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"
try
do.call(rbind, yourByList)
hth,
Kingsford Jones
On Wed, Jun 24, 2009 at 9:34 PM, Stephan Lindner wrote:
> Dear all,
>
>
> I have a code where I subset a data frame to match entries within
> levels of an factor (actually, the full script uses three difference
> factors do do that). I'm very
On Jun 24, 2009, at 8:01 PM, Roslina Zakaria wrote:
Hi r-users,
If I have 10 data generated from multi skew-t distribution, how
do I find the tail probability of alpha=.01?
You have given us the tail probability, so I assume you are asking for
the the value of x above which the tail
On Jun 24, 2009, at 9:26 PM, R_help Help wrote:
Hi,
I do not understand why after I called apply on a function that
returns an
xts (getIdvAdjSeries) it returns a matrix whose columns are just
numeric
value of time series in xts instead of a list of xts objects.
Basically, I called the fo
On Jun 24, 2009, at 9:29 PM, R_help Help wrote:
I do not want to use histogram because I have more control over base
graphs
that R provide. I'm wondering if hist function provides a parameter
that can
set y axis as % of count instead of raw count?
So you want 100 times the proportions t
Good Day,
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
>write.table(planned.meal, file="plannedMeal1.txt",
+ quote=FA
Dear all,
I have a code where I subset a data frame to match entries within
levels of an factor (actually, the full script uses three difference
factors do do that). I'm very happy with the precision with which I can
work with R, but since I loop over factor levels, and the data frame is
big, the
Hi,
I wonder whether there is a way to generate a polygon (a triangle in
my case) with color gradient using grid.polygon() in package grid?
I tried something like
library(grid)
grid.polygon(x=c(0, 0.5, 1), y=c(0.5, 1, 0.5), gp=gpar(col=NA,
fill=colorRampPalette(c("green", "lightgray"),
I do not want to use histogram because I have more control over base graphs
that R provide. I'm wondering if hist function provides a parameter that can
set y axis as % of count instead of raw count?
If not, I'd like to know how I can add text or line on histogram function? I
don't like the idea o
Hi,
I do not understand why after I called apply on a function that returns an
xts (getIdvAdjSeries) it returns a matrix whose columns are just numeric
value of time series in xts instead of a list of xts objects.
Basically, I called the following:
apply(matrix(tickers,ncol=1),1,FUN=getDivAdjSer
We’re very pleased to announce an update to http://crantastic.org/,
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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 have just
started to
learn R
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 have just started to
learn R, and my initial euphoria of the "I can do anything with
Hi r-users,
If I have 10 data generated from multi skew-t distribution, how do I find
the tail probability of alpha=.01?
Thank you.
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Hi Duncan,
Thanks for this useful answer. See below.
Duncan Temple Lang a écrit :
Hi Olivier
Olivier Cailloux wrote:
Hello,
I am trying to reach a web service using the SOAP package. I
succeeded calling the web service, but not sending parameters to it.
After much research and tries, I th
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), c(2,2,2
> a
[[1]]
[[1]][[1]]
[1] 1 1 1
[[1]][[2]]
[,1] [,2] [,3]
[1,]100
[2,]010
[3,]0
Hi,
perhaps the question I have is more a statistical question than a
question about R.
But I hope nevertheless someone could help me.
If I have the following anova table;
Analysis of Variance Table
Response: aktiv
Df Sum Sq Mean Sq F value Pr(>F
Perhaps:
x <- 1:26
xyplot(x ~ x, groups = x, pch = letters, col = rainbow(26))
On Wed, Jun 24, 2009 at 3:28 PM, Katharina
May wrote:
> Hi,
>
> I've got the following problem which I cannot think of a solution right now:
>
> if got a lattice xyplot in black and white and a grouping variable
> wit
my earlier comment is probably irrelevant since you are fitting only
one qss component and have no other covariates.
A word of warning though when you go back to this on your new machine
-- you are almost surely going to want to specify
a large lambda for the qss component in the rqss call.
