On Fri, Mar 20, 2009 at 10:02 PM, johnhj wrote:
>
> Hii,
>
> Is it possible, to use the plot() funktion and the boxplot() funktion
> together ?
> I will plot a simple graph and additionally to the graph on certain places
> boxplots. I have imagined to plot the graph a little bit transparency and
>
On Fri, Mar 20, 2009 at 8:03 PM, t c wrote:
> Dear r list,
>
> I am using glm.nb in the MASS package to fit negative binomial models to data
> on manta ray abundance, and AICctab in the bbmle package to compare model
> IC. However, I need to test for the goodness of fit of the full model, and
On Fri, Mar 20, 2009 at 7:18 PM, science! wrote:
>
> I am aware that it is easily possible to create var names on the fly. e.g.
> assign(paste("m",i,sep=""),j)
> but is it possible to assign dataframes to variables created on the fly?
>
> e.g.
> If I have a dataframe called master and I wanted to
Now that I have my list of flags with theri respective values (thanks to all
those who posted their suggestions):
Flags Values
1 TrendOff 0
2 MOdwt 1
3ZeroPadding 1
4 Step1HSOff 1
5 Step1NumHSOff 4
6 Step1NumHSOff 1
7Step1ExtOff 1
Many thanks to Bill and Thomas for both insight and this nice solution.
Kathy Gerber
William Dunlap wrote:
If you diff the desired series you get all 2^n
possible n-long sequences of 0's and 1's. Hence
another solution is to convert the numbers in
0:(2^n-1) to their binary representations, put
I am aware that it is easily possible to create var names on the fly. e.g.
assign(paste("m",i,sep=""),j)
but is it possible to assign dataframes to variables created on the fly?
e.g.
If I have a dataframe called master and I wanted to subset parts of those
data into separate dataframes, I could
Hii,
Is it possible, to use the plot() funktion and the boxplot() funktion
together ?
I will plot a simple graph and additionally to the graph on certain places
boxplots. I have imagined to plot the graph a little bit transparency and
show in the same graph on certain places boxplots
Is it p
Thanks for the tip. I'll start with rJava, and see if that easier
approach works.
On Mar 20, 2009, at 10:50 PM, Martin Morgan wrote:
Michael Kubovy writes:
Dear R-helpers,
I run R version 2.8.1 (2008-12-22) on i386-apple-darwin8.11.1
I would like to call R from Java. I downloaded SJava_0
Michael Kubovy writes:
> Dear R-helpers,
>
> I run R version 2.8.1 (2008-12-22) on i386-apple-darwin8.11.1
>
> I would like to call R from Java. I downloaded SJava_0.69-0.tar.gz,
> and then issued the command suggested on http://www.omegahat.org/
> RSJava/ :
Others will point you to rJava
Dear R-helpers,
I run R version 2.8.1 (2008-12-22) on i386-apple-darwin8.11.1
I would like to call R from Java. I downloaded SJava_0.69-0.tar.gz,
and then issued the command suggested on http://www.omegahat.org/
RSJava/ :
% R INSTALL -c SJava_0.69-0.tar.gz
ARGUMENT 'INSTALL' __ignored__
WARNI
Aha, I get it now. I was under the mistaken, intuitive impression that
the subset condition was evaluated element-by-element... I guess it
must actually work out to a vector of booleans, each element of which
gets compared to the corresponding element of the data to be
subsetted. That is, in hindsi
If you use Jim's example and use grep() with ordinary and and then
negative indexing, you get these results:
> x[grep("her", x$input),]
input output corpusFreq pvolOT pvolRatioOT
2 donate(her,thebook) P 48.7 68928 0.1899471
6give(her,it) P 100.0 1
grep and regexpr return different values. regexpr returns a vector of
the same length as the input and this can be used to construct a
logical subscript. grep return a vector of only the matches, in which
case you can have a length of zero if there are no matches. Makes it
harder to create the s
Thanks, Jim (and Mark, who replied off-list) -- that does the trick. I
had tried using an index expression with grep, but that failed in the
same way as the subset method. It is still rather mysterious why this
works with regexpr but not with grep :)
-Max
On Fri, Mar 20, 2009 at 7:57 PM, jim holt
Dear r list,
I am using glm.nb in the MASS package to fit negative binomial models to data
on manta ray abundance, and AICctab in the bbmle package to compare model IC.
