This is not the appropriate list for C-level questions. Please study the
posting guide and repost on the right list.
Also, you might want to also show how you actually call this from R. Are
you by any chance using .External instead of .Call ?
Romain
Le 15/06/10 23:20, Fabian Zäpernick a éc
Hi R,
I have a vector with all positive values and have fit the GEV
distribution (shape parameter=0) to the data and I get the negative
value of the location parameter. Is this possible?
library(evd)
r=read.table("clipboard",header=F)[,1]
> length(r)
[1] 2087
> sum(r>=0) #All data point
Duncan,
thanks, if the header is skipped, then it is much better.
It seems that I will be able to run the tests without actually running
'R CMD check' and extract the results, so it all looks good now.
Best Regards,
Gabor
On Tue, Jun 15, 2010 at 7:25 PM, Duncan Murdoch
wrote:
> Gábor Csárdi wr
Hello colleagues,
I have tried to use the package bigmemory, biganalytics and biglm. I
want to specify a multivariate regression with a weight.
I have imported a large dataset with the library(bigmemory). I load the
library (biglm) and specified a regression with a weight. But I get
every
Hi
I implemented the age-structure model in Gove et al (2002) in R, which is a
nonlinear statistical model. However running the model in R was very slow.
So Dave Fournier suggested to use the AD Model Builder Software package and
helped me implement the model there.
ADMB was incr
Beautiful! Thanks Peter and Phil for your kind help-
sherri
Quoting Peter Alspach :
Tena koe Sherri
You could turn on graphics recording (if you are using Windows click
History on the graphics window and choose recording). Then you can page
up/down to view them. If you save.image on exit,
David and Marc,
thanks, both your answers work for me.
all.vars() is ideal for me as I can get a vector of characters which I
can use as argument of deriv() to get a vector of derivatives of an
expresssion for all its variables.
On Wed, Jun 16, 2010 at 12:28 AM, Marc Schwartz wrote:
> On Jun 15
Use predict(my.gbm,type="response")
If you use bag.fraction=1.0 then all observations are guaranteed to be
used. For other choices, and assuming that you are doing a reasonable
number of iterations, like 1000, then there is only a negligible chance
that an observation won't be selected.
_
Tena koe Sherri
You could turn on graphics recording (if you are using Windows click
History on the graphics window and choose recording). Then you can page
up/down to view them. If you save.image on exit, you'll probably want
to rm(.SavedPlots) (I think that's the correct name. ls(all=TRUE) wi
Thanks for the reply, Phil.
My computer gets hung up on the par(ask=TRUE) call. I just tried
readline("Hit to proceed.")
and, it works well going forward. However, I cannot go backwards. Any
thoughts?
thanks again-
sherri
Quoting Phil Spector :
Sherri -
Perhaps calling
par(ask=TRU
You can use one-sided limits like ylim = c(0, NA), xlim = c(NA, 1) in
ggplot2 and lattice >= 0.18 (approximately; the version with R 2.11.0
will do it).
-Felix
On 16 June 2010 01:49, Ted Harding wrote:
> Greetings!
> I would like to be able to specify a fixed (say) lower limit
> for plotting, w
Sherri -
Perhaps calling
par(ask=TRUE)
before plotting would be useful. (You'll be prompted to
hit Return to see the next plot in the series.)
- Phil Spector
Statistical Computing Facility
Dear All-
I am trying to plot over one hundred figures. I do not want to save
them, just walk through them to take a look. If I run the code as it
is below, then the plots just rapidly run through. I tried adding a
new device, but I reached the device number limit. I have gone through
the
Do you have comment characters in your data (#)? Are there unbalanced
quotes in your data? How may rows does it read in? If you look at
the last line read, can you see a problem in your data? These are
problems that you will have with your data and try:
comment.char='', quotes=''
to see what ha
Hi all,
I have imported csv file for processing and exported as txt file (using *
"write.table"* with sep = "\t", row.names = FALSE options)
However, when I import this txt file in R (*"read.delim"* with header =
TRUE, sep = "\t" options) the dimension (i.e number of rows) is inconsistent
with th
Dear R-helpers,
I am working on an optimization problem, and I tried to use nlminb function
to solve it. But now I have an equality constrain, the Langrange multiplier
can solve it. However I would like to know if there is any existing function
or package solving this kind of problem. Any suggest
Hello everybody
After doing a MANOVA on a bunch of data, I want to be able to make some comment
on the amount of variation in the data that is explained by the factor of
interest. I want to say this in the following way: XX% of the data is explained
by A.