On Wed, Jun 24, 2009 at 2:05 PM, Katharina
May wrote:
> That's a point. I justed wanted to provide an overview for myself to
> see the tendencies in a direct comparement
> and with an easy way to distinct them, but maybe the text panel can
> help me with that...
>
> Well anyway, is it right that a
Hi
Kexin Ji wrote:
Hi,
I want to produce two parallel rotated strips with color gradient. So
far, the sample strip is something produced by this:
pushViewport(viewport(x = unit(0.638, "npc"), y =unit(0.386, "npc"),
width=.62, height=0.006, angle=137.2))
grid.rect(y=100:1/100, just="top",
Yep, its looking like a memory issue -- we have 6GB RAM and 1GB swap --
I did notice that the analysis takes far less memory (and runs) if I:
tahoe_rq <-
rqss(ltbmu_4_stemsha_30m_exp.img~ltbmu_eto_annual_mm.img,tau=.99,data=boundary_data)
(which I assume fits a line to the quantiles)
vs.
ta
On 24 June 2009 at 14:07, Jonathan Greenberg wrote:
| I installed R 2.9.0 from the Debian package manager on our amd64
| system that currently has 6GB of RAM -- my first question is whether
| this installation is a true 64-bit installation (should R have access to
| > 4GB of RAM?) I suspe
Jonathan,
Take a look at the output of sessionInfo(), it should say x86-64 if
you have a 64bit installation, or at least I think this is the case.
Regarding rqss(), my experience is that (usually) memory problems are
due to the fact that early on the processing there is
a call to model.mat
Perhaps what you do should depend on what you want to see. If all the lines
lie near one another that says one thing. If all but one or two "agree" but
the mavericks are in obvious disagreement, then that begs to ask "why?" If
the entire lot looks like spaghetti, then that is informative also.
That's a point. I justed wanted to provide an overview for myself to
see the tendencies in a direct comparement
and with an easy way to distinct them, but maybe the text panel can
help me with that...
Well anyway, is it right that a grouped black and white plot can
contain a maxinum of 8 distingui
Rers:
I installed R 2.9.0 from the Debian package manager on our amd64
system that currently has 6GB of RAM -- my first question is whether
this installation is a true 64-bit installation (should R have access to
> 4GB of RAM?) I suspect so, because I was running an rqss() (package
quantr
Dear Tu,
Use library(agricolae)
LSD.test(), It is necessary to aov()
comparison <- LSD.test(yield,virus,df,MSerror,group=F)
Comparison between treatments means
tr.i tr.j diff pvalue
112 11.533 0.0176
213 11.933 0.0151
314 12.500 0.0121
423 23.
On Wed, Jun 24, 2009 at 2:55 PM, Duncan Murdoch wrote:
> On 6/24/2009 2:41 PM, Sean Davis wrote:
>
>> A simple question, I hope I have a binary file format that I want to
>> slurp into R using readBin. There are a couple of fields in the file that
>> are 4-byte floats. Since R has no "float
As I said, the example works. So for simulated data it works, which means
that for correctly provided data, it should work as well.
year%in%c(1) means for all years in the vector. In this case, it is
identical to year==1. However, the option I use is extendable since you
could write year%in%c(1,3,
Thanks!
