I am about to evaluate the outlier detection mode of random forests.
At
[1]http://www.stat.berkeley.edu/~breiman/RandomForests/cc_home.htm#outliers
the raw outlier measure for a case n is defined as
nsample/AverageProximity
I wasn't able to figure out, what "nsample" stands for
I can use gdata to successfully read in the example Excel file, but not any
other excel files. Why might this be the case? It seems that the problem
has something to do with opening the database but no indication as to what
the problem is. So i'm at a loss of how to fix it.
> library(gdata)
gdata
Pablo,
Check the qqPlot function in "car":
require(car)
qqPlot(x, dist = "gamma", shape = 1.7918012, rate = 0.9458022)
Best,
Jorge.-
On Tue, Apr 2, 2013 at 4:41 AM, pablo.castano <> wrote:
> Hi,
>
> I want to create upper and lower 95% confidence intervals for a p-p plot of
> an empirical di
There's probably a better way to do it, but here's one way:
require(stringr)
windows()
file_list <- list.files(pattern=".txt")
for (file in file_list)
{
dataset <- read.table(file, header=TRUE, sep="\t")
plot(dataset)
savePlot(filename=str_replace(file,".txt",""),type="png")
Hi,
May be this helps.
list.files()
#[1] "file1.txt" "file2.txt" "file3.txt"
lapply(list.files(),function(x)
{x1<-read.table(x,header=TRUE);x2<-gsub("txt","png",x);png(x2);plot(col2~col1,data=x1,type="l");dev.off()})
A.K.
- Original Message -
From: David Lyon
To: "r-help@r-project.or
I have many files in 1 directory, file names end in .txt.
Each file has 2 columns
col1 col2
2 3
3 4
4 5
5 6
I want to make a list of the file names and iterate through each plotting them
in a separate file $filename\.png with the png swapped for txt.
So far I have this, can someone help
Hi Kamil,
(I have replied to the forum but my posts as replies never seem to get
accepted, so here is the text in an email also. Apologies if it comes
through twice.)
Thanks so much for the prompt reply. You've solved it! Actually I had the
latest version of the package, but an older version of R
Hello,
My question is related to ctree() function from the library 'party'.
Is there a way to force ctree() to use a specific variable in the first
split? I am asking because the first split contains two variables with
very similar scores, and choosing the alternative variable would induce
a t
On 2013-04-01 19:23, arun wrote:
gsub("\\,.*","",x)
#[1] "foo" "bar" "qux"
A.K.
No big deal, but does "," have to be escaped?
sub(",.*", "", x)
Peter Ehlers
- Original Message -
From: Gundala Viswanath
To: "r-h...@stat.math.ethz.ch"
Cc:
Sent: Monday, April 1, 2013 10:13 PM
Subjec
On 04/02/2013 03:13 PM, Gundala Viswanath wrote:
I have the following list of strings:
x <- c("foo, foo2, foo3", "bar", "qux, qux1")
what I want to do is to obtain
foo, bar qux
Namely for each element in the vector obtain only string
before the first comma.
What's the way to do it?
This se
gsub("\\,.*","",x)
#[1] "foo" "bar" "qux"
A.K.
- Original Message -
From: Gundala Viswanath
To: "r-h...@stat.math.ethz.ch"
Cc:
Sent: Monday, April 1, 2013 10:13 PM
Subject: [R] How to remove all characters after comma in R
I have the following list of strings:
x <- c("foo, foo2, foo
I have the following list of strings:
x <- c("foo, foo2, foo3", "bar", "qux, qux1")
what I want to do is to obtain
foo, bar qux
Namely for each element in the vector obtain only string
before the first comma.
What's the way to do it?
- G.V.
[[alternative HTML version deleted]]
_
On Apr 1, 2013, at 7:50 AM, Eder Paulo wrote:
> Hi,
>
> I work with
> remote sensing in Brazil. I would like to know if there is any algorithm or
> package that does image segmentation by region growing in R.
>
It's possible that there is a domain of investigation where the phrase " image
seg
On 2013-04-01 15:06, Ted Harding wrote:
On 01-Apr-2013 21:26:07 Robert Baer wrote:
On 4/1/2013 4:08 PM, Peter Ehlers wrote:
On 2013-04-01 13:37, Ted Harding wrote:
Greetings All.