I can acheive something like wh
On Jun 15, 2010, at 6:02 PM, Josh B wrote:
Dear listserv,
I am trying to perform a principal components analysis and create an
output table of the eigenvalues for the dependent variables. What I
want is to see which variables are driving each principal components
axis, so I can make stat
Dear listserv,
I am trying to perform a principal components analysis and create an output
table of the eigenvalues for the dependent variables. What I want is to see
which variables are driving each principal components axis, so I can make
statements like, "PC1 mostly refers to seed size" or s
Hi
when I call the function below in R, i get the error: Object 'pairlist'
can't be converted to 'double'.
#include
#include
#include
SEXP CSimPoisson(SEXP lambda, SEXP tgrid, SEXP T2M, SEXP Ni, SEXP NT)
{
double sign, EVar;
double *xlambda, *xtgrid, *xT2M, *xNi, *xNT, *xtau;
Hello:)
I have 4 points in a 2-dimension coordinate. how can I do the regression
analysis and calculate the p-value£¿
thanks a lot,
Liang
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/li
hi,
could you please tell me what kind of cross validation that SVM of e1071 uses?
Cheers,
Amy
_
View photos of singles in your area! Looking for a hot date?
[[alternative HTML
Same for me on Mac OS X:
> x <- c(0.1819711,0.4811463,0.1935151,0.1433675)
> 1-sum(x)
[1] 0
> version$version.string
[1] "R version 2.11.1 Patched (2010-05-31 r52180)"
Regards,
Jorge
On Tue, Jun 15, 2010 at 6:40 PM, Gabor Grothendieck <> wrote:
> On Tue, Jun 15, 2010 at 6:00 PM, Yen Lee <> wr
Hello, everyone,
Thank you for all your kindness.
I've solved the problem through your help with the function all.equal.
Thank you very much!
Yen
-Original Message-
From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Sent: Wednesday, June 16, 2010 6:40 AM
To: Yen Lee
Cc: r-help@
On Tue, Jun 15, 2010 at 6:00 PM, Yen Lee wrote:
> Hello, everyone,
>
> There's a problem about zero in R and I really need your help.
>
>
>
> I have a vector shown as x=c(0.1819711,0.4811463,0.1935151,0.1433675),
>
> The sum of this vector is shown as 1 in R, but when I type 1-sum(x), the
> value
On Jun 15, 2010, at 5:07 PM, David Winsemius wrote:
>
> On Jun 15, 2010, at 5:45 PM, Alberto Lusiani wrote:
>
>> I would like to get the list of variables of an expression: is that
>> possible? I searched the help and the web without success. In
>> practice, given an expression like
>>
>> exp
On Tue, 2010-06-15 at 06:28 -0700, Farhad Shokoohi wrote:
> Dear All,
> I revise my question about the problem I have.
> Take a look at the article
> http://www.jstatsoft.org/v09/i01
> and download the attached code.
> try to run one of the codes for example section 2.1 in R
> here is the code
Th
Hello Yen,
You may find ?zapsmall helpful. If your algorithm expects 1-sum(x) to
have a lower bound of zero, then also simply setting a cutoff point
seems reasonable.
Josh
On Tue, Jun 15, 2010 at 3:00 PM, Yen Lee wrote:
> Hello, everyone,
>
> There's a problem about zero in R and I really need
On Jun 15, 2010, at 5:45 PM, Alberto Lusiani wrote:
I would like to get the list of variables of an expression: is that
possible? I searched the help and the web without success. In
practice, given an expression like
expr = parse(text="x^2+y^3")
I would like to have a function such as:
var
Hi Yen,
See
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
HTH,
Jorge
On Tue, Jun 15, 2010 at 6:00 PM, Yen Lee <> wrote:
> Hello, everyone,
>
> There's a problem about zero in R and I really need your help.
>
>
>
> I have a vector shown as x=c
Hello, everyone,
There's a problem about zero in R and I really need your help.