jholtman wrote:
>
> Try this:
>
>> fdata
> state statenum yearincome popul
> 1 ALABAMA11 9.703193 3973
> 2 ALABAMA12 9.745950 3992
> 3 ALABAMA13 9.762092 4015
> 4 ALASKA21 10.221640 532
> 5 ALASKA22 10.16960
Don't be silly. They can't be made "distinguishable" by any number of line
types and/or colors. The brain can't keep that many different symbol
representations straight. Referring back and forth to a legend is also
similarly useless. You need to think more creatively about how to make a
more meanin
On 6/24/2009 3:28 PM, Katharina May wrote:
> Hi,
>
> I've got the following problem which I cannot think of a solution right now:
>
> if got a lattice xyplot in black and white and a grouping variable
> with many (more than 8
> values) and I plot it as regression lines (type="r"), just like this
Hi,
I've got the following problem which I cannot think of a solution right now:
if got a lattice xyplot in black and white and a grouping variable
with many (more than 8
values) and I plot it as regression lines (type="r"), just like this
one (not reproducable but that's
I guess not the point he
Thanks! I tried these, but I got the following messages:Warning message:In
getDependencies(pkgs, dependencies, available, lib) : package âurootâ is
not available
Error in library(uroot) : there is no package called 'uroot'
I download a package called "uroot" and put it into: C:\Program
F
The subplot function (at least the one in the TeachingDemos package and the
copy in Hmisc) only does rectangles. Clipping regions are also limited to
rectangles currently. There are external tools (gimp, imagemagick) that may me
of use or triangles and other shapes can be approximated by enoug
Thanks David, for trying to replicate. May be it is a 2.9 problem? I
should have attached session info:
> agrep("Staatssekretar im Bundeskanzleramt","Bundeskanzler",max.distance=.9)
integer(0)
> agrep("Staatssekretar im Bundeskanzleramt","Bundeskanzler",max.distance=.7)
[1] 1
sessionInfo()
R vers
Since you provided no code, the following is just a guess, Try:
install.packages("uroot")
library(uroot)
Then try your analysis again.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of DongHongwei
Sent: Wednesday, June 24, 2009 3:
Greetings!
I'm trying to use R to test the unit root for a univariate data. By this link:
http://rss.acs.unt.edu/Rdoc/library/uroot/html/ADF.test.html
it tells me that I can use the function ADF.test(). However, when I tried this
in R, I got this message:
"Error: could not find function "ADF.te
On 6/24/2009 2:02 PM, Mike Beddo wrote:
Greetings!
Can someone provide a simple script for a R function that recursively builds a
binary tree. I am most familiar with C and pass by reference, but I think R is
like Fortran and pass by value.
Here's a tree with a random depth:
makeTree <- fu
Unable to reproduce:
> agrep("Staatssekretar im
Bundeskanzleramt","Bundeskanzler",max.distance=.6)
[1] 1
>
> agrep("Staatssekretar im
Bundeskanzleramt","Bundeskanzler",max.distance=.89)
[1] 1
> agrep("Staatssekretar im
Bundeskanzleramt","Bundeskanzler",max.distance=.9)
[1] 1
> agrep("Sta
On 6/24/2009 2:41 PM, Sean Davis wrote:
A simple question, I hope I have a binary file format that I want to
slurp into R using readBin. There are a couple of fields in the file that
are 4-byte floats. Since R has no "float" data type (4-byte), I am
wondering how to go about this task. I
On Wed, Jun 24, 2009 at 12:34 PM, Mark Na wrote:
> The problem is that after running the ifelse statement, data$SOCIAL_STATUS
> is converted from a factor to a character.
> Is there some way I can avoid this conversion?
I'm afraid that ifelse has very bizarre semantics when the yes and no
argument
You might also want to look at the plyr package,
http://had.co.nz/plyr. In particular, ddply + transform makes these
tasks very easy.
library(plyr)
ddply(mtcars, "cyl", transform, pos = seq_along(cyl), mpg_avg = mean(mpg))
Hadley
On Wed, Jun 24, 2009 at 11:48 AM, David
Hugh-Jones wrote:
> That
A simple question, I hope I have a binary file format that I want to
slurp into R using readBin. There are a couple of fields in the file that
are 4-byte floats. Since R has no "float" data type (4-byte), I am
wondering how to go about this task. I could simply do all the work in C,
but the
Hi,
I'm trying to get a color-gradient triangle out of a large plot.
Wonder can subplot() generate a triangle?
So far, I'm only getting rectangles.
Great appreciation for any reply!!
