup values from a 2-d table using the row names and
column
names as indices. I was wondering if there's a way to do this
without an
explicit loop.
Example:
#x is the 2-d table that holds the values
x -
On 6/25/2009 7:20 AM, John Sorkin wrote:
R 2.8
Windows XP
Excel 2003
I am trying to read an Excel spread sheet. I have looked at the RODBC help
pages and am having trouble setting up code that will work. My code and the
results are pasted below:
On Jun 25, 2009, at 10:24 AM, Rama Ramakrishnan wrote:
Follow-on question: is there a way to do this for higher-dimensional
(i.e.
more than 2 dimensions) arrays?
The apply method I just posted generalizes to higher dimensional arrays.
--
DW
On Thu, Jun 25, 2009 at 10:17 AM, Rama
Hello,
Is there a way to access function's slots from inside the function? I
want to make functions slot dependent without recurring to generic
function mechanism.
Probably this goes a bit against R philosophy, but otherwise I don't
really see the use of extending functions in R.
Try this (shown for stated problem but generalizes by just adding
additional arguments):
mapply([, list(x), ltrs, mnths)
On Thu, Jun 25, 2009 at 10:24 AM, Rama Ramakrishnanr...@alum.mit.edu wrote:
Follow-on question: is there a way to do this for higher-dimensional (i.e.
more than 2
Hello everybody,
I have programed a function to translate the R object structure into a XML
data structure but I haven't found a function in the XML package to export
this data in an XML file!
my data look like that:
FactorAssessor:codec/Factor
Sum Sq 33.98159/Sum Sq
Df 28/Df
F
I get this error while computing partial correlation.
*Error in solve.default(Szz) :
system is computationally singular: reciprocal condition number =
4.90109e-18*
Why is it?Can anyone give me some idea ,how do i get rid it it?
This is the function i use for calculating partial correlation.
See the StatDataML package.
On Thu, Jun 25, 2009 at 10:54 AM, guillaume Le
Rayleray.guilla...@gmail.com wrote:
Hello everybody,
I have programed a function to translate the R object structure into a XML
data structure but I haven't found a function in the XML package to export
this data in
Hi Guillaume
Once you have created an XML representation using any of the various
ways to represent XML in R via the XML package (e.g. internal nodes,
lists of lists, hash trees), you can use the saveXML() function:
saveXML(myXML, fileName.xml)
Personally, I use newXMLNode() and friends
Thanks, David, that works too!
On Thu, Jun 25, 2009 at 10:30 AM, David Winsemius dwinsem...@comcast.netwrote:
On Jun 25, 2009, at 10:24 AM, Rama Ramakrishnan wrote:
Follow-on question: is there a way to do this for higher-dimensional (i.e.
more than 2 dimensions) arrays?
The apply
The updated package has been submitted to CRAN and will propagate to
mirrors over the next day or so.
It is maintained on R-Forge at http://r-forge.r-project.org/projects/writexls
, where downloads will be available as well. There is a transient
problem at the moment with R-Forge and the
BTW, the XML you show is not legal XML.
For instance, the element
Pr(F)NA/Pr(F)
is not a legal XML element name.
Similarly
/anova:2
is not a legal name since it is 2.
And anova:2 means an element with name 2 with (XML) name space prefix
anova (which must be declared previously)
As Gabor
Hi All...
I¹m trying to build a small demo using gWidgets which permits interactive
scaling and selection among different things to plot. I can get the widgets
for scaling to work just fine. I am using gcheckboxgroup to make the
(possibly multiple) selections. However, I can¹t seem to figure
Thanks, Gabor. Works great!
On Thu, Jun 25, 2009 at 10:38 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
Try this (shown for stated problem but generalizes by just adding
additional arguments):
mapply([, list(x), ltrs, mnths)
On Thu, Jun 25, 2009 at 10:24 AM, Rama
I am trying to estimate models with subsets using the NLME package. However, I
am getting an error in the case below (among others):
subset - c(rep(TRUE, 107), FALSE)
fm2 - lme(distance ~ age + Sex, data = Orthodont, random = ~ 1,
subset=subset)
Error in xj[i] : invalid subscript type
Dear R-help,
I'm very sorry to ask 2 questions in a week. I am using the package
'crr' and it does exactly what I need it to when I use the dataset a.