I have a vector shown as x=c(0.1819711,0.4811463,0.1935151,0.1433675),
The sum of this vector is shown as 1 in R, but when I type 1-sum(x), the
value is not zero, but -2.220446e-16.
I can accept that this value i
I would like to get the list of variables of an expression: is that
possible? I searched the help and the web without success. In
practice, given an expression like
expr = parse(text="x^2+y^3")
I would like to have a function such as:
vars(expr)
returns
c("x", "y") or a list like the one ret
On Jun 15, 2010, at 4:39 PM, GL wrote:
Have the following function that is called by the statement below.
Trying to
return the two dataframes, but instead get one large list including
both
tables.
ReadInputDataFrames <- function() {
dbs.this= read.delim("this.txt", header = TRUE, sep =
Have the following function that is called by the statement below. Trying to
return the two dataframes, but instead get one large list including both
tables.
ReadInputDataFrames <- function() {
dbs.this= read.delim("this.txt", header = TRUE, sep = "\t", quote="\"",
dec=".")
dbs.that= read.
You need to constrain your parameters such that 0 < vector[1] < 1 and
vector[2] > 0.
You could try the `spg' function in package "BB" or the `L-BFGS-B' function
in optim to perform this box-constrained optimization. There are other
options as well including the `nlminb' function.
Ravi.
-Orig
On 15-Jun-10 19:35:54, Marc Schwartz wrote:
> On Jun 15, 2010, at 12:11 PM, Ted Harding wrote:
>
>> On 15-Jun-10 16:01:24, William Dunlap wrote:
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of
ted.hard...@manche
Dear Alain,
> -Original Message-
> From: Alain Guillet [mailto:alain.guil...@uclouvain.be]
> Sent: June-15-10 12:25 PM
> To: John Fox
> Cc: r-help@r-project.org
> Subject: Re: [R] Problem with the recode function
>
> I found out what the problem is: when I start R Commander, some plug-ins
On Jun 15, 2010, at 12:11 PM, Ted Harding wrote:
> On 15-Jun-10 16:01:24, William Dunlap wrote:
>>> -Original Message-
>>> From: r-help-boun...@r-project.org
>>> [mailto:r-help-boun...@r-project.org] On Behalf Of
>>> ted.hard...@manchester.ac.uk
>>> Sent: Tuesday, June 15, 2010 8:49 AM
>
George Coyle wrote:
Hi All,
I am trying to turn a Matrix into a vector for analysis purposes. I need to
select only certain columns from the entire matrix for the vector (intraday
time intervals). Also I need to transpose the Matrix (so times are in rows)
stack each successive new column on
How about:
?subset
?t
?as.vector
I'm not sure transposing is really needed, but without a workable example
as requested in the posting guide (hint, hint), it's hard to say for certain.
Sarah
On Tue, Jun 15, 2010 at 1:51 PM, George Coyle wrote:
> Hi All,
>
> I am trying to turn a Matrix into a v
apply(test, 1, paste, collapse = ",")
On Tue, Jun 15, 2010 at 11:27 AM, Jonathan Greenberg
wrote:
> Folks:
>
> Say I have a matrix:
>
> test=matrix(c(1,2,3),nrow=10,ncol=3)
>
> I would like to have an output character vector where each line is
> row's values delimited by commas, e.g.:
>
> "1,2,3
> apply(test, 1, paste, collapse=",")
[1] "1,2,3" "2,3,1" "3,1,2" "1,2,3" "2,3,1" "3,1,2" "1,2,3" "2,3,1" "3,1,2"
[10] "1,2,3"
On Tue, Jun 15, 2010 at 2:27 PM, Jonathan Greenberg
wrote:
> Folks:
>
> Say I have a matrix:
>
> test=matrix(c(1,2,3),nrow=10,ncol=3)
>
> I would like to have an output
Trying to get 2.11.1 built on a solaris 10 machine. This particular
machine I do not have root access so I have just created an area under
my home to use. I am currently having issues in the src/library area
most specifically with getting the datasets area built. Here is the
output form my most rec
Hi Gabor,
Not sure where to report this, but
Mac 10.5.8
R: 11.1
When you examine the zoo vignette and hit the back button, you get a hang.
