Error originates in the customized function ode. When IN!=0, You did
not assign a value to dIN which is required in list(c(dP1,dP2,dIN)).
On 2010-9-21 2:30, Tianchan Niu wrote:
Dear All,
I am trying to use ode solver rk4 to solve an ODE system, however, it
keeps saying: Error in eval(expr,
I don't see a data frame name in the rlm call...
Dennis
On Mon, Sep 20, 2010 at 7:08 PM, uttara_n nilawar.utt...@gmail.com wrote:
I am absolutely new to R and I am aware of only a few basic command lines.
I
was running a robust regression in R, using the following command line
library
On 09/21/2010 05:02 AM, Gregory Ryslik wrote:
Hi,
I think I've found away around that issue. The following works. If
this method is inefficient and one has something faster, I'll appreciate
it though!
lapply(mylist, function(x) as.numeric(as.character(x)))
You could avoid making them
Hi guys,
Im new to R and am having a bit of trouble with what should be a simple loop.
It sprobably something very fundamental that im doing wrong.
for(i in c(1:520))
{
tmp1- ((1-samp.pct[i])^2)*(log(1-theor.pct[i])-log(1-theor.pct[i+1]))
tmp2-
I created a EM algorithm for Generalized hyperbolic distribution.
I want to estimate mutheldaplus, sigmatheldaplus, betasigmaplus in my code.
After getting use these value , then my iteration have to be begin of this code.
But I can not to do iteration part.
Can you help me use my code and get
Oh, I am sorry I did not include the data.frame command line since it loaded
the data correctly without any errors. The following is the command line I
used to load the data:
data - read.table
(C:/MUP/Internship/Distance_Statistics/Excel_data/ForModelling_Rev_2.csv ,
sep= ,, header = TRUE)
Hi!
Is there an R-package with which I can calculate bandwidths via cross
validation in a data set??
Greetz,
Valentin
--
View this message in context:
http://r.789695.n4.nabble.com/Package-for-calculating-bandwidths-tp2548091p2548091.html
Sent from the R help mailing list archive at
What are you doing at iteration 520 with the objects corresponding to the i
+ 1 index? And there's a reason why you get one value for tmp1 and tmp2 -
you replace their values every iteration. Maybe you want tmp1[i] and
tmp2[i]?
If this is not a recursive calculation, you could easily replace the
First up, you've got a problem if your vectors are 520 elements in
length because you accessing element i+1 in your loop and when i is
520 you won't have a valid value.
Next, you don't really need a for loop at all :) You can do those
operations on the whole vectors...
e.g. tmp1 - (1 -
Dear list,
I'm seeking some advice regarding a particular numerical integration I
wish to perform.
The integrand f takes two real arguments x and y and returns a vector
of constant length N. The range of integration is [0, infty) for x and
[a,b] (finite) for y. Since the integrand has values in
I'm accessing around 95 tar files on an FTP server ranging in size between
10 and 40MB a piece.
while certainly can click on them and download them outside of R, I'd like
to have my script do it.
Retrieving the ftp directory with RCurl works fine (about 90% of the time)
but downloading the
Hi Ralf,
I am sure there is a function to do exactly that but I am not sure what is the
name and the package (Try plotrix or gregmisc...). To help you with your
problem we would need a reproducible example with your data, not the example
which obviously works.
As a guess, when you run
info
Dear R list
I have a problem with NA, which should be a string, but R seems that it
doesn't recognize it. What I do is first give the format command to my data
frame:
format.data.frame(mydata,big.mark= )
so I give a blank as thousand separator. All my records in my data frame
become strings,
On 09/21/2010 05:02 AM, David Winsemius wrote:
...
Not all university libraries have access via that link and efforts at
identifying the citation in Pubmed failed, so could I request a more
complete citation, please? Oh never mind I got it with Google. If anyone
else needs the ISSN of JASA in
On Tue, Sep 21, 2010 at 9:39 AM, steven mosher mosherste...@gmail.com wrote:
I'm accessing around 95 tar files on an FTP server ranging in size between
10 and 40MB a piece.
while certainly can click on them and download them outside of R, I'd like
to have my script do it.
