rc-list(c(
123,321,234,543,654,768,986,987,246,284),c(Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec))
# the matrix has rownames that are used as identifiers and columns
# of time. 1 years worth of data. Thats the native format
test-matrix(seq(1,120, by=1), nrow=10,dimnames=rc)
test
Jan
On Tue, 10 Aug 2010, Carrie Li wrote:
Thank you both!
I found that the model with both x and y have measurement errors should be
pretty common in practice. But it seems to me that there is no a simple
solution for it...(I mean, a ready-to-use package or program handing this
model fitting
Hey guys,
I'm trying to run a Canonical Correlation Analysis (CCA) between two data
sets. But for my case I am dealing with data sets(which are stored as two
matrices X and Y) in which the number of experimental units is greater than
the number of variables, so I want to use a sample from my
Trevor,
Thanks for your reply.
That doesnot help
Any other suggestions?
Anamika
On Tue, Aug 10, 2010 at 11:58 AM, Trevor Davies tdav...@mathstat.dal.cawrote:
I think you need to load is R2WinBUGS again.
require(R2WinBUGS)
Trevor
Hi All:
I am getting an error while trying to plot in
Hi All,
I am trying to run the following script but I am getting and error message
one=read.table(sample.txt,sep=\t)
bitmap(file=sample.JPG,type=jpeg,width=5,height=5,res=300,pointsize=10)
“Error in (st + 1):(en - 1) : argument of length 0”
plot(one$V1,one$V2)
I tried to google the error but
How do I get the std. errors of correlation coefficients? When I use the cor
function it only prints the correlation.
Thanks, Fusun.
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The labels on the x-axis are defaulting to scientific notation no matter
how small cex.axis is. How can I override scientific notation to get the
labels to print out as specified? Here is the code (UNIT here is
0.0105):
plot(xm,yv,log=xy,ylim=c(0.1,20)/UNIT,xlim=c(0.004,20)*UNIT,xaxt=n,t
ype=n)
Thank you both!
I found that the model with both x and y have measurement errors should be
pretty common in practice. But it seems to me that there is no a simple
solution for it...(I mean, a ready-to-use package or program handing this
model fitting problem. )
One more question, is any reference
The types of bitmap available depend on the ghostscript you have installed.
Have you checked that yours handles jpeg?
What happens if you use the jpeg driver directly?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of kayj
Sent:
On Aug 10, 2010, at 8:23 PM, Michal Figurski wrote:
Peter,
Since in PB the procedure is to calculate a whole list of slopes and
intercepts, wouldn't it be a solution to determine the correlation and go
from there? How do I do it?
You can't. (Perhaps someone could, but it looks like a
On Wed, 11 Aug 2010, bill.venab...@csiro.au wrote:
The types of bitmap available depend on the ghostscript you have
installed. Have you checked that yours handles jpeg?
That's true, but bitmap() is failing in its check of the available
types.
As so very often, we lack the 'at a minimum'
Thanks Ravi.
Gildas
Ravi Varadhan a écrit :
I think the problem is because the the Hessian of the augmented Lagrangian
iis singular at c(0,0).
Try this:
require(alabama)
heq - function(x) {
x[1]^2+x[2]^2 - 1
}
constrOptim.nl(par=c(0,0), fn=f, heq=heq,
Hi,
Not really a request for help, but I thought it would be useful to others
who use Gentoo (http://www.gentoo.org/) to know that you can now install
CRAN and bioconductor packages using Gentoo's package management systme
Portage
Hello!
er.. do I need to calculate the values into a grid, or is there a way to
plot it without that? E.g cloud works, but I can't seem to get a surface
like that.
eg I made a test csv with some data
library(lattice)
mp - read.csv(myPolinom.csv, sep=;, header=TRUE)
cloud(zz~xx*yy, data=mp) #
Hello!
Is there a simplier way in R to get a nonlinear regression (like nls) for a
surface? I have 3D data, and it is definitely not a linear surface with
which it would fit the best. Rather sg like z = a + f(x) + g(y) where
probably both f and g are polinomes (hopefully quadratic).
Szilvia
--
Thanks.
On the other hand,
I try to obtain the same result but from this list :
x - list()
x$i - 5
x$j - 9
x$k - 15
names(x$i) - a
names(x$j) - b
names(x$k) - b
Thanks in advance,
Carlos
2010/8/10 Wu Gong w...@mtmail.mtsu.edu:
Hi Carlos,
I give a handmade code, hope it helps.
y - list()
Hi everyone,
I am writing a code for cointegration between n (n=6 initially) pairs.