I havent tested with other vignettes and cant imagine that is is specific to
yours
FWIW.
Did I mention that zoo is great. Thx for your work on it.
On Tue,
Hi All,
I am trying to turn a Matrix into a vector for analysis purposes. I need to
select only certain columns from the entire matrix for the vector (intraday
time intervals). Also I need to transpose the Matrix (so times are in rows)
stack each successive new column on top of each other (lates
Folks:
Say I have a matrix:
test=matrix(c(1,2,3),nrow=10,ncol=3)
I would like to have an output character vector where each line is
row's values delimited by commas, e.g.:
"1,2,3"
"2,3,1"
"3,1,2"
...
"1,2,3"
What is the fastest way of doing this? I can paste() row-by-row but
this seems an ine
Dear R Users:
What I need to do is to plot two ellipses in the same graph (one for
each group). You can use either X1 or X2 as the X-axis and the other
variable as the Y-axis.
Also I need to show the 95% C.I. or P.I. of the overlapping part of the two
ellipses, and for each ellipse separately.
Hello again,
You are right: it corresponds to the DST time transition in France (I think
you have guessed it was France from my "Frenglish" style - sorry about it).
To get around this problem, I now use :
as.numeric(format(strptime("2010-03-28 02:00:00", format="%Y-%m-%d
%H:%M:%S"), format="%H")
Hello colleagues,
I have tried to use the package biglm. I want to specify a
multivariate regression with a weight.
I have imported a large dataset with the library(bigmemory). I load
the library (biglm) and specified a regression with a weight. But I
get everytime a error message like “object n
Hi. I am trying to do a nonlinear regression on a set of data with Monod
kinetics and Haldane inhibition. I am using the following commands to do
the nonlinear regression:
dce<-read.delim("data.txt", header = TRUE, sep = "\t", quote="\"", dec=".",
fill = TRUE, comment.char="")
dce.m1<-nls(rate~k
Hey.
Looking at EuStockMarkets[1:1860] -- the DAX.
ar() estimates an AR(1) model with \alpha_1 ~= 1 and high \sigma^2.
But forecasting from a given X_t_0 according to an AR(1) model with \alpha_1 =
1 should be X_t_0 for all X_t, t > t_0 . ?
How do you do this forecasting in R?
predict(ar(),n.
Dear All,
I revise my question about the problem I have.
Take a look at the article
http://www.jstatsoft.org/v09/i01
and download the attached code.
try to run one of the codes for example section 2.1 in R
here is the code
fossil <- read.table("fossil.dat",header=T)
x <- fossil$age
y <- 10*f
You can define a function that does just that: sum the 1s in Acc and divide
by the length of Acc. Then use tapply to apply the function for each
subject.
f=function(x){sum(as.numeric(as.character(x)))/length(x)}
tapply(d$Acc,list(d$S),f)
HTH,
Daniel
--
View this message in context:
http://r.78
On Jun 15, 2010, at 1:34 PM, Federico Calboli wrote:
Hi everyone,
I'm running a cox ph model on a dataset with a number of variables.
Each variable has a different number of missing data, so that
coxph() drops the individuals who are missing data at one or more
variables. Because of this
HI, Dear Greg and R community,
I have one question about the output of gbm package. the output of Boosting
should be f(x), from it , how to calculate the probability for each
observations in data set?
SInce it is stochastic, how can guarantee that each observation in training
data are selected at
On 15 Jun 2010, at 18:34, Federico Calboli wrote:
> I'm running a cox ph model on a dataset with a number of variables. Each
> variable has a different number of missing data, so that coxph() drops the
> individuals who are missing data at one or more variables. Because of this
> dropping (total
On Tue, Jun 15, 2010 at 1:54 PM, steven mosher wrote:
> Hi Gabor,
> Not sure where to report this, but
> Mac 10.5.8
> R: 11.1
> When you examine the zoo vignette and hit the back button, you get a hang.
> I havent tested with other vignettes and cant imagine that is is specific to
> yours
> FWIW.
