Try read.S in package foreign. It was written back in the days of that
S-PLUS format.
On 16/02/2012 06:06, Patrick Connolly wrote:
I have a small file of the type that S-PLUS produced and stored in
.Data/ directories back in the days before we used R. I'm fairly sure
it would become a datafram
On Wed, Feb 15, 2012 at 10:22 PM, arunkumar wrote:
> Hi,
>
> My input is
>
> x=c("D","A","B")
> Y=c("A","B","C","D")
>
> varorder=which(Y %in% X)
>
> I'm getting result as 1 2 4
> But i need the result to be 4 1 2
Here's a hint: lookup the function match.
Peter
Are there any result parameters which can be used to compare these
algorithms?
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Sent from the R help mailing list archive at Nabble.c
Hi,
My input is
x=c("D","A","B")
Y=c("A","B","C","D")
varorder=which(Y %in% X)
I'm getting result as 1 2 4
But i need the result to be 4 1 2
can any one help
-
Thanks in Advance
Arun
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On Wed, Feb 15, 2012 at 11:58 PM, Ben quant wrote:
> Thank you Duncan. Interesting. I find it strange that you can't get a list
> of the environments. But I'll deal with it...
>
> Anyway, I'm about to start a new R dev project for my company. I'm thinking
> about architecture, organization, and go
I have a small file of the type that S-PLUS produced and stored in
.Data/ directories back in the days before we used R. I'm fairly sure
it would become a dataframe if read into Ver 3.4 of that august
predecessor to what we use now. (It was Solaris in case that makes
any difference.)
AFAIK, the f
Thank you Duncan. Interesting. I find it strange that you can't get a list
of the environments. But I'll deal with it...
Anyway, I'm about to start a new R dev project for my company. I'm thinking
about architecture, organization, and gotchas. I went through much of the
documentation you sent me.
Thank you for your replies. That can be quite tricky indeed.
The data consists of monthly unemployment rates, which I get as a text
file along with the information: monthly, period of beginning and end.
So it does not come with some timestamp as I specify the time inside R.
This tweak time(X2
On Wed, Feb 15, 2012 at 10:25 PM, P.B. Lecavalier wrote:
> Greetings,
>
> I would like to subset observations in a time series using xts, after
> converting from ts to xts.
>
> X=ts(1:100, frequency=12, start=c(1976))
> X2=as.xts(X)
> X2["1984"]
>
> The output:
>
> Feb 1984 98
> Mar 1984 99
>
Time zones strike again!
In short, your problem is that your machine is probably not in UTC for
its timezone (and it probably shouldn't be) so 12:00am Jan1 in London
is still Dec31 for you. To see this directly, look at
as.POSIXct(time(X2)) -- you'll see that item 97 which "should be" Jan
1 is act
thanks! this was very helpful-- ultimately changed the name although it was
a little confusing to do so...
On Wed, Feb 15, 2012 at 6:12 PM, David Winsemius wrote:
>
> On Feb 15, 2012, at 5:30 PM, Yihui Xie wrote:
>
> Man, the original reply told you to use backticks `` rather than
>> double quot
Greetings,
I would like to subset observations in a time series using xts, after
converting from ts to xts.
X=ts(1:100, frequency=12, start=c(1976))
X2=as.xts(X)
X2["1984"]
The output:
Feb 1984 98
Mar 1984 99
Apr 1984 100
What happened to January? The index is always one month off, wit
It looks like a bug in bxp (which plot(factor,numeric)
calls) -- it tries to make a nGroup*nGroup long vector out of boxwex:
> f <- rep(factor(as.character(1:25000)), len=5)
> x <- sin(1:5)
> boxplot(split(x, f))
Error in rep.int(boxwex, n) : cannot allocate vector of length 62500
For
Tom Roche Wed, 15 Feb 2012 17:43:05 -0500
>> 1 I'm plotting to PDF, so everytime I dev.off() creates a new file,
>> and I want everything in one file (as does my boss :-)
>> 2 I'm doing the work on a cluster, where I very much do not have
>> root, and which has a fairly minimal set of install
Hi,
I got a little lost in your verbal description, but if I understand
your code correctly (what is n?), I think I'd implement it with the
split function: step i) make a single (hopefully vectorized) pass to
get whatever split points you need (the offset is in this step); ii)
use split() to break
This is certainly not a reproducible example, but this works fine for me.
lets = factor(sample(letters[1:3], 10, TRUE))
plot(1:10, 1:10, col = lets)
legend("topleft", legend = levels(lets), col = seq.int(length(lets)), lty = 1)
Michael
On Wed, Feb 15, 2012 at 9:35 AM, agent dunham wrote:
> Dear
Can you give minimal reproducible code? I'm worried about using
strings for the x axis but it'll be easier to diagnose with data and
actual code. Use dput() to put your data sample (not your whole data
set) in email-friendly plain text.