Retrieving the ftp
Short answer: don't do that.
The format function is for preparing data for output. Do your data
manipulations on a data frame you keep for such use, and only use format to
prepare for output.
n.via...@libero.it n.via...@libero.it wrote:
Dear R list
I have a problem with NA, which should be a
All the solutions in this thread so far use the lapply(split(...)) paradigm
either directly or indirectly. That paradigm doesn't scale. That's the
likely
source of quite a few 'out of memory' errors and performance issues in R.
data.table doesn't do that internally, and it's syntax is pretty
Sorry Ralf,
I was in a hurry when I replied to your plot. The function you are looking for
is:
dx=density(x)
plot(dx)
That works with the code you send in your mail. Just adjust the plot limits and
change x and y for the density plot of y and it works. A good reference can be
found here:
Thanks Frank,
I have one small question regarding this, understand you are very busy and if
you cant answer i would greatly appreciate any thoughts from the list.
Split-sample validation is not reliable unless you have say 10,000 samples to
begin with
I am a little confused. When i ran
I would like to thank you very much for making this clear. It seems that the
solution you suggested is right one as the second attempt does find all the
cells that are touched.
Now I ll try to find out how much the line gets into one of this cell as every
cell affects acts like a weight. The
Dear All,
Consider a simple example
a-c(1,4,3,0,4,5,6,9,3,4)
b-c(0,4,5)
c-c(5,4,0)
I would like to be able to tell whether a sequence is contained (the
order of the elements does matter) in another one e.g. in the example
above, b is a subsequence of a, whereas c is not. Since the order
Convert to strings and use grep functions.
using c for variable is a bad idea.
a - paste(a, collapse=)
b - paste(b, collapse=)
d - paste(d, collapse=)
grepl(b,a)
grepl(d,a)
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
nikhil.l...@gmail.com
On Sep 21,
On 20/09/2010 8:36 PM, Hey Sky wrote:
Hey, R Users
my windows is 64 bit windows 7. I am trying to install the package ucminf into
my 64 bit version R but cannot. the package I downloaded is from
http://cran.r-project.org/web/packages/ucminf/index.html and I installed it with
the install
Hi
I got a barplot that has values between 0-150 and the y-axis shows the steps
0 50 100 and 150 but I would like to change it to 0 10 20 30... ...130 140
150
Dont really know the word in english so sry about it beeing abit confusing
:)
Thx for your help
Joel
--
View this message in context:
On Tue, 2010-09-21 at 03:55 -0700, Joel wrote:
Hi
I got a barplot that has values between 0-150 and the y-axis shows the steps
0 50 100 and 150 but I would like to change it to 0 10 20 30... ...130 140
150
Dont really know the word in english so sry about it beeing abit confusing
:)
The urgency and the vague description of your problem strongly suggest
that this is homework. This list is not for homework---see the posting
guide at the bottom of every message. Nonetheless since I know this
problem reasonably well I will offer some comments.
QRMlib is a package created to
On 09/21/2010 08:55 PM, Joel wrote:
Hi
I got a barplot that has values between 0-150 and the y-axis shows the steps
0 50 100 and 150 but I would like to change it to 0 10 20 30... ...130 140
150
Hi Joel,
This is a combination of the pretty calculation of axis tick intervals
and the silent
Hey,
I have a dataset where two columns are factors and another column consists
of values. Each combination of factors can only have a single value assigned
to it.
I'd like to represent this as a matrix or table where the rows are the first
column factors and the columns the second column
I'm confused, hope someone can point out what is not obvious to me.
I thought I was creating a new data frame by 'deleting' rows from an
existing dataframe - I've tried 2 methods.
But this new data frame seems to remember values from its parent - even
though there are no occurences.
Where
example(factor)
iris1$Species - factor(iris1$Species, drop=T)
will get you what you need.