Kexin
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Dear list -
I am a bit puzzled by the behavior of agrep:
The following command finds a match:
agrep("Staatssekretar im Bundeskanzleramt","Bundeskanzler",max.distance=.6)
But if I _increase_ the maximum distance to .9 it fails:
agrep("Staatssekretar im Bundeskanzleramt","Bundeskanzler",max.dist
Greetings!
Can someone provide a simple script for a R function that recursively builds a
binary tree. I am most familiar with C and pass by reference, but I think R is
like Fortran and pass by value.
- Mike Beddo
__
R-help@r-project.org mailing list
Hi
I have been using a Zero-Inflated negative binomial model fitted using
the pscl zeroinfl command but I would like to extract a goodness of fit
measure are there any suitable pseudo R^2 measures available for this
type of analysis to try and assess the amount of variation in the data
explain
aliceduggan wrote:
>
> Sorry I meant bwlabel on this message
> [...]
> I am using R-Image in an MSc statistics dissertation and have found, in
> various EBImage manuals a functions - bwimage, which I would like to use,
> however I can't find it within my version of R.
> Could you tell me if the
Hi Olivier
Olivier Cailloux wrote:
Hello,
I am trying to reach a web service using the SOAP package. I succeeded
calling the web service, but not sending parameters to it. After much
research and tries, I think I found that the problem lies in the
namespace including the parameters in the S
Hello,
I am currently trying to simulate data and analyze it using the frailty option
in the coxph function. I am working with recurrent event data, using counting
process notation. Occasionally, (about 1 in every 100 simulations) I get the
following warning:
Error in coxph(Surv(start, end,
Luis,
On 24 June 2009 at 16:35, Luis Ridao Cruz wrote:
| R-help,
|
| I'm now on the process of installing RODBC on UBUNTU
| but an error occured. I copy-paste the output:
You can simply do
$ sudo apt-get install r-cran-rodbc
As for the error:
| > install.packages("RODBC")
[...]
| co
Explicitly convert it to a factor...
data$SOCIAL_STATUS<-factor(ifelse(data$SOCIAL_STATUS=="B" & data$MALE>4,
"C", data$SOCIAL_STATUS))
However, note that this would, in general, change the levels attribute to
the levels actually present in the converted vector. If you wish to (and it
makes sense
That seems to work. I should add that to make "ave" work like "by" one can
do:
mydata$newvar <- ave(1:nrow(mydata), mydata$some_factor, FUN= function (x) {
x <- ds[x,]
# ... etc...
})
Thanks!
David
[[alternative HTML version deleted]]
__
R-h
You can work with levels of factor:
levels(DF$SOCIAL_STATUS)[DF$MALE > 4 & DF$SOCIAL_STATUS == "B"] <- "C"
On Wed, Jun 24, 2009 at 1:34 PM, Mark Na wrote:
> Hi R-helpers,
>
> Please see the below R output.
> The problem is that after running the ifelse statement, data$SOCIAL_STATUS
> is convert
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 space between
R-help,
I'm now on the process of installing RODBC on UBUNTU
but an error occured. I copy-paste the output:
> install.packages("RODBC")
Warning in install.packages("RODBC") :
argument 'lib' is missing: using
'/home/luisridaocruz/R/x86_64-pc-linux-gnu-library/2.8'
trying URL 'http://cran.ii.uib
Cecilia, this is impossible to tell from the information you provide. The
simulated data below...
##START
year=rep(1:10,10)
year.fe=rep(rgamma(10,10,2),10)
firm.fe=rep(rnorm(10,0,3),each=10)
x=rnorm(100,0,3)
e=rnorm(100,0,3)
y=10+firm.fe+year.fe+x+e
newdata=data.frame(y,x,e,firm,year,firm.fe,year.
Hi R-helpers,
Please see the below R output.
The problem is that after running the ifelse statement, data$SOCIAL_STATUS
is converted from a factor to a character.
Is there some way I can avoid this conversion?
Thanks in advance, Mark Na
> str(data)
'data.frame': 2100 obs. of 11 variables:
$ D
Thanks a lot!