However, when I use dataset b I get the following error message:
Error in drop(.Call(La_dgesv, a, as.matrix(b), tol, PACKAGE = base)) :
system
On Jun 25, 2009, at 11:35 AM, Rebecca Sela wrote:
I am trying to estimate models with subsets using the NLME package.
However, I am getting an error in the case below (among others):
subset - c(rep(TRUE, 107), FALSE)
fm2 - lme(distance ~ age + Sex, data = Orthodont, random = ~ 1,
Rebecca,
I think the problem is that subset is a nume of an R function. If you do
something like
subs - c(rep(TRUE, 107), FALSE)
fm2 - lme(distance ~ age + Sex, data = Orthodont, random = ~ 1,
subset=subs)
everything works fine.
Hope this helps,
Andy
__
This means that your design matrix or model matrix is rank deficient, i.e it
does not have linearly independent columns. Your predictors are collinear!
Just take your design matrices covaea or covaeb with 17 predcitors and
compute their rank or try to invert them. You will see the problem.
On Wed, Jun 24, 2009 at 9:04 PM, Rolf Turnerr.tur...@auckland.ac.nz wrote:
Do not get your knickers in a twist. R works simply and straightforwardly
in simple straightforward situations.
Though I find R an incredibly useful tool, alas, it is simply not true
that R works simply and
Your covariance matrix Szz is not positive definite. It is singular. The
following test that you are doing is neither necessary nor useful:
zz.ev - eigen(Szz)$values
if(min(zz.ev)[1]0){
stop(\'Szz\' is not positive definite!\n)
}
You may want to use Moore-Penrose
Erratum:
ifelse(TRUE,dd,dd) = 1230786000 (class numeric)
should be
ifelse(TRUE,tt,tt) = 1230786000 (class numeric)
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Hello useRs:
Does anyone have thoughts on the lifecycle of older releases of R? I
know that currently the 2.8.x and 2.9.x releases seem to be actively
supported on the mailing lists, but what about older releases, say
2.4.x? Curious to hear when people think older versions of R become
obsolete
Hi,
How can I obtain the residuals of my long memory model:
x.fd = fracdiff(dif, nar=1, nma=2, M=30)
There is no function as acf() as for arima or garch models...
Many thanks
Ana
[[alternative HTML version deleted]]
__
R-help@r-project.org
Dear Rolf,
Rolf Turner wrote:
On 25/06/2009, at 12:27 PM, Craig P. Pyrame wrote:
Dear Stavros,
What you discuss below is somewhat scary to me as an R newbie. Is this
just an incident, a bug perhaps, or rather the way things typically go
in R, as your Welcome to R! seems to suggest? I
On Thu, 25 Jun 2009, Rowe, Brian Lee Yung (Portfolio Analytics) wrote:
Does anyone have thoughts on the lifecycle of older releases of R? I
know that currently the 2.8.x and 2.9.x releases seem to be actively
supported on the mailing lists, but what about older releases, say
2.4.x? Curious to
On Thu, Jun 25, 2009 at 9:38 AM, Rowe, Brian Lee Yung (Portfolio
Analytics)b_r...@ml.com wrote:
Hello useRs:
Does anyone have thoughts on the lifecycle of older releases of R? I
know that currently the 2.8.x and 2.9.x releases seem to be actively
supported on the mailing lists, but what about
There is an archive for all packages for older versions of R, but if
you want up-to-date functionality of packages then you need the
newest versions.
my 2 cents
stephen
On Thu, Jun 25, 2009 at 12:38 PM, Rowe, Brian Lee Yung (Portfolio
Analytics)b_r...@ml.com wrote:
Hello useRs:
Does anyone
Hi,
I was trying to read some help of functions and in all functions I try to see
is giving me this error:
?write.table
Erro em
print.help_files_with_topic(C:/ARQUIV~1/R/R-29~1.0/library/utils/chm/write.table)
:
CHM file could not be displayed
Anybody knows what is happening?
Good to know. I know that other software projects (whether languages,
OSes, applications) tend to keep recent versions in maintenance mode for
a certain period of time prior to retiring them. I wonder if that
would happen with R, either by design or out of necessity of an
increasing user base.