I can't resist
http://www.lmgtfy.com/?q=set+windows+PATH+variable
On 6/15/2010 8:37 AM, yi li wrote:
i've just installed the package"xlsx",which is used for read,write,format
Excel 2007 files, as i loaded the package ,i was told to intall the package
rJava, the problem was that i didn't h
Hi everyone,
I'm running a cox ph model on a dataset with a number of variables. Each
variable has a different number of missing data, so that coxph() drops the
individuals who are missing data at one or more variables. Because of this
dropping (totally fine btw) I want to know how many events
Hello,
I am trying to compute MLE for non-Gaussian AR(1). The error term follows a
difference poisson distribution. This distribution has one parameter
(vector[2]).
So in total I want to estimate two parameters: the AR(1) paramter (vector[1])
and the distribution parameter.
My function is the
Hello,
I am trying to compute MLE for non-Gaussian AR(1). The error term follows a
difference poisson distribution. This distribution has one parameter
(vector[2]).
So in total I want to estimate two parameters: the AR(1) paramter (vector[1])
and the distribution parameter.
My function is
I find that sometimes the recode function in car is easier to use, while at
other times it's easier to use the recode in Hmisc. If both packages are
loaded, you can always use car::recode or Hmisc::recode to specify which
function you want.
hth,
David Freedman, CDC
--
View this message in conte
Gábor Csárdi wrote:
Dear all,
I would like to write some tests for my R package, and the usual
'tests' directory seemed like a good solution, but there is something
I cannot understand.
It is possible to supply .Rout.save files with the expected output for
the tests, which is great. But since t
On 15-Jun-10 16:01:24, William Dunlap wrote:
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] On Behalf Of
>> ted.hard...@manchester.ac.uk
>> Sent: Tuesday, June 15, 2010 8:49 AM
>> To: r-h...@stat.math.ethz.ch
>> Subject: [R] Unspecifie
Hello Hadley, Tormod and every one else.
I just published a post on my blog, giving the code and presenting an
example of it's use (on the Iris data set)
http://www.r-statistics.com/2010/06/clustergram-a-graph-for-visualizing-cluster-analyses-r-code/
I welcome any comments (pitfalls, suggestions
I found out what the problem is: when I start R Commander, some plug-ins
are automatically loaded and it seems that the problem comes from the
RcmdrPlugin.Export, more precisely from the Hmisc package (the plug-in
depends on it) which contains a recode() function too with the following
document
Dear Natalie,
recode() does work for factors; is it possible that you haven't put the
level names in quotes? Example:
> (f <- factor(rep(letters[1:3], 3)))
[1] a b c a b c a b c
Levels: a b c
> recode(f, " c('a', 'b') = 'A'; else='B' ")
[1] A A B A A B A A B
Levels: A B
Best,
John
---
On Mon, 14 Jun 2010, david hilton shanabrook wrote:
basically I need to create a sliding window in a string. a way to explain this
is:
v <-
c("a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y")
window <- 5
shift <- 2
I want a matrix of char
This solution also seems to be the fastest of the proposed options for
this data set:
library("rbenchmark")
benchmark(columns = c("test", "elapsed", "relative"), order = "elapsed",
apply =apply(iris[, -5], 2, tapply, iris$Species, mean),
with = with(iris, rowsum(iris[, -5], S
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of
> ted.hard...@manchester.ac.uk
> Sent: Tuesday, June 15, 2010 8:49 AM
> To: r-h...@stat.math.ethz.ch
> Subject: [R] Unspecified [upper] xlim/ylim?
>
> Greetings!
> I would like
Dear Alain,
Taking a stab in the dark here but I find that if my original column is
a factor then recode won't work for me. Would a simple ifelse statement
not do the same thing?
test$variable <- ifelse(test$x>=1 & test$x<=5,0,1)
On 15/06/2010 16:53, John Fox wrote:
Dear Alain,
I'm afraid
Dear Alain,
I'm afraid that I can't duplicate your problem. First, there is no recode
function in the Rcmdr package; it uses recode from car.
Here's a record of my Rcmdr session, using the recode dialog to generate the
recode command:
> test$variable <- recode(test$x, '1:5=0; else=1; ', as.fact
Greetings!
