Dear Rohit
On 3 May 2011 22:53, Rohit Pandey wrote:
> Hello R community,
>
> I have been using R's inbuilt maximum likelihood functions, for the
> different methods (NR, BFGS, etc).
>
> I have figured out how to use all of them except the maxBHHH function. This
> one is different from the others
Accelerated Failure Time Model to Proportional Hazard Form
Greetings R users:
I have been working on a problem for a while and can't seem to get any result.
I am trying to convert accelerated failure time estimates to proportional
form.
I keep getting an error that I can't understand and do
Why the mean value " h" is not changing as the slider moves from 0 to 25 ?.
It remains always constant.
library(manipulate)
example <- function(x.max){
plot(cars, xlim=c(0,x.max))
abline(h=mean(cars$dist),col="blue",lty=2) }
manipulate(
example(x.max), x.max=slider(0,25, step=5) )
veepsirtt
the original data are
V2 =c(371000,285000 ,156000, 20600, 4420, 3870, 5500 )
T2=c( 0.3403 ,0.4181 ,0.4986 ,0.7451 ,1.0069 ,1.553)
nls2=nls(V2~v0*(1-epi+epi*exp(-cl*(T2-t0))),start=list(v0=10^7,epi=0.9,cl=6.2,t0=8.7))
after execution error occurs as below
Hi Steven,
Thanks for the quick reply. i have tried but
its giving me error--->Error in optim(x = c(38.1815173696765,
-12.7988197976440, -3.88212459045077, :
initial value in 'vmmin' is not finite
i have tried something like this:
library(MASS)
x<-rnorm(n=100,mean=10,sd=20);
fitdistr(x,dbeta,
Hi,
How i can install the package "adapt" in some version of R for mac?
i try in 2.13, 2.9,2.7 and other previous versions... and nothing happens.
and another question: There are some packages that do the same but that it
is implemented for mac? (calculate integrals in 2 or more dimmensions).
h
On May 3, 2011, at 21:18 , Kalicin, Sarah wrote:
>
> I have a work around for this, but can someone explain why the first example
> does not work properly? I believed it worked in the previous version of R, by
> selecting just the rows=200525 and omitting the na's. I just upgraded to
> 2.13.
On May 3, 2011, at 10:03 PM, Usha wrote:
Thanks for the help.
I would like to explain my problem.
I have sample of scores from tests which varies form 0 to 35.
Now, i want to find out the best fit distribution for this data. I
need to
order the distributions based on their best fit.
For this
Thanks for the help.
I would like to explain my problem.
I have sample of scores from tests which varies form 0 to 35.
Now, i want to find out the best fit distribution for this data. I need to
order the distributions based on their best fit.
For this i am using the function fitdistr(). [One of th
Hi all ,
1. In Laavan the package when you have two variables a and b which have a
direct effect on one another for example like this: (a â b), To write this in
the model we must write how:a ~~b Or we must write two equations:a ~ bb ~ a
in this case there is some information to add for the res
Kalicin, Sarah wrote:
\begin{quote}
I have a work around for this, but can someone explain
why the first example does not work properly?
I believed it worked in the previous version of R,
by selecting just the rows=200525 and omitting the na's.
\end{quote}
You can prove this statement by providin
maxBHHH is *not* an in-built R function. It is in a distributed package called
"maxLik". Always tell us which package is being used so that it is easier for
us to help you.
The error message says that the gradient function is returning a 10 x 2 matrix,
whereas you say that you have 1000's of
At 10:15 04/05/2011, you wrote:
Content-Type: text/plain
Content-Disposition: inline
Content-length: 537
Hi everyone,
I would like to improve my plot and I was
wondering if someone can help me whith it. I'm
trying this plot using two groups, but I want to
choice the colors (the black and whi
Hi,
I have no familiarity with these functions --- I see that they are not
in base R --- so I suggest that at very least you identify the package
that you are using. Better would be to contact the package maintainer
directly. Sometimes maintainers do not read R-help.