Yet is there a way to incorporate the lifting score into Cross
Validation, not just a plot?
Thanks again!
On Wed, Jun 24, 2009 at 9:07 AM, Tobias Sing wrote:
> Michael,
>
> a lift chart for evaluating binary scoring classifiers, as I
> understand it, plots...
>
> lift score: P(Yhat
I removed all of the objects that were from the examples I tried, saved,
and quit.
R starts up cleanly now. Yeah!
That was annoying behavior, and I would recommend that the authors
change it, please.
Thank you for your to-the-point advice, Professor Ripley!
-- David
ps - Sorry for the lengthy dis
How about something like this:
> x
id data
1 1 0.7773207
2 3 0.9606180
3 2 0.4346595
4 3 0.7125147
5 2 0.344
6 2 0.3253522
7 2 0.7570871
8 3 0.2026923
9 3 0.7111212
10 2 0.1216919
> # compute running sum for each ID
> x$run <- ave(x$data, x$id, FUN=cumsum)
> x
id
That was unintentional: the mail server crashed and apparently sent
the just-started reply.
It is nothing to do with 'Autoloads'. What is most likely if that you
have saved a reference to a namespace such as 'distrMod' in your
.RData file. If that it is the case, when .RData is loaded it wil
Hello all
I have a big data frame and I regularly want to break it down into subsets,
calculate some new data, and add it back to the data frame.
At the moment my technique seems a bit ugly and embarrassing. Something
like:
result <- by(mydata, mydata$some_factor, function (x) {
# do something
Michael,
a lift chart for evaluating binary scoring classifiers, as I
understand it, plots...
lift score: P(Yhat = + | Y = +)/P(Yhat = +)
against
rate of rate of positive predictions: P(Yhat = +).
...across the continuum of possible cutoffs. If you want to do this,
here is how you would do this
Hi
I am trying to explore the use of random forests for regression to
identify the important environmental/microclimate variables involved in
predicting the abundance of a species in different habitats, there are
approx 40 variable and between 200 and 500 data points depending on the
dataset. I
Maybe the packages caret,RWeka and ROCR are usefuel starting points.
Cheers, Christian
Hi all,
Could anybody give me some pointers to Cross Validation using Lifting
Score as error function, as commonly used in data-mining and
classification field in marketing and e-commerce research?
Thanks!
On Wed, 24 Jun 2009, dav...@rhotrading.com wrote:
I wanted to try out package distrMod, so I did
install.packages('distrMod')
library(distrMod)
and played around, saved and quit.
Now whenever I start up in this directory, I get distr and lots of other
stuff loaded and lots of messages.
How
This "one liner" works great!
Thanks (for all replies)
>
> Try this:
>
> subset(d, eval(parse(text = paste(paste(columns, values, sep = "=="),
> collapse = " & "
>
>>
>> Hello
>>
>> I have a data frame d with columns "var1", "var2", "var3"
>>
>> Then I have two vectors:
>> columns <- c("var2
Karl,
I have written a function to compute the gradient of a multivariate normal
density (derivatives at a given point, with respect to the parameters of the
density). It is in the atttached file. I would appreciate any feedback.
There are couple of interesting points:
1. Note is that there
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 missi
I wanted to try out package distrMod, so I did
> install.packages('distrMod')
> library(distrMod)
and played around, saved and quit.
Now whenever I start up in this directory, I get distr and lots of other
stuff loaded and lots of messages.
How do I keep it from automatically loading, other tha
Try this
Rich
tmp <- data.frame(
y=rnorm(100),
category=rep(factor(letters[1:5]),each=20),
level=rep(factor(0:1), length=100))
tmp
table(tmp[,2:3])
tmp$y[with(tmp, category=="a" & level=0)] <- NA
tmp$y[with(tmp, category=="a" & level==0)] <- NA
tmp$y[with(tmp, category=="e" & level==1)] <- NA
t
Hi all,
Could anybody give me some pointers to Cross Validation using Lifting
Score as error function, as commonly used in data-mining and
classification field in marketing and e-commerce research?