I have done the whole thing for 2 variables. There is a function coint(x1,x2)
that takes 2 inputs x1 and x2 and does the following:
coint-function (x1,x2)
{
adfdata(x1)
adfdata(x2)
A beginning of solution...
n - sapply(x, function(i) names(i))
tapply(x, n, c)
2010/8/11 Carlos Petti carlos.pe...@gmail.com:
Thanks.
On the other hand,
I try to obtain the same result but from this list :
x - list()
x$i - 5
x$j - 9
x$k - 15
names(x$i) - a
names(x$j) - b
names(x$k) -
Or rather :
n - sapply(x, function(i) names(i))
tapply(x, n, names)
2010/8/11 Carlos Petti carlos.pe...@gmail.com:
A beginning of solution...
n - sapply(x, function(i) names(i))
tapply(x, n, c)
2010/8/11 Carlos Petti carlos.pe...@gmail.com:
Thanks.
On the other hand,
I try to obtain the
Dear R People,
Im trying to link R and TK. Im using R 2.11.1 with Tinn-R 1.17.2.4.
For simple examples like the following one its working (see the
initiation of the variables, the handler and the link to the function
which multiply Zahl, Exponent and Factor) :
require(tcltk)
On 11/08/2010 6:15 AM, szisziszilvi wrote:
Hello!
Is there a simplier way in R to get a nonlinear regression (like nls) for a
surface? I have 3D data, and it is definitely not a linear surface with
which it would fit the best. Rather sg like z = a + f(x) + g(y) where
probably both f and g are
On 08/11/2010 06:18 AM, Watkins, Janice wrote:
The labels on the x-axis are defaulting to scientific notation no matter
how small cex.axis is. How can I override scientific notation to get the
labels to print out as specified? Here is the code (UNIT here is
0.0105):
Dear R community,
I have a question concerning the generation of an experimental design for
discrete choice experiments (=choice based conjoint analysis, CBC). I
understand that there is an article Design and Analysis of Choice Experiments
Using R by Aizaki and Nishimura (available at
Hi,
I would like to extract the coefficients of a logistic regression
(estimates and standard error as well) in lrm as in glm with
summary(fit.glm)$coef
Thanks
David
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
How about this?
untested since no data is provided.
a - paste(x,1:6,sep=)
b- combn(a,2)
coint-function (x)
{
x1 - get(x[1])
x2 - get(x[2])
adfdata(x1)
adfdata(x2)
engle-lm(x1~x2)
residual-resid(engle)
adfresd(residual, k=1)
On 08/09/2010 06:51 PM, JH wrote:
I want to edit a function in the nlme package. I download the package source
nlme_3.1-96.tar.gz from the link below then edit it one of the scripts
inside of it. From this stage how can I turn it into a package that I can
use in R?
well, you can directly use the coef() and vcov() generics, e.g.,
library(rms)
y - rbinom(100, 1, 0.5)
x - runif(100, -3, 3)
Fit - lrm(y ~ x)
Fit
coef(Fit)
sqrt(diag(vcov(Fit)))
I hope it helps.
Best,
Dimitris
On 8/11/2010 12:56 PM, david dav wrote:
Hi,
I would like to extract the
On Aug 11, 2010, at 6:56 AM, david dav wrote:
Hi,
I would like to extract the coefficients of a logistic regression
(estimates and standard error as well) in lrm as in glm with
summary(fit.glm)$coef
?coef
Try instead:
coef(fit.glm)
?vcov
The accessor function for the covariance matrix
On Wed, 11 Aug 2010, david dav wrote:
Hi,
I would like to extract the coefficients of a logistic regression
(estimates and standard error as well) in lrm as in glm with
summary(fit.glm)$coef
Thanks
David
coef(fit)
sqrt(diag(vcov(fit)))
But these will not be very helpful except in the
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
On Tue, 10 Aug 2010, Mark Seeto wrote:
Hi Frank (and others),
Thank you for your comments in reply to my questions. I had not
encountered contrast
Has someone implemented a Computable General Equilibrium (CGE) model in R?
Usually such models are coded in GAMS ( a general algebraic solver) or GEMPACK
(a solver specifically for CGE models). As it is possible to implement CGE
models in the econometric software EViews (Essama-Nssah, B.
Hello,
I use the earth package to obtain a approximate models to my data.
Now, I want to get the gradiente from this model...
Is there an automatized form to get this?
The model has a hundred of the terms...