This is a somewhat generic query (I'm really asking on behalf
of a friend who uses R on Windows, whereas I'm on Lin
On 01-Apr-2013 21:26:07 Robert Baer wrote:
> On 4/1/2013 4:08 PM, Peter Ehlers wrote:
>> On 2013-04-01 13:37, Ted Harding wrote:
>>> Greetings All.
>>> This is a somewhat generic query (I'm really asking on behalf
>>> of a friend who uses R on Windows, whereas I'm on Linux, but
>>> the same phenome
I have never used the data.table package. I am trying to do the following
SQL left join in R
create table all as select a.*
from dates b left outerjoin activitycount a on
a.tdate=b.tdate
and a.activity=b.activity
On 2013-04-01 13:08, Matthew Lundberg wrote:
Note the edited subject line! I don't know why I typed it as it was before.
This says that as.numeric(as.character(f)) will work regardless of the
implementation, and I agree.
It's the recommendation to use as.numeric(levels(f))[f] that has me
wonde
On 4/1/2013 4:08 PM, Peter Ehlers wrote:
On 2013-04-01 13:37, Ted Harding wrote:
Greetings All.
This is a somewhat generic query (I'm really asking on behalf
of a friend who uses R on Windows, whereas I'm on Linux, but
the same phenomenon appears on both).
Say one has a largish dataframe -- cal
Your example is not reproducible [1]. We don't know what device you are writing
to, and we don't have your data or even a subset of it.
However, facet_wrap is not used for generating separate graphs. You will need
to make some kind of loop construct (for or lapply) that opens the device,
prints
On 2013-04-01 12:46, Jose Narillos de Santos wrote:
Hi all,
I´m using
titt<- expression(paste("R con ventanas de 200 ", italic(datos)))
And it works properly on a plot(...,main=titt,..)
But if I want to add an improvement to avoid typing n on the plot so that
n<-200
titt<- expression(paste
On 2013-04-01 13:37, Ted Harding wrote:
Greetings All.
This is a somewhat generic query (I'm really asking on behalf
of a friend who uses R on Windows, whereas I'm on Linux, but
the same phenomenon appears on both).
Say one has a largish dataframe -- call it "G" -- which in the
case under discus
Use bquote(), as in
n <- 200
titt <- bquote(paste("R con ventanas de ", .(n), " ", italic(datos)))
or, using ~ instead of paste(),
titt <- bquote("R con ventanas de " ~ .(n) ~ italic(datos))
plot(1, 1, main=titt)
The notation .(xxx) means to replace xxx by the value of the variable called
library(ggplot2)
a<- read.table("data", header=T)
b = na.omit(a)
ggplot(data=b) + geom_line(aes(x=timepoint, y=value,group=sample, colour=
factor(sample))) + geom_point(aes(x=timepoint, y=value, group=s
ample)) + facet_wrap(~bio, scales = "free",ncol = 5) +theme_bw() +
opts(legend.direction = "h
Greetings All.
This is a somewhat generic query (I'm really asking on behalf
of a friend who uses R on Windows, whereas I'm on Linux, but
the same phenomenon appears on both).
Say one has a largish dataframe -- call it "G" -- which in the
case under discussion has 592 rows and 41 columns. The inte
Today it's 16 years ago and 367,496 messages later since MartinMächler
started the R-help (321,119 msgs), R-devel (45,830 msgs) and
R-announce (547 msgs) mailing lists [1] - a great benefit to all of
us. Special thanks to Martin and also thanks to everyone else
contributing to these forums.
[1] h
Note the edited subject line! I don't know why I typed it as it was before.
This says that as.numeric(as.character(f)) will work regardless of the
implementation, and I agree.
It's the recommendation to use as.numeric(levels(f))[f] that has me
wondering about section 2.3.1 of the language defini
?plotmath
tells you how.