Thanks!
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https://s
$ sudo R
> install.packages(rgdal)
worked for me on my Ubuntu machine
but, I already had GDAL library installed with
$ sudo apt-get install libgdal1-1.5.0
--
B. H. Braswell (Rob)
University of New Hampshire
On Jun 24, 2009, at 9:24 AM, Luis Ridao Cruz wrote:
R-help,
I'm trying to install
How about this:
> x[order(x$One,-x$Two,decreasing=T),]
One Two
8 5 3
7 4 3
6 3 3
4 2 2
5 2 3
3 1 1
1 1 2
2 1 3
2009/6/24 Daniel Brewer :
> Hello,
>
> I have a data.frame which I would like to sort with the primary key
> decreasing while the secondry key is increa
Hello,
I am trying to reach a web service using the SOAP package. I succeeded
calling the web service, but not sending parameters to it. After much
research and tries, I think I found that the problem lies in the
namespace including the parameters in the SOAP body.
In short, my question is:
Hello,
I have a data.frame which I would like to sort with the primary key
decreasing while the secondry key is increasing e.g.
x <- data.frame(One=c(1,1,1,2,2,3,4,5),Two=c(2,3,1,2,3,3,3,3))
I would like to order it so it looks like this:
One Two
8 5 3
7 4 3
6 3 3
4 2 2
5 2
reproducible code.
On Wed, Jun 24, 2009 at 8:53 AM, Santosh wrote:
> Dear R-sians..
>
> I am trying to plot boxplots with side-by-side option.. I tried some of the
> posted suggestions and could not make it work due to unequal sizes of
> categories...
>
> e.g.
> weekly measured water depth values
R-help,
I'm trying to install the rgdal package under Ubuntu
but I get the following warning:
> install.packages("rgdal")
Warning in install.packages("rgdal", configure.args =
"--with-gdal-modules=/usr/local/lib") :
argument 'lib' is missing: using
'/home/luisridaocruz/R/x86_64-pc-linux-gnu-
Hello,
Mango is pleased to announce that the next LondonR Group meeting will be
held on 21st July. At this meeting we plan to have a feedback session
from people who attended useR! in Rennes. To help us plan this, if you
are attending at Rennes next month and would be interested in giving
some
See the example in plot.zoo labelled Fancy X Axis:
library(zoo)
example(plot.zoo)
?plot.zoo
On Wed, Jun 24, 2009 at 10:11 AM, Thomas Levine wrote:
> One last tiny problem: How do I add months to the scale? It currently just
> has years
> http://school.thomaslevine.org/mywall.png
>
> Thanks again
Do you mean something like source? Type ?source for help
--- On Tue, 6/23/09, Derek Lacoursiere wrote:
> From: Derek Lacoursiere
> Subject: [R] driver file
> To: r-help@r-project.org
> Received: Tuesday, June 23, 2009, 3:05 PM
>
> Hi,
>
> How can I, from a single "driver" file, source othe
I get:
> unit(-1, "line")
Error in valid.units(units) : Invalid unit
> unit(-1, "lines")
[1] -1lines
I have an older version of R (2.8.1) and of grid (2.8.1). The source
for grid 2.8.1 does not have pseudonyms.
http://svn.r-project.org/R/branches/R-2-8-branch/src/library/grid/src/unit.c
--
Da
One last tiny problem: How do I add months to the scale? It currently just
has years
http://school.thomaslevine.org/mywall.png
Thanks again
Tom
On Sat, Jun 20, 2009 at 12:14 PM, Thomas Levine wrote:
> I wasn't really thinking that far ahead; plot tries to do something, so I
> figured I'd try th
Dear R-sians..