Michael
On Wed, Feb 15, 2012 at 10:46 AM, carol white wrote
It looks like period is a factor, so you can reorder the factor levels
however you wish and then plot them the same way. I believe recode()
in library(car) is a popular function for doing this, but I haven't
used it myself.
Michael
On Wed, Feb 15, 2012 at 11:34 AM, vd3000 wrote:
> Hi, all,
>
> I
I've never used neuralnet, but if this is like other modelling
functions in R, you need to put the column names corresponding to the
outputs, not the name of the object they came from. They might have
been changed in creating the data frame so take a look at
colnames(dt)
Hope that helps,
Michael
On Tue, Feb 14, 2012 at 11:17 PM, Redding, Matthew
wrote:
> I've been trawling through the documentation and listserv archives on this
> topic -- but
> as yet have not found a solution. I'm sure this is pretty simple with R, but
> I cannot work out how without
> resorting to ugly nested loops.
yet another solution (I think)
> patrn<- c(1,2,3,4)
> exmpl<- c(3,3,4,2,3,1,2,3,4,8,8,23,1,2,3,4,4,34,4,3,2,1,1,2,3,4)
> indx <- embed(rev(seq_along(exmpl)), length(patrn))
> matches <- apply(indx, 1, function(.indx){
+ all(exmpl[.indx] == patrn)
+ })
> indx[matches, 1L]
[1] 23 13 6
>
On We
On 12-02-15 5:26 PM, wpcmba wrote:
Issue: Connecting to a computer in a network.
I would like to connect to a file on a computer on my internal network that
has a different password. I have tried download.file and read.csv to no
avail.
When use this link in explorer it will prompt me for a user
We've recently installed a Netezza server and I am trying to integrate the R
GUI with this server and run functions using the server itself. I am able
to successfully connect to the server, but cannot get any of the
Netezza-specific R functions to work that exist in the package called "nza".
Spec
Thankyou all for your great, and creative solutions.
There is definately more than one way to skin a cat.
A colleague alerted me to another solution:
seq.strt <- which( sapply( 1:(length(exmpl)-length(patrn)+1), function(i)
isTRUE( all.equal( patrn, exmpl[ i + 0:(length(patrn)-1) ] ) ) ) )
Ap
Hi dear all,
I'm wondering about the question that; Does the influence.measures(model)
for linear models valid for general linear models such as logistic
regression models?
That is;
If I fit the model like
model <- glm( y~X1+X2, family=binomial)
Then, if i apply the function "influence.measur
Issue: Connecting to a computer in a network.
I would like to connect to a file on a computer on my internal network that
has a different password. I have tried download.file and read.csv to no
avail.
When use this link in explorer it will prompt me for a username/password:
file://hs-907/D$/
If you don't dev.off(), all plots will be sent to the open graphical
device. That usually doesn't impact behavior of other output types:
pdf(file='fooout.pdf')
hist(x <- rnorm(100))
y <- sin(x)
print(str(y))
cat(y,file='fooout.txt')
plot(x,y)
dev.off()
Hope this helps
On Wed, Feb 15, 2012 at 3:4
On Feb 15, 2012, at 5:30 PM, Yihui Xie wrote:
Man, the original reply told you to use backticks `` rather than
double quotes "". They are very different in R; see ?Quotes
And GSE19829-GPL570_eset is really a bad name. I recommend you to
rename it.
Regards,
Yihui
--
Yihui Xie
Phone: 515-29
I'm attempting to refactor an R script that does a lot of plotting,
among other things. Ideally I'd like to do something like
setup # does pdf(...)
for each part of input {
plot(process(part))
}
cleanup # does dev.off()
but have problems:
1 I'm plotting to PDF, so everytime I dev.off()
Man, the original reply told you to use backticks `` rather than
double quotes "". They are very different in R; see ?Quotes
And GSE19829-GPL570_eset is really a bad name. I recommend you to rename it.