However, I need to test for the goodness of fit of the full model, and have not
been able to find a Pearson's Chi Squared st
Try using regexpr instead:
> x <- read.table(textConnection("input output corpusFreq pvolOT pvolRatioOT
+ give(mysister,theoldbook) P 47.0 56016 0.1543651
+ donate(her,thebook) P 48.7 68928 0.1899471
+ give(mysister,thebook) P 73.4 80136 0.2208333
+ donate
Kevin E. Thorpe wrote:
Ravi Varadhan wrote:
Good try, Kevin. But that doesn't seem to do it.
set.seed(123)
x <- sort(runif(100))
y <- sin(4*pi*x) + rnorm(100, sd=0.2)
ans.lo2 <- loess(y ~ x, degree=2, span=0.75)
ans.gam2 <- gam(y ~ lo(x, degree=2, span=0.75))
summary(ans.lo2$fitted - ans.g
Ravi Varadhan wrote:
Good try, Kevin. But that doesn't seem to do it.
set.seed(123)
x <- sort(runif(100))
y <- sin(4*pi*x) + rnorm(100, sd=0.2)
ans.lo2 <- loess(y ~ x, degree=2, span=0.75)
ans.gam2 <- gam(y ~ lo(x, degree=2, span=0.75))
summary(ans.lo2$fitted - ans.gam2$fitted) # larger d
I have some data that looks like this:
> dataP
input output corpusFreq pvolOT pvolRatioOT
1 give(my sister, the old book) P 47.0 56016 0.1543651
5 donate(her, the book) P 48.7 68928 0.1899471
9 give(my sister
Really belongs o R-SIG-Mac list and have copied it there. I cannot
reproduce on my Mac. I get the same colors on the pdf file with that
code as I do on the screen device. I have in in the past needed to set
up a device before plotting:
trellis.device(device="postscript", color = TRUE) ... bu
Hi! I am calling ismev rlarg.diag() multiple times inside a "for
loop" in a script I've written. I don't want to have to hit the Return
key many times, so I am using the following command at the top of my
script so I am not prompted to press for the Return key.
options(device.ask.default=FALSE)
T
I lose control on colors when I plot points with pch=1 or 16
(empty or solid circle) or any letter, say "A". For example:
Â
x=runif(5)
y=runif(5)
pdf("plot.pdf")
plot(x, y,
type='p', pch=1, col = 1:5) #just black points
plot(x, y,
type='p', pch=0, col = 1:5) #points with different colors
I lose control on colors when I plot points with pch=1 or 16
(empty or solid circle) or any letter, say "A". For example:
Â
x=runif(5)
y=runif(5)
pdf("plot.pdf")
plot(x, y,
type='p', pch=1, col = 1:5) #just black points
plot(x, y,
type='p', pch=0, col = 1:5) #points with different colors
Dear R community,
Is this sample size large enough to study differences between two groups of the
populations?
Q1: do the body temperatures differ between the two groups of the overwintering
turtles juveniles and adults?
One group (adults) has 6 turtles
Second group (juveniles) has 1 tu
Hi there,
Is there any such package? I searched but found none. Thanks in advance.
Xin Zheng
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PLEASE do read the posting gui
On 21/03/2009, at 12:50 AM, Fredrik Karlsson wrote:
Dear list,
Sorry for posting a borderline statistical question on the list,
but hte
SPSS people around me just stares at me blankly when refering to
tests with
any term other than ANOVA and post-hoc. I would appreciate any
insight on
h
Hi all,
I would like to use an object and to add some definition to the [
function, but I do not manage...
My objet is a "trajectories", a matrix whose columne name does contain
information :
-
trajectories <- function(traj,varName,time){
colnames(traj) <- paste(varName,time,sep="")
I don't think it necessary to prune trees in RF, per brieman's paper.
On 3/20/09, Liaw, Andy wrote:
> The way the trees are structured in randomForest, there's no way to stop
> tree growth by depth (what you called level).
>
> (If anyone has ideas, I'm all ears.)
