On Wed, 27 Apr 2011, Joshua Wiley wrote:
Hi,
I am not incredibly knowledgeable about gamma distributions, but
looking at your data, you have a tiny mean:variance ratio, which, I
believe, means that the bulk of the distribution will be near 0 and
you may run into computational problems (again I
It seems like the 'manipulate' package comes as part of the RStudio
distribution. It's not part of the standard R and not on CRAN.
You're more likely to get help on this via RStudio's website/forums.
/Henrik
On Wed, Apr 27, 2011 at 8:51 PM, veepsirtt wrote:
> require(quantmod)
> getSymbols("GLD
Hi,
I am not incredibly knowledgeable about gamma distributions, but
looking at your data, you have a tiny mean:variance ratio, which, I
believe, means that the bulk of the distribution will be near 0 and
you may run into computational problems (again I think. I would
gladly be corrected). This
require(quantmod)
getSymbols("GLD")
library(manipulate)
manipulate(
plot(as.matrix(Cl(GLD)), xlim=c(0,x.max)),
x.max=slider(10,1000))
I am not getting the plot of closing price of GLD.
V.Periasamy M.E
Associate Professor
Dept of Mechanical Engineering,
IRTT-ERODE-638316
INDIA
--
View thi
There was a small error in the data creation step and have fixed it as below:
test <- c(895.1358,2915.7447,335.5472,1470.4022,194.5461,1814.2328,
1056.3067,3110.0783,11441.8656,142.1714,2136.0964,1958.9022,
891.89,352.6939,1341.7042,167.4883,2502.0528,1742.1306,
837.1481,867.8533,3590.4308,1125
Hi,
I have following problem when trying to feed an CSV file to quantmod
using following command:
> require (quantmod)
> getSymbols("test1",src="csv")
Error in charToDate(x) :
character string is not in a standard unambiguous format
The sample test1.csv file contents:
Symbol, Date, Open, High
What exactly are you trying to accomplish? A little debugging might
show you what the problem is. I put the following statement in the
loop right after the assignment to 'newrow':
print(newrow)
What I saw is that you never reset 'iter' and once it reached a
maximum length, you were continuing t
On Apr 27, 2011, at 9:29 PM, Dale wrote:
Yes my example was not the best, it is an oversimplified example. I
tried
using your method and didn't seem to work.
And to whom might you be addressing this complaint?
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Hello,
I'm trying to follow the documentation of how to use gridBase, and I've reached
the minimal code example below as my best effort. Can someone explain how to
keep the column of boxplots on the same page as the rectangles (even though
I've tried new = TRUE) ? Also, would it be hard / possi
Yes my example was not the best, it is an oversimplified example. I tried
using your method and didn't seem to work.
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Sent from the R help mailing list archive
I was wondering if there exists a function that automatically tries to detect
the format of a datafile. E.g. if it is an ascii datafile, that it can
detect appropriate defaults for the read.table() parameters. One could for
example read the first 10 lines of the file and analyze the format of the
f
Hi,
I have a problem with Kolmogorov-Smirnov test fit. I try fit distribution to
my data. Actualy I create two test:
- # First Kolmogorov-Smirnov Tests fit
- # Second Kolmogorov-Smirnov Tests fit
see below. This two test return difrent result and i don't know which is
properly. Which result is prop
Dear list
I have 5 markers that can be used to detect an infection in combination. Could
you please advise me how to use functions in ROCR/ other package to produce the
ROC curve for a combination of markers?
I have used the following to get ROC statistics for each marker.
pred <- prediction(
The 'findCorrelation' function in the caret package may be helpful.
On Tue, Apr 19, 2011 at 3:10 PM, Rita Carreira wrote:
>
> Hello R Users!
> I have a data frame that has many variables, some with missing observations,
> and some that are correlated with each other. I would like to subset the
Grateful for any hints as to why I'm not getting the inner loop to cycle the
expected number of times.
Code and one run's results below.