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistic
Hello Duncan,
Thanks for the reply. I tried your command, and I got "raw(0)" in responseâ¦
However, your response that I was on the right track was in fact very helpful,
since I went back to my "header" file and found that the problem was the
transfer from the Ubuntu instrument to my Mac. No
Hi Martin,
Thanks! You were right. Quotes seemed to work, except I kept running into
a problem on the following line:
time <- "GSE19829-GPL570_eset$days_to_death" / 365
The error is that R atomic vector operator invalid. It's actually much
easier if I could use quotes rather than re-name-- so
Brilliant! Problem solved and you have just saved me many hours of
frustration.
Thanks so much.
On Wed, Feb 15, 2012 at 1:12 PM, Martin Morgan wrote:
> On 02/15/2012 12:33 PM, Jonathan Callahan wrote:
>
>> Greetings,
>>
>> I am creating a standalone executable script using Rscript and have run
On Feb 15, 2012, at 5:41 AM, Duncan Murdoch wrote:
On 12-02-14 10:58 PM, Jooil Kim wrote:
Hello,
I'm wondering if I can get some help with reading Fortran binary
"unformatted" output files into R.
The Fortran output files were generated in Ubuntu 10.04 LTS using
gfortran4.4, on a 32bit
This post https://stat.ethz.ch/pipermail/r-sig-mixed-models/2009q1/001819.html
may help you understand why the standard p-values in some cases are
not the right thing to do and what one alternative is.
On Tue, Feb 14, 2012 at 3:36 PM, Xiang Gao wrote:
> Hi
>
> I am working on a Nested one-way ANO
I saw I made a little mistake in the loop,
in the line
test.number$numberb[y-Year[1]+1]<-length(which(test.starty==1 &
test.f[(n+1)]== 1 ))
it is n+1 instead of n-1 (like I wrote in the beginning)
But the question I have about it is still the same.
My excuzes
Year<-data.frame(Date=c(1980,19
Ciao Etienne, thank you.
Today I tried to understand something more. Here's what I did (The file
names are a bit different):
*training <- data.frame(cbind(TL_training_2006_id,
AST_L1B_2008_05_2009_area_giusta_1, AST_L1B_2008_05_2009_area_giusta_2,
AST_L1B_2008_05_2009_area_giusta_3N))*
Then ...
*x
On Feb 15, 2012, at 17:29 , gianni lavaredo wrote:
> Dear Reasearchers,
>
> I am writing a report and i need (and wish) cite R. somebody know the
> citation of R for the 2012? or the more actual?
There's no released R version from 2012 -- yet. Wait two weeks and there will
be one.
-pd
--
Pe
On 16/02/12 03:59, Marion Wenty wrote:
Thank you for your answers!
Jeff, thanks very much for the tip with the<<- instead of using<- to save
the objects outside of the function, as well! I find this very usefull and
convenient, also if I decide to choose more objects.
This practice is to be
On 02/15/2012 12:18 PM, Ben Ganzfried wrote:
Hey,
I keep getting the following warning:
data(GSE19829-GPL570_eset)
Warning in data(GSE19829 - GPL570_eset) :
data set ‘GSE19829 - GPL570_eset’ not found
## maybe str(GSE19829-GPL570_eset) ; plot(GSE19829-GPL570_eset) ...
Hi Ben --
'-' is
On 02/15/2012 12:33 PM, Jonathan Callahan wrote:
Greetings,
I am creating a standalone executable script using Rscript and have run
into problems when I define a new S4 class. Here is a small script that
demonstrates my problem:
Hi --
Check out the difference between
R --vanilla -e "sessio
Hello List !
I'm a bright new R user, and I encounter a problem when trying to use the
neuralnet package.