Cheers
Andrew
On Tue, May
Hi everyone,
I would like to improve my plot and I was wondering if someone can help me
whith it. I'm trying this plot using two groups, but I want to choice the
colors (the black and white circles) but I don't know how to change it from
here. These are my sentences:
myplot3d<- scatterplot3d(
On 05/03/2011 03:23 PM, Ben Rhelp wrote:
Hi all,
I am trying to compile Rgraphiz on Windows 7 64bit with R-2.13.0. I have
installed
Rtools213.exe from [1]. The 64bit packages in [2] provided me with the 64 bit
version
of graphviz. After intalling the binary version Rgraphviz 1.30 (in 32bit) it
Does anyone know how to write the code for an anova for a generalized
randomized block design? I have two blocks (random) and three treatments
(fixed). Each treatment has two reps at each site. I know that for a RCBD
with no replication that an anova can be run using aov(object~Block *Trt),
but
Hi Sarah,
I'm not sure that I understand your problem. You have shown us three
ways to try to omit missing values, and one of them seems to work.
But you're concerned because some aspect of it doesn't match the ones
that don't work? But they don't work!
I wonder if you could send an example i
Try the function
rownames()
Andrew
On Tue, May 03, 2011 at 03:29:37AM -0700, agent dunham wrote:
> Dear community,
>
> I uploaded an excel with read.xls. My xls file actually have a column which
> is an id, ("plot" is the id) :
>
> plot height area
> 347.6 5.4
> 853.2 4.1
> 8
I suggest that you provide some commented, minimal, self-contained,
reproducible code.
Cheers
Andrew
On Wed, May 04, 2011 at 02:23:29AM +0530, Rohit Pandey wrote:
> Hello R community,
>
> I have been using R's inbuilt maximum likelihood functions, for the
> different methods (NR, BFGS, etc).
>
Your interpretation of what the output is supposed to look like is
actually correct. Take a look at the estimates of the bias in the
BootStrap Statistics. You will see that they are the same as the
difference between the location of colMeans of t and t0.
I hope that this helps,
Andrew
On Tue,
Uwe,
It think the problem turned out to be with my "profile.site" file.
As I recall, temporarily deleting / renaming it was the simple fix. I think
it had to do with something I use to parse the commandArgs
Thank you for your reply, and sorry for not responding sooner. I didn't
think anyone re
Hi all,
I am trying to compile Rgraphiz on Windows 7 64bit with R-2.13.0. I have
installed
Rtools213.exe from [1]. The 64bit packages in [2] provided me with the 64 bit
version
of graphviz. After intalling the binary version Rgraphviz 1.30 (in 32bit) it
complains (as
expected) that:
> libra
I'm not clear on what you're looking for here. Your x-axis is numeric, why are
you converting it to a factor? If you keep it numeric, the labels don't
overlap. Or perhaps you don't want it to be numeric, in which case why not
just change the aspect ratio of the plot until they no longer overl
Hi Richard,
Thanks for your advice.
I think that your suggestion is that I run the ANOVA with Combined.Plot as a
factor. I have tried that does not alleviate the problem.
Did I understand you properly?
Do you have another idea?
Thanks,
David
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View this message in context:
http://r.789695.
Hi R users
I apologise in advance for this question as I suspect it is simple and
perhaps others have had this problem.
I am struggling to sort out how to fix the x axes so that the labels
don't overlap.
I have put the following example together to show my problem.
library(lattice)
titre <- as
On 03/05/2011 4:43 PM, Jun Shen wrote:
Dear list,
This may sound silly. What is the right way to change the names of a
dataframe? Let's say I have this data frame (dose) with four columns with
names "ID", "DOSE", "TIME" "CMT". I want to change "DOSE" to "AMT". So I did
names(dose[2])<-'AMT'
Bu
Hi,
The package maintainer is aware of this feature request. In the
meantime, I've used Currying,
require(cubature)
f <- function(x, a) cos(2*pi*x*a) # a simple test function
adaptIntegrate(roxygen::Curry(f, a=0.2), lower=0, upper=2)
HTH,
baptiste
On 4 May 2011 05:57, Ravi Varadhan wrote
Hello R community,
I have been using R's inbuilt maximum likelihood functions, for the
different methods (NR, BFGS, etc).