Thanks,
Galen
> # source("looptest.r")
> sp<-numeric()
> iter<-numeric()
> rn<-numeric()
> ds<-data.frame(sp, iter, rn)
>
> for (sp in
Hello all,
I keep on getting the following error message when I try downloading
statmod:
> install.packages("statmod")
Installing package(s) into C:\Users\Isaac\Documents/R/win-library/2.12
(as lib is unspecified)
trying URL '
http://www.revolution-computing.com/cran/bin/windows/contrib/2.12/s
I don't know the answer to your question, but I avoid these problems by saving
my data as csv and avoiding direct interaction with Excel files. Excel is NOT a
database, even though it has supposed support through ODBC. I find this holds
true regardless of the programming environment from which I
On Wed, Apr 27, 2011 at 2:03 PM, mathijsdevaan wrote:
> Hi,
>
> Is there an alternative to "z <- read.zoo(DF, split = 2, index = 3, FUN =
> identity)" and "r <- rollapply(z, 3, sum.na, align = "right", partial =
> TRUE)"? I am trying to use the following script in which the split data (B)
> conta
Hi:
Here's one way to do this with plyr and data.table.
# plyr
As Hadley inferred, when using ddply(), it's convenient to write a
function for a generic (sub-)data frame and have it return a data
frame. Here's my function and call:
f <- function(d) {
require(MASS)
est <- fitdistr(d$win
Hi all,
I am trying to read Excel file usingthe follwoing commnad
library(RODBC)
data=odbcConnectExcel(file.choose())
sqlTables(data)
Bdat=sqlFetch(data, "test")
odbcClose(data)
head(Bdat)
1. The above script works if the Excel file is opened. If the excel file is
not
On 28/04/11 06:12, David Winsemius wrote:
On Apr 27, 2011, at 11:54 AM, Dale wrote:
If I have a vector of n elements, e.g. a vector of length 4 with
elements 10,
20, 30, 40 and want to find the different values of x such that x^2=10,
x^2=20, x^30 and x^2=40, how could I do this in R? I'm thin
RExcel has its own mailing list (the included documentation says so!)
Please subscribe at rcom.univie.ac.at and post your question on that list.
On 4/27/2011 8:00 PM, wwreith wrote:
> I have columns of data in Excel 2007, A2:A196, B2:B196...ET2:ET196 that I
> would like to place into arrays in R
On 28/04/11 06:09, Jonathan Daily wrote:
Try ?scan or ?readLines.
I think she might actually want readline() rather than either of the above.
cheers,
Rolf Turner
On Wed, Apr 27, 2011 at 11:42 AM, Lisa wrote:
Dear all,
I am trying to write a script to pause the execution of a f
> but it's a little clumsy, because
>
> with_connection(file("myfile.txt"), {do stuff...})
>
> isn't very useful because you have no way to reference the connection
> that you're using. Ruby's blocks have arguments which would require
> big changes to R's syntax. One option would to use pronouns:
Peter, I have indeed worked with Gregory-Newton and divided differences in my
very first numerical analysis course a couple of decades ago! However, I am
perplexed by the particular form of this matrix where the differences are
stored along the diagonals. I know that this is not the *same* as t
> From: h.wick...@gmail.com [mailto:h.wick...@gmail.com] On
> Behalf Of Hadley Wickham
> Sent: Wednesday, April 27, 2011 2:21 PM
> To: luke-tier...@uiowa.edu
> Cc: William Dunlap; r-help@r-project.org
> Subject: Re: [R] setting options only inside functions
>
> > Put together a list and we can se
> require(MASS)
> out <- with(weib.test.too, tapply(wind_speed, site, function(x) fitdistr(x,
> 'weibull')))
> estimates <- t(sapply(out, "[[", 1))
> SDs <- t(sapply(out, "[[", 2))
> estimates
> SDs
>
> HTH,
> Jorge
That'll do nicely and is much more elegant than my current method. I was sure
t
> Put together a list and we can see what might make sense. If we did
> take this on it would be good to think about providing a reasonable
> mechanism for addressing the small flaw in this function as it is
> defined here.