I would like to be able to specify a fixed (say) lower limit
for plotting, while leaving the upper limit "floating, when
plotting. The context is that the maximum in the data to be
plotted is unpredictable, being the consequence of a simulation,
whereas I know that it cannot be less than
Hello,
I am reading "Using The foreach Package" document and I have tried the
following:
-
> sessionInfo()
R version 2.10.1 (2009-12-14)
i386-pc-mingw32
locale:
[1] LC_COLLATE=German_Switzerland.1252
LC_CTYPE=German_Switzerland
Hello,
I am using the recode() function in Rcmdr and the result is not what I
expect so I am almost sure I did something wrong but what...
> test <- data.frame(x=1:10)
> library(car)
> recode(test$x,'1:5=0 ; else=1', as.factor.result=TRUE)
[1] 0 0 0 0 0 1 1 1 1 1
Levels: 0 1
BUT
> library(R
Almost continuously I get some variant of the following message
fillingb up my R console:
> bubble(data1, "logvol")
> ?variogram
** (evince:2799): WARNING **: DBus error
org.freedesktop.DBus.Error.Failed: Unable to set metadata key
** (evince:2799): WARNING **: DBus error
org.freedesktop.DBus.Err
On Jun 15, 2010, at 9:05 AM, Frankvb wrote:
Dear all,
Currently I am trying to integrate a function which depends on four
variables, two of which are given, one is given in the integrate
function,
so there is one variable to integrate on.
The code is as follows:
Pmatrix =
function(th) {
Hi Hadley,
I wrapped the code into a function.
I made it so all the lines would always start from the cluster mean.
And I tried to give more meaning to the colors by giving the
color according the the order of the first principal component of that
observation.
What do you think ?
Tal
# -
On Jun 15, 2010, at 7:45 AM, PtitBleu wrote:
Hello,
In one of my functions, I need to extract the hour from a date.
For example:
as.POSIXlt("2010-03-27 02:00:00")$hour gives 2 as expected.
It works for all the dates I've tested except the following one
which is in
my list of dates:
as.POSI
I use two PCs for my R programming, one Win32 and one Linux (Ubuntu 10).
In Windows I use Tinn-R to save my code and occasionally I am unable to
open a Tinn-R saved file in Ubuntu. If I try to open in gedit I get a
message that "gedit has not been able to detect the character encoding."
To correct
Frank,
The trouble is that your integrand function `int' is not vectorized. You
can use `Vectorize' or `sapply' to rectify this.
int.vec = function(theta, s, a, lambda) {
sapply(theta, function(t,s,a,lambda) int(t,s,a,lambda), s=s, a=a,
lambda=lambda)
}
integrate(int.vec,lower=0.0001,u
In context below:
On Jun 15, 2010, at 4:20 AM, Yogesh Tiwari wrote:
Dear John,
Great thanks,
I would appreciate if any expert can throw more light on it.
Thanks in advance,
Best Regards,
Yogesh
On 6/14/10, John Kane wrote:
Hi Yogesh,
I think you accidentelly replied only to me and not to
> The glitches are the cases where you would have a bundle of lines belonging
> to a specific cluster, but had spaces between them (because the place of one
> of the lines was saved for another line that in the meantime moved to
> another cluster).
I think that display looked just fine!
> I just
Dear all,
I would like to write some tests for my R package, and the usual
'tests' directory seemed like a good solution, but there is something
I cannot understand.
It is possible to supply .Rout.save files with the expected output for
the tests, which is great. But since the tests are not run w
On Tue, Jun 15, 2010 at 8:27 AM, skan wrote:
>
> Hello
>
> Where could I find examples on how to work with the time index in a
> timeseries or zoo series?
>
> Let say I've got this series
>
> DATA
> 1990-01-01 10:00:00 0.900
> 1990-01-01 10:01:00 0.910
> 1990-01-01 10:03:00 0.905
> 1990-01-
On 15.06.2010 13:45, PtitBleu wrote:
Hello,
In one of my functions, I need to extract the hour from a date.
For example:
as.POSIXlt("2010-03-27 02:00:00")$hour gives 2 as expected.
It works for all the dates I've tested except the following one which is in
my list of dates:
as.POSIXlt("2010-0
Dear All,
I revise my question about the problem I have.
