?do.call
/H
On Sun, Nov 1, 2009 at 10:58 PM, Charlie Sharpsteen
wrote:
> On Sun, Nov 1, 2009 at 8:07 PM, Ning Ma wrote:
>>
>> Hi, everybody
>>
>> Is there any way to execute a function, which name is stored in a string.
>> such as:
>> a <- "ls()"
>> foo(a) ## same as ls() itself.
>
> One way to
Can someone provide me a good way to circumvent the lack of calling
Stored Procedures from RMySQL?
I can not rewrite those stored procedures on R because there a lot
more folks here that only understands SQL.
The stored procedure returns a resultset.
Thanks
Caveman
_
On Sun, Nov 1, 2009 at 8:07 PM, Ning Ma wrote:
>
> Hi, everybody
>
> Is there any way to execute a function, which name is stored in a string.
> such as:
> a <- "ls()"
> foo(a) ## same as ls() itself.
One way to accomplish this by using get() to search for a function
that matches your string. You
parkbomee writes:
> Thank you all.
>
> What Chuck has suggested might not be applicable since the number of
> different times is around 40,000.
>
> The object of optimization in my function is the varying "value",
> which is basically data * parameter, of which "parameter" is the
> object of opti
Dear R users,
I wish to utilise processed and saved objects as arguments of a function.
Specifically, I have created objects using *"assign"* & *"paste"* functions
with an incremental index i, the names of the objects are:
fund1, fund2, fund3,., fund80,. (where the numerical value
in
On Sun, Nov 1, 2009 at 9:18 PM, David Winsemius wrote:
>
> On Nov 1, 2009, at 11:28 PM, Henrik Bengtsson wrote:
>
>> On Sun, Nov 1, 2009 at 7:48 PM, David Winsemius
>> wrote:
>>>
>>> On Nov 1, 2009, at 10:16 PM, Henrik Bengtsson wrote:
>>>
path <- "data";
dir.create(path);
for
On Nov 1, 2009, at 11:28 PM, Henrik Bengtsson wrote:
On Sun, Nov 1, 2009 at 7:48 PM, David Winsemius > wrote:
On Nov 1, 2009, at 10:16 PM, Henrik Bengtsson wrote:
path <- "data";
dir.create(path);
for (i in 1:10) {
m <- i:5;
filename <- sprintf("m%02d.Rbin", i);
pathname <- file.path(pat
Erin Hodgess schrieb:
xy
[1] 0.7305081 2.4224211
str(xy)
num [1:2] 0.73 2.42
any(xy) > 1
[1] FALSE
Warning message:
In any(xy) : coercing argument of type 'double' to logical
What am I doing wrong please?
xy > 1 should return TRUE FALSE, and you want to appl
Dear R People:
Sorry to be so pesky tonight. I figured it out.
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
__
R-help@r-project.org mailing lis
Dear R People:
I am working with a numeric vector and trying to use the "any"
function. However, I'm getting some odd results, as shown below:
> xy
[1] 0.7305081 2.4224211
> str(xy)
num [1:2] 0.73 2.42
> any(xy) > 1
[1] FALSE
Warning message:
In any(xy) : coercing argument of type 'double' to l
On Nov 1, 2009, at 11:07 PM, Ning Ma wrote:
Hi, everybody
Is there any way to execute a function, which name is stored in a
string.
such as:
a <- "ls()"
foo(a) ## same as ls() itself.
Need to leave the "()" off.
> fstr <- "sum"
> eval(parse(text=fstr))(1:5)
[1] 15
Or, to execute a R c
Got it!
There is an xxx$model$theta that does the trick.
Thanks,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
__
R-help@r-project.org maili
On Sun, Nov 1, 2009 at 7:48 PM, David Winsemius wrote:
>
> On Nov 1, 2009, at 10:16 PM, Henrik Bengtsson wrote:
>
>> path <- "data";
>> dir.create(path);
>>
>> for (i in 1:10) {
>> m <- i:5;
>> filename <- sprintf("m%02d.Rbin", i);
>> pathname <- file.path(path, filename);
>> save(m, file=path
Dear R People:
I have the output from an arima model fit in an object xxx.