In devtools, I have:
#' Evaluate code in specified locale.
with_locale <
Hi Justin,
One way of doing it is using a combination of tapply() and sapply() as
follows:
# data
set.seed(144)
weib.dist<-rweibull(1,shape=3,scale=8)
weib.test.too<-data.frame(cbind(1:10,weib.dist))
names(weib.test.too)<-c('site','wind_speed')
# results
require(MASS)
out <- with(weib.test.t
On Wed, Apr 27, 2011 at 3:55 PM, Justin Haynes wrote:
> I am trying to extract the shape and scale parameters of a wind speed
> distribution for different sites. I can do this in a clunky way, but
> I was hoping to find a way using data.table or plyr. However, when I
> try I am met with the foll
On Apr 27, 2011, at 21:34 , Ravi Varadhan wrote:
> My apologies in advance for being a bit off-topic, but I could not quell my
> curiosity.
>
> What might one do with a matrix of all order finite differences? It seems
> that such a matrix might be related to the Wronskian (its discrete analo
I am trying to extract the shape and scale parameters of a wind speed
distribution for different sites. I can do this in a clunky way, but
I was hoping to find a way using data.table or plyr. However, when I
try I am met with the following:
set.seed(144)
weib.dist<-rweibull(1,shape=3,scale=8
On Wed, Apr 27, 2011 at 10:12 PM, Kenn Konstabel wrote:
> On Wed, Apr 27, 2011 at 12:58 PM, Nick Sabbe wrote:
>> No, that does not work.
>> You cannot do assignment within (l)apply.
>> Nor in any other function for that matter.
>
> Yes that may work if you want to.
> You can do non-local assignme
Put together a list and we can see what might make sense. If we did
take this on it would be good to think about providing a reasonable
mechanism for addressing the small flaw in this function as it is
defined here.
Best,
luke
On Wed, 27 Apr 2011, Hadley Wickham wrote:
This has the side effe
I wanted to use an F-statistic to get p-values for treatment differences. I
have parameter estimates and standard errors. I posted about combining the
parameters in a previous post, but here I would just like to test the
parameter differences by treatments (it is a nonlinear function). I
calculated
There is an extensive statistical literature on how to correct for boundary
bias in kernel density estimates. See, for example,
An Improved Estimator of the Density Function at the Boundary
S. Zhang, R. J. Karunamuni and M. C. Jones
Journal of the American Statistical Association
Vol. 94, No. 4
Cubist is a rule-based machine learning model for regression. Parts of the
Cubist model are described in:
Quinlan. Learning with continuous classes. Proceedings
of the 5th Australian Joint Conference On Artificial
Intelligence (1992) pp. 343-348
Quinlan. Combining instance-based and m
Thanks. I will try them.
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On Wed, 27-Apr-2011 at 11:00AM -0700, wwreith wrote:
|> I have columns of data in Excel 2007, A2:A196, B2:B196...ET2:ET196 that I
|> would like to place into arrays in R. I have been trying to write a macro
|> that would automatically create all of my arrays for me with a array names
|> coming fro
Forgot to mention that the ctree command is from the party library.
//M
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented,
Thanks for your two suggestions.
I am too much of a beginner in R and R-forum to
be able to do either of them.
How do I/
Sorry.
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My apologies in advance for being a bit off-topic, but I could not quell my
curiosity.