Thaks for any help
Cleber
--
O bom senso é a coisa do mundo mais bem distribuída:
Hi Allan,
Thank you for your response. Finally I succeeded. These were the
commands to compile Atlas and R:
Atlas:
../configure -t 4 -Fa alg -fPIC
make
make install
(-fPIC is needed to wrap some of objects of lapack.a as into a shared lib)
R:
../configure
Hi,
Look at ?cor.test, it does not give you the standard error, but it
does give you a confidence interval around the correlation coefficient
and a significance test (which is my best guess of what you are
probably looking to do with the standard error). You could also get
it out of a simple
Hello:
I try to use function bms to select the best model with the biggest
posterior probability for different quantile. I have one return and 16
variables. However, I can just select for the linear model, how to change
the code for quantile:
bma1=bms(rq(y~.,tau=0.25,data=EQT), burn =
Hi, I'm trying to put a document through odfWeave that has some R code for
coursework in it:
library(odfWeave)
Loading required package: lattice
Loading required package: XML
inFile = ~/Documents/Squirrel.odf
outFile = ~/Documents/ Squirrel Out.odt
inFile = ~/Documents/Squirrel.odt
odfWeave
Peter, Frank, David and others,
Thank you all for your ideas. I understand your lack of trust in PB
method. Setting that aside (it's beyond me anyways), please see below
what I have finally came up with to calculate the CI boundaries. Given
the slope and intercept with their 05% 95% CIs, and
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt University
On Wed, 11 Aug 2010, Michal Figurski wrote:
Peter, Frank, David and others,
Thank you all for your ideas. I understand your lack of trust in PB
Given a long zoo matrix, the goal is to sweep out a statistic from the
entire length of the
sequences.
longzoomatrix-zoo(matrix(rnorm(720),ncol=6),as.yearmon(outer(1900,seq(0,length=120)/12,+)))
cnames-c(12345,23456,34567,45678,56789,67890)
colnames(longzoomatrix)-cnames
longzoomatrix[1:24,]
What does this mean?
It's impossible to tell. Read the posting guide and figure out all the
details that you left out. If we don't have more information, you
should have low expectations about the quality of any replies to might
get.
--
Max
__
Hello Everyone!
Here's what I'm trying to do. I'm working on generating occurrences of
precipitation based upon precipitation occurrence for a station during the
previous day and two stations that have already been generated by joint
probablities and 1st order Markov chains or by the same
Hello!
I have something like this:
test1 - data.frame(intx=c(4,3,1,1,2,2,3),
status=c(1,1,1,0,1,1,0),
x1=c(0,2,1,1,1,0,0),
x2=c(1,1,0,0,2,2,0),
sex=c(0,0,0,0,1,1,1))
and I can easily fit a cox
Are you for some reason against writing your function to accept a single
argument, a formula, that you simply pass on to coxph?
Mendolia, Franco wrote:
Hello!
I have something like this:
test1 - data.frame(intx=c(4,3,1,1,2,2,3),
status=c(1,1,1,0,1,1,0),
Dear R users,
I am trying to simulate some multitrait-multimethod models using the
packages sem and psych but whatever I do to deal with models which do not
converge I always get stuck and get error messages such as these:
Error in summary.sem(M1) : coefficient covariances cannot be computed
Dear R users,
I am trying to simulate some multitrait-multimethod models using the packages
sem and psych but whatever I do to deal with models which do not converge I
always get stuck and get error messages such as these:
Error in summary.sem(M1) : coefficient covariances cannot be computed
Frank Harrell f.harr...@vanderbilt.edu 11/08/2010 17:02:03
This problem seems to cry out for one of the many available robust
regression methods in R.
Not sure that would be much more appropriate, although it would
_appear_ to work. The PB method is a sort of nonparametric/robust
approach
Dear R-help list users,
I have a ts object and I would like to make a new one that contain mean by year.
Is there a direct way of do it?
Thks,
Sebastián.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
Hi,
Is it possible to do a Matrix Plot and in the cell perform a linear
regression also adding to the cell the r2 and the equation. If so, how?
Thanks,
As hz
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Hi As hz,
The following might get you started:
http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=137
Also, see ?pairs.
HTH,
Jorge
On Wed, Aug 11, 2010 at 1:34 PM, ashz wrote:
Hi,
Is it possible to do a Matrix Plot and in the cell perform a linear
regression also adding to
I could do that. However, the function f that I mentioned below is part of a
bigger program and is nested inside another function, say function A. In
function A I determine the covariates that I want to use and then call my
function f. So even if I use a formula as single argument, I would
Hi. I need to convert data to time series.
