I have a list of 5 stocks consisting of IBM, AXP, XOM, BA, GE. I am able to
solve the optimal portfolio without long only constraints along with group
weight constraints. However, I am not able to add one more constraint
regarding Beta of the portfolio with upper bound of 1.1 and lower bound of
.9.
Thanks everyone for you help with my last question, and now I have one last
one...
Here is a sample of my data in .csv format
site,time_local,time_utc,reef_type_code,sensor_type,sensor_depth_m,temperature_c
06,2006-04-09 10:20:00,2006-04-09 20:20:00,BAK,sb39, 2, 29.63
06,2006-04-09 10:40:00,2006
I have a spatial weight file in csv that I want as listw object in R.
The file has the following 3 variables (left to right in the file) -- OID_, NID
and WEIGHTS. NID stands for the neighbors and OID_ as the origins. There are
217 origins with 4 neighbors each.
I have been able to read the csv
Dear all:
Could anyone please provide some R codes to plot the below survival data to
compare two groups (0 vs 1) after 2 simulations (TRL)? need 95% prediction
interval on the plot from these 2 trials. I would like to simulate 1000 trials
later. Thanks a lot for your great help and consid
Hello,
I used glsm.mcmc and likfit.glsm to create model. Now I want to predict at
different locations, but I can't get glsm.krige to work. I keep getting the
error that trend.d and trend.l must have similar specifications. I have tried
numerous ways to include the covariates in the glsm.krige
Hi, I have two questions regarding plotting of this dataset:
Product ColorStoreA StoreB StoreC Price
A Red 4 2378
Blue 5 22 78
Greem4 3 2 80
B Red 3 ..
sorry!
vols=read.csv(file="C:/Documents and Settings/Hugh/My
Documents/PhD/Quick&DirtyVols.csv"
+ , header=TRUE, sep=",")
> x<-ts(vols[,2])
> x
so I have an ts object of 500 elements.
I want to be able to split the array into say two parts of 250?
How can I do it?
I want to keep the elements
Sorry, I should have really started a new thread with this because really it
is a new question only loosely related to the first Q.
Thanks for the assist.
As suggested I switched over to sweave. I have a lot of .tex tables that I
> have already created that I was previously inserting into my tex
As suggested I switched over to sweave. Not to bore you with the details
but I have a lot of .tex tables that I have already created that I was
previously inserting into my tex document (using \input). The journal I
plan on submitting to eventually wants the final .tex file so I thought it
would
require(Hmisc)
?areg.boot
Frank
Mishra, Srikanta wrote:
>
> I am using ACE/AVAS to fit a non-parametric regression model to a
> multivariate data set. Is there a way to add confidence intervals to the
> fit and predictions by using bootstrap? Thanks for any suggestions.
>
>
> [[alterna
On 10/29/2011 09:23 AM, Hurr wrote:
Can I also label selected reciprocal numbers easily either
alongside or without the automatic ones, and have them
automatically in their correct spots.
Sorry, I really don't know R now.
Hi Hurr,
What you want is probably achievable with the "axis" function. T
On Oct 28, 2011, at 6:23 PM, Hurr wrote:
Can I also label selected reciprocal numbers easily either
alongside or without the automatic ones, and have them
automatically in their correct spots.
Sorry, I really don't know R now.
I am unable to parse your request into standard English. Please po
Hi Michelle,
In addition to Jeff's advice, can you now reproduce the original
error? What you are describing is highly improbable if everything is
properly executed. It becomes more likely if more was chagned between
running than just inserting browser(). For example, if you reran/had
made chan
On 11-10-28 11:34 AM, playballa23 wrote:
Hello,
I'm not getting a great looking 3d graph when plotting my z values:
x<- seq(0.0, max(d1), max(d1)/20)
y<- seq(0.0, max(d2), max(d2)/20)
z<- matrix(NA, length(x), length(y))
persp3d(x, y, z, phi=30, theta=30, scale=TRUE,shade = 0
[ perhaps?