I want to verify that the ma1 coefficient is there, so I did the following:
> xxx$coef
ar1ar2ma1 intercept
1.3841297 -0.4985667 -0.996 -0.1091657
> str(xxx$coef)
Named num [1:4] 1.384 -0.499 -
Hi, everybody
Is there any way to execute a function, which name is stored in a string.
such as:
a <- "ls()"
foo(a) ## same as ls() itself.
Or, to execute a R command, which is stored in a string
such as:
a <- "m1 <- matrix(1:9,3,3)"
foo(a) ## same as the assignment itself
__
On Nov 1, 2009, at 10:16 PM, Henrik Bengtsson wrote:
path <- "data";
dir.create(path);
for (i in 1:10) {
m <- i:5;
filename <- sprintf("m%02d.Rbin", i);
pathname <- file.path(path, filename);
save(m, file=pathname);
}
That would result in each of the ten files containing an object with
There is no package that does this, but you can do it yourself pretty
easily (although I think that it is a waste of time to bag this
particular model). You can use the function bagEarth in the caret
package as a prototype.
Max
On Sun, Nov 1, 2009 at 8:32 PM, 柯洁 wrote:
>
> Dear sir,
> If I want
path <- "data";
dir.create(path);
for (i in 1:10) {
m <- i:5;
filename <- sprintf("m%02d.Rbin", i);
pathname <- file.path(path, filename);
save(m, file=pathname);
}
/H
On Sun, Nov 1, 2009 at 6:53 PM, jeffc wrote:
>
> Hi,
>
> I would like to save a few dynamically created objects to disk
Dear sir,
If I want to use bagging with SVM, which package should I choose?Thanks!
Best wishes,Jie
_
[[alternative HTML version deleted]]
__
Hi,
I would like to save a few dynamically created objects to disk. The
following is the basic flow of the code segment
for(i = 1:10) {
m = i:5
save(m, file = ...) ## ???
}
To distinguish different objects to be saved, I would like to save m as m1,
m2, m3 ..., to file /home/data/m1, /home/
Can you at least provide an 'str' of the 'task' object (not sure if it
is a dataframe; you said a 'list') so that we know what it looks like.
It would also be helpful if you would provide a subset of the data
that we could try out a script on it. The best way would be to post
the first 10 or so r
it appears that what you really want is to use:
task[[i]]
instead of task[i]
b
On Nov 1, 2009, at 11:04 PM, dadrivr wrote:
I would like to preface this by saying that I am new to R, so I
would ask
that you be patient and thorough, so that I'm not completely
clueless. I am
trying to con
Katja Seis wrote:
>
> Hi,
> my data frame consist of 8 Variables and 120 000 observations. With those
> datas I am running a PCA and after I want to calculate the Volume of the
> PCA-cloud of certain subsets of my data. Does anyone have an idea about a
> function that can do this?
>
>
Not re
The manual on Writing R Extensions says that one should ``only
document a
***single*** data object per Rd file''. My recollection is that in the
past this commandment (which I occasionally found to be inconvenient)
was enforced.
I recently saw a post to R-help which appeared to indicate that
I would like to preface this by saying that I am new to R, so I would ask
that you be patient and thorough, so that I'm not completely clueless. I am
trying to convert a list to numeric so that I can perform computations on it
(specifically mean-center the variable), but I am running into proble
introduce a factor variable with the months and then use tapply?
Kjetil
On Sun, Nov 1, 2009 at 9:07 PM, Tim Bean wrote:
> Hello, I have a quick question about time series methodology. If I want to
> display a boxplot of time series data, sorted by period, I can type:
>
> boxplot(data ~ cycle(dat
Hello, I have a quick question about time series methodology. If I want to
display a boxplot of time series data, sorted by period, I can type:
boxplot(data ~ cycle(data));
where data is of class "ts"
Is there a similar method for calculating, say, the median value of each
time step within the s
On Nov 1, 2009, at 3:12 PM, Charles C. Berry wrote:
On Sun, 1 Nov 2009, David Winsemius wrote:
On Nov 1, 2009, at 1:46 PM, spencerg wrote:
Hi, Chuck:
Thanks very much, but why do I get "package 'expm' is not
available"
from install.packages("expm",repos="http://R-Forge.R-
project
Jim and John,
Thanks for your replies. I ended up using both suggestions to plot the full
tide series and then overlay the averages for rise and fall in different
colours, which illustrated very well how such parameters can
be misleadings in some cases.
