Make sure to look at
as.character(FnO_Data$Date[m:l])
My $.02
/Henrik
On Sat, Sep 18, 2010 at 10:31 PM, Santosh Srinivas
wrote:
> I tried this and it works too (For most part) strangely for certain
> dates (20090831) it is giving NA ...
>
>> FnO_Data$Date[m:l]
> [1] 20090828 20090828 200
I tried this and it works too (For most part) strangely for certain
dates (20090831) it is giving NA ...
> FnO_Data$Date[m:l]
[1] 20090828 20090828 20090828 20090828 20090828 20090828 20090828 20090828
20090828 20090828 20090828 20090828
[13] 20090828 20090828 20090828 20090828 20090828 2009
> have <- list(a=7,b=3,c=1)
> have2 <- lapply(have, rep, 2)
> have2
$a
[1] 7 7
$b
[1] 3 3
$c
[1] 1 1
>
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Derek Ogle
Sent: Sunday, 19 September 2010 11:20 AM
To: R (r-help@R-project.
Ahoy maties,
As I'm sure everyone is aware, today (ISOdate(2010,9,19)) is International
Talk Like A Pirate Day.
I hope Arrr-help and Arrr-devel, and the whole of the Arrr community will be
observing the day properly.
__
R-help@r-project.org mailin
On 2010-09-07 13:58, Cable, Samuel B Civ USAF AFMC AFRL/RVBXI wrote:
I would like to open an existing netCDF file and add a variable to it.
I am using the ncdf package. This test code gives the idea of what I am
trying to do:
library(ncdf)
print('here we go')
print('first, construct netCDF fi
On Fri, 17 Sep 2010, Greg Snow wrote:
I think what is going on (and someone is likely to correct me otherwise) is
that formulas have an associated environment that gets passed along with them
while character strings do not.
Yes.
model.frame(), which is called from most modelling functions,
Thanks to Dennis for solving my first question and also pointing me in the
right direction.
To complete the thread ... there are times when I need to have different
numbers of repeats for each object in the list. This can be accomplished with
mapply() as follows ...
> mapply(rep,x=have,c(1,1,
Hi:
How about
> have <- list(a=7,b=3,c=1)
> lapply(have, rep, 2)
$a
[1] 7 7
$b
[1] 3 3
$c
[1] 1 1
HTH,
Dennis
On Sat, Sep 18, 2010 at 6:19 PM, Derek Ogle wrote:
> I have a list that looks like this ...
>
> > have <- list(a=7,b=3,c=1)
> > have
> $a
> [1] 7
>
> $b
> [1] 3
>
> $c
> [1] 1
>
>
I have a list that looks like this ...
> have <- list(a=7,b=3,c=1)
> have
$a
[1] 7
$b
[1] 3
$c
[1] 1
and I want to have a simple way to change it to the following without re-typing
the values ...
> desire <- list(a=c(7,7),b=c(3,3),c=c(1,1))
> desire
$a
[1] 7 7
$b
[1] 3 3
$c
[1] 1 1
In othe
Dear Ravi
thanks for your reply. I think the number of obs may do be the problem for a
zero hessian matrix. since it is simulated data, I have increased the sample
size to 500 obs and tried for around 20 times, the zero hessian did not appear,
compared with the fact that it happens too often wi
Hi, Kenneth
It is not clear if you mean that your pdf output usually works, but it
does not in this special case, or that this is a first effort with
pdf. The answer might depend on which is the case case.
If you are just getting started, can I refer you to some lecture notes
I made about saving
as, silly me.
clearG()
this now works!
--
View this message in context:
http://r.789695.n4.nabble.com/getting-a-function-to-do-something-tp2545594p2545596.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org maili
Hi,
I want to repeatedly do a task, so thought I could put it in a function and
then just call the function.
The task is just clearing all the graphics devices and then opening a new
one of a specified size.
Now, when I call the function below, nothing appears to happen. But when I
run the 2 lin
When saving a data frame with long character variable, approximately longer
than 3 characters, to a text field in PostgreSQL, using RODBC on Windows
XP, I get a timestamp saved to the database instead. Is there any way to
extend the number of characters that a text variable can receive?
Here a
On 09/18/2010 03:32 AM, Graham Smith wrote:
...
What I am hoping for is:
an x-axis labelled "Distance" with tick marks at 0, 5, 10,15,20 and 25.
And, an y-axis labelled "Species" and tick marks labelled A, B and C.
So I would appreciate some help on how the data should be prepared for
kiteCha
Hello all-
Lets say I have my dependent variable Y_ij and a time varying covariate
X_ij. I want to formulate some regression model to check if X_ij is
dependent on X_i1,X_i2,,X_i(j-1) and Y_i1,Y_i2,,Y_i(j-1). I want to
test no dependence on the Y component.
I'm thinking a simple linear r
On 2010-09-18 13:14, Joshua Wiley wrote:
Hello Ben,
I believe the problem is not strictly update() in the for loop, but
the formula being passed to update(). The value of "i" changes, but
its symbol does not, so the second time through its like you overwrite
the first. Here is one way around i
I was able to get proper convergence in "BFGS", when I use the starting value
from Nelder-Mead with 5000 iterations.