You really need to say more about what you are looking to do if you want a more
informative answer
M
On Oct 28, 2011, at 3:11 PM, Bazman76 wrote:
> Hi there,
>
> I have a ts object that I would like wo split into arbirary sizes.
>
> Can'tfind how to do this?
>
> I realise its p
Why don't you try substituting your vector of values and see what comes
out...once you figure out what happened, the sum() command will solve your
problems.
Michael
On Oct 28, 2011, at 5:10 PM, djbanana wrote:
> I understand that the likelihood function is a product and hence the log
> likel
Are you sure your variables are categorical or numeric? Of course, glm
differentiates these two kinds of variables. For example, I ran the
same variable with different modes, the results are very different.
> dat<-data.frame(y=rpois(100,5),xf=as.factor(sample(1:4,100,replace=T)))
> glm(y~xf,data=d
Can I also label selected reciprocal numbers easily either
alongside or without the automatic ones, and have them
automatically in their correct spots.
Sorry, I really don't know R now.
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Ask a psychic?
Perhaps show your real problem (not this vague handwaving) to a statistician?
If you are in a statistics course, talk to your instructor or teaching
assistant?
This is R-help, not a statistics helpline, or a homework helpline. You are
expected to know at least the basic statisti
I understand that the likelihood function is a product and hence the log
likelihood function is a sum. However I can't figure out what the problem
is.
Here's the likelihood function:
[(alpha1*beta2*gamma)^v1 exp^(-alpha1*beta2*gamma)]/v1! * [(alpha2*beta1)^v2
exp^(-alpha2*beta1)]/v2!
Isn't the l
Short answer: I have not.
Long answer: Read the posting guide. Provide a reproducible example. Show
actual error messages. Identify your operating system and version of R. Read
the posting guide. Then try posting again.
---
Daniel,
I want to use the contrast library because I want to be able to specify any
arbitrary post-hoc contrast, e.g. Given a 3x2 table describing smoking (never,
former, current) by sex (male,female), I can use a post-hoc contrast to compare
the fraction of female former smokers to the fraction
Hi there,
I have a ts object that I would like wo split into arbirary sizes.
Can'tfind how to do this?
I realise its probaly very simple buy I can't sem to find the right
function?
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Dear R community:
I am working on a model frame problem which is important for my data
analysis. What I am trying to get is to get data set d3 through the model
formula "ff=y~a+b+a*b" and data set d1 to generate a new data set d3 which
includes a interaction column between a and b rather than d2.
Hey all,
I am attempting to replicate my results achieved in another program within
R (so I can expand my options for methods). I am trying to run a GLM (Family
= Poisson) for count data in R. Some of my variables are factors and I am
under the impression that the function glm() cannot run a mode
Dear R community:
I am working on a model frame problem which is important for my data
analysis. What I am trying to get is to get data set d3 through the model
formula "ff=y~a+b+a*b" and data set d1 to generate a new data set d3 rather
than d2. I have tried several ways but did not get that done.
The likelihood function is a product. Thus, the log likelihood function is a
sum. Your log.lik statement, however, fails to compute the sum, which it
should minimize. Hence your optim statement does not know what to optimize
because log.lik is a vector of the length of the number of observations in
Hello. I have this error:
> install.packages("rJava")
Installing package(s) into ‘C:/PROGRA~1/R/R-212~1.0/library’
(as ‘lib’ is unspecified)
Warning in install.packages :
package ‘rJava’ is not available
Does this mean that rJava is not available at all?
--
View this message in context:
http://
> Hello,
>
> I'm not getting a great looking 3d graph when plotting my z values:
>
> x<- seq(0.0, max(d1), max(d1)/20)
> y<- seq(0.0, max(d2), max(d2)/20)
> z<- matrix(NA, length(x), length(y))
>
> persp3d(x, y, z, phi=30, theta=30, scale=TRUE,shade = 0.75,
> col=pal[col.ind], xlab=v
I am trying to run this code and obtain the MLEs for my parameters. However I
am getting this error at the end.