I have a training set with 8 inputs, and there are 3 outputs (I need 3 distinct
neurones as output). Although I read the examples, and the package article, I
don't know how to tell R there
Greetings,
I am creating a standalone executable script using Rscript and have run
into problems when I define a new S4 class. Here is a small script that
demonstrates my problem:
#!/usr/bin/env Rscript
#
# toy.Rscript -- testing S4 class creation in an Rscript
setClass("toy",
representation(n
I'm not sure what the "official description" of R is beyond that given
in citation(), but the R FAQ gives this: "[R] consists of a language
plus a run-time environment." Part of R's personality, as far as I can
tell, comes from the fact it straddles the general purpose
language/domain specific lang
Hey,
I keep getting the following warning:
> data(GSE19829-GPL570_eset)
Warning in data(GSE19829 - GPL570_eset) :
data set GSE19829 - GPL570_eset not found
> ## maybe str(GSE19829-GPL570_eset) ; plot(GSE19829-GPL570_eset) ...
>
> if(require(affy)){
+ summary(GSE19829-GPL570_eset$vital_statu
. you don't really consider this a well formed question do you?
(Among others:)
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
http://www.gerv.net/hacking/how-to-ask-good-questions/
What immediately comes to mind:
i) Plotting engine? (base, lattice, or
On Feb 15, 2012, at 15:17 , R. Michael Weylandt wrote:
> But that's dropping unused factor levels, not the drop of `[`
Is there a good reason that nobody has suggested looking at aggregate()? There
would seem to be a little red tape to sort out, but, I mean, this is generally
speaking what agg
I have some data set which is available from 2005-2010 . I would like plot
monthly maps out of it.
So how I should write loop that can plot this yearly data to for every month
and write title for individual months too.
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First, in R there is no need to declare the dimensions of your objects
before they are populated so couldn't you reduce some run time by not
going through the double data.frame step ?
> df<- data.frame()
> df
data frame with 0 columns and 0 rows
> for(i in 1:100) for(j in 1:3) df[i,j]<- runif(1)
>
On Wed, 15 Feb 2012, dlofaro wrote:
Dear R-Users,
I'm a R beginner and I have a similar problem:
I fitted a SurvivalTree with rpart and I'm tryng to plot with the partykit
package by the funcion as.party but it sends me this error:
plot(as.party(rpartSurv))
Error in plot.constparty(as.party(r
Dear All,
I am using package fPortfolio to run minimum variance portfolio
optimizations in R. I already know how to set portfolioSpecs, portfolio
objects and constraints. Unfortunately I am not able to set the following
type of constraints.
I have a timeSeries object with returns data for roug
Hi Tony,
?nls
Cheers,
Tsjerk
On Feb 15, 2012 8:03 PM, "Anthony Fristachi" wrote:
Greetings,
Any suggestions for approaching the fitting of the function
y = b/exp(a*x) + c*x + y0
where a, b, c, and y0 are unknown constants and y and x are variables in a
give dataset.
Thanks
Tony
Greetings,
Any suggestions for approaching the fitting of the function
y = b/exp(a*x) + c*x + y0
where a, b, c, and y0 are unknown constants and y and x are variables in a
give dataset.
Thanks
Tony
[[alternative HTML version deleted]]
environment[[names[j]]][i]<-I does seem to take longer than
environment$name[i]<-I when looping over the length of the
vector involved. Using a list instead of an environment seems
to work better.
Below are 4 functions that populate a data.frame row by row.
f0 : the naïve approach
f1 : use
Thank you,
Its looking like your package will work for me. I have two questions.
First, how do I rotate the plot 90 degrees so the group labels are on
the x axis and the response value on the y axis?
Second, I'm having trouble with the group labels. I need to order my
groups into meaningful group
Dear R Community,
I am happy to introduce the next release of the *dclone* package (see
R Journal, 2(2): 29-37, 2010) which now includes a suite of functions
for parallel computations:
- see functions parJagsModel, parUpdate, parCodaSamples using 'snow'
type clusters; and the
- jags.parfit functi
Dear list,
The new package 'multivator' is now available on CRAN. This presents
a multivariate generalization of the emulator package.
The corresponding JSS article is:
Robin K. S. Hankin (2012), "Introducing multivator: A Multivariate Emulator",
Journal of Statistical Software, 46(8), 1-20.
Hello,
>
> I'm trying to build a lower triangular matrix (with zeros in the diagonal)
> from a particular dataframe.
>
This example constructs a lower triangular matrix from a vector, not a
data.frame.