What might one do with a matrix of all order finite differences? It seems that
such a matrix might be related to the Wronskian (its discrete analogue,
perhaps).
http://en.wikipedia.org/wiki/Wronskian
Ravi
Dear all,
I was intrigued by the ctree command and wanted to check it out. I first ran
the demo with example(ctree) and did get the survival graphs in the end. Upon
doing this with my own data and yielding a "Invalid operation on a survival
time" I tried to rerun example(ctree) and now I also g
> This has the side effect of ignoring errors
> and even hiding the error messages. If you
> are concerned about multiple calls to on.exit()
> in one function you could define a new function
> like
> withOptions <- function(optionList, expr) {
> oldOpts <- options(optionList)
> on.exit(option
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Jannis
> Sent: Wednesday, April 27, 2011 11:16 AM
> To: Jonathan Daily
> Cc: r-help@r-project.org
> Subject: Re: [R] setting options only inside functions
>
> Thanks to all who s
Thanks to all who supplied suggestions. All of them worked. The best solution,
however, was to wrap the stuff inside dummy() after the options(...) into a
try() command. That way it also worked with the following setup:
dummy=function()
{
old.options=options(error=quote{dummy1()})
try(...
Have you tried:
pen <- rep(1,ncol(X))
pen[c(1,5,7)] <- 0
fit <- glmnet(X,y,penalty.factor=pen)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Brian H. Chen
Sent: Wednesday, April 27, 2011 12:01 PM
To: r-help@r-project.org
Subject
I have columns of data in Excel 2007, A2:A196, B2:B196...ET2:ET196 that I
would like to place into arrays in R. I have been trying to write a macro
that would automatically create all of my arrays for me with a array names
coming from the cells A1, B1, etc.
I can manually create an array using REx
Hi,
Is there an alternative to "z <- read.zoo(DF, split = 2, index = 3, FUN =
identity)" and "r <- rollapply(z, 3, sum.na, align = "right", partial =
TRUE)"? I am trying to use the following script in which the split data (B)
contains about 30 unique cases and obviously I am getting an alloca
On Apr 27, 2011, at 11:54 AM, Dale wrote:
If I have a vector of n elements, e.g. a vector of length 4 with
elements 10,
20, 30, 40 and want to find the different values of x such that
x^2=10,
x^2=20, x^30 and x^2=40, how could I do this in R? I'm thinking of
using the
uniroot function, but
Try ?scan or ?readLines.
On Wed, Apr 27, 2011 at 11:42 AM, Lisa wrote:
> Dear all,
>
> I am trying to write a script to pause the execution of a function and
> provide some additional commands to the function and then continue execution
> of the function. For example, when my function detects a w
Look at TkListView in the TeachingDemos package as one option (though for large
lists it is noticeably slower than str). It creates a tree structure
representing your list that you can then expand or collapse branches in.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain He
I need to create a binary matrix with all node of a phylogenetic tree and the
presence of each taxo in their respective node.
Example:
require(ape)
y<-read.tree(text="(E,((H,I)D,(F,G)C)B)A;")
y
plot(y, show.node=TRUE)
I need to create a binary matrix as follows:
A B C
I am trying to fit a large Cox model with many predictors. Because
there are many predictors, I would like to use the ridge() function to
get penalized ml estimates for all coefficients. The problems are that:
1. When I include a factor (like race) in the ridge() function, dummy
variable
Try this:
sapply(listObj, '[', 1:max(sapply(listObj, length)))
On Wed, Apr 27, 2011 at 2:23 PM, Bogaso Christofer
wrote:
> Dear all, let say, I have following list object:
>
>
>
> listObj <- vector("list", length = 3)
>
> listObj[[1]] <- rnorm(3)
>
> listObj[[2]] <- rnorm(4)
>
> listObj[[3]] <-
If I have a vector of n elements, e.g. a vector of length 4 with elements 10,
20, 30, 40 and want to find the different values of x such that x^2=10,
x^2=20, x^30 and x^2=40, how could I do this in R? I'm thinking of using the
uniroot function, but am finding difficult applying it to a vector. Than
Dear all,
I am trying to write a script to pause the execution of a function and
provide some additional commands to the function and then continue execution
of the function. For example, when my function detects a wrong number in a
dataset, the function pauses automatically and returns informatio
Rich,
thanks a lot, i will definitly check it out. however, since the analysis
mentioned above is already implemented could you or anyone tell me whether
it contains any statistical flaws?
best
Lisa
__
Von:
Two problems with the code below.