Nikhil Kaza
Asst. Professor,
City and Regional Planning
University of North Carolina
nikhil.l...@gmail.com
On Sep 21, 2010, at 7:41 AM, pdb wrote:
I'm confused, hope someone can point out what is
From: jlu...@ria.buffalo.edu
Clearly inferior treatments are unethical.
The Big Question is: What constitute clearly? Who or How to decide
what is clearly? I'm sure there are plenty of people who don't
understand much Statistics and are perfectly willing to say the results
on the two
Thx for all your help!
--
View this message in context:
http://r.789695.n4.nabble.com/change-y-axis-distance-tp2548363p2548461.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Removing elements from a factor does not change the levels of the
factor.
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium
Research
I would like to thank you very much for your reply.
Actually I would like to ask you if there is
a small list called fred:
fred - list(happy = 1:10, name = squash)
and a big list called bigfred that included fred list 5 times
bigfred - rep(fred,5)
Is it possible somehow to index all these
Baptiste;
You should see if this meets your requirements:
help(adaptIntegrate, package=cubature)
(I got errors when I ran the code and NaN's when I looked at the
output of test function, f.)
vAverage(mixedrule, -4, 4, 0.0, 1, 20, f) - c(pi, pi/2, 2*pi)
Error: object 'mixedrule' not found
baptiste auguie baptiste.auguie at googlemail.com writes:
Dear list,
I'm seeking some advice regarding a particular numerical integration I
wish to perform.
The integrand f takes two real arguments x and y and returns a vector
of constant length N. The range of integration is [0,
This function might be helpful:
bleh - function(a, b) {
where - list()
matches - 0
first - which(a == b[1])
for (i in first) {
seq.to.match - seq(i, length = length(b))
if (identical(a[seq.to.match], b)) {
matches - matches + 1
where[[matches]] - seq.to.match
}
}
Dear list,
I'm calculating the integral of a Gaussian function from 0 to
infinity. I understand from ?integrate that it's usually better to
specify Inf explicitly as a limit rather than an arbitrary large
number, as in this case integrate() performs a trick to do the
integration better.
However,
Thanks, but that was what I just discovered myself the hard way.
What I really wanted to know was how to solve this issue.
--
View this message in context:
http://r.789695.n4.nabble.com/removed-data-is-still-there-tp2548440p2548527.html
Sent from the R help mailing list archive at Nabble.com.
On Sep 21, 2010, at 1:34 AM, Ralf B wrote:
Hi,
in order to save space for a publication, it would be nice to have a
combined scatter and density plot similar to what is shows on
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=78
I wonder if anybody perhaps has already
Hi,
thanks for the pointer to cubature (which i had probably dismissed too
quickly). Your tests with f should not work: the domain of f(x,.) is
restricted to positive reals, but this domain of integration is then
transformed in mixedrule() to map the semi-infinite range to a more
reasonable
On Sep 21, 2010, at 6:52 AM, Liaw, Andy wrote:
From: jlu...@ria.buffalo.edu
Clearly inferior treatments are unethical.
The Big Question is: What constitute clearly? Who or How to decide
what is clearly? I'm sure there are plenty of people who don't
understand much Statistics and are
On Sep 21, 2010, at 6:31 AM, Lorenzo Isella wrote:
Dear All,
Consider a simple example
a-c(1,4,3,0,4,5,6,9,3,4)
b-c(0,4,5)
c-c(5,4,0)
I would like to be able to tell whether a sequence is contained (the
order of the elements does matter) in another one e.g. in the
example above, b is a
Hi Chuan,
I'm forwarding your question to the list because I haven't used the
rimage package... It's best if you post questions to the list anyway
because you are more likely to get a fast and useful answer.
On 20 September 2010 23:03, chuan zun liang wrote:
Dear Michael:
I am so sorry,disturb
Thanks Frank,
I have one small question regarding this, understand you are very busy and if
you cant answer i would greatly appreciate any thoughts from the list.