My problem is that when I converted my data to time series the data comes
with time. I just want the date because doesn't work with blockMaxima().
data(bmwRet)
BMW=as.timeSeries(bmwRet)
head(BMW)
GMT
BMW.RET
1973-01-02 0.047704097
1973-01-03
On Aug 11, 2010, at 1:53 PM, Mendolia, Franco wrote:
I could do that. However, the function f that I mentioned below is
part of a bigger program and is nested inside another function, say
function A. In function A I determine the covariates that I want to
use and then call my function
I may be missing something but I don't see how PB handles errors in
variables any differently than other regression methods that ignore
this problem.
Frank
Frank E Harrell Jr Professor and ChairmanSchool of Medicine
Department of Biostatistics Vanderbilt
Thanks for your hint. This works:
form - formula(paste(Surv( time, event,) ~,paste(covar,
collapse=+),+strata(stratum), sep= ) )
Franco
From: David Winsemius [dwinsem...@comcast.net]
Sent: Wednesday, August 11, 2010 1:05 PM
To: Mendolia, Franco
Cc: Erik
On Wed, Aug 11, 2010 at 12:22 PM, steven mosher mosherste...@gmail.com wrote:
Given a long zoo matrix, the goal is to sweep out a statistic from the
entire length of the
sequences.
longzoomatrix-zoo(matrix(rnorm(720),ncol=6),as.yearmon(outer(1900,seq(0,length=120)/12,+)))
This shoul work:
coxph(Surv(intx, status) ~ . + strata(sex) - sex, test1)
On Wed, Aug 11, 2010 at 1:54 PM, Mendolia, Franco fmendo...@mcw.edu wrote:
Hello!
I have something like this:
test1 - data.frame(intx=c(4,3,1,1,2,2,3),
status=c(1,1,1,0,1,1,0),
Hi Adrienne,
I guess apply should be better than for loop.
Code like this:
event.gen2 = function(genmat,use1,use2,num,ortho_obs_used){
onerow.gen - function(one.row, use1){
one.row[num] - ifelse(...}
genmat[,num] - NA ##Add one row with NA values
suppose covnames is a vector containing your covariate names (e.g. as
character strings)
Then (warning: untested)
rhs - paste(covnames, collapse=+)
## makes them into a single string separated by +-es and
form - formula(paste(y, rhs, sep=~))
## creates your formula.
?substitute can also be
The colMeans comes closest,
for a single series the assume you have 100 years of monthly data.
The mean you want to scale by is the mean for a restricted period in the
center
of the series.. say 1950-1960
for this period you have the average jan (1950-1960) average feb, ect.
your final series
On Wed, Aug 11, 2010 at 2:38 PM, steven mosher mosherste...@gmail.com wrote:
The colMeans comes closest,
for a single series the assume you have 100 years of monthly data.
The mean you want to scale by is the mean for a restricted period in the
center
of the series.. say 1950-1960
for this
Dear R-help list:
I have want seems to be a simple task, but can't find a solution.
I tried querying the archives, but did not succeed.
My goal is to set up two (or more) graphs in one graph window, and
then during execution of a sequential algorithm add points
sequentially to these graphs. The
What is wrong with using a loop?
It used to be that loops were much slower than some of the alternatives, but
now days a well crafted loop runs almost as fast (sometime faster) than the
apply functions. So if the loop is working for you, use it and don't worry
about it (though there may be
Dear all,
I have some growth curve data from an experiment that I try to fit using
lm and lmer. The curves describe the growth of classification accuracy
with the amount of training data t, so basically
y ~ 0 + t (there is no intercept because y=0 at t0)
Since the growth is somewhat nonlinear
If it were just one loop by itself and I was doing the calculation for just
one month for all of my 317 stations, I would agree with. However, this
function itself is inside another loop which goes through each month and
year that I need the calculation for each of the stations.
If you have any
The simplest thing to do is just recreate the plot(s) from scratch at each
iteration with all the accumulated data. Most graphics devices are fast enough
now that you will not notice the redraw. If you really just want to add to the
plots, you can save and restore the graphics parameters,
Hi, sorry I've gotten around the issue, I there was an error with the
document format itself, I managed to correct by making a new file.
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Sent from the R help mailing list archive at Nabble.com.