However, the hessian is not positive-definite. This indicates that you have a
problem in your model. It seems to me that the model is over-parametrized.
You have 20-odd parame
Hello Ben,
I believe the problem is not strictly update() in the for loop, but
the formula being passed to update(). The value of "i" changes, but
its symbol does not, so the second time through its like you overwrite
the first. Here is one way around it that works for me:
mdat <- matrix(c(1,2
Hi,
First let me say I am a big fan of R and appreciate all your time and
effort.
The update() function does not seem to work in a for loop. Consider the
following:
mdat <- matrix(c(1,2,3, 11,23,13, 12,4,8), nrow = 3, ncol=3, byrow=TRUE)
reg <- lm(mdat[7:9]~1)
for(i in 1:2) {
reg <- update(reg,
On Sep 18, 2010, at 11:36 AM, David Winsemius wrote:
On Sep 18, 2010, at 11:25 AM, Santosh Srinivas wrote:
Strangely this is not working ... what am I doing wrong here?
tDate <- FnO_Data$Date[1]
tDate
[1] 20090101
as.Date(c(tDate),format="%Y%m%d")
[1] NA
?sasDate
as.Date does not tak
Thanks.
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: 18 September 2010 21:07
To: Santosh Srinivas
Cc: r-help@r-project.org
Subject: Re: [R] Date issues
On Sep 18, 2010, at 11:25 AM, Santosh Srinivas wrote:
> Strangely this is not working ... what am I
On Sat, Sep 18, 2010 at 11:25 AM, Santosh Srinivas
wrote:
> Strangely this is not working ... what am I doing wrong here?
>
>> tDate <- FnO_Data$Date[1]
>> tDate
> [1] 20090101
>> as.Date(c(tDate),format="%Y%m%d")
> [1] NA
>
Do you have zoo loaded? If you do then a minimal reproducible example
(
On Sep 18, 2010, at 10:25 AM, Santosh Srinivas wrote:
> Strangely this is not working ... what am I doing wrong here?
>
>> tDate <- FnO_Data$Date[1]
>> tDate
> [1] 20090101
>> as.Date(c(tDate),format="%Y%m%d")
> [1] NA
What version of R are you running?
What is the output of:
str(FnO_Data$D
Hey, R Users
a few days ago I have met a zero hessian with optim command. I reproduced it
with simulated data. plz check the code and data at the bottom of the post. I
also attachment them with this email. hope it can reduce some workload as
copying and pasting.
I have simunated data many time
On Sep 18, 2010, at 11:25 AM, Santosh Srinivas wrote:
Strangely this is not working ... what am I doing wrong here?
tDate <- FnO_Data$Date[1]
tDate
[1] 20090101
as.Date(c(tDate),format="%Y%m%d")
[1] NA
?sasDate
as.Date does not take numeric arguments. Try:
> as.Date(as.character(tDate)
Strangely this is not working ... what am I doing wrong here?
> tDate <- FnO_Data$Date[1]
> tDate
[1] 20090101
> as.Date(c(tDate),format="%Y%m%d")
[1] NA
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read th
On Sep 18, 2010, at 12:28 AM, Stephen Liu wrote:
Hi Erik,
I try to find out whether Emacs/ESS is available on R repo. What is
its
versiohn?
There is no R repository that distributes Emacs. Emacs and the ESS
package (for Emacs) are not part of R. You choose your version of
Emacs and
See if rowttests is any faster.
library(genefilter)
?rowttests
You have to install Bioconductor. I've used this on large datasets,
but I haven't compared
timings.
On Mon, Sep 13, 2010 at 4:26 PM, Alexey Ush wrote:
> Hello,
>
> I have a question regarding how to speed up the t.test on large dat
Thanks Michael,
Quite an effort! I failed to explain myself sufficiently clearly, what I need
is for this if possible, to work with more than 4 treats, but still produce a
matrix with all possible combinations of treatments (minus whichever treatment
is in c1) in c2 and c3 but there would only
Hello Petr,
thank you for your ideas. The split() looks most realistic.
What about this idea:
1. Define three functions Refun1, Refun2, Refun3 for the three different
sections of the calculations (same as you suggested)
2. lambda = (Re <= 2320) * Refun1(Re) + ((Re > 2320) && (Re < 65 * dk))
*
Hello all,
After many e-mails and comments, I made corrections to my post on the topic
of R and funding.
I hope this does a better service to the R community:
http://www.r-statistics.com/2010/09/open-source-and-money-%E2%80%93-why-paying-r-developers-might-not-always-help-the-project/
Sorry for h
On 09/17/2010 10:50 PM, array chip wrote:
>
>
> Thank you Peter. Actually 3 people from mixed model mailing list tried my
> code
> using lmer(). They got the same results as what I got from lme4(). So they
> couldn't replicate my lmer() results:
>
> Random effects:
> Groups NameVari
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