Split-sample validation is not reliable unless you have say 10,000 samples to
begin with
I am a little confused. When i ran the
Hello there,
I started to use rcom package and there were no problems until I tried to
call some external function (method) which returns safearray type of data
where either me or rcom or something else fails. Specifically I connected to
a database and called a method doing something like this
You can get exactly the plot you want (except with relative frequency
polygons instead of the histograms on the marginals, but they are
equivalent) with the package
chplot
(which is available from CRAN)!
Use just one group (class, sample, category -- i.e., omit the conditioning
variable as
Try this:
d - data.frame(A = letters[1:10], B = sample(letters[11:20]), C =
sample(10))
xtabs(C ~ A + B, d)
On Tue, Sep 21, 2010 at 8:39 AM, ZeMajik zema...@gmail.com wrote:
Hey,
I have a dataset where two columns are factors and another column consists
of values. Each combination of
On Sep 21, 2010, at 5:28 AM, Jeff Newmiller wrote:
Short answer: don't do that.
The format function is for preparing data for output. Do your data
manipulations on a data frame you keep for such use, and only use
format to prepare for output.
That is excellent advice. But to answer the
Thanks, adaptIntegrate() seems perfectly suited, I'll just need to
figure a transformation rule for the infinite limit. The suggestion of
x-1/x does not seem to work here because it also transforms 0 into
-infinity. I think exp(pi* sinh(x)) could be a better choice,
according to Numerical Recipes.
http://r.789695.n4.nabble.com/file/n2548152/25jfmyx.jpg
HOW to create image like this?
**tp://i55.tinypic.com/25jfmyx.jpg[/IMG]
I don't known how to create the image above or which function can create
this image?
--
View this message in context:
On Sep 21, 2010, at 11:28 , Jeff Newmiller wrote:
Short answer: don't do that.
The format function is for preparing data for output. Do your data
manipulations on a data frame you keep for such use, and only use format to
prepare for output.
But isn't that what the OP is doing? It is
Would nobody like to answer me?
--
View this message in context:
http://r.789695.n4.nabble.com/Package-for-calculating-bandwidths-tp2548091p2548509.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Hi
I am trying to determine the mean of a Weibull function that has been fit to
a data set, adjusted for a categorical covariate , gender (0=male,1=female).
Here is my code:
library(survival)
survdata-read.csv(data.csv)
##Fit Weibull model to data
On Sep 21, 2010, at 8:39 AM, pdb wrote:
Thanks, but that was what I just discovered myself the hard way.
What I really wanted to know was how to solve this issue.
Although that was _not_ what you requested in your first post.
2 options:
?table
?factor
iris1$Species
On Tue, Sep 21, 2010 at 8:58 AM, zcrself zcrs...@gmail.com wrote:
http://r.789695.n4.nabble.com/file/n2548152/25jfmyx.jpg
HOW to create image like this?
**tp://i55.tinypic.com/25jfmyx.jpg[/IMG]
My first response is On an empty stomach, with a handy supply of
anti-migraine tablets.
I don't
On Sep 21, 2010, at 9:04 AM, David Winsemius wrote:
On Sep 21, 2010, at 8:39 AM, pdb wrote:
Thanks, but that was what I just discovered myself the hard way.
What I really wanted to know was how to solve this issue.
Although that was _not_ what you requested in your first post.
2
Bandwidth of what? Internet traffic/some sort of smoother?
Your question is not clear to me.
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il
Hi,
I knew about that way already, with factor(). Isn't there another
possibility, directly at the subsetting step? That would be of great help
iris1 - iris[iris$Species == 'setosa',] ## I mean here
Ivan
Le 9/21/2010 15:14, David Winsemius a écrit :
On Sep 21, 2010, at 9:04 AM, David
This looks like it has been created in Circos: http://mkweb.bcgsc.ca/circos/
HTH
Peter
On Tue, Sep 21, 2010 at 9:58 AM, zcrself zcrs...@gmail.com wrote:
http://r.789695.n4.nabble.com/file/n2548152/25jfmyx.jpg
HOW to create image like this?