Error in optim(c(1.4, 1.1, 0.8, 0.92, 0.4), poisson.lik, v = v) :
objective function in *optim evaluates to length 100 not 1*
Code:
poisson.lik <- function(theta,v){
v=matrix(c(1,3,2
Dear All
I have a program that breaks at the following lines of code:
bigfunction =
{
...
object1 = myfunction(x)
object2 = strsplit(object1, ",")[[1]]
...
}
where myfunction is defined elsewhere outside of bigfunction.
The error I get is "error in strsplit() -- object1 not found".
Howeve
Thanks for your assistance David, I'll run your code and see how it goes.
Take care.
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Hi. I have generated a heatplot and also added a colour bar (using
"classvec") under the dendrogram that clusters the column subjects. These
subjects are divided into four groups, each with a different colour. I have
been trying to find a way to reorder the dendrogram by rotating some of the
branch
On Oct 28, 2011, at 3:57 PM, Hurr wrote:
Does R graphics have a way to easily label the horizontal axis by the
reciprocal of the scaled value?
plot(x<-1:10,y<-1:10, xaxt="n")
axis(1, at=1:10, labels=round(1/x, 3) )
It is a bit more difficult to do it as a fractional label but possible:
A correspondant suggests it would help if I said the OS,
it's Intel Mac OS X 10.4...yes I should update but I'm
trying to hold off a little longer due to numerous other
things I don't want to have to reinstall just yet...
On 10/28/11 10:54 AM, Nick Matzke wrote:
Hi all,
I am attempting to in
Thanks!
On Fri, Oct 28, 2011 at 10:59 AM, Marc Schwartz wrote:
> On Oct 28, 2011, at 9:49 AM, Ben Ganzfried wrote:
>
> > Hey,
> >
> > I'm trying to match patient identifiers from two separate input files,
> and
> > then add information from one of the input files to the corresponding
> output
> >
Is there a specific reason why you insist on using the contrast library? If
not:
# Create 2x2 contingency table.
counts=c(50,50,30,70)
row <-gl(2,2,4)
column <- gl(2,1,4)
mydata <- data.frame(row,column,counts)
print(mydata)
#Create contrasts
row<-factor(row)
column<-factor(column)
contrasts
Does R graphics have a way to easily label the horizontal axis by the
reciprocal
of the scaled value?
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Hi
I have a table with four columns and 3 rows. I calculated the average for
each columns as average (A), average (B), average (C) and average (D), then
calculate the relative ratio as average (A)/average (B), and average
(C)/average (D). I would like to know how can I calculate the significance
b
> dat
[,1] [,2] [,3]
[1,]001
[2,]010
[3,]011
[4,] NA11
1 - is.na(dat)
[,1] [,2] [,3]
[1,]111
[2,]111
[3,]111
[4,]011
D.
On Fri, Oct 28, 2011 at 11:40 AM, Evgenia wrote:
> I have matrix data
(!is.na(data)) + 0
-- Bert
On Fri, Oct 28, 2011 at 11:40 AM, Evgenia wrote:
> I have matrix data
> data<-matrix(cbind(c(0,0,0),c(NA,0,1),c(1,1,1),c(0,1,1)),ncol=3)
> and I want to create a new matrix by checking each element of the data and
> put value 0 if i have NA and 1 otherwise.
> For this
On 28/10/2011 2:40 PM, Trevor Davies wrote:
I have found that I like having my captions and labels in my latex document
rather than having them contained in my xtable output file (I haven't fully
gone to sweave yet).
The remark in the parens is the problem here. Use Sweave.
Duncan Murdoch
I have found that I like having my captions and labels in my latex document
rather than having them contained in my xtable output file (I haven't fully
gone to sweave yet). I know I can do something like this by using the
'only.contents' argument in xtable. Unfortunately, the only.contents
argume
I have matrix data
data<-matrix(cbind(c(0,0,0),c(NA,0,1),c(1,1,1),c(0,1,1)),ncol=3)
and I want to create a new matrix by checking each element of the data and
put value 0 if i have NA and 1 otherwise.