I have figured out how to use all of them except the maxBHHH function. This
one is different from the others as it requires an observation level
gradient.
I am using the foll
In the first case you create a new data frame consisting of the 2nd column of
the original, then change the name of the only column in that new data frame,
then since nothing is done with that data frame it gets thrown away. So it is
not that nothing happened, but just that nothing useful happe
Hello!
I have data that contain, among other things the date for the
beginning and for the end of a (daily) time series (see example below
- "mydata")
mystring1<-c("String 1", "String 2")
mystring2<-c("String a", "String b")
starts<-c(as.Date("2011-02-01"),as.Date("2011-03-02"))
ends<-c(as.Date("
Dear list,
This may sound silly. What is the right way to change the names of a
dataframe? Let's say I have this data frame (dose) with four columns with
names "ID", "DOSE", "TIME" "CMT". I want to change "DOSE" to "AMT". So I did
names(dose[2])<-'AMT'
But nothing happened. The name of the secon
library(MASS)
fitdistr(x,"beta",list(shape1=1,shape2=1))
On Tue, May 3, 2011 at 9:44 PM, Shekhar wrote:
>
> Hi,
> I have some random data and i want to find out the parameters of Beta
> distribution ( a and b) such that this data approximately fits into
> this distribution. I have tried by plot
hi all - i'm trying to 'R CMD build' a package, but i have what
appears to be a bootstrapping problem:
i've included a vignette in my package, with R code interwoven (and
built using Sweave), but in this documentation i have a code line:
> library(MyPackage)
now, when trying to build a .tar.gz ins
Most likely your combined.trt is linearly dependent on the combined.plot
factor. Try
Anova.Trt.D.M.T.Pr.Model <- aov(Combined.Rs ~ as.factor(Combined.Plot)
.
and see if combined.plot now has the 11 df you are anticipating.
Rich
On Tue, May 3, 2011 at 3:37 PM, Rovinpiper wrote:
> I'm runn
NEW SECTION BEGINS JUNE 1 (11AM-2PM ET) AND MAY 27 (6PM-9PM ET)
The non-profit organization Information Institute (
http://www.information-institute.org) is offering a live, interactive,
synchronous online course entitled Fundamentals of Using R. The registration
cost for the 14-hour, 5 week cou
I'm running an ANOVA on some data for respiration in a forest. I am having a
problem with my degrees of freedom. For one of my variables I get one fewer
degrees of freedom than I should.
I have 12 plots and I therefore expected 11 degrees of freedom, but instead
I got 10.
Any ideas?
I have some
I am attempting to use package boot to summarize and compare the performance
of three models. I'm using R 2.13.0 in a Win32 environment.
My statistic function returns a vector of 6 values, 3 of which are error
rates for different models, and 3 are pairwise differences between those
error rates.
On Apr 28, 2011, at 15:18 , JP wrote:
>
>
> I have found that when doing a wilcoxon signed ranked test you should report:
>
> - The median value (and not the mean or sd, presumably because of the
> underlying potential non normal distribution)
> - The Z score (or value)
> - r
> - p value
>
.
On Mon, May 2, 2011 at 5:38 PM, Mike Harwood wrote:
> Hello,
>
> I am apparently confused about the use of an id parameter for an event
> history/survival model, and why the EHA documentation for aftreg does
> not specify one. All assistance and insights are appreciated.
Which version of eha are
Hi:
Here are two more candidates, using packages plyr and data.table. Your
toy data frame is called dd below.
library(plyr)
ddply(dd, .(Site, Prof), transform, Hadj = H - min(H))
Site Prof H Hadj
111 248
211 160
311 67 51
412 230
512 56 3
I have a work around for this, but can someone explain why the first example
does not work properly? I believed it worked in the previous version of R, by
selecting just the rows=200525 and omitting the na's. I just upgraded to 2.13.