A. It produces empty JPEGs. When the 'bwplot' line alone is submitted, the
plot duly shows up.
B. When the 'bwplot' line alone is submitted, y labels are values 1 to 6,
not actual distinct values of y$maxthreads.
(C. I would, of course, prefer to produce plots fo
Anyone have experience specifying the "penalty.factor" option in the
"glmnet" command?
I have 3 variables (out of a million genotype variables) that I want to
force into the model (i.e., set penalty factor to 0), but I can't figure out
how to do that.
[[alternative HTML version deleted]]
You might want to use the logspline package instead of the density function, it
allows you to specify bounds on a distribution.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@
Absence of evidence is not evidence of absence. Perhaps you are not getting
answers for a good reason.
>From a previous email:
1) reading the source code of packages that use rJava, such as RWeka is the
best way to understand how things work. If you are asked to do so is for a
reason; expl
I don't know who to contact in the management of this R-forum,
and there are a few things I cannot figure out.
As I understand it, to participate and learn on the R-forum,
I must receive all the emails even concerning topics I have no
interest in currently. It clogs up my email, takes a long
time
Here's one way:
uselen = max(sapply(listObj,length))
do.call(rbind,lapply(listObj,function(x){length(x) = uselen;x}))
[,1] [,2] [,3] [,4] [,5]
[1,] 0.0702225 -1.143031 1.6437560 NANA
[2,] -0.6100869 2.657910 -0.6028418 -0.7739858NA
[3,
Hi Christofer,
You might try
sapply(listObj, function(l) l[1:max(sapply(listObj, length))] )
HTH,
Jorge
On Wed, Apr 27, 2011 at 1:23 PM, Bogaso Christofer <> wrote:
> Dear all, let say, I have following list object:
>
>
>
> listObj <- vector("list", length = 3)
>
> listObj[[1]] <- rnorm(3)
>
Dear all, let say, I have following list object:
listObj <- vector("list", length = 3)
listObj[[1]] <- rnorm(3)
listObj[[2]] <- rnorm(4)
listObj[[3]] <- rnorm(5)
Now I want to convert above list into a Matrix. Ofcourse I can do it using
"Reduce("rbind", listObj)". However as you notice t
Dear Julian,
Though it's not exactly what you're looking for, you might take a look at
the heplots package (on CRAN).
I hope this helps,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton,
Hi. Comments below
On Wed, Apr 27, 2011 at 2:32 AM, agent dunham wrote:
> Hi, thanks, I think I've changed the previous as you told me but I'm having
> this error, what does it mean?
>
>
> model<- lm(log(v1)~log(v2)+v3, data=dat)
>
> newax<- expand.grid(
> v2 = seq(min(log(dat$v2)), max(log(da
Massimiliano -
Here's one way, assuming you wanted b to be the same
length as a:
df = transform(df,b=ave(as.character(df$a),df$ID,
FUN=function(a)paste(a,collapse='')))
If you just want one observation for each value of ID, you
could use
aggregate(df$a,list(ID=df$ID),
Thanks Claudia,
Meanwhile I implemented a simple function to evaluate the Youden-Index and
subsequently all other parameters. This is sufficient for my purpose.
Cheers,
Christian
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On Wed, Apr 27, 2011 at 11:25:42AM +, Hans W Borchers wrote:
> Jeroen Ooms gmail.com> writes:
>
> >
> > Is there an easy way to turn a vector of length n into an n by n matrix, in
> > which the diagonal equals the vector, the first off diagonal equals the
> > first order differences, the sec
Please ask about Rgraphviz (and particularly binary distributions of
it) on the list of those providing it (which is not R!).
And do use a sensible subject line (which really does need to include
'Rgraphviz'), as the posting guide asked you to.