(In your example, you also use a vector, the last column of the DF.)
x <- runif(15)
y <- matri
Dear R-Users,
I'm a R beginner and I have a similar problem:
I fitted a SurvivalTree with rpart and I'm tryng to plot with the partykit
package by the funcion as.party but it sends me this error:
> plot(as.party(rpartSurv))
Error in plot.constparty(as.party(rpartSurv)) :
node_surv not yet imple
Please read the very last two lines in the footer of your email below.
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Wed, Feb 15, 2012 at 8:55 AM, Bos, Roger wrote:
> I am completely new t
runif's min and max argument may be vectors:
> x=c(0,0,0,0,0)
> y=c(10,20,30,40,50)
> runif(length(y), min=x, max=y)
[1] 7.005180 14.035296 9.388089 22.094211 43.624606
This is equivalent to x + (y-x)*runif(length(y)) but
such a transformation is not always so easy for nonuniform
random
Inline
On Wed, Feb 15, 2012 at 3:04 AM, uday wrote:
> Hi ,
>
> Thanks for reply
>
> My latitude and longitude contains 9-10 observations per file
> when I run coords <- expand.grid(lat=1:5,long=1:5) then my computer
>
You don't have to run this part. As your original post did not provide
On 15/02/2012 15:00, Nicholas Reich wrote:
Hello.
I'm getting an unexpected result when running smooth.spline(). Here
is a simple example that replicates the error I'm getting:
aa<- c(1, 2, 3, 8, 8, 8, 8, 8, 8, 8, 8, 8, 12, 13, 14) bb<-
1:length(aa) plot(aa, bb) smooth.spline(aa, bb)
Error i
Sorry
I don't know the citation inside the report is corret
The Analysis was done using a script written in the statistical computing
environment of R (R Development Core Team, 2011)
is It correct the form "statistical computing environment" ?
Gianni
On Wed, Feb 15, 2012 at 5:38 PM, Sarah Gos
On Feb 15, 2012, at 11:47 AM, nymphita wrote:
Hi Tsjerk!
Thanks for your quick reply!
It's a nice way to built a lower triangular matrix with zeros in the
diagonal, but what I can't work out is *how to include the values of
the
third column of the dataframe inside the matrix*.
I just reali
On 15.02.2012 17:33, Dimitris.Kapetanakis wrote:
Dear all,
I am trying to use the integrate function in R but it seems that it does not
work in my example and I cannot figure out why. I create a function Mu1
(which works fine) and try to integrate by the code:
n <- 100
Ctrl<
Integrate works on functions that are vectorized (i.e., the algorithm
puts in N inputs and expects N outputs) -- your function is not
vectorized (and I'm not sure what integrating it means, but I'm not
looking too closely) but you can make it "look vectorized" with the
Vectorize() HOF. Note that th
Perhaps ignore the lower-triangularity for a moment and do something like this:
x <- matrix(NA, ncol = max(j), nrow = max(i))
x[i, j] <- k
Your code will be clearer if you use with() rather than df$i constructs.
Hope this helps,
Michael
On Wed, Feb 15, 2012 at 11:47 AM, nymphita
wrote:
> Hi T
Typo below, 'latter' -> 'former'. namedVector[[i]]
drops the names and namedVector[i] does not.
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 William Dunlap
> Sent: W
Sorry, I just realized that it's not a lower triangualr matrix, but an upper
triangular matrix!
But still the solution/s should be rather similar in both cases.
http://r.789695.n4.nabble.com/file/n4391127/matrix2.png
I apologize for creating confusion...
Nymphita
--
View this message in context
You could use [[ instead of [, as the latter drops
the names. E.g., after
> v <- c("a"=3,"a"=4,"c"=5,"d"=6)
change your
> c("n"=v["a"]+v["c"],"n1"=v["d"]*3) # "n.a"=8 "n1.d"=18
n.a n1.d
8 18
to
> c("n"=v[["a"]]+v[["c"]],"n1"=v[["d"]]*3) # "n"=8 "n1"=18
n n1
8 18
Bill Dunl
On 15/02/2012 14:18, Yogs wrote:
Hi,
I am trying to estiamte parameters for gamma distribution using mle for
below data using fitdist& fitdistr functions which are from "fitdistrplus"
& "MASS"packages . I am getting errors for both functions. Can someone
please let me know how to overcome th
Hi, all,
I am currently trying to learn this example.
http://learnr.wordpress.com/2009/03/17/ggplot2-barplots/
I created the stack bar easily.