I am also concern with the row numbers being different in the
William, you are right. Thanks for clarification.
Andrija
On Tue, May 3, 2011 at 9:04 PM, William Dunlap wrote:
>
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
> > -Original Message-
> > From: r-help-boun...@r-project.org
> > [mailto:r-help-boun...@r-project.org] On B
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of andrija djurovic
> Sent: Tuesday, May 03, 2011 11:28 AM
> To: Woida71
> Cc: r-help@r-project.org
> Subject: Re: [R] Sim
Hi,
I've noticed that the usual "break", "next" commands do not work in a
foreach loop, is there a nice way to do that?
A little more detail: I am using foreach to conduct a very time
consuming (may take several days if done sequentially) simulation
study. The number of simulations is set to 100
I have a FORTRAN DLL file obtained from Compaq Visual Fortran and when I try
to load the DLL into the R environment I get an error.
> dyn.load("my_function.dll")
"This application has failed to start because MSCVRTD.dll was not found.
Re-installing this application may fix the problem."
When I
Dear Stuart,
See ?bcPower and ?powerTransform in the car package, the latter for
univariate and multivariate conditional and unconditional ML Box-Cox.
I hope this helps,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department of Sociol
Hi.
There is no need to do this in a for loop.
Here is one approach:
x <- read.table(textConnection("Site Prof H
1 1 24
1 1 16
1 1 67
1 2 23
1 2 56
1 2 45
2 1 67
2 1 46"), header = TRUE)
closeAllConnections()
x
cbind(x,newCo
It is actually possible and preferable to do this with no loops.
Assuming your data is in a dataframe called dat:
idx <- with(dat, Site == 1 & Prof == 1)
dat <- within(dat, { new = H - ifelse(Site == 1 & Prof == 1,
min(H[idx]), min(H[!idx])) })
dat
which also serves to illuminate the difference b
On 03/05/2011 1:26 PM, Ryan Utz wrote:
Well... that could work. Problem is in the actual graphs I'm making, there
are to be>30 lines per graph (as many as 60 in some cases). Any way I could
use the lines command without having to write out 60 lines of code per
figure? That's why I like ablines; y
Ok, I get it.
require(cubature)
f <- function(x, a) cos(2*pi*x*a) # a simple test function
# this works
a <- 0.2
adaptIntegrate(function(x, argA=a) f(x, a=argA), lower=0, upper=2)
# but this doesn't work
rm(a)
adaptIntegrate(function(x, argA=a) f(x, a=argA), lower=0, upper=2, a=0.2)
Ravi.
__
?barplot
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Caitlin
> Sent: Tuesday, May 03, 2011 11:36 AM
> To:
On Tue, May 3, 2011 at 10:40 AM, Joshua Wiley wrote:
> The issue is the plotting region is slightly padded. The easiest
> option, I think, would be to clip() it. I have a general sense that
> one of the par() options would let you adjust the padding to 0, but I
> could just be imagining that (an
Hi Ryan,
The issue is the plotting region is slightly padded. The easiest
option, I think, would be to clip() it. I have a general sense that
one of the par() options would let you adjust the padding to 0, but I
could just be imagining that (anyone else??). Anyway, here are some
options:
###
p
There is the bct function in the TeachingDemos package that does Box-Cox
transforms (though you could also write your own fairly simply). The
lappy/sapply functions will apply a function to each column of a data frame.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healt
Hi all.
I need to construct a plot showing words on the x-axis and how many times
each word was given as a verbal response on the y-axis as solid bar
(frequency). Is there a convenient function to do this in R? I considered
hist(), but I'm not sure how to construct the text file. Example:
apple,
Hi again,
Now that I have the data.frame as ordered factors, when I try to transpose
it, I lose the factor orders.
> datfact<-data.frame(c1,c2,c96)
> sapply(datfact, class)
c1c2c96
[1,] "ordered" "ordered" "ordered"
[2,] "factor" "factor" "factor"
>
> dafacT<-as.data.