For this reason i made the function below
pdata<-matrix(NA,ncol=ncol(data),nrow=nrow(data))
pdat
Forgive my resending this post. To data I have received only one response
(thank you Bert Gunter), and I still do not have an answer to my question.
Respectfully,
John
Windows XP
R 2.12.1
contrast package.
I am trying to understand how to create contrasts for a model that contatains
an intera
The various suggestions seem kind of complex to me, at least on a
unix-like system (including Mac OS X).
This is what I do:
sink('tmp.txt')
cat('This is the body of the message\n')
sink()
system('cat tmp.txt | mail -s "A test email" macque...@llnl.gov')
One could probably avoid the tempor
Hi all,
I am attempting to install a package called phylobase from
source directory. It all seems to work until the end, at
which point it looks like the last compile command fails
because the line is too long...perhaps because the g++
command line also includes "Loading ~/.Rprofile...", lik
On 10/28/2011 01:27 AM, diyanah wrote:
Yes, I'm using windows. Thank you. I made some improvement, I think, but then
why did I receive this?
utils:::menuInstallLocal()
package 'EBImage' successfully unpacked and MD5 sums checked
library("EBImage")
Error: package 'EBImage' was built before R
And to further the example, length() of matrix is not equal to the number
of rows either.
> mm <- matrix(1:6, ncol=2)
> length(mm)
[1] 6
> dim(mm)
[1] 3 2
Also, NROW() and nrow() are different; I'd be cautious about using NROW
without making sure I understood the difference.
> NROW
function (x)
I am using ACE/AVAS to fit a non-parametric regression model to a multivariate
data set. Is there a way to add confidence intervals to the fit and
predictions by using bootstrap? Thanks for any suggestions.
[[alternative HTML version deleted]]
I would think you could also do so with lines() directly, though you'll have to
choose an appropriate mesh of points to do so. A little wrapper function could
be written to do so pretty easily if you want to combine it with the
scatterplot, but since the fitted values are already in the output o
On Fri, 28 Oct 2011, Rich Shepard wrote:
May not be until Monday that I return with results on this effort as I'm
going to be carfully checking and documenting each step and its results.
Dan, et al.:
Got it working properly now. Took a different approach after re-reading
the source data in
ethan.shepherd gmail.com> writes:
>
> I'm getting errors when running what seems to be a simple Weibull
> distribution function:
>
[snip]
> If I change the data to this:
>
[snip]
> I get the error "Error in fitdistr(x, "weibull"): optimization failed"
>
> I can run a Weibull distribut
Hi,
On Oct 28, 2011, at 11:49 AM, Fernando Andreacci wrote:
> The cubic regression of my model is significant and I want to plot a line
> that best fits. It's not abline() function, because it has a curve. Please,
> how can I plot it?
>
Will curve() do it for you?
?curve
Cheers,
Ben
>
>
On Oct 28, 2011, at 11:26 AM, William Dunlap wrote:
Are you looking for something like the following? ifactor()
is like factor but assumes that x is integral and that levels
should be {1, 2, ..., max(x)} with no gaps.
x <- c(1,3,4,9,1,9,1,5,4,5,2,1,1,1,6)
ifactor <- function(x, levels=seq_le
I will. Thanks
On Fri, Oct 28, 2011 at 10:18 AM, David Winsemius
wrote:
>
> On Oct 28, 2011, at 10:32 AM, Joe Stuart wrote:
>
>> Hi,
>> I have a time series ts.score that gets created based on a function.
>> When I print the time series out it prints two rows, first the date
>> then the value.
>>
The cubic regression of my model is significant and I want to plot a line
that best fits. It's not abline() function, because it has a curve. Please,
how can I plot it?
---
Fernando Andreacci
Biólogo
Fone +55 47 9921 4015
fandrea...@gmail.com
[[alternative HTML version deleted]]
__
You need to use either the lme4 or nlme packages for mixed models.