>
> Andy
>
> From: Anirudh Kondav
On Mar 20, 2009, at 3:55 PM, Altaweel, Mark R. wrote:
I have a problem where I have two columns of data that I can simply
plot using:
plot(wV[0:15,3],wY[0:15,3]).
Perhaps:
plot(wV[0:15,3],wY[0:15,3], col = ifelse(wY[0:15,3]>0, "blue","red") )
This produces my desired plot.
Now, say I ha
I have a problem where I have two columns of data that I can simply plot using:
plot(wV[0:15,3],wY[0:15,3]).
This produces my desired plot.
Now, say I have a third variable that I would like to introduce and use that
variable to set different colors in the plot
In this case, say I wanted val
Do you only want the HH:MM:SS of the data? What are you going to do
with it? You can 'extract' it with
format(hora, "%H%:%M:%S")
If you don't want the date and are only working with the hours in a
single day, then convert the time to a numeric value of hours (or
minutes), whichever is appropria
Dear all,
Im having an awkward problem in R. When I write the command
Fisher.test(school.data,workspace=2e+07), where school.data is the matrix
corresponding to the data set,
I get the error message:
FEXACT error 7.
LDSTP is too small for this problem.
Try increasing the size o
The way the trees are structured in randomForest, there's no way to stop
tree growth by depth (what you called level).
(If anyone has ideas, I'm all ears.)
Andy
From: Anirudh Kondaveeti
>
> Hi all!
>
> The randomForest in R enables us to prune the trees using the nodesize
> feature where we c
There is probably a greater density of potential users on the SIG-
Finance mailing list.
I still think you attribute special powers to that function that the
authors never build into it. It does not predict the future. It
doesn't even simulate the future. It just reformats into a matrix the
We were hoping to use RSPerl to interface various R scripts we have to
a Perl-based web interface. I have been able to get some simple code
working; however, I can't seem to return a dataframe from R to Perl,
and I can't seem to determine from the documentation whether this is
possible or not.
I
Hi all!
The randomForest in R enables us to prune the trees using the nodesize
feature where we can stop splitting a node if it contains less than the
specified no.of of records/entities at that node.
However is there a way to stop the tree growing after a specified number of
levels. To be more c
mau...@alice.it wrote:
> I would like to create a vector of pairlist (flag, binary_value) like:
>
> (variable ="TrendOff", value = 0)
> (variable ="MOdwt",value = 1)
> (variable ? "ZeroPadding", value =1)
>
>
> I tried the followin
Thanks a lot. That should help!
Anirudh Kondaveeti
2009/3/20 Uwe Ligges
>
>
> Uwe Ligges wrote:
>
>>
>>
>> Anirudh Kondaveeti wrote:
>>
>>> To be more clear,
>>>
>>> My data set contains two classes.. Class 1 and Class 2
>>> Class 1 has original data with 300 rows
Uwe Ligges wrote:
Anirudh Kondaveeti wrote:
To be more clear,
My data set contains two classes.. Class 1 and Class 2
Class 1 has original data with 300 rows
Class 2 is randomly generated data with 1500 rows.
I want to sample a new data set with
Class 1 - all the rows
Class 2 - only 300 row
Anirudh Kondaveeti wrote:
To be more clear,
My data set contains two classes.. Class 1 and Class 2
Class 1 has original data with 300 rows
Class 2 is randomly generated data with 1500 rows.
I want to sample a new data set with
Class 1 - all the rows
Class 2 - only 300 rows out of 1500 rows
a
Uwe had been right all along. I don't understand what you don't
understand from the documentation.
You can use sampsize=c(300, 300) and replace=FALSE to make sure that all
300 class 1 rows are used, but be warned that that leaves no rows for
OOB estimate.
Andy
From: Anirudh Kondaveeti
>
> To
> I was including the settings in my document, but I also tried setting it
> from the command line. This resulted in more of a change, but only in that
> the ttBlue now being defined as black rather than blue. So, some effect, but
> not all the way.
You will have to define the styles before callin
Actually I was looking for someone who has some experience in using
the package. Anyway, thanks for your input.
On Fri, Mar 20, 2009 at 10:51 PM, David Winsemius
wrote:
> How could I (or anyone for that matter) possibly answer that question?