-- Bert
On Mon, Apr 1, 2013 at 12:46 PM, Jose Narillos de Santos
wrote:
> Hi all,
>
> I惴 using
>
>
> titt<- expression(paste("R con ventanas de 200 ", italic(datos)))
>
> And it works properly on a plot(...,main=titt,..)
>
> But if I want to add an improvement to avoid
Yup. Note also:
> as.character.factor
function (x, ...)
levels(x)[x]
But of course this is OK, since this can change if the implementation
does. Which is the whole point, of course.
-- Bert
On Mon, Apr 1, 2013 at 12:16 PM, Matthew Lundberg
wrote:
>
> When used as an index, the factor is impl
Hi all,
I´m using
titt<- expression(paste("R con ventanas de 200 ", italic(datos)))
And it works properly on a plot(...,main=titt,..)
But if I want to add an improvement to avoid typing n on the plot so that
n<-200
titt<- expression(paste("R con ventanas de ",n, italic(datos)))
It doesn´t
When used as an index, the factor is implicitly converted to integer. In
the expression as.numeric(levels(f))[f], the vector as.numeric(levels(f))
is indexed by as.integer(f).
This appears to rely on the current implementation, as mentioned in section
2.3.1 of the language definition.
On Mon, A
On Apr 1, 2013, at 10:22 AM, Yolanda wrote:
> I need to use the next robust functions: t1way, box1way and lincon.
> I don't find these funcions in any library.
Try:
pkg:WRS
install.packages("WRS", repos="http://R-Forge.R-project.org";, type="source"))
> What libraries do I have to install?
On 2013-04-01 10:48, Matthew Lundberg wrote:
These two seem to be at odds. Is this the case?
From help(factor) - section Warning:
To transform a factor f to approximately its original numeric values,
as.numeric(levels(f))[f] is recommended and slightly more efficient than
as.numeric(as.chara
Hi Mathew,
In what way are they at odds?
On Mon, Apr 1, 2013 at 1:48 PM, Matthew Lundberg
wrote:
> These two seem to be at odds. Is this the case?
>
> >From help(factor) - section Warning:
>
> To transform a factor f to approximately its original numeric values,
> as.numeric(levels(f))[f] is re
Hi,
I want to create upper and lower 95% confidence intervals for a p-p plot of
an empirical distribution with a theoretical gamma distribution.
This is my code:
x<-rgamma(100,shape=2, rate=1) # empirical data
fitdistr(x,"gamma") # fit a gamma distribution
dist<-pgamma(x,shape=1.9884256 ,rate=0
Hi Many Thanks to all.
dev.print(jpeg, file="test.jpeg", height=800,width=800)
Works perfectly it saves in the default directory the jpeg file.
I use RStudio.
Many Thanks.
2013/4/1 David Winsemius
>
> On Apr 1, 2013, at 9:26 AM, R. Michael Weylandt wrote:
>
> > On Mon, Apr 1, 2013 at 5:05
These two seem to be at odds. Is this the case?
>From help(factor) - section Warning:
To transform a factor f to approximately its original numeric values,
as.numeric(levels(f))[f] is recommended and slightly more efficient than
as.numeric(as.character(f)).
>From the language definition - secti
I need to use the next robust functions: t1way, box1way and lincon.
I don't find these funcions in any library.
What libraries do I have to install?
Thanks.
Yolanda
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLE
Hi, I am trying to run a panel regression using 88 observations and 9
variables. In-built Hausman Test did not work, then I found a code for
auxiliary regression method for the Hausman test.
The panel models are:
fe=plm(gd ~ l+g+o+c+g1+h+n+r, model = "within", data = new.frame,index =
c("id"))
r
On Mon, Apr 1, 2013 at 12:42 PM, Gabor Grothendieck wrote:
> On Mon, Apr 1, 2013 at 1:20 PM, Paul Johnson wrote:
> > Here's my little discussion example for a quadratic regression:
> >
> > http://pj.freefaculty.org/R/WorkingExamples/regression-quadratic-1.R
> >
> > Students press me to know the
I certainly second all Jeff's comments.
**HOWEVER** :
http://www.tandfonline.com/doi/pdf/10.1080/00401706.1978.10489610
IIRC, DUD's provenance is old, being originally a BMDP feature.