(There are some other possibilities as well). See
http://glmm.wikidot.com/faq for MUCH more detail
On 10/27/2011 7:19 PM, Molly Hanlon wrote:
> Hi All,
>
> I'm working with some SAS code to analyze an experiment set up as follows
On Oct 28, 2011, at 4:08 AM, Samir Benzerfa wrote:
Hi David,
In the general case, there is still a gap in your solution
>sum( tbl["1",
2:ncol(tbl)] ). This solution refers to a specific column number
(here:
column number 2) and not to the actual length of the run, doesn't it?
That is
On 28/10/2011 11:04 AM, playballa23 wrote:
Hello,
I'm not getting a great looking 3d graph when plotting my z values:
x<- seq(0.0, max(d1), max(d1)/20)
y<- seq(0.0, max(d2), max(d2)/20)
z<- matrix(NA, length(x), length(y))
persp3d(x, y, z, phi=30, theta=30, scale=TRUE,shade = 0.
Are you looking for something like the following? ifactor()
is like factor but assumes that x is integral and that levels
should be {1, 2, ..., max(x)} with no gaps.
> x <- c(1,3,4,9,1,9,1,5,4,5,2,1,1,1,6)
> ifactor <- function(x, levels=seq_len(max(0, x, na.rm=TRUE))) factor(x,
> levels=levels)
On Oct 28, 2011, at 10:32 AM, Joe Stuart wrote:
Hi,
I have a time series ts.score that gets created based on a function.
When I print the time series out it prints two rows, first the date
then the value.
2011-06-14
-1.25947868
The function gets called multiple times in my script and I'm tryi
Hello,
I'm not getting a great looking 3d graph when plotting my z values:
x <- seq(0.0, max(d1), max(d1)/20)
y <- seq(0.0, max(d2), max(d2)/20)
z <- matrix(NA, length(x), length(y))
persp3d(x, y, z, phi=30, theta=30, scale=TRUE,shade = 0.75,
col=pal[col.ind], xlab=var.name[1], ylab
the part of the question dawned on me now is, should I try to do the parallel
processing of the full code or only the iteration part? if it is full code
then I am at the complete mercy of the R help community or I giveup on this
and let the computation run the serial way, which is continuing from p
I'm getting errors when running what seems to be a simple Weibull
distribution function:
This works:
x <-
c(23,19,37,38,40,36,172,48,113,90,54,104,90,54,157,51,77,78,144,34,29,45,16,15,37,218,170,44,121)
rate <- c(.01,.02,.04,.05,.1,.2,.3,.4,.5,.8,.9)
year <- c(100,50,25,20,10,5,3.3,2.5,2,1.2,1.1)
Thanks for posting this. If anyone is interested in a short extension to
include an attachment, try the function below.
I make no guarantees though. Also, note that winDialog() is used in a
couple places, so it may need some therapy before working outside of
Windows.
send.email <- function(to,
Hi All,
sorry but its probably another beginners question. I have run my data
through both the 'lars' and 'glmnet' packages in R in an effort to reduce
the number of predictors going into my regression equation.
Is there a way to view the final (best) equation output from the function?
Hopeful
Yes, I'm using windows. Thank you. I made some improvement, I think, but then
why did I receive this?
> utils:::menuInstallLocal()
package 'EBImage' successfully unpacked and MD5 sums checked
> library("EBImage")
Error: package 'EBImage' was built before R 2.10.0: please re-install it
which one d
Though I have not found the solution anywhere I have come up with a trick
that I will post here for future reference:
As long as you only want to include AR terms of the external variables (even
the dependant variable can be used as an external variable) it is possible
to manually lag the variable
Hi David,
In the general case, there is still a gap in your solution >sum( tbl["1",
2:ncol(tbl)] ). This solution refers to a specific column number (here:
column number 2) and not to the actual length of the run, doesn't it? That
is, in this simple example the column number 2 actually correspo
The (maximum) limit for shape is set on 10 for the function garchfit. Hence
it is impossible to estimate data for a student-t distributed error term
with a higher degrees of freedom than 10. I wonder how to change this
parameter. I hope someone could give me the answer. Thank you verry much in
adva
In the general case, there is still a gap in your solution >sum( tbl["1",
2:ncol(tbl)] ). This solution refers to a specific column number (here:
column number 2) and not to the actual length of the run, doesn't it? That
is, in this simple example the column number 2 actually corresponds to the
len
I am using Package survey version 3.20 and R 2.11.1.