I've had the same problem and ended up using the xlsReadWrite package. It
takes more time to import a sheet but does have the colClasses command.
Following your example:
library(xlsReadWrite)
read.xls("testtable", sheet = "sheet1", colClasses="character")
should worked, it did for me
--
View t
Dr. Ligges,
Thanks a lot for providing syntax for passing additional parameters. It
worked for me and has solved my problem.
Many thanks for your quick help.
HC
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View this message in context:
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Dear R-helpers,
I used the step.gam function (package gam, Trevor Hastie) on a data frame
without problems. Then I created a list of several bootstrap samples from this
data frame. Now I want to use the step.gam function on this list using a
for-loop. The code is working well until the step.gam
Check your par() settings, specifically "xpd". For more control see ?clip. If
that does not do enough for you then use lines or segments for complete control.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Mes
Hello everybody,
I am beginning with loops and functions and would be glad to have help in
the following question:
If i have a dataframe like this
Site Prof H
1 1 24
1 1 16
1 1 67
1 2 23
1 2 56
1 2 45
2 1 67
2 1 46
An
Hi
Could any one please help how I can trnasform data based on Box-Cox
Transformations. I have massive data set with many variables. If
possible someone can write few lines so I can read in all data set
once and transform it.
g1 g2 g2
97.03703704 89.25925926 4.4
24.9074
Well... that could work. Problem is in the actual graphs I'm making, there
are to be >30 lines per graph (as many as 60 in some cases). Any way I could
use the lines command without having to write out 60 lines of code per
figure? That's why I like ablines; you just have to specify a single value
a
Hi,
g.data is perhaps a interesting package for you.
HTH,
Christian
Dear List members,
I would like to load R objects saved as RData file but ran into the problem
that these objects are too large for my RAM ('Can not allocate vactor of size
XX...'). Switching to a Linux machine is no opti
On Tue, 2011-05-03 at 10:36 -0600, Ryan Utz wrote:
> Hi all,
>
> I'm attempting to make a quite-specific plot where the axes cross at the
> origin and with gridlines for guidance. I've been using ablines to create
> the reference lines because I want a lot of control as to where they are
> placed
Hi all,
I'm attempting to make a quite-specific plot where the axes cross at the
origin and with gridlines for guidance. I've been using ablines to create
the reference lines because I want a lot of control as to where they are
placed on the axis. This command works very well for such control.
How
?cumsum
On Tue, May 3, 2011 at 11:32 AM, Alaios wrote:
> Dear all,
> I would like to know what is the most time efficient way to calculate the
> following in a huge vector.
>
> Let's say that I have the vector
>
> 1,2,3,4,5,6
> and I want to return a vector of the same length which every cell co
Thanks, Phil.
I could have sworn that I tried that (several times).
It works perfectly, of course.
Thanks again,
Robert
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Sent from the R help mailing list archive
Dear all,
I would like to know what is the most time efficient way to calculate the
following in a huge vector.
Let's say that I have the vector
1,2,3,4,5,6
and I want to return a vector of the same length which every cell containing
the sum of the previous cells like this
1, 1+2, 1+2+3, 1+2+3
I saw the format of the caret data some days ago. It is possible to convert
my csv data with the same data a format as the caret dataset. My idea is to
use firstly the same scripts as caret tutorial, then i want to remove
problems related with data formats and incompatibilities.
Thanks for your t
Dear List members,
I would like to load R objects saved as RData file but ran into the problem
that these objects are too large for my RAM ('Can not allocate vactor of size
XX...'). Switching to a Linux machine is no option, neither is raising the
memory limit.
I am now wondering whether it i
Your simulation example is bad. You cannot fit a beta distribution to a data
that is not in [0,1], leave alone negative data.
x <- runif(1007)
fitdistr(x, "beta", start=list(shape1=0.5, shape2=0.5))
But try this instead:
x <- runif(100, 1, 27)
fitdistr(x, "beta", start=list(shape1=0.5, shap
On Tue, 3 May 2011, Usha wrote:
Please guide me through to resolve the error message that I get
this is what i have done.
x1<- rnorm(100,2,1)
x1fitbeta<-fitdistr(x1,"beta")
Error in fitdistr(x1, "beta") : 'start' must be a named list
You have many errors, starting with not reading the post
See the examples at the end of:
http://cran.r-project.org/web/packages/caret/vignettes/caretTrain.pdf
for a QSAR data set for modeling the log blood-brain barrier
concentration. SVMs are not used there but, if you use train(), the
syntax is very similar.