**tp://i55.tinypic.com/25jfmyx.jpg[/IMG]
I don't
There is nothing mysterious. You need to increase the accuracy of
quadrature by decreasing the error tolerance:
# I scaled your function to a proper Gaussian density
shiftedGauss - function(x0=500){
integrate(function(x) 1/sqrt(2*pi * 100^2) * exp(-(x-x0)^2/(2*100^2)), 0,
Inf,
On Sep 21, 2010, at 3:58 AM, zcrself wrote:
http://r.789695.n4.nabble.com/file/n2548152/25jfmyx.jpg
HOW to create image like this?
**tp://i55.tinypic.com/25jfmyx.jpg[/IMG]
I don't known how to create the image above or which function can
create
this image?
THe corresponding author in
mark.fisher123 marko.fisher2008 at gmail.com writes:
[snip]
library(survival)
survdata-read.csv(data.csv)
##Fit Weibull model to data
WeiModel-survreg(Surv(survdata$Time,survdata$Status)~survdata$gender)
summary(WeiModel)
P-pweibull(n, scale=exp(WeiModel$coef[1]),
Hi Everyone,
I am interested in taking the mode over several thousand matrices. I show an
example below. For the [1,1] entry of my mode matrix that I want to create I
would like to have a 2. For the [1,2] entry I would want a 2. For the [2,2]
entry it would be 4 and so forth. Earlier, I was
Ivan Calandra ivan.calandra at uni-hamburg.de writes:
Hi,
I knew about that way already, with factor(). Isn't there another
possibility, directly at the subsetting step? That would be of great help
iris1 - iris[iris$Species == 'setosa',] ## I mean here
Ivan
Not as far as I know.
David Winsemius wrote:
On Sep 19, 2010, at 5:59 PM, zozio32 wrote:
Thanks for you're long answer.
I have to say, I am not fully sure of what you're meaning
everywhere. As I
said, I am merely following a recipe book, and when things depart
from it I
am a bit lost.
I'll try to
Hi,
Thanks for the tip, but it's still mysterious to me. Reading ?integrate did not
give me much hint as to what relative accuracy means in this context. I
looked at the source of integrate.c but it's still not clear to me how the
default value of rel.tol (10^-4 for me) is not enough to
On Sep 21, 2010, at 15:43 , Ben Bolker wrote:
mark.fisher123 marko.fisher2008 at gmail.com writes:
[snip]
library(survival)
survdata-read.csv(data.csv)
##Fit Weibull model to data
WeiModel-survreg(Surv(survdata$Time,survdata$Status)~survdata$gender)
summary(WeiModel)
You could try pnorm also:
shiftedGaussR - function(x0 = 500) {
sd - 100/sqrt(2)
int - pnorm(0, x0, sd, lower.tail=FALSE, log.p=TRUE)
exp(int + log(sd) + 0.5 * log(2*pi))
}
shiftedGaussR(500)
[1] 177.2454
shiftedGauss(500)
[1] 177.2454
-Matt
On Tue, 2010-09-21 at 09:38 -0400,
You are dealing with functions that are non-zero over a very small interval,
so you have to be very careful. There is no method that is going to be
totally foolproof. Having said that, I have always felt that the default
tolerance in integrate is too liberal (i.e. too large). I always use
Try this:
mode - function(x, ...)
as.numeric(names(which.max(table(x
apply(array(unlist(mymats), dim = c(length(mymats), dim(mymats[[1]]))), 1:2,
mode)
On Tue, Sep 21, 2010 at 10:47 AM, Gregory Ryslik rsa...@comcast.net wrote:
Hi Everyone,
I am interested in taking the mode over
On Tue, Sep 21, 2010 at 8:34 AM, Ralf B ralf.bie...@gmail.com wrote:
in order to save space for a publication, it would be nice to have a
combined scatter and density plot similar to what is shows on
Not quite the same thing, but I like the scatterplots in Rcmdr, which
feature boxplots instead
Thanks, I'll do that too from now on.