--Amitava
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R-he
Thank you, it was exactly what I was looking for!
Regards,
Xavier
Le 28. 10. 11 15:45, Prof Brian Ripley a écrit :
> See the help for postscript ... especially the 'Printing' section.
>
>
> On Fri, 28 Oct 2011, Xavier Robin wrote:
>
>> Hi!
>>
>> In Windows the "win.print" function allows plott
On Oct 28, 2011, at 9:49 AM, Ben Ganzfried wrote:
> Hey,
>
> I'm trying to match patient identifiers from two separate input files, and
> then add information from one of the input files to the corresponding output
> file. I'd greatly appreciate any help!
>
> More specifically,
> Input_File_1 h
Looks like a job for merge().
On Fri, Oct 28, 2011 at 10:49 AM, Ben Ganzfried wrote:
> Hey,
>
> I'm trying to match patient identifiers from two separate input files, and
> then add information from one of the input files to the corresponding output
> file. I'd greatly appreciate any help!
>
> M
Hey,
I'm trying to match patient identifiers from two separate input files, and
then add information from one of the input files to the corresponding output
file. I'd greatly appreciate any help!
More specifically,
Input_File_1 has a column header "bcr_patient_barcode"
Input_File_2 has a column
Hi,
I have a time series ts.score that gets created based on a function.
When I print the time series out it prints two rows, first the date
then the value.
2011-06-14
-1.25947868
The function gets called multiple times in my script and I'm trying to
append the time series to an object called ts.
Thanks to all,
>a <- a[order(a[[4]]),]
gives desired result
>sort(a[,4])
sorts the data a[4]
regards,.
On Fri, Oct 28, 2011 at 7:42 PM, John Sorkin wrote:
> I am not sure (I can't test at the moment) but will this work?
> sort(a[,4])
>
>
> John David Sorkin M.D., Ph.D.
> Chief, Biostatistics a
I am not sure (I can't test at the moment) but will this work?
sort(a[,4])
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1
a <- a[order(a[[4]]),]
On Fri, Oct 28, 2011 at 10:01 AM, nandan amar wrote:
> I want to sort my following data set according to value in 4th column
>
> 0 1 0 27877.3044386212 15.8733019973557 2640.42407064348
> 0 1 1 27470.1699006254 35.4182473588807 2303.26461260826
> 0 1 2 27468.0314985496 3
I want to sort my following data set according to value in 4th column
0 1 0 27877.3044386212 15.8733019973557 2640.42407064348
0 1 1 27470.1699006254 35.4182473588807 2303.26461260826
0 1 2 27468.0314985496 38.4400363878507 2300.05521684593
0 1 3 27469.1130543141 40.7540672493746 2299.32564458
Thank you Marc and David for your suggestions. I think Marc's is going to
work nicely for what I need but I'll also try David's.
Many thanks both
Fiona
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See the help for postscript ... especially the 'Printing' section.
On Fri, 28 Oct 2011, Xavier Robin wrote:
Hi!
In Windows the "win.print" function allows plotting directly to a
printer (or copying an open device to the printer). This is very
convenient to quickly print a plot once it looks g
Thanks Raphael,
dev.print will create a copy of the device into another device, but it
won't actually push the data to a paper printer unless that other device
represents a printer device (as win.print does).
Typically in Windows I use it together with win.print as
dev.print(win.print, printer=bl
Windows XP
R 2.12.1
contrast package.
I am trying to understand how to create contrasts for a model that contatains
an interaction. I can get contrasts to work for a model without interaction,
but not after adding the interaction. Please see code below. The last two
contrast statements show th
I have never used, but take a look on ?dev.print
On Fri, Oct 28, 2011 at 11:16 AM, Xavier Robin wrote:
> Hi!