On Tue, May 3, 2011 at 9:38 AM, yprive
On May 3, 2011, at 7:10 AM, David Winsemius wrote:
On May 3, 2011, at 6:12 AM, swaraj basu wrote:
Hello Everyone,
I am having problem in defining specific
axis for
plotting a vactor.
vecAVG <- c(0.2, 0.4, 0.6, 0.2, 0.4)
names(vecAVG)<-c("br
On May 3, 2011, at 6:12 AM, swaraj basu wrote:
Hello Everyone,
I am having problem in defining specific
axis for
plotting a vactor.
vecAVG <- c(0.2, 0.4, 0.6, 0.2, 0.4)
names(vecAVG)<-c("brain","heart","kidney","lung","blood")
well, first of all thank for your answer. I need some example that works with
Support Vector Regression. This is the format of my data:
VDP V1V2
9.15 1234.5 10
9.15 2345.6 15
6.7789.0 12
6.7234.6 11
3.2 123.6 5
3.2 235.7 8
VDP is t
The expression
0:g_range[2]
is not meaningful. The : operator is for integers, while your data is
continuous. Likely, you want something along the lines of
axis(2, las=1, at=pretty(vecAVG))
___
Patrick Breheny
Assistant Professor
Department of Biostatistics
Department of
I've changed to the hard path in the PATH environment variable in Windows and
it works.
If you are talking about the search path in R, I do not have a clue on how to
test it.
Regards,
Stefan
-Oprindelig meddelelse-
Fra: Jonathan Daily [mailto:biomathjda...@gmail.com]
Sendt: 3. maj 2011
I am fairly new to R and have a (for me) slightly complicated set of data to
analyse. It contains several continuous and categorical variables for a
group of individuals – e.g;
ID Sex Age Familysize Phone Education
1 M 23 3 Yes Primary
Dear community,
I uploaded an excel with read.xls. My xls file actually have a column which
is an id, ("plot" is the id) :
plot height area
347.6 5.4
853.2 4.1
895.4 8.4
121 6.76.2
...
1325 2.11.5
However R uses another id, this way:
r id plot height are
I set the path with no spaces and run as administrator, but the problem is
not fixed. I'm not quite shure, but I can't remember the problem bevore R
version 2.13. Could it be, that the new R version causes that problem?
Frank Lehmann
-Ursprüngliche Nachricht-
Von: Uwe Ligges [mailto:lig..
Hi,
I have a erdos-renyi game with 6000 nodes and probability 0.003.
g1 = erdos.renyi.game(6000, 0.003)
How to create a Watts Strogatz game with the same probability.
g1 = watts.strogatz.game(1, 6000, ?, ?)
What should be the third and fourth parameter to this argument.
--
View this message i
On 03.05.2011 15:05, Richard Wang wrote:
Thanks. I didn't know that. I just found it in Brian Ripley's page. Is this
the cran extras?
Right, and under Windows it is a default repository.
Uwe Ligges
Thanks
Richard
On 3 May 2011, at 13:19, Uwe Ligges wrote:
On 03.05.2011 14:09,
So I have been playing with bubble plot and I was able to create bubble plot
with relatively simple data. I am no having problems when I increase the
complexity of my data. Below is a example of my data:
phytochemicalMainClassFitValues
Name A 0.5
Thanks for all your help and I apologize for not being clear in the
beginning. I will try the "group lasso" packages. From the paper, it
seems like that is what I want to do. Thanks again!