Regards,
Tony
2009/10/31 Jim Lemon
> O
You just need to extract the list component named "value":
I1$value
Ravi.
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of Geriatric Medicine and Gerontology
School of Medicine
Johns Hopkins University
Ph. (410) 502-2619
Peter Dalgaard wrote:
Kevin E. Thorpe wrote:
I'm sure this is simple enough, but an R site search on my subject
terms did suggest a solution. I have a numeric vector with many
values that I wish to create a factor from having only a few levels.
Here is a toy example.
> x <- 1:10
> x <-
fact
Kevin E. Thorpe wrote:
I'm sure this is simple enough, but an R site search on my subject
terms did suggest a solution. I have a numeric vector with many
values that I wish to create a factor from having only a few levels.
Here is a toy example.
> x <- 1:10
> x <-
factor(x,levels=1:10,labels
On Nov 1, 2009, at 3:51 PM, Kevin E. Thorpe wrote:
I'm sure this is simple enough, but an R site search on my subject
terms did suggest a solution. I have a numeric vector with many
values that I wish to create a factor from having only a few levels.
Here is a toy example.
> x <- 1:10
> x <-
Thank you all.
What Chuck has suggested might not be applicable since the number of different
times is around 40,000.
The object of optimization in my function is the varying "value", which is
basically data * parameter, of which "parameter" is the object of optimization..
And from the r prof
Hi Kevin,
Here are two suggestions:
# Combination of levels() and table()
table(levels(x))
# A B C
# 3 3 4
# Or defining a function
mysummary <- function(x) table(levels(x)) # you can easily improve it :-)
mysummary(x)
# A B C
# 3 3 4
HTH,
Jorge
On Sun, Nov 1, 2009 at 3:51 PM, Kevin E. Thorpe
I'm sure this is simple enough, but an R site search on my subject
terms did suggest a solution. I have a numeric vector with many
values that I wish to create a factor from having only a few levels.
Here is a toy example.
> x <- 1:10
> x <-
factor(x,levels=1:10,labels=c("A","A","A","B","B","B"
On Tue, Oct 20, 2009 at 8:14 AM, Ted Harding
wrote:
> On 20-Oct-09 13:34:49, Peng Yu wrote:
>> fisher.test() gives a very small p-value, which is underflow on my
>> machine. However, the log of it should not be underflow. I'm wondering
>> if there is a way get log() of a small p-value. An approxim
On Sun, 1 Nov 2009, David Winsemius wrote:
On Nov 1, 2009, at 1:46 PM, spencerg wrote:
Hi, Chuck:
Thanks very much, but why do I get "package 'expm' is not available"
from install.packages("expm",repos="http://R-Forge.R-project.org";)?
In my case I think it was it is because there
Hi,
my data frame consist of 8 Variables and 120 000 observations. With those datas
I am running a PCA and after I want to calculate the Volume of the PCA-cloud of
certain subsets of my data. Does anyone have an idea about a function that can
do this?
Thanks
[[alternative HTML
Hi Laila,
Here is a suggestion:
res <- integrate(dnorm, -1.96, 1.96)
res
# 0.9500042 with absolute error < 1.0e-11
res[[1]]
# [1] 0.9500042
res$value
# [1] 0.9500042
HTH,
Jorge
On Sun, Nov 1, 2009 at 3:02 PM, Laila Alkhalfan <> wrote:
> Hi
> Can we get the result of an intigration without
Hi
Can we get the result of an intigration without the absolute error?
for example
f1<-function(x1){(1/gamma(alpha))*x1^(alpha-1)*exp(-x1)*log(x1)}
I1<-integrate(f1, 0, (max(cc)-tau1+(theta2/theta1)*tau1)/theta2)
I1
0.08007414 with absolute error < 7.2e-05
I need the answer 0.08007414 withou the
What you need to do is to understand how to use Rprof so that you can
determine where the time is being spent. It probably indicates that
this is not the source of slowness in your optimization function. How
much time are we talking about? You may spent more time trying to
optimize the function
On 2/11/2009, at 2:49 AM, Laura Saltyte wrote:
Hello,
I have daily wind speed data and need
``need'' is probably not the operative word!
to fit seasonal ARIMA model, problem
is that my period is 365. But when I use arima(...) function, with
period
365, I’m getting error message:
On Nov 1, 2009, at 1:46 PM, spencerg wrote:
Hi, Chuck:
Thanks very much, but why do I get "package 'expm' is not
available" from install.packages("expm",repos="http://R-Forge.R-project.org
")?