>
> In Windows the "win.print" function allows plotting directly to a
> printer (or copying an open device to the printer). This is very
> convenient to quickly print a plot once it looks
Upgrade R and try again. Preferably use the release candidate version of
the upcoming R-2.14.0.
Uwe Ligges
On 28.10.2011 11:10, bdeep...@ibab.ac.in wrote:
Hi,
I am trying to install qvalue, however its giving installation error ->
Error : package 'tcltk' does not have a name space
ERRO
Hi!
In Windows the "win.print" function allows plotting directly to a
printer (or copying an open device to the printer). This is very
convenient to quickly print a plot once it looks good.
Is there an equivalent function under Linux? For example through CUPS,
IPP, LPD or other ?
Obviously with
On Thu, 27 Oct 2011, Rich Shepard wrote:
Nope. Can't.
Dan et al.:
I apologize for the certainty; anything can happen to data as un-intended
consequences of processing. I will write that there _should_ be no
duplicates.
What I'm going to do is re-read the data frame from the source text
Thank you all very much, specially Ellison.
I have been looking at some codes for simulation and I am trying to recreate
them to my needs, but I understand I'm probably not doing it in the most
efficient way.
What you said was very useful.
I really want to create a series of vector of random nu
On 27.10.2011 18:34, Baris Demiral wrote:
Hi guys,
I have 64-bit Windows 7 machine. I need to execute the "latex" and
"pdflatex" commands from R to print out tables in pdf.
I use R 2.13 64-bit, and I have GSView, GSScript and TexWorks
installed. I remember i did this once but I forgot the ne
Hi Picohan
I had a really similar error messages recently when trying to install packages
from sources and it was related to my version of windows not being in english.
See this for a discussion:
http://r.789695.n4.nabble.com/Building-package-DESCRIPTION-file-not-existing-td3935084.html
Chee
Thanks,
i have it now:
library(lattice)
mein.panel <- function(x, y){
panel.xyplot(x, y)
panel.abline(a=0, b=1, lwd=2, col="red")}
x<-c(1,2,3,4,5,6)
y<-x
xyplot(x ~ y, main =
"xyplot", xlab="Sequenz 2",
ylab="Sequenz 1", las=1,
panel=mein.panel)
2011/10/28 S Ellison :
>
>
>> From: Jörg
> From: Jörg Reuter
>
> Why
> is the line with xyplot() not always in the middle of the
> dots like plot()?
Because you used the base graphics command abline() on a lattice plot?
They don't mix. The plot regions are different for lattice and base graphics -
notice that the second abline
Hi,
I draw two Plots, one with xyplot() and one with plot(). Why is the
line with xyplot() not always in the middle of the dots like plot()?
library(lattice)
x<-c(1,2,3,4,5,6)
y<-x
plot.new()
plot(x ~ y, main = "Sequenz 1 und Sequenz 2", xlab="Sequenz 2",
ylab="Sequenz 1", las=1)
abline(a=0, b=1,c
Hello All,
I have successfully installed rJava. I have written a sample code in java
to check how it runs. the code successfully compiled.
to compile i used javac --classpath
/home/deepthi/R/x86_64-unknown-linux-gnu-library/2.13/rJava/jri/JRI.jar:/home/deepthi/R/x86_64-unknown-linux-gnu-library/
Thanks for the tip, but I am not on win 7, I am on win xp.
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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R-help@r-project.o
Hi,
I am trying to install qvalue, however its giving installation error ->
Error : package 'tcltk' does not have a name space
ERROR: lazy loading failed for package qvalue
* removing /home/sbw/R/x86_64-unknown-linux-gnu-library/2.12/qvalue
The downloaded packages are in
/tmp/Rtm
On 10/28/2011 07:21 AM, Enrico Schumann wrote:
> A simple way may be not to plot all data points. But it will depend on
> your data if that is a good idea.
>
> Regards,
> Enrico
>
> ## EXAMPLE
>
> require("zoo")
>
> ## random data
> dates <- seq(from = as.Date("2000-01-01"),
> to = a
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