On Jun 25, 2009, at 11:38 AM, Rowe, Brian Lee Yung (Portfolio
Analytics) wrote:
Hello useRs:
Does anyone have thoughts on the lifecycle of older releases of R? I
know that currently the 2.8.x and 2.9.x releases seem to be actively
supported on the mailing lists, but what about older
Hi,
I'm trying to fit a binary logistic regression model, and would like to
consider certain characteristics B and C only for people with variable A=1 and
not for those with variable A=0, so I'm trying to do the following:
model- lrm(formula= y ~ A: (B+C) + D + E +...)
I've had no
Alejandra Solis Herrera wrote:
Hi,
I'm trying to fit a binary logistic regression model, and would like to
consider certain characteristics B and C only for people with variable A=1 and
not for those with variable A=0, so I'm trying to do the following:
model- lrm(formula= y ~ A:
Hello all,
How to draw a line in plot when I know the start point(x1,y1) and end
point(x2,y2)? I need make this as additional information in the graph:
plot(wl2[[1]],wl2[[2]])
I think that is possible make this with the function abline(), is possible? I
looked the function lines() too, but
Dear Lesandro,
Take a look at
?segments
HTH,
Jorge
On Thu, Jun 25, 2009 at 2:30 PM, Lesandro lesand...@yahoo.com.br wrote:
Hello all,
How to draw a line in plot when I know the start point(x1,y1) and end
point(x2,y2)? I need make this as additional information in the graph:
On Jun 25, 2009, at 1:30 PM, Lesandro wrote:
Hello all,
How to draw a line in plot when I know the start point(x1,y1) and
end point(x2,y2)? I need make this as additional information in the
graph:
plot(wl2[[1]],wl2[[2]])
I think that is possible make this with the function abline(), is
On 25-Jun-09 18:38:37, Marc Schwartz wrote:
On Jun 25, 2009, at 1:30 PM, Lesandro wrote:
Hello all,
How to draw a line in plot when I know the start point(x1,y1)
and end point(x2,y2)? I need make this as additional information
in the graph:
plot(wl2[[1]],wl2[[2]])
I think that is possible
I have sent a note to Peter Ruckdeschel who wrote the excellent
SweaveListingUtils package but find myself up against a deadline in
preparing a handout for useR! 2009.
The following is supposed to work:
=
plot(x, y) # Figure `\ref{myfig}`
@
Where the back tick ` is an escape character and
On Jun 25, 2009, at 1:51 PM, Ted Harding wrote:
On 25-Jun-09 18:38:37, Marc Schwartz wrote:
On Jun 25, 2009, at 1:30 PM, Lesandro wrote:
Hello all,
How to draw a line in plot when I know the start point(x1,y1)
and end point(x2,y2)? I need make this as additional information
in the graph:
Thanks Jorge and Marc,
I drew the line using the function:
segments(x0, y0, x1, y1)
Lesandro
--- Em qui, 25/6/09, Marc Schwartz marc_schwa...@me.com escreveu:
De: Marc Schwartz marc_schwa...@me.com
Assunto: Re: [R]
A naive question: what happened to the xlsReadWrite package?
http://cran.r-project.org/web/packages/xlsReadWrite/
It says that it was removed from the CRAN repository. Are there any plans
for it be available again?
Thanks,
Andrew
[[alternative HTML version deleted]]
Glad if it helps.
check out this page of examples for tikz,
http://www.texample.net/tikz/examples/feature/shadings/
If you do choose this route, you could perhaps read the new wiki page on
importing graphics in a R plot,
Hi, Jim,
Thank you for your reply. I just want to increase the height of y axis in
the second plot in order to show all the indices (x1, x2, ...). Can you help
me? Thank you again.
Legen
Jim Lemon-2 wrote:
legen wrote:
Hallo, All,
I have a question about changing the height or scale of
Hi, Dieter Menne,
Thank you for your help. I tried par(las=1,cex=0.5), but it changed only
the size of indices on the y axis in the second plot relative to the
default. I really want to increase the height of y axis in order to show all
the indices (x1, x2, ...). In the genetic study, we aften
Hello,
I've been trying to calculate home range sizes (for Icelandic geese!) using
minimum convex polygons with the adehabitat package. I've tried to use the R
code shown by demo(homerange) in adehabitat and when that didn't work I've
fiddled around with it but to no avail...Below is the output
I have many S-plus project folders that I need to convert to R workspaces.
For the smaller project folders ( 200MB), using data.dump with oldStyle
= T and data.restore (in the foreign package) within R seems to work
fine. However, I have several project folders that are quite large (~ 4GB).
When
1 - 100 of 117 matches
Mail list logo