In my case I think it was it is because there is no 2.10 branch to
either the:
http://r-
Hi,
Here is one approach to solve your likelihood maximization subject to
constraints. I have written a function called `constrOptim.nl' that can solve
constrained optimization problems. I will demonstrate this with a simulation.
# 1. Simulate the data (x,y).
a <- 4
b <- 1
c <- 2.5
p <- 0.
Hi, Chuck:
Thanks very much, but why do I get "package 'expm' is not
available" from
install.packages("expm",repos="http://R-Forge.R-project.org";)?
Best Wishes,
Spencer Graves
Charles C. Berry wrote:
On Sun, 1 Nov 2009, spencerg wrote:
A question, a comment, and an a
Confirmed with recent Zelig and R-2.10.0.
CCing Kosuke Imai, the Zelig maintainer:
Please can you take a look at this one and additionally fix the warnings
given on your package's overview page at
http://cran.r-project.org/web/checks/check_results_Zelig.html
Please upload a fixed version with
Hello,
I have daily wind speed data and need to fit seasonal ARIMA model, problem
is that my period is 365. But when I use arima(...) function, with period
365, Im getting error message: Error in makeARIMA(trarma[[1]],
trarma[[2]], Delta, kappa) : maximum supported lag is 350. Can someone
hel
On Sat, 31 Oct 2009, parkbomee wrote:
Thank you both.
However, using tapply() instead of a loop does not seem to improve my code much.
I am using this inside of an optimization function,
and it still takes more than it needs...
Well, you haven't given us much to work with.
The optimizatio
Arne Schulz wrote:
Dear list,
my problem seems to be primarily a statistical one, but maybe there is a
misspecification within R (and hopefully a solution).
I have two groups with two measured variables as training data. According to
the variables, the groups differ totally. I know that this i
On Sun, 1 Nov 2009, spencerg wrote:
A question, a comment, and an alternative answer to matrix^(-1/2):
QUESTION:
What's the status of the "expm" package, mentioned in the email you cited
from Martin Maechler, dated Apr 5 19:52:09 CEST 2008? I tried both
install.packages('expm') and
instal
A question, a comment, and an alternative answer to matrix^(-1/2):
QUESTION:
What's the status of the "expm" package, mentioned in the email you
cited from Martin Maechler, dated Apr 5 19:52:09 CEST 2008? I tried both
install.packages('expm') and
install.packages("expm",repos="http://R-For
Duh, thought of that after I'd left for dinner :(
--- On Sat, 10/31/09, David Winsemius wrote:
> From: David Winsemius
> Subject: Re: [R] how to loop thru a matrix or data frame , and append
> calculations to a new data frame?
> To: "John Kane"
> Cc: r-help@r-project.org, "Robert Wilkins"
>
On Oct 31, 2009, at 9:33 PM, David Winsemius wrote:
On Oct 31, 2009, at 4:39 PM, Kajan Saied wrote:
Dear R-Help Team,
as a R novice I have a (maybe for you very simple question), how do
I get
the following solved in R:
Let R be a n x n matrix:
\mid R\mid^{-\frac{1}{2}}
solve(A) gives
Dear Stefan,
See two comments inserted below.
Stefan Grosse wrote:
On Sun, 1 Nov 2009 00:47:50 -0700 (PDT) jomni wrote:
J> So do I write the function as wilcox.test(original, test,
J> alternative="l")? or wlcox.test(original, test, alternative = "g")?
J> or wilcox.test(test, original, alterna
Hi Jason,
If I understand correctly, you are looking for something along the lines of
with(X, tapply(author, articleID, function(x) length(unique(x
# 1 2 3
# 1 2 2
with X your data frame.
HTH,
Jorge
On Sun, Nov 1, 2009 at 1:20 AM, Jason Priem <> wrote:
> I've got a data frame describing
Dear R Experts,
I have a query concerning SEMs in a multilevel context. I have an unbalanced
panel where children are nested in families, which in turn are nested in
districts. The problem is the following: the outcome variable of interest is
measured at the child level but the (explanatory) la
Hi R Users,
When I use package lme4 for mixed model analysis, I can't distinguish
the significant and insignificant variables from all random independent
variables.