I am trying to plot boxplots with side-by-side option.. I tried some of the
posted suggestions and could not make it work due to unequal sizes of
categories...
e.g.
weekly measured water depth values are categorized into 5 levels based on
their values
such measurement is again cate
You're right, I meant to write "lines"not "line". The strange thing is,
although "line" isn't listed in ?unit, it doesn't return an error on my
machine,
> unit(-1, "line")
[1] -1line
> unit(-1, "lines")
[1] -1lines
Reading from
http://svn.r-project.org/R/trunk/src/library/grid/src/unit.c "l
Try this:
dotplot(value ~ factor(le, levels = le),data=df)
On Wed, Jun 24, 2009 at 10:24 AM, Christian Schulz wrote:
> Thanks, how is it possible that the x-axis labels getting the row sorting
> instead the alphabetically one?
> Could i just printed only every n (i.e. 3rd) label?
>
> Christi
Thanks, how is it possible that the x-axis labels getting the row
sorting instead the alphabetically one?
Could i just printed only every n (i.e. 3rd) label?
Christian
df <- data.frame(value=runif(26),le=letters)
df <- df[order(df[,1]),]
dotplot(value ~ le,data=df)
Maybe you want to have
Hi,
I want to produce two parallel rotated strips with color gradient. So
far, the sample strip is something produced by this:
pushViewport(viewport(x = unit(0.638, "npc"), y =unit(0.386, "npc"),
width=.62, height=0.006, angle=137.2))
grid.rect(y=100:1/100, just="top",
gp=gpar(col=NA
Hi
r-help-boun...@r-project.org napsal dne 24.06.2009 14:16:15:
> Hello
>
> I have a data frame d with columns "var1", "var2", "var3"
>
> Then I have two vectors:
> columns <- c("var2", "var3")
> values <- c(0, 1)
>
>
> Is there a compact way to subset the data frame
> using these two vectors
Dear Alice,
EBImage is a general-purpose image processing package for R, able to
read, write, process and analyze images, but is part of the Bioconductor
project. Please post your future questions on the BioC mailing list (see
http://www.bioconductor.org/docs/mailList.html).
There is no func
Hi,
I have data collected from PTT-GPS devices tagged to birds. These data
typically are delivery in time slots, comprising the day-light period for
example.
I'm interested in testing the temporal correlation of observations (points)
and eventually exclude locations falling below a given corre
Try this:
subset(d, eval(parse(text = paste(paste(columns, values, sep = "=="),
collapse = " & "
On Wed, Jun 24, 2009 at 9:16 AM, Blazej Krzeminski wrote:
> Hello
>
> I have a data frame d with columns "var1", "var2", "var3"
>
> Then I have two vectors:
> columns <- c("var2", "var3")
> valu
Hello,
Thanks! Join is not possible, as the database is simply way to large. When
we exactly now, what we are looking for maybe, but not yet…
By now, I figured the paste command - not familiar with R…, however, it only
seems to work for one - the last - entry I am using. Unfortunately, the help
Sorry I meant bwlabel on this message
Alice
--- On Wed, 24/6/09, alicedug...@btopenworld.com
wrote:
From: alicedug...@btopenworld.com
Subject: bwimage
To: r-help@r-project.org
Date: Wednesday, 24 June, 2009, 12:34 PM
Dear Sir/Madam,
I am using R-Image in an MSc statistics dissertation
I am using R-Image in an MSc statistics dissertation and have found, in
various EBImage manuals a functions - bwlabel, which I would like to use,
however I can't find it within my version of R.
Could you tell me if the function has changed or what I need to do to be
able to use it.
--
View this
i am a new user of R working on "Cost sensitivity for Classical classification
functions"
i have codes for cross validation for both linear and quadratic classifier but
i cant not very cost of missclassification. please any body with help should
help me
thanks
ARIYO,OLUDARE SAMUEL
Statis
Dear Sir/Madam,
I am using R-Image in an MSc statistics dissertation and have found, in various
EBImage manuals a functions - bwimage, which I would like to use, however I
can't find it within my version of R.
Could you tell me if the function has changed or what I need to do to be able
to use i
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