If I would like to rank the stack bar from the highest on the right,
shortest on the left and eventually
I could show the data "1991-00" on the left
Hi Tsjerk!
Thanks for your quick reply!
It's a nice way to built a lower triangular matrix with zeros in the
diagonal, but what I can't work out is *how to include the values of the
third column of the dataframe inside the matrix*.
I just realized that I forgot to explain something about the data
Dear all,
I am trying to use the integrate function in R but it seems that it does not
work in my example and I cannot figure out why. I create a function Mu1
(which works fine) and try to integrate by the code:
n <- 100
Ctrl<- as.matrix(cbind(runif(n, -30, 30)))
W
Dear all,
I am currently running an experiment using quantile regression. In order to get
more accurate results for a hypothesis test, I need to run a bootstrapping
version of quantile regression and I need to find the estimated covariance
matrix among all the coefficients for several quantiles
Typing
citation()
at an R prompt will provide you with complete citation information for
the version of
R you are using. Same goes for packages, with citation("pkgname").
Sarah
On Wed, Feb 15, 2012 at 11:29 AM, gianni lavaredo
wrote:
> Dear Reasearchers,
>
> I am writing a report and i need (and
Hi gianni,
Type "citation()" without quotes in the R console. You will get something
like
To cite R in publications use:
R Development Core Team (2011). R: A language and environment for
statistical computing. R Foundation for Statistical Computing, Vienna,
Austria. ISBN 3-900051-07-0,
Dear Reasearchers,
I am writing a report and i need (and wish) cite R. somebody know the
citation of R for the 2012? or the more actual?
thanks in advance
Gianni
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https:/
Thanks for your help. I already considered using list. But as I want to put
some constraints on the entries and the form of the vector, I thought it
would be easier to verify them by using a vector instead of a list.
--
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Hi Nymphita,
?upper.tri
x <- as.data.frame(matrix(1:6,6,6))
x[upper.tri(x,diag=TRUE)] <- 0
x
Cheers,
Tsjerk
On Wed, Feb 15, 2012 at 4:33 PM, nymphita wrote:
> Hello!
>
> I'm trying to build a lower triangular matrix (with zeros in the diagonal)
> from a particular dataframe.
>
> The matrix I
Hi all,
I was wondering if there is a function kind of similar that splits a
dataframe, applies a function to each row and returns in a data frame. I
know ddply but this one isn’t useful in this situation.
I have a dataframe with values for each day (rows) for different objects
(columns). I have
Hello!
I'm trying to build a lower triangular matrix (with zeros in the diagonal)
from a particular dataframe.
The matrix I have to construct has 203 rows and 203 columns and that makes
20503 values to be included within (that's why I can't do it manually).
To illustrate the dataframe I have, I
Hello.
I'm getting an unexpected result when running smooth.spline(). Here is a
simple example that replicates the error I'm getting:
> aa <- c(1, 2, 3, 8, 8, 8, 8, 8, 8, 8, 8, 8, 12, 13, 14)
> bb <- 1:length(aa)
> plot(aa, bb)
> smooth.spline(aa, bb)
Error in smooth.spline(aa, bb) : need at l
Hi,
I am trying to estiamte parameters for gamma distribution using mle for
below data using fitdist & fitdistr functions which are from "fitdistrplus"
& "MASS"packages . I am getting errors for both functions. Can someone
please let me know how to overcome this issue??
data
y1<-
c(256656, 7637
Sorry for the error: the first column is of the string type and the second
column of float, respectively.
- Forwarded Message -
From: carol white
To: "r-h...@stat.math.ethz.ch"
Cc:
Sent: Wednesday, February 15, 2012 4:46 PM
Subject: Error in rep.int(boxwex, n) : negative length vector
Hi
> Hi, I am working under R2.11.1 Windows.