>
> David Winsemius, MD
> Heritage Laboratories
> West
To be more clear,
My data set contains two classes.. Class 1 and Class 2
Class 1 has original data with 300 rows
Class 2 is randomly generated data with 1500 rows.
I want to sample a new data set with
Class 1 - all the rows
Class 2 - only 300 rows out of 1500 rows
and then use it in random fores
Anirudh Kondaveeti wrote:
sampsize uses the same sample for all the trees in the random Forest.
No.
Uwe Ligges
But I want to use different sample for each tree of the 500 trees in the
random Forest. Thanks!
Anirudh Kondaveeti
2009/3/20 Uwe Ligges
Ani
sampsize uses the same sample for all the trees in the random Forest.
But I want to use different sample for each tree of the 500 trees in the
random Forest. Thanks!
Anirudh Kondaveeti
2009/3/20 Uwe Ligges
>
>
> Anirudh Kondaveeti wrote:
>
>> Hi!
>>
>> I am deali
Agreed --- if you provided a working example it might be as simple as,
library(reshape)
recast(res, Id~Tick, fun="mean", id.var=c("Id", "Tick"),
measure.var="X")
which i had to test with this,
res <- list(read.table(textConnection(
"IdX Tick
1 2.2 1
2 3.1 1
1
I say again:
Build a simple example;
--
David Winsemius
On Mar 20, 2009, at 1:00 PM, Altaweel, Mark R. wrote:
I will try to be clearer with what I wanted.
What I am trying to do is take a list, say defined as res, that has
a size of 1000
For example:
res[[1]]
re
Hi,
see ?par
You are probably interested in par("usr")
Mark Borowsky schrieb:
I would like to query an existing plot to learn the actual x and y
limits of the plottable area. Is this possible? I can discover the
extreme tick mark locations but not the actual plot area limits.
Setting the
How could I (or anyone for that matter) possibly answer that question?
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
On Mar 20, 2009, at 1:07 PM, Chirantan Kundu wrote:
Thanks for the help. RATING & NAME are not required for this call as
par documentation. About the CASHFLOWS I'
I'd do this with recursion, I think
* A valid (n+1) number sequence is a valid n-number sequence followed either by
the last number of the sequence or the last number plus one.
* A valid 1-number sequence is 0
-thomas
On Fri, 20 Mar 2009, Kathy Gerber wrote:
I am trying to print out
Thanks for the help. RATING & NAME are not required for this call as
par documentation. About the CASHFLOWS I'm not sure as I'm going to
call create_cashflows_matrix which is supposed to return me the
cashflow matrix. Do I still need to have the CASHFLOWSlist in the
structure?
On Fri, Mar 20, 2009
I will try to be clearer with what I wanted.
What I am trying to do is take a list, say defined as res, that has a size of
1000
For example:
res[[1]] res[[2]] .until
res[[1000]]
--
IdX Tick
Assunto: Extract Time in POSIXct
I have a dataframe with a column [hora] with this format:
hora: POSIXct, format: "1899-12-30 14:30:00"
It was obtained reading a access2007 database table.
Can I extract the time from each entry, keeping the POSIXct format?
Paulo E. Cardoso
Hi Thomas,
Thanks for responding.
You are right about bioconductor, however their solution was to
implement RHDF5 package for linux only, which allows selecting fields,
but not ranges of the data. Any idea about ranges of data? My
datafiles are as big as 20GB.
thx,
Dani
On Fri, Mar 20, 2009 at 1
I am trying to print out a list of strings of length 11 based on
integers 0 through 10. The rules as given to me for the ordering are:
The first digit must be 0.
The 2nd digit must be 0 or 1.
The 3rd digit must equal the 2nd digit or the 2nd digit +1.
...
Given the final digit, n, all digits 0 t
I would like to query an existing plot to learn the actual x and y
limits of the plottable area. Is this possible? I can discover the
extreme tick mark locations but not the actual plot area limits.
Setting the limits with ylim will not help in this case because I am
overplotting an existing
It was not clear what you were asking for. Did you want a 'list' of
'lists' pairs? What are you going to use the data for? How are you
expecting to access it? This will create a list similar to:
z <- list(list(variable="TrendOff", value=0), list(...