It strikes me that in a case such as this one it may be safer to use a
truncated, finite interval around the region where the integrand is non-zero,
rather than following the advice of ?integrate to use Inf as integration limit.
At least one wouldn't risk
Dear users,
I would like all the ticks on a boxplot (x and y) to be labeled
I have checked all the par() arguments but couldn't find what I'm
looking for
Here is an example to show it:
df - structure(list(SPECSHOR = structure(c(1L, 1L, 1L, 3L, 3L, 3L, 3L,
3L, 4L, 4L), .Label = c(cotau,
On Tue, Sep 21, 2010 at 3:09 AM, Matthew Dowle mdo...@mdowle.plus.com wrote:
All the solutions in this thread so far use the lapply(split(...)) paradigm
either directly or indirectly. That paradigm doesn't scale. That's the
likely
source of quite a few 'out of memory' errors and performance
Should anyone feel like reinventing that coloured wheel in R, the arcTextGrob()
function in gridExtra answered a more basic query on R-help a few months ago:
draw text labels on a circle and connect them with arcs. It might be a starting
point.
baptiste
On Sep 21, 2010, at 3:14 PM, Barry
Sorry, bandwidth via cross validation for a kernel smoothing.
--
View this message in context:
http://r.789695.n4.nabble.com/Package-for-calculating-bandwidths-tp2548091p2548630.html
Sent from the R help mailing list archive at Nabble.com.
__
Hello,
I want to make an 2D plot of .xyz data. So plot on x-axis and y-axis and
use a color scale for the z.
For every x-location along a cross section of the soil at several depths
a resistivity must be displayed. This must result in a picture/graph
which shows the resistivity for that cross
hi i'm francesco, i'am a new r user.
I have a dataset with these variables
[1] timestamp categoria latlong
ammontare_euro
[6] provincia risoluzione
and i want to make an analjsy on the data of variable 'categoria'.
now the variable 'categoria' has 98
Hadley Wickham
London: 1st - 2nd November 2010
Data Visualisation in R: Harnessing the power of ggplot2 to produce elegant
data graphics
Mango Solutions is delighted to offer a one-off 2 day training course with Dr.
Hadley Wickham, R Project Data Visualisation Guru and creator of ggplot 2. The
On Tue, Sep 21, 2010 at 6:14 AM, Barry Rowlingson
b.rowling...@lancaster.ac.uk wrote:
On Tue, Sep 21, 2010 at 8:58 AM, zcrself zcrs...@gmail.com wrote:
http://r.789695.n4.nabble.com/file/n2548152/25jfmyx.jpg
HOW to create image like this?
**tp://i55.tinypic.com/25jfmyx.jpg[/IMG]
My first
I am trying to use the cca/rda/capscale functions in vegan to analyse
genetic distance data ( provided as a dist object calculated using
dist.genpop in package adegenet) with geographic distance partialled out
( provided as a distance object using dist function in veganthis method
is
Dear group,
I have recognized a strange behaviour of palette(). I tried to find any
explanation but failed so far (or even didnt understood the idea behind
- what is most probable).
My original plan was to define a palette, save it in a variable and use
it later for an image-plot. One
Dear Jacob,
Do you want something like this? Using dummy data instead of yours.
Copy-paste the output of dput() if you want to pass your data in a
format that is easy to use.
library(ggplot2)
dataset - expand.grid(X = -3:3, Depth = seq(0, -1, by = -0.25))
dataset$Resistivity -
On Sep 21, 2010, at 9:49 AM, Jacob Oosterwijk wrote:
Hello,
I want to make an 2D plot of .xyz data. So plot on x-axis and y-axis
and
use a color scale for the z.
For every x-location along a cross section of the soil at several
depths
a resistivity must be displayed. This must result
Hello,
This is my first post to the help request list, so I'm going to err on the
side of giving too much information.
I'm working on writing a simulation in which agents will make repeated
production and exchange decisions with randomly chosen partners.