On Wed, 27 Apr 2011, Fang, Yongxiang wrote:
De
Dear all,
I have the following R dataframe:
set.seed(11)
(df <- data.frame(ID=rep(1:10,1:10),a=factor(sample(1:4,55,rep=T))))
where ID is an identification code.
I need to create a new variable "b" in which I would paste the full group
of "a" variable according to ID variable.
F
hi all.
I have a matrix of data with 5 different groups and 20 individual
response per group, and about 12 variables collected for each. I want to
represent the result in a 2D plot. PCA is not so good because the
difference between the groups is not obvious. I have seen, in a recent
paper, pe
On 27/04/2011 9:43 AM, Cormac Long wrote:
Hello R-help,
I am wondering if anyone can help me with this:
I want to access data in a list which has been passed
into a C function, but I cannot work out how to access
the values. How do I move from the given SEXP pointer
to the next object in the li
Lisa,
Please look at some of the demos in the HH package.
These are built on the capabilities of the glht function in the multcomp
package.
## install.packages("HH") ## if necessary
library(HH)
demo("MMC.WoodEnergy-aov", package="HH") ## first
demo("MMC.WoodEnergy", package="HH") ## second
Ri
It is not a human growth curve. The parameter estimates are about:
a = .5
b = 7
k = 1
It is not a sigmoidal curve as there is never a concave segment.
From: ml-node+3478241-1447170361-211...@n4.nabble.com
[mailto:ml-node+3478241-1447170361-211...@n4.nabble.com]
Sent: Wednesday, April 27, 2011 09:
Thanks a lot Jim, Dennis.
Jim, actually where "dots" were I meant "etcetera" or "similar data". ;)
My output is close to that one Dennis proposed with function aggregate, but
i would like to keep the column var2 and the rest of columns with the unique
value corresponding to var1 greatest value, So
On Apr 27, 2011, at 13:29 , Alaios wrote:
> That was great :)
> REgards
You may need to turn your sarcasm detector back on. Beware:
> x <- matrix(rnorm(1e6), 1000,1000)
> y <- solve(solve(x))
> identical(x,y)
[1] FALSE
> all.equal(x,y)
[1] TRUE
> summary(c(x-y))
Min.1st Qu. Median
> Date: Wed, 27 Apr 2011 14:40:23 +0200
> From: jonat...@k-m-p.nl
> To: r-help@r-project.org
> Subject: Re: [R] Speed up plotting to MSWindows graphics window
>
>
> On 27/04/2011 13:18, Mike Marchywka wrote:
> >
> >> > Date: Wed, 27 Apr 2011 11:
Hello R-help,
I am wondering if anyone can help me with this:
I want to access data in a list which has been passed
into a C function, but I cannot work out how to access
the values. How do I move from the given SEXP pointer
to the next object in the list? I have tried to use CDR,
but to no avail
On Wed, Apr 27, 2011 at 4:58 AM, Dennis Murphy wrote:
> Hi:
>
> You could try something like
>
> df <- data.frame( expand.grid( Week = 1:52, Year = 2002:2011 ))
expand.grid already returns a data frame... You might want
KEEP.OUT.ATTRS = F though. Even it feels like you are yelling at R.
Hadley
Dear All,
I run R on a windows 7 machine and it has been worked very well. I installed
Graphvis 2.20.3 and Rgraphviz.
recently, however, I cannot load the Rgraphviz package and error message popped
up
The message shown on the pop up window with the title: R Consol: Rgui.exe -
Sysytem error
Th
Many thanks for your messages.
I will take a look at the survey package.
I was concerned with the issues raised by Cramer (1999) in "Predictive
performance of the binary logit model in unbalanced samples".
In this particular case, misclassification costs are much higher for
the smaller group (def
Hi Dennis,
My replies are in-line.
