I am using RExcel and the R Commander and so far find it easy to use.
However, I am looking for a DFA and it is not clear which function that
might be in RExcel. Can anyone point me to where I can find it?
Thank you.
[[alternative HTML version deleted]]
I was only just now able to investigate and give this a try (not sure how I
missed it previously when I looked at the page). It worked perfectly! Thanks
to you both. =)
On Mar 31, 2013, at 8:28 PM, Nicole Ford wrote:
John,
Thanks for your reply! I will check again I must have missed
I got this erro
Error in c(red, blue, green)[r] : invalid subscript type 'S4'
--
View this message in context:
http://r.789695.n4.nabble.com/How-to-represent-certain-values-in-a-file-as-we-want-tp4662925p4662935.html
Sent from the R help mailing list archive at Nabble.com.
Hi, All
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]]
Hi all,
My replies within the forum aren't getting approved, though my emails
always go through, so here is my reply to a question I previously
posted (all questions and answers shown). Thanks, Cat
I'm having trouble with the model generating 'dredge' function in the MuMIn
'Multi-model
On Mar 31, 2013, at 12:37 PM, Jonsson wrote:
I got this erro
Error in c(red, blue, green)[r] : invalid subscript type 'S4'
I don't have your data so testing is not possible. I assumed a plot call was
base graphics and that 'r' was a vector. I was wrong. It's an S4 object. I
would follow
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
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
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 1 100.00
On Apr 1, 2013, at 5:18 AM, Hin-Tak Leung wrote:
--- On Sat, 30/3/13, Hin-Tak Leung ht...@users.sourceforge.net 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 ...
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
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
--- On Sat, 30/3/13, Hin-Tak Leung ht...@users.sourceforge.net 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 ...
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
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)+
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.
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]]
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)+
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/
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 ramine.mossad...@finra.org wrote:
Hello,
I have 2 data frames: activity and dates. Activity
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 =
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 not
See
?jpeg
Sarah
On Mon, Apr 1, 2013 at 12:05 PM, Trying To learn again
tryingtolearnag...@gmail.com 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
On Mon, Apr 1, 2013 at 5:05 PM, Trying To learn again
tryingtolearnag...@gmail.com 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
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 Apr 1, 2013, at 9:26 AM, R. Michael Weylandt wrote:
On Mon, Apr 1, 2013 at 5:05 PM, Trying To learn again
tryingtolearnag...@gmail.com 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
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 2013-02-04
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
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
On Mon, Apr 1, 2013 at 1:20 PM, Paul Johnson pauljoh...@gmail.com 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
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
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 -
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 jdnew...@dcn.davis.ca.uswrote:
a) Read the Posting
On Mon, Apr 1, 2013 at 12:42 PM, Gabor Grothendieck ggrothendi...@gmail.com
wrote:
On Mon, Apr 1, 2013 at 1:20 PM, Paul Johnson pauljoh...@gmail.com wrote:
Here's my little discussion example for a quadratic regression:
http://pj.freefaculty.org/R/WorkingExamples/regression-quadratic-1.R
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))
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
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 -
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 dwinsem...@comcast.net
On Apr 1, 2013, at 9:26 AM, R. Michael Weylandt wrote:
On Mon, Apr
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
Hi Mathew,
In what way are they at odds?
On Mon, Apr 1, 2013 at 1:48 PM, Matthew Lundberg
matthew.k.lundb...@gmail.com 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,
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
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?
There
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,
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
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
matthew.k.lundb...@gmail.com wrote:
When used as an
?plotmath
tells you how.
-- Bert
On Mon, Apr 1, 2013 at 12:46 PM, Jose Narillos de Santos
narillosdesan...@gmail.com 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
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
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]
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
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 =
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
xxx.
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
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(R
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,
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 --
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
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 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 phenomenon appears on
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
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
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]]
gsub(\\,.*,,x)
#[1] foo bar qux
A.K.
- Original Message -
From: Gundala Viswanath gunda...@gmail.com
To: r-h...@stat.math.ethz.ch 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
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 seems to
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 gunda...@gmail.com
To: r-h...@stat.math.ethz.ch r-h...@stat.math.ethz.ch
Cc:
Sent: Monday,
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
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
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
67 matches
Mail list logo