> flags_val <- c( 0,1,1,1,4,1,1,1,4,1,
I have found the gdata library quite helpful:
library(gdata)
ll()
ll(dimensions=TRUE)
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PLEASE do read the posting guide http:/
Dear Max,
I was including the settings in my document, but I also tried setting it
from the command line. This resulted in more of a change, but only in that
the ttBlue now being defined as black rather than blue. So, some effect, but
not all the way.
Here is the session info:
> sessionInfo()
R
Besides needing a grouped structure and not calling that function with
a group name, your sub-structure does not seem complete when compared
with the AAA group within the example dataset;
> str(allbonds) # your structure
List of 1
$ mybonds:List of 7
..$ ISIN: chr [1:2] "IN0020080
I would like to create a vector of pairlist (flag, binary_value) like:
(variable ="TrendOff", value = 0)
(variable ="MOdwt",value = 1)
(variable ? "ZeroPadding", value =1)
I tried the following syntax but the emcompassing list (t
Does this do what you want:
> x<-data.frame(id=c(1,2,3), snp1=c("AA","GG",
+ "AG"),snp2=c("GG","AG","GG"),snp3=c("GG","AG","AA"))
> do.call(rbind, apply(x, 1, function(.row){
+ data.frame(c(.row[1], .row[1]),
+strsplit(.row[2], ''),
+strsplit(.row[3], ''),
+
Did you do this within the document, or prior to running odfWeave?
Also, please provide the output of sessionInfo() after starting odfWeave.
Max
On Fri, Mar 20, 2009 at 11:03 AM, Fredrik Karlsson wrote:
> Dear list,
>
> The output of my embedded code sections come out in the style "ttBlue" wich
On Mar 20, 2009, at 10:15 AM, Chirantan Kundu wrote:
Here it goes -
library(termstrc)
ISIN <- vector()
ISIN[1]<-"IN0020080019"
ISIN[2]<-"IN0020020163"
MATURITYDATE<-as.Date(c("20081231","20101231"),"%Y%m%d")
STARTDATE<-as.Date(c("20010101","20020101"),format="%Y%m%d")
COUPONRATE<-c(8.24,7.95)
Anirudh Kondaveeti wrote:
Hi!
I am dealing with random forest using R.
Is there a way to sample a fixed no.of rows from a dataset for use with
different trees in random Forest.
To be more clear, my data set contains 1500 rows, and I am growing 500 trees
in Random Forest
Is it possible to samp
Dear list,
The output of my embedded code sections come out in the style "ttBlue" wich
is Times New Roman and Blue when I open document. I would like to change
this to a fixed witdth font, smaller and less blue by using style
definitions in the document, but it really does not affect the output.
The argument 'x' of adf.test should not only be a vector but also a
numeric vector (see ?adf.test). Have a closer look at your data and
how they are stored. What does is.numeric(x) give? I guess that
originally your data are stored as a data frame. Converting a
data.frame to a vector is not d
Hello,
is it possible to make so that once I have installed and loaded the
necessary packages in R, I don't need to install and load them for a
new session in R?
Up to now I install packages, for example, like this:
install.packages("package_name")
and then load them like this:
library("package
This is an eigenvalue problem with 0 on the main diagonal.
It is almost always inefficient to find the determinant
as an intermediate step. The original poster is looking for the
ngative of the eigenvalues of the matrix with the x replaced by zeros.
> tmp <- matrix(c(0,1,0,0,1,0,1,0,0,1,0,1,0,0,1
Hi,
I have a large dataset on which I would like to do the following:
x<-data.frame(id=c(1,2,3), snp1=c("AA","GG",
"AG"),snp2=c("GG","AG","GG"),snp3=c("GG","AG","AA"))
> x
id snp1 snp2 snp3
1 1 AA GG GG
2 2 GG AG AG
3 3 AG GG AA
And
On Fri, Mar 20, 2009 at 9:07 AM, Etches Jacob wrote:
> I am trying to specify a legend title to be other than the variable name,
> but I find that the legend splits because scale_shape() takes effect but
> scale_colour() does not. Can someone spot my error? Here's some toy code
> that produces t
Hi,
I have a large dataset on which I would like to do the following:
x<-data.frame(id=c(1,2,3), snp1=c("AA","GG",
"AG"),snp2=c("GG","AG","GG"),snp3=c("GG","AG","AA"))
> x
id snp1 snp2 snp3
1 1 AA GG GG
2 2 GG AG AG
3 3 AG GG AA
And
I am trying to specify a legend title to be other than the variable
name, but I find that the legend splits because scale_shape() takes
effect but scale_colour() does not. Can someone spot my error?