On Wed, Aug 11, 2010 at 2:44 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Wed, Aug 11, 2010 at 2:38 PM, steven mosher mosherste...@gmail.com wrote:
The colMeans comes closest,
for a single series the assume you have 100 years of monthly data.
The mean you want to scale by is the
Classification accuracy is an improper scoring rule, and one of the
problems with it is that the proportional classified correctly can be
quite good even if the model uses no predictors. [Hence omitting the
intercept is also potentially problematic.]
Frank E Harrell Jr Professor and
Dear Gui,
Because you're wrapping the calls to summary() in try(), the errors there
shouldn't stop your simulation. It probably makes sense to test whether the
object returned by try() is of class try-error, and possibly to specify
the argument silent=TRUE, examining the error messages in your
Hi. I need to convert data to time series.
My problem is that when I converted my data to time series the data comes
with time. I just want the date because doesn't work with blockMaxima().
data(bmwRet)
BMW=as.timeSeries(bmwRet)
head(BMW)
GMT
BMW.RET
1973-01-02 0.047704097
1973-01-03
*What I want to do:
*Create a windows shortcut that will start the R gui **and**
simultaneously source a file
*What I have already tried:
*This almost works, but it's not the interactive R GUI:
R --no-save --sdi -file=C:\SomePath\example.R
These open the R GUI, but doesn't recognize -f --f
Gene Leynes wrote:
*What I want to do:
*Create a windows shortcut that will start the R gui **and**
simultaneously source a file
*What I have already tried:
*This almost works, but it's not the interactive R GUI:
R --no-save --sdi -file=C:\SomePath\example.R
These open the R GUI, but
On Aug 11, 2010, at 4:13 PM, Gene Leynes wrote:
*What I want to do:
*Create a windows shortcut that will start the R gui **and**
simultaneously source a file
Can't you set the file associations to start up RGUI when you launch
an .R file?
*What I have already tried:
*This almost
On 11/08/2010 4:13 PM, Gene Leynes wrote:
*What I want to do:
*Create a windows shortcut that will start the R gui **and**
simultaneously source a file
*What I have already tried:
*This almost works, but it's not the interactive R GUI:
R --no-save --sdi -file=C:\SomePath\example.R
These
Hi,
?plot will help you.
plot(0:18,col=white)
points(0:18,0:18,pch=0:18,col=rainbow(19),cex=2)
Turn amp; Fall wrote:
How do you change the standard plotting function so that points are solid
and can have a pretty colour?
-
A R learner.
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View this message in context:
It seems very likely you are working on a 32-bit version of R, but it's a
little surprising still that you would have a problem with any single year.
Please tell us the operating system and version of R. Did you preprocess
the airline CSV file using the utilities provided on bigmemory.org? If
Hi,
Try ?unique please.
x - c(2,2,9,4,6,2,4,4,6,8,6) # Original vector
unique(x) #New vector only has unique elements
sort(unique(x)) # Ordered
Regards,
Wu
-
A R learner.
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Sent
Hi, folks,
##
food=c('fruit','fruit','fruit','drink','drink','drink')
type=c('apple','apple','orange','water','soda','soda')
value=c(2,3,1,5,7,6)
data=data.frame(food,type,value)
share=c((2+3)/(2+3+1),5/6,1/6,5/(5+7+6),13/18,13/18)
I'll respond to everyone, but Duncan's response worked, so I'll put that
first. Thank you *everyone* for responding, I really appreciate it
Duncan:
Great idea, it's simple and works. Using the --args causes the final flags
to be ignored by R, but you can still retreive them using commandArgs()
Hi listers,
I made some search, but i didn`t find in the forum.
I have a data set.
I would like to make a search (conditon) on my data set.
x-c(1,2,3,4,5,6,7,8,9,10)
count-0
if (CONDITON){count-1}else{count-0}
My CONDITION would be: is there number 5 in my data set?
Thanks in advance,
Marcio
?match, look at the %in% operator.
Mestat wrote:
Hi listers,
I made some search, but i didn`t find in the forum.
I have a data set.
I would like to make a search (conditon) on my data set.
x-c(1,2,3,4,5,6,7,8,9,10)
count-0
if (CONDITON){count-1}else{count-0}
My CONDITION would be: is there
I have formulae with coefficents that I would like to update. However, I get
some strange results. For example, see the following:
For the formula y ~ d+ 3*r+t I want to add a variable p, so
update(y~d+0*r+t, .~.+p)
produces
y ~ d + t + p - 1
If the coefficient is not 0, but rather,
On Aug 11, 2010, at 6:03 PM, Jarrett Byrnes wrote:
I have formulae with coefficents that I would like to update.