On Tue, May 3, 2011 at 2:40 AM, Nick Sabbe wrote:
> For performance reasons, I advise on using the followi
Hi,
I have some random data and i want to find out the parameters of Beta
distribution ( a and b) such that this data approximately fits into
this distribution. I have tried by plot the histograms and graph, but
it requires lot of tuning and i am unable to do that. can anyone tell
me how to do it p
Please guide me through to resolve the error message that I get
this is what i have done.
>x1<- rnorm(100,2,1)
>x1fitbeta<-fitdistr(x1,"beta")
Error in fitdistr(x1, "beta") : 'start' must be a named list
Yes, I do understand that sometime for the distribution to converge to the
given set of data
Hello Everyone,
I am having problem in defining specific axis for
plotting a vactor.
vecAVG <- c(0.2, 0.4, 0.6, 0.2, 0.4)
names(vecAVG)<-c("brain","heart","kidney","lung","blood")
par(mar=c(12,4.1,4.1, 2.1))
plot(sort(
On May 3, 2011, at 4:24 AM, Uwe Ligges wrote:
On 03.05.2011 13:16, nuncio m wrote:
Hi list,
I have a matrix with all elements of some columns are zeroes. Is it
possible to remove these columns:
Xnew <- X[ , as.logical(colSums(X)), drop=FALSE]
A counter-example:
X <- matrix(c(1,1,1,-1
On May 3, 2011, at 2:10 AM, Albert-Jan Roskam wrote:
Hi Jeff,
Ah, thanks a lot! Yes, meanwhile I also switched to csv. This still
requires
knowledge about the regional settings (Sys.getlocale), but it's a
lot more
transparent.
I'm quite new to R and I must say that stuff like this is ea
On 11-05-03 09:23 AM, Jurgens de Bruin wrote:
> So I have been playing with bubble plot and I was able to create bubble
> plot with relatively simple data. I am no having problems when I
> increase the complexity of my data. Below is a example of my data:
>
> phytochemicalMainClassFitValue
Thanks. I didn't know that. I just found it in Brian Ripley's page. Is this
the cran extras?
Thanks
Richard
On 3 May 2011, at 13:19, Uwe Ligges wrote:
>
>
> On 03.05.2011 14:09, Richard Wang wrote:
>> Thanks. One more question. If I use install.packsges, do I need to install
>> Rtool
On 04/13/2011 05:06 PM, Ben Bolker wrote:
> Thomas Lumley uw.edu> writes:
>
>>
>> On Thu, Apr 14, 2011 at 5:30 AM,
>> Michael Friendly yorku.ca> wrote:
>>> I have a diagram to be included in latex, where all my figures are .eps
>>> graphics (so pdflatex is not an option)
>>
>> You could use
On 03.05.2011 14:09, Richard Wang wrote:
Thanks. One more question. If I use install.packsges, do I need to install
Rtool or utils package is sufficient?
It depends on how demanding the package is. For the one you mentioned,
you will need the Rtools, since C/C++ sources are to be compiled
Thanks. One more question. If I use install.packsges, do I need to install
Rtool or utils package is sufficient?
Thanks,
Richard
On 3 May 2011, at 12:26, Uwe Ligges wrote:
>
>
> On 02.05.2011 23:48, Richard Wang wrote:
>> Hi,
>>
>> I'd like to ask a installation question. I want to insta
A safe way out of this mess is to install R somewhere else.
For example, create a directory c:\Programs and install R there.
Regards
Søren
-Oprindelig meddelelse-
Fra: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] På
vegne af Jonathan Daily
Sendt: 3. maj 2011 13:
Ah ok. I suppose the fix is to get the hard path (C:/Program
Files/...) on the search path and remove the symlink from the search
path. Does that work?
2011/5/3 Stefan McKinnon Høj-Edwards :
> Yes, the message is pretty clear, but it has nothing to do with running as
> admin.
> I have just tried
On 02.05.2011 23:48, Richard Wang wrote:
Hi,
I'd like to ask a installation question. I want to install a source code
through the following command,
R CMD INSTALL RDCOMClient
This is intended to be used in the shell of your OS (assuming Windows
given the package), not in R.
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