Here's some toy code that produces the problem on my system (R2.8.1,
windows, today's CRAN
Good try, Kevin. But that doesn't seem to do it.
set.seed(123)
x <- sort(runif(100))
y <- sin(4*pi*x) + rnorm(100, sd=0.2)
ans.lo2 <- loess(y ~ x, degree=2, span=0.75)
ans.gam2 <- gam(y ~ lo(x, degree=2, span=0.75))
summary(ans.lo2$fitted - ans.gam2$fitted) # larger differences, about 10%
Here it goes -
library(termstrc)
ISIN <- vector()
ISIN[1]<-"IN0020080019"
ISIN[2]<-"IN0020020163"
MATURITYDATE<-as.Date(c("20081231","20101231"),"%Y%m%d")
STARTDATE<-as.Date(c("20010101","20020101"),format="%Y%m%d")
COUPONRATE<-c(8.24,7.95)
PRICE<-(50,100)
ACCRUED<-c(5,2)
TODAY<-c(Sys.Date(),Sys.D
Dear Dirk,
As a Ubuntu newbie I was going by the information on the following
couple of webpages which talk about the need to turn the i386 binaries
into binaries for lpia architecture:
http://mydellmini.com/forum/i386-packages-on-lpia--t3540s75.html
http://ubuntuforums.org/showthread.php?p=6
On Mar 20, 2009, at 9:09 AM, Bob Gotwals wrote:
I need to find the determinant of this matrix
x 1 0 0
1 x 1 0
0 1 x 1
0 0 1 x
det yields x^4-3x^2+1
I can then use polyroot to find the roots of the coefficients.
The question is about the use of "x", which is what I'm solving for.
If you ar
Sorry for bothering you - but I have a little problem. I`m completely
new in R and trying to use lmer. That works, but I m not sure if my code
is right, so I would just like to get an opinion from the experts.
I have a Growthrate of plants (RGR
Fixed Factors: Fertilizer, Status (alien or native
Hi,
Just to add to this thread (with my first ever R list post). I got a
Dell Mini 9 (I think it is called the Inspirion 910 in the US?)
yesterday running Ubuntu with 2GB and a 32 GB SSD.
While the machine came with Ubuntu 8.04 LTS it is a version which has
been compiled by Dell to suit the
Got code?
On Mar 20, 2009, at 8:15 AM, Chirantan Kundu wrote:
I'm trying to use the package termstrc. However I cannot figure out
how to invoke helper functions like create_cashflows_matrix &
create_maturities_matrix. Even when I try to invoke those with the
data supplied with the package (say,
The following site is your friend:
http://www.rseek.org/
-Christos
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of
> meenu.s...@ingim.com
> Sent: Friday, March 20, 2009 5:47 AM
> To: r-help@r-project.org
> Subject: [R] Sta
Dear Members
Is there an R code for the following statistical tests ?
1. marquering_verbeek
2.HenrikssonMerton
3.cumby&modest
Regards
Meenu
-
ATTENTION:
The information in this electronic mail message is priva...{{dropped:18}}
Thanks,
now it works!
Max
Romain Francois-2 wrote:
>
> Ooops, I meant this :
>
> con.voidEval("sink('test.txt')");
> con.voidEval("cat(1+1)");
> con.voidEval("sink()");
>
> Romain Francois wrote:
>> Hi,
>>
>> try this instead:
>>
>> con.voidEval("sink(test.txt)");
>> con.voidEval("cat(1+1)"
Thank you. I will try the approacch you suggested.
Here is the list of 20 flags whose concurrent values values select the program
branch.
"TRUE" / "FALSE" can be replaced by 1 / 0
All the seperating lines "$$" and "..." and comments (in capital letters) can
be removed (this is a print-out).