-- Bert
On Mon, Apr 1, 2013 at 10:59 AM, Jeff Newmiller wrote:
> a) Read the Posting Guide. It will tell you t
Here is an example of the numerical instability that poly() can help with.
Imagine a 100-minute experiment where you chose to record the times as POSIXt
objects (stored as seconds since 1970). Using poly() gives the right predicted
values, but using I(x^2) does not:
> d <- data.frame(Date=as.
a) Read the Posting Guide. It will tell you things like:
1) Don't post in HTML format. Adjust your email client appropriately.
2) Don't repost. Your emails are all there, looking silly next to each other.
3) Use the search facilities yourself, such as RSiteSearch().
b) Regarding your "DUD" algorit
On Mon, Apr 1, 2013 at 1:20 PM, Paul Johnson wrote:
> Here's my little discussion example for a quadratic regression:
>
> http://pj.freefaculty.org/R/WorkingExamples/regression-quadratic-1.R
>
> Students press me to know the benefits of poly() over the more obvious
> regression formulas.
>
> I thi
Dear All,
wondering if someine can access the link to the randsamp code referenced in the
R-help archive here:
http://www.mail-archive.com/r-help@stat.math.ethz.ch/msg75645.html ? I have
tried but for whatever reason I can not get trough. My problem seems to be
similar to what the author orig
Here's my little discussion example for a quadratic regression:
http://pj.freefaculty.org/R/WorkingExamples/regression-quadratic-1.R
Students press me to know the benefits of poly() over the more obvious
regression formulas.
I think I understand the theory on why poly() should be more numericall
Hi,
Not sure if this is what you wanted:
activity<-
data.frame(Name=paste0("activity",LETTERS[1:5]),stringsAsFactors=FALSE)
dates1<-
data.frame(dat=as.Date(c("2013-02-01","2013-02-04","2013-02-05"),format="%Y-%m-%d"))
merge(dates1,activity)
# dat Name
#1 2013-02-01 activityA
#2 2
On Apr 1, 2013, at 9:26 AM, R. Michael Weylandt wrote:
> On Mon, Apr 1, 2013 at 5:05 PM, Trying To learn again
> wrote:
>> Hi all if I plot a graph on R, I press on the plot Export/Save Plot as
>> Image
>
> I assume this means you're using R-Studio?
>
>>
>> I change name on "File name:"
>
SAS has DUD (Does not Use Derivatives)/Secant Method for nonlinear
regression, does R offer this option for nonlinear regression?
I have read the helpfile for nls() and could not find such option, any
suggestion?
Thanks,
Derek
[[alternative HTML version deleted]]
__
On Mon, Apr 1, 2013 at 5:05 PM, Trying To learn again
wrote:
> Hi all if I plot a graph on R, I press on the plot Export/Save Plot as
> Image
I assume this means you're using R-Studio?
>
> I change name on "File name:"
>
> I select DirectoryBibliotecas\Documentos
>
> And select Width 800
See
?jpeg
Sarah
On Mon, Apr 1, 2013 at 12:05 PM, Trying To learn again
wrote:
> Hi all if I plot a graph on R, I press on the plot Export/Save Plot as
> Image
>
> I change name on "File name:"
>
> I select DirectoryBibliotecas\Documentos
>
> And select Width 800 and Height 800
>
> And fi
Hi all if I plot a graph on R, I press on the plot Export/Save Plot as
Image
I change name on "File name:"
I select DirectoryBibliotecas\Documentos
And select Width 800 and Height 800
And finally save in format JPEG
It is posible to "type" code so that I can run my function and it is n
I have a problem to plot label (Year) only for significant values (in this
case spoz and sneg).
I use this code, but don't work with labels.
library(ggplot2)
ggplot(data1, aes(x = Year, y = value,fill=type,width=1))+
geom_bar(stat="identity",position="identity")+
scale_y_continuous(breaks = rou
That sounds like a job for merge().
If you provide an actual reproducible example using dput(), then you
will likely get some actual runnable code.
Sarah
On Mon, Apr 1, 2013 at 11:54 AM, ramoss wrote:
> Hello,
>
> I have 2 data frames: activity and dates. Activity contains a l variable
> list
Hello,
I have 2 data frames: activity and dates. Activity contains a l variable
listing all activities: activityA, activityB etc.