Take a look at the article
http://www.jstatsoft.org/v09/i01
and download the attached code.
try to run one of the codes for example section 2.1 in R
here is the code
fossil <- read.table("fossil.dat",header=T)
x <- fossil$age
y <- 10*fo
Hi Hadley,
Thanks for replying.
The glitches are the cases where you would have a bundle of lines belonging
to a specific cluster, but had spaces between them (because the place of one
of the lines was saved for another line that in the meantime moved to
another cluster).
I just came up with a so
Dear all,
Currently I am trying to integrate a function which depends on four
variables, two of which are given, one is given in the integrate function,
so there is one variable to integrate on.
The code is as follows:
Pmatrix =
function(th) {
P = matrix(nrow=6, ncol=6, data=0)
> showMethods("apply")
Function: apply (package base)
X="ANY"
X="missing"
(inherited from: X="ANY")
X="timeSeries"
On Tue, Jun 15, 2010 at 15:10, Gavin Simpson wrote:
> On Tue, 2010-06-15 at 14:56 +0200, Sergey Goriatchev wrote:
>> Maybe I have to much stuff loaded in the workspace, Gavin, yo
On Tue, 2010-06-15 at 14:56 +0200, Sergey Goriatchev wrote:
> Maybe I have to much stuff loaded in the workspace, Gavin, you are right:
OK, so now do
showMethods("apply")
And R should list out the available methods. See which package
(re)defines apply.
But it is likely going to be simpler to st
Yes I had a data.frame but did not named the categorical variable "class" and
the numerical one "var". It is working now.
However I would like to do a comparison with several classes and variables.
Is there a way of doing this? Or do I have to do a Wilcox.test for every
variable?
Thanks
Marie.
-
Hello,
In one of my functions, I need to extract the hour from a date.
For example:
as.POSIXlt("2010-03-27 02:00:00")$hour gives 2 as expected.
It works for all the dates I've tested except the following one which is in
my list of dates:
as.POSIXlt("2010-03-28 02:00:00")$hour which gives 0.
I do
Hi Daniel, thanks for your reply. Unfortunately, that is not doing what I
need. In the example I sent, there are three subjects (S1, S2 & S3). Each
subject has 3 trials worth of data and each trial has 10 samples. What I
want to return is the accuracy rate for each subject. The answer is 66.6
Dear John,
Great thanks,
I would appreciate if any expert can throw more light on it.
Thanks in advance,
Best Regards,
Yogesh
On 6/14/10, John Kane wrote:
> Hi Yogesh,
>
> I think you accidentelly replied only to me and not to the mailing list.
> I'd suggest that you copy your post to me to the
Hello
Where could I find examples on how to work with the time index in a
timeseries or zoo series?
Let say I've got this series
DATA
1990-01-01 10:00:00 0.900
1990-01-01 10:01:00 0.910
1990-01-01 10:03:00 0.905
1990-01-01 10:04:00 0.905
1990-01-01 10:05:00 0.890
...
Maybe I have to much stuff loaded in the workspace, Gavin, you are right:
> sessionInfo()
R version 2.10.1 (2009-12-14)
i386-pc-mingw32
locale:
[1] LC_COLLATE=German_Switzerland.1252
LC_CTYPE=German_Switzerland.1252
LC_MONETARY=German_Switzerland.1252
[4] LC_NUMERIC=CLC_TI
This is also an interesting discussion to read in this respect :
http://tolstoy.newcastle.edu.au/R/e8/help/09/10/1228.html
Cheers
Joris
On Mon, Jun 14, 2010 at 5:55 PM, Karl-Dieter Crisman wrote:
>> Looking for a recommended package that handles prime number computations.
>
> I'm not sure whethe
On Tue, 2010-06-15 at 14:38 +0200, Sergey Goriatchev wrote:
> Erik, I see the following when I type "apply" at the prompt:
>
> > apply
> standardGeneric for "apply" defined from package "base"
Looks like you have something loaded in your workspace (or have created
something) that has altered the
Sergey Goriatchev wrote:
Erik, I see the following when I type "apply" at the prompt:
apply
standardGeneric for "apply" defined from package "base"
function (X, MARGIN, FUN, ...)
standardGeneric("apply")
Methods may be defined for arguments: X, MARGIN, FUN
Use showMethods("apply") for curr
1 - 100 of 120 matches
Mail list logo