Just
Hi,
I'm trying to use the package termstrc. However I cannot figure out
how to invoke helper functions like create_cashflows_matrix &
create_maturities_matrix. Even when I try to invoke those with the
data supplied with the package (say, corpbonds), it throws error
saying "Error in as.vector(x, mo
R friends,
I need to find the determinant of this matrix
x 1 0 0
1 x 1 0
0 1 x 1
0 0 1 x
det yields x^4-3x^2+1
I can then use polyroot to find the roots of the coefficients.
The question is about the use of "x", which is what I'm solving for.
thanks in advance, and this is a back-burner ques
On 3/20/2009 8:40 AM, Irina Foss wrote:
Dear,
I am trying to plot contours (with filled.contour function) inside UK area, so
that no contours are plotted over the sea. I was wondering if anyone can help
me with this task and if I can get any suggestions how to do that.
If you have a grid of
Dear,
I am trying to plot contours (with filled.contour function) inside UK area, so
that no contours are plotted over the sea. I was wondering if anyone can help
me with this task and if I can get any suggestions how to do that.
Thank you,
Irina
Irina Foss
Environmental Research Institute
N
Hi,
it would help if you provided a minimal example.
Here is one approach I often use with the plotting package ggplot2,
parameters <- expand.grid(m=c(0, 1), s=seq(0.1, 1,length=10))
x <- seq(-5, 5, length=300)
foo <- function(m, s){
data.frame(x=x, y=dnorm(x, m, s), m=factor(m), s=fa
Hi Mary,
To print without [1,] use cat() instead of print(). To get rid of the
row names when printing a data.frame use:
print(data.frame, row.names = FALSE)
To get nice Latex tables, you can use xtable (as the other reply
suggested). If you want to include it in Word, you can export your da
Dear list,
Sorry for posting a borderline statistical question on the list, but hte
SPSS people around me just stares at me blankly when refering to tests with
any term other than ANOVA and post-hoc. I would appreciate any insight on
how this all is possible:
I have a model fitted by aov() stored
Hi Tom,
it would have been nice (and it is in fact requested by the posting
guide) to give a running example instead of letting us construct your data.
Anyway, it wasn't too hard with
Xa<-array(1:12,dim=c(2,2,3))
Xa[c(2,6,8)]<-NA
#so next, create a vector for the colMeans with length 2*2*3, rep
I am running a simulation many times changing one parameter each time and
recording the outcome.
I have to produce a plot to make a sense of the bunch of numbers I get from
every run.
My problem is to insert a multi-line text to keep track of which result
correspond to which parameter values.
A
Dear R-users,
Can anyone tell me how to convert point coordinates from WSG84 (google
maps coordinates) format to UTM HUSO-31 ED-50. And if I have a map
(shp format) with UTM HUSO-31 ED-50 coordinates, how can I convert to
WSG84 (google maps coordinates) coordinates. Should I install some
external
of course! as some might suggest, i should have used times new sarcastic ;)
vQ
PS beware of
vector(vector)
and
function(function)
Bert Gunter wrote:
> Not surprising at all -- expected!
>
> is.function(vector) ##TRUE
>
> -- Bert
>
> -Original Message-
> From: Wacek Kusnie
On Fri, 20 Mar 2009, Pradheep K E wrote:
When I run predict on the model output on a new data (for testing) or on the
same data, I get only NA's. I'm able to run predict with some other models
constructed with biglm. One reason I suspect is that the model itself has a
few undefined terms (NA's).
Hi,
The paper
@ARTICLE{AlmironSilvaMM:2009,
author = {Almiron, M. and Almeida, E. S. and Miranda, M.},
title = {The Reliability of Statistical Functions in Four Software Packages
Freely used in Numerical Computation},
journal = {Brazilian Journal of Probability and Statistics},
ye
Dear Gavin, Gad and all,
Thank you very much for the reply!
Indeed my problem resembles Gavin's previously encountered one, and the
suggested method of package brglm() and profilemodel() seems solve the
problem right on target.
Right now I get stabilized AIC and realistic prediction, I'll check
mtmor...@fhcrc.org wrote:
>> setClass("foo",representation(x="numeric"))
>
> This creates an *S4* class which HAS A 'slot' x that is of type numeric.
>
>>> setMethod("plot","foo",function(x,y,...)boxplot(x,...))
>>> x <- rnorm(100)
>>> class(x) <- "foo"
>
> uh oh, this is creating an *S3* class tha
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