The dates contain all the valid business dates. I need to combine the 2 so
that I get a single data frame activitydat that contains the activity name
along w/ evevr
Dear R users,
I have a problem to plot label (Year) only for significant values (in this
case spoz and sneg).
I use this code, but don't work with labels.
library(ggplot2)
ggplot(data1, aes(x = Year, y = value,fill=type,width=1))+
geom_bar(stat="identity",position="identity")+
scale_y_continuou
Hi,
I work with
remote sensing in Brazil. I would like to know if there is any algorithm or
package that does image segmentation by region growing in R.
Thanks,
Eder.
[[alternative HTML version deleted]]
_
Hi: Google for koyck distributed lag. Based on what you wrote, I think
that's what you're looking for or something close to it.
There is tons of literature on that model and if you read enough about it,
you'll see that
through a transformation, reduces to something that much simpler to
estimate.
Hello all!
I have a problem to plot label (Year) only for significant values (in this
case spoz and sneg).
I use this code, but don't work with labels.
library(ggplot2)
ggplot(data1, aes(x = Year, y = value,fill=type,width=1))+
geom_bar(stat="identity",position="identity")+
scale_y_continuous(
Hello all!
I want to make a barplot with ggplot2.
I want to view in the same chart the semn values (significant values
(pointer over 50)). I try this code, but only for pointer values.
ggplot(data, aes(x = Year, y = pointer)) + geom_bar(stat="identity")
please help me with this problem.
I use
--- On Sat, 30/3/13, Hin-Tak Leung wrote:
> "... was committed to freetype in January and will form the
> next release (2.4.12)".
It is perhaps worth repeating the quote: 'The official R binaries for windows
... are compiled against static libraries of cairo 1.10.2 ... are firmly in the
"do
Sorry for this message it's just a test.
Thank you!
--
---
Catalin-Constantin ROIBU
Lecturer PhD, Forestry engineer
Forestry Faculty of Suceava
Str. Universitatii no. 13, Suceava, 720229, Romania
office phone +4 0230 52 29 78, ext. 531
mobile phone +4 0745 53 18 01
+
Hi Cat,
are you using some very old version of MuMIn? That would explain the
missing 'QAICc'.
As for the error message about 'logLik', it usually occurs when there
are some misspelled arguments (that go into "..." and are passed to the
rank function, 'AICc' in your case). Check if there is some ar
On Apr 1, 2013, at 5:18 AM, Hin-Tak Leung wrote:
> --- On Sat, 30/3/13, Hin-Tak Leung wrote:
>
>> "... was committed to freetype in January and will form the
>> next release (2.4.12)".
>
> It is perhaps worth repeating the quote: 'The official R binaries for
> windows ... are compiled again
On Apr 1, 2013, at 10:56 , catalin roibu wrote:
> Hello all!
>
> I have a problem to draw a polygon with R. My data is like this>
> Year Nb.series Perc.pos Perc.neg Nature RGV_mean RGV_sd neg poz
> 1 1901 1 0.00 0.00 0 4.29 NA 0.00 0.00
> 2 1902 1 100.00 0.00 1 16.47 NA 0.00 100.00
> 3 1903 1
Hello all!
I have a problem to draw a polygon with R. My data is like this>
Year Nb.series Perc.pos Perc.neg Nature RGV_mean RGV_sd neg poz
1 1901 1 0.00 0.00 0 4.29 NA 0.00 0.00
2 1902 1 100.00 0.00 1 16.47 NA 0.00 100.00
3 1903 1 100.00 0.00 1 31.31 NA 0.00 100.00
4 1904 1 0.00 0.00 0 -9.62 N
Hi guys,
I am afraid I am stuck with an estimation problem.
I have two variables, X and Y. Y is explained by the weighted sum of n
lagged values of X. My aim is to estimate the two parameters
c(alpha0,alpha1) in:
Yt = Sum from j=1 to n of ( ( alpha0 + alpha1 * j ) * Xt-j )
Where Xt-j denotes th
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