Rather old one, consider upgrading
> I work with a set of parameters which is save under a vector. I think it
is
> easier to understand, which parameter is called in the vector by naming
all
> vector elements. For example, we have a vector which t
Dear All,
I tried to invoke plot for a matrix of 196475 x2 dimension. I loaded the data
with read.table. On one dimension, there are strings (names) and on the other
dimension, the floats between 0 and 1. When I invoked plot by setting x to
names and y to floats and ylim = c(0,1), I got the foll
Tom Roche Fri, Feb 3, 2012 at 11:16 AM
>> summary: image.plot-ing two sets of netCDF data, with the second
>> derived from the first.
Should have noted that the data (atmospheric emissions) in
* the second dataset is heavily positively-skewed. (This leads to a
problem with the legend when plot
On 15-02-2012, at 15:27, Martin Morgan wrote:
> On 02/14/2012 11:45 PM, Petr Savicky wrote:
>> On Wed, Feb 15, 2012 at 02:17:35PM +1000, Redding, Matthew wrote:
>>> Hi All,
>>>
>>>
>>> I've been trawling through the documentation and listserv archives on this
>>> topic -- but
>>> as yet have n
On Wed, Feb 15, 2012 at 06:27:01AM -0800, Martin Morgan wrote:
> On 02/14/2012 11:45 PM, Petr Savicky wrote:
> >On Wed, Feb 15, 2012 at 02:17:35PM +1000, Redding, Matthew wrote:
> >>Hi All,
> >>
> >>
> >>I've been trawling through the documentation and listserv archives on
> >>this topic -- but
>
s_elternmz <- function(Var="balt")
{
Dg_a<-mz[,c("asbhh","apkz",Var)]
colnames(Dg_a)[colnames(Dg_a)=="apkz"]<-"bpkzm"
colnames(Dg_a)[colnames(Dg_a)==Var]<-paste(Var,"_m",sep="")
mz_int<-merge(mz,Dg_a,by=c("asbhh","bpkzm"),all.x=T)
colnames(Dg_a)[colnames(Dg_a)=="bpkzm"]<-"bpkzv"
colnames
I am completely new to LaTeX and Sweave, so I am trying to follow the
tutorial in the following link:
http://www.r-bloggers.com/using-latex-r-and-sweave-to-create-reports-in-
windows/
I installed MiKTeX 2.9 and TeXnicCenter, but I get a build error when
compiler the example file:
The two errors a
Gab,
Make sure you have variables for each training.
training <- data.frame(Training_2006, AST_L1B_1, AST_L1B_2, AST_L1B_3N)
If you can't do that, then you don't have as many training observations
than you have predictive informations. Make sure to create a line for each
set of predictive pixels
I want to use spplot function for plotting data on wold map
spplot.points(obj, zcol = names(obj), ..., names.attr,
scales = list(draw = FALSE), xlab = NULL, ylab = NULL, aspect =
mapasp(obj,xlim,ylim),
panel = panel.pointsplot, sp.layout = NULL, identify = FALSE, formula,
Hello to everybody !!! I am a student in the Mathematics department of the
university of Milano.
I am interested in an r -package or an available code for fitting two
component mixtures of generalized linear mixed effects models for cluster
correlated data. In particular with normal random effec
Dear community,
I've plotted data and coloured depending on the factor variable v3.
In the legend, I'd like to assign properly the same colors than in the
factor (the factor has 5 levels).
I've been trying this but it doesn't work.
plot(var1, var2, xlab = "var1", ylab = "var2", col =var3 , b
On 02/14/2012 11:45 PM, Petr Savicky wrote:
On Wed, Feb 15, 2012 at 02:17:35PM +1000, Redding, Matthew wrote:
Hi All,
I've been trawling through the documentation and listserv archives on this
topic -- but
as yet have not found a solution. I'm sure this is pretty simple with R, but I
cannot
On Wed, Feb 15, 2012 at 7:24 AM, aajit75 wrote:
>
> Hi All,
> This might be simple question, I need to retrive data for modelling from the
> databases. Eveytime date values changes so I countnot fix date value in the
> code, it is required to pass as parameter.
> When I pass the date as parameter,
But that's dropping unused factor levels, not the drop of `[`
Michael
On Wed, Feb 15, 2012 at 9:16 AM, David Winsemius wrote:
>
> On Feb 15, 2012, at 7:19 AM, R. Michael Weylandt wrote:
>
>> Sorry about that -- forgot the case where you might only have a
>> certain entry once: try this,
>>
>> sa
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