On Tue, Apr 26, 2011 at 9:15 PM, Dennis Murphy wrote:
> Hi:
>
> My view, which may well be narrow, is that techniques like PLS and PCR
> are useful fit procedures, but I would be very leery about using them
> as prediction machines. With new data, why should a
On Apr 27, 2011, at 9:28 AM, Schatzi wrote:
Here is more information on the equation. It is a growth function:
Growth = a + b*(1-exp(-k*time))
where a, b and k are parameters. I wanted to test the difference in
total growth between treatments and the parameters a + b represent
total growt
On Apr 27, 2011, at 6:49 AM, Dr. Meesters, Christian wrote:
Hi,
I have some questions for the wireframe function of the lattice
package. My dataset's "x-data" are sampled logarithmically and as
such I would like to have a semilogarithmic 3D plot when plotting a
time series. Does anyone k
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On Apr 27, 2011, at 7:25 AM, Hans W Borchers wrote:
Jeroen Ooms gmail.com> writes:
Is there an easy way to turn a vector of length n into an n by n
matrix, in
which the diagonal equals the vector, the first off diagonal equals
the
first order differences, the second... etc. I.e. to do
Thanks, David! That is another interesting perspective to (sub/super)
diagonal story! For now I was looking only at block sizes of lower triangle
submatrices as Dennis suggested.
Regards,
Santosh
On Wed, Apr 27, 2011 at 5:57 AM, David Winsemius wrote:
>
> On Apr 27, 2011, at 12:07 AM, Dennis Mur
On Apr 27, 2011, at 5:28 AM, Gary Nobles wrote:
hi I am still tryingto do this, I have been working on this for a
year but i
have remained stuck...
I have points at regular intervals but within an irregular window
I need to make a nb weights matrix, then nb2wlist
Not sure I understand tho
There is probably a more elegant way to do this, but you could write
it into dummy1():
dummy1 <- function()
{
...original function
options(old.options)
}
Alternatively, you could use ?tryCatch with the finally argument as a
call to options.
HTH,
Jon
On Wed, Apr 27, 2011 at 9:16 AM, Jannis wrot
Here is more information on the equation. It is a growth function:
Growth = a + b*(1-exp(-k*time))
where a, b and k are parameters. I wanted to test the difference in total
growth between treatments and the parameters a + b represent total growth.
Thus, I figured that I could add the parameters
On 27.04.2011 15:16, Jannis wrote:
Dear list members,
is it possible to set some options only inside a function so that the original
options are restored once the function is finished or aborted due to an error?
Until now I do something like:
dummy=function()
{
old.options=options(err
See ?on.exit
Jeremy
On Wednesday, April 27, 2011 9:16:13 AM UTC-4, Jannis wrote:
>
> Dear list members,
>
>
> is it possible to set some options only inside a function so that the
> original options are restored once the function is finished or aborted due
> to an error? Until now I do someth
Dear list members,
is it possible to set some options only inside a function so that the original
options are restored once the function is finished or aborted due to an error?
Until now I do something like:
dummy=function()
{
old.options=options(error=dummy1())
options(old.opti
Grr,
and 1) do not cross post to several lists (as the posting guide says)!
2) what is your name, "Stat Consult"? Don't you feel the name "Stat
Consult" somewhat ridiculous when asking such elementary question that
could have been sorted out if you had read the posting guide and the
manual
On 27.04.2011 13:46, Stat Consult wrote:
Dear ALL
I want to load "HTSanalyzeR", It 's necessary to load "igraph" package.
This time I see this error:
library(igraph)
library(HTSanalyzeR)
Loading required package: GSEABase
Loading required package: Biobase
Error: package 'Biobase' is not ins
On Apr 27, 2011, at 12:00 AM, kparamas wrote:
Thanks for the info.
I have 2 degree distributions that have different degrees.
Do you mean a different range of values?
I want both these barplots to have the same axes. Is this possible?
I have used xlim and ylim. ylim works fine for both pl
On Apr 27, 2011, at 12:07 AM, Dennis Murphy wrote:
Hi:
Maybe this can help get you started. Reading your data into a matrix
m,
m <- structure(c(1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1,
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