Here is my data and result:
Data:
Rice<-data.frame(Yield=c(8,7,4,9,7,6,9,8,8,8,7,5,9,9,5,7,7,8,8,8,4,8,6,4,8,8,9),
I did this on the source files which were semi-colon delimted (to delimit the
fields, I am not sure what character denotes the new tweet)
After loading the tm package
> txt <- system.file("texts", "txt", package = "tm")
> (twitter <- Corpus(DirSource(txt),
+ readerControl = list(language = "lat"
On Nov 1, 2009, at 8:24 AM, onyourmark wrote:
Hello. The "fields" are separated by a ';'. I think that the data is
"rectangular" in the sense that there are about 15 fields for each
row.
There either are 15 fields or there aren't. You can't make a dataframe
with an approximate number of
Dear All,
I'm doing an imputation on the missing data and is looking for a decomposition
method in R. Could somone advice me the way to do this?
Thank you
Fir
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
On 01/11/2009 7:43 AM, onyourmark wrote:
Hi. I have a huge list called twitter:
It's a list, but more importantly it's a VCorpus and a Corpus. You
should use the functions appropriate to those classes to extract the
strings making up the data, declare their encoding properly (or convert
the
Hello. The "fields" are separated by a ';'. I think that the data is
"rectangular" in the sense that there are about 15 fields for each row. Some
of the fields are empty. In the dput() display below, it seems that the rows
are delimited by ' " ' .
Any idea from this?
Here is the end of the output
Three suggestions:
-- drop the idea of using a dataframe. It's only appropriate when the
data is rectangular.
-- look at strsplit for separating at "@" characters.
-- post the output of dput() on your sample, since email is probably
not capable of rendering this data without creating distort
On Nov 1, 2009, at 1:59 AM, Patrick Connolly wrote:
On Sun, 01-Nov-2009 at 01:20AM -0500, Jason Priem wrote:
I've got a data frame describing comments on an electronic journal,
wherein each row is a unique comment, like so:
commentID author articleID
1 1 smith 2
2
On Fri, Oct 30, 2009 at 8:33 PM, Ottorino-Luca Pantani
wrote:
> Dear R users,
> this is a follow up of this message
> http://tolstoy.newcastle.edu.au/R/e6/help/09/05/13897.html
> I'm reproducing the core of it for convenience.
>
>> //
>> / data(Oats, package = "MEMSS") /
>> / tp1.oats <- xyplot(yi
Hi. I have a huge list called twitter:
> dim(twitter)
NULL
> str(twitter)
List of 1
$ :Classes 'PlainTextDocument', 'TextDocument', 'character' atomic
[1:35575] 11999;10:47:14;20;10;2009;ObamaLouverture;Trails Mixed Lessons For
Governance From Campaigner-in-chief: President obama jumps campaig
As from subject: is it possible to use the rect.hclust (or something
equivalent) with horizontal dendrograms?
thanks
nico
--
View this message in context:
http://old.nabble.com/rect.hclust-and-horizontal-dendrograms-tp26148886p26148886.html
Sent from the R help mailing list archive at Nabble.co
On Sun, 1 Nov 2009 00:47:50 -0700 (PDT) jomni wrote:
J> So do I write the function as wilcox.test(original, test,
J> alternative="l")? or wlcox.test(original, test, alternative = "g")?
J> or wilcox.test(test, original, alternative="g")?
J> or wilcox.test(test, original, alternative="l")?
J> How
Hi, I am very confused with constructing the wilcox.test in R.
I have two populations 'original' and 'test'.
I want to know if the 'test' is generally 'lower' than original.
I use alpha of 0.05.
So do I write the function as wilcox.test(original, test, alternative="l")?
or wlcox.test(original, t
On Sun, 01-Nov-2009 at 01:20AM -0500, Jason Priem wrote:
> I've got a data frame describing comments on an electronic journal,
> wherein each row is a unique comment, like so:
>
> commentID author articleID
> 1 1 smith 2
> 2 2 jones 3
> 3 3 andrews
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