However, I get some strange results. For example, see the following:
For the formula y ~ d+ 3*r+t
Did you really get meaningful results from that formula? Care to
provide
Hi Yi,
It would be helpful for others to provide a solution if you give your
formulas that calculating the value of share and concentration.
?apply will helps.
Regards,
Wu
-
A R learner.
--
View this message in context:
Thank you for reminding me.
##
food=c('fruit','fruit','fruit','drink','drink','drink')
type=c('apple','apple','orange','water','soda','soda')
value=c(2,3,1,5,7,6)
data=data.frame(food,type,value)
share=c((2+3)/(2+3+1),5/6,1/6,5/(5+7+6),13/18,13/18)
On Wed, 11 Aug 2010, David Winsemius wrote:
On Aug 11, 2010, at 6:03 PM, Jarrett Byrnes wrote:
I have formulae with coefficents that I would like to update. However, I
get some strange results. For example, see the following:
For the formula y ~ d+ 3*r+t
Did you really get meaningful
On Aug 11, 2010, at 6:45 PM, Charles C. Berry wrote:
On Wed, 11 Aug 2010, David Winsemius wrote:
On Aug 11, 2010, at 6:03 PM, Jarrett Byrnes wrote:
I have formulae with coefficents that I would like to update.
However, I get some strange results. For example, see the
following:
For
oh, god, please don't tell anybody...
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R-help@r-project.org mailing
# Two test for a class like this:
setClass(XXX,
representation=representation(
matrix
)
)
i-new(XXX);
m=matrix();
colnames(m)-c(colA);
i...@.data=m;
# i
# An object of class “XXX”
# colA
#[1,] NA
#
#First Test
Hi,
I wanted to generate a plot which is almost like the plot generated by the
following codes.
category - paste(Geographical Category, 1:10)
grp1 - rnorm(10, mean=10, sd=10)
grp2 - rnorm(10, mean=20, sd=10)
grp3 - rnorm(10, mean=15, sd=10)
grp4 - rnorm(10, mean=12, sd=10)
mydat -
Dear All,
I tried to plot contour lines using R function contour, but got the results
which are not expected.
require(RTOMO)
x - seq(-1,1,0.1)
y - seq(-1,1,0.1)
xy - meshgrid(x,y)
z - xy$x^2+ 3*xy$y^2
contour(x,y,z,col=blue,xlab=x,ylab=y)
The above code gave me the contour graph for
I didn't really know what to post as the topic subject, but I have a vector,
for instance (2,2,4,6,2,4,4,6,8,6) and I want to create another vector which
is just numbers from 1 to 4 since there are only 4 unique numbers in my
vector, so for instance 2 would be 1, 4 would be 2, 6 would be 3, and 8
Simple question.
How do you change the standard plotting function so that points are solid
and can have a pretty colour?
Thanks for the help
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an error in the document format? Good job figuring it out. Did you submit a
bug report?
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Hi,
i'm searching for an R library able to calculate the Regression Error
Characteristic Curves
as explained in:
On Wed, 11 Aug 2010, David Winsemius wrote:
On Aug 11, 2010, at 6:45 PM, Charles C. Berry wrote:
On Wed, 11 Aug 2010, David Winsemius wrote:
On Aug 11, 2010, at 6:03 PM, Jarrett Byrnes wrote:
I have formulae with coefficents that I would like to update. However,
I get some
Hi,
if I use the jpeg driver directly like
one - read.table(sample.txt,sep=\t)
jpeg(file=sample.JPG)
plot(V2 ~ V1,one)
and it did not plot anything! why is that?
the version of ghostscript that I installed i from the following link
http://pages.cs.wisc.edu/~ghost/doc/GPL/gpl871.htm
and
Hi,
The code seems complicate, but it's understandable.
##
food=c('fruit','fruit','fruit','drink','drink','drink')
type=c('apple','apple','orange','water','soda','soda')
value=c(2,3,1,5,7,6)
data=data.frame(food,type,value)
share=c((2+3)/(2+3+1),5/6,1/6,5/(5+7+6),13/18,13/18)
I found some amazing package called lqa
one question!!!
Can I use this package in linaer model(not GLM)?
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Hey folks,
I have a graphic of dispersion...
I would like to color some specific points (x,y) according to another
indicator variable, for example:
x-rnorm(5)
y-rnorm(5)
ind-(1,0,0,1,0)
plot(x,y)
Any suggestion, thanks...
Marcio
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