The idea is, all agents can produce two
On Sep 21, 2010, at 10:50 AM, Daniel Stepputtis R wrote:
Dear group,
I have recognized a strange behaviour of palette(). I tried to find
any explanation but failed so far (or even didnt understood the idea
behind - what is most probable).
My original plan was to define a palette, save
I used the following command line:
Regression - read.table
(C:/MUP/Internship/Distance_Statistics/Excel_data/Robust_Regression/ForModelling_Rev_Rob.csv
, sep= ,, header = TRUE)
attach (Regression)
summary (Regression)
library (MASS)
rlm(TotalEmployment_2004 ~ MISSISSIPPI + LOUISIANA +
I am trying to compute the smallest control group size using Fleiss
formula. Assuming that my population is 5,000, the smallest expected
lift (smallest detectable change) is 5%, and the highest likely rate
in the control group is 50%, what is the minimum control group size
assuming a 95%
baptiste auguie baptiste.auguie at googlemail.com writes:
Thanks, adaptIntegrate() seems perfectly suited, I'll just need to
figure a transformation rule for the infinite limit. The suggestion of
x-1/x does not seem to work here because it also transforms 0 into
-infinity. I think exp(pi*
Hello Uttara,
What happens if you try something like this?
Regression -
read.table(C:/MUP/Internship/Distance_Statistics/Excel_data/Robust_Regression/ForModelling_Rev_Rob.csv,
sep= ,, header = TRUE)
library (MASS)
rfdmodel1 - rlm(TotalEmployment_2004 ~ MISSISSIPPI + LOUISIANA +
Probably true, thats cunning, but look at base::match. The
first thing it does is coerce factor to character (an allocate
and copy needed internally). data.table doesn't do that
either, see data.table:::sortedmatch.
I made first basic steps towards a proper reproducible test
suite (timings.Rnw).
Hello,
this might be little off-topic, but still... I have not found any official web
forum for R users. Did I look good? If not, I'm sorry and shame on me! :-)
Such forums are very probably the most common way to share information about
all sorts of problems, to ask and get answer. See
I will take a look and get back to you.
Ravi.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Erik Kimbrough
Sent: Tuesday, September 21, 2010 11:01 AM
To: R-help@r-project.org
Subject: [R] Trouble with Optimization in Alabama
Hello,
I'm running into the following error:
C:\Program Files\R\Rtools\bin\sh.exe: *** fatal error - couldn't allocate heap,
Win32 error 487, base 0x7A, top 0x7b, reserve_size 61440, allocsize
65536, page_const 4096
I've re-installed Rtools and R (2.10.1) and followed all the
On 21/09/10 17:40 PM, Nevil Amos nevil.a...@gmail.com wrote:
I am trying to use the cca/rda/capscale functions in vegan to analyse
genetic distance data ( provided as a dist object calculated using
dist.genpop in package adegenet) with geographic distance partialled out
( provided as a
Hello,
I am using maptools for plooting geographical data.
The colour of the region indicates some region dependent value
(population for example).
I pass the colours of the regions to the plot.Map function by defining
the foreground colour:
jet.colors = colorRampPalette(c(#7F, blue,
Hey, Duncan
thanks for your reply.
I am not sure which version i have installed but I downloaded it from
http://cran.skazkaforyou.com/. when I check the R installed, it says 2.11.1.
I do not know I answered your question or not. if not, where I can find them?
(in fact, I did not notice/find
Thanks, Bill and Michael, you have answered the question I asked, but not
the one I wished to ask
I want to obtain the maximum in each group of variables, so I could scale
each variable by the maximum for its group. If I use tapply, as in the
example below, there's a mismatch in dimensions of
Error in data.frame(times = NonCensored.data, censor = Censored.data) :
arguments imply differing number of rows: 14, 6
Anan Halabi
Reliability Eng, RD
HP Scitex
Tel: 972-9-8924648
mobil: 972-52-6624231
__
R-help@r-project.org mailing list
1 - 100 of 207 matches
Mail list logo
