Sorry, typo. It should be names(e.hp)
Pascal
On Sun, Mar 16, 2014 at 3:42 PM, Pascal Oettli wrote:
> Dear Sabina,
>
> If you carefully read the help page (always a good idea), examples
> show you that it is "object$cycle", not "object$Cycle". Also, you can
> use names, i.e. names(e.hp$).
>
> HTH
Dear Sabina,
If you carefully read the help page (always a good idea), examples
show you that it is "object$cycle", not "object$Cycle". Also, you can
use names, i.e. names(e.hp$).
HTH,
Pascal
On Sun, Mar 16, 2014 at 12:56 AM, Liana-Sabina Luncasu
wrote:
> Dear all,
>
> I am working with some ti
On Sat, 15 Mar 2014, peter dalgaard wrote:
On 15 Mar 2014, at 20:54 , Mike Miller wrote:
$ cat data1.txt
0.005
0.00489
I don't know why it shows 17 digits and doesn't round to 15, but it is showing
that the numbers are different, for some reason.
Aiding my weakening eyesight
It is not order that you are having difficulty with... it is name scope. Your
"z" is not a standalone variable, but a column in your "t" data frame.
Try
t[order(t$z), ]
Note that "t" is the name of a commonly-used function in R that transposes
matrices. It is generally not a good idea to reuse
Indeed you are -- the syntax for selecting columns of a data frame (or
components from a list).
Advice: Read "An Introduction to R" or online tutorial of your choice
to learn proper syntax.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. In
Did you not notice:
"Residual standard error: 22.85 on 5876 degrees of freedom
(36 observations deleted due to missingness)
??
(No residuals for missings...)
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And kno
Evidently, I'm overlooking something simple. I'm trying to used order with
data.frame.
For example:
t = data.frame(x = c(11,12,14), y = c(19,20,21), z = c(10,9,7))
t[order(z), ]
Error in order(z) : object 'z' not found
Thank you for any insights and advice provided.
[[alternative
Dear all,
I am working with some time series data and I applied an HP filter to it
using the following code:
e.hp = hpfilter(empl)
Now I am trying to access the last column (Cycle), that is, to save it in a
separate vector. I tried with both e.hp$Cycle and e.hp[ , 3] but none of
them works. I ge
I am trying to plot the residuals from a linear model and I get the following
error message: Error in xy.coords(x, y, xlabel, ylabel, log) : 'x' and 'y'
lengths differ. The outcome is a continuous variable and the explanatory
variable is ordinal. My immediate suspicion was that it had something
On Sat, 15 Mar 2014, Rui Barradas wrote:
I haven't followed this thread since its start but I think you now have
a case for FAQ 7.31. See inline below.
Try
(1-0.995) - 0.005
[1] 4.336809e-18
(2-1.995) - 0.005
[1] -1.066855e-16
Hope this helps,
Yes, that does show the problem well, but
On 15 Mar 2014, at 20:54 , Mike Miller wrote:
> On Sat, 15 Mar 2014, peter dalgaard wrote:
>
>> I don't think so. I think some of your numbers differ sufficiently from
>> numbers with only a few digits to the right of the decimal that write.table
>> needs to write them with increased precisio
Having just learned a few tricks from you guys, this might be the neatest
way to show the issue:
write.table(c(1-0.995, 2-1.995), row.names=F, col.names=F)
0.005
0.00489
options(digits) only works with write(), and not with write.table():
options(digits=7)
write(c(1-0.995, 2-1.9
Hello,
I haven't followed this thread since its start but I think you now have
a case for FAQ 7.31. See inline below.
Em 15-03-2014 19:54, Mike Miller escreveu:
On Sat, 15 Mar 2014, peter dalgaard wrote:
I don't think so. I think some of your numbers differ sufficiently
from numbers with on
On Sat, 15 Mar 2014, peter dalgaard wrote:
I don't think so. I think some of your numbers differ sufficiently from
numbers with only a few digits to the right of the decimal that
write.table needs to write them with increased precision. You didn't
read them like that, didn't you? You did some
Thanks for the ideas. It is great to have such skilled assistance with
this issue. That said, I don't think we've solved this one, yet.
Looking back at where my numbers came from, I found that I had read in
integers from a file, divided by 1000, then (critically) subtracted those
numbers fro
On Sat, 15 Mar 2014, Tim Marcella wrote:
Hi,
I am using a two part hurdle model to account for zero inflation and
overdispersion in my count data. I would like to account for a segmented or
breakpoint relationship in the binomial logistic hurdle model and pass
these results onto the count model
Hi,
I am using a two part hurdle model to account for zero inflation and
overdispersion in my count data. I would like to account for a segmented or
breakpoint relationship in the binomial logistic hurdle model and pass
these results onto the count model (negative binomial).
Using the segemented
Installing the openbals library may help.
Shige
On Sat, Mar 15, 2014 at 12:00 PM, Jeff Newmiller
wrote:
> Comparing with an unspecified benchmark makes answering this too hard.
> Following instructions in the Posting Guide will lead to more accurate Q
> and A.
>
> Note that you may not need to
Comparing with an unspecified benchmark makes answering this too hard.
Following instructions in the Posting Guide will lead to more accurate Q and A.
Note that you may not need to compile if you have not as yet followed the
recommendations: http://cran.r-project.org/bin/linux/ubuntu/README. The
Dear Thomas,
I cannot really answer because this is not a reproducible example; but your
traceback() output already gives a hint: try changing the random.method to
something different from default. In fact, as the singular matrix problem
happens during estimation of variance components, using
My guess is that maybe the default Ubuntu binaries aren't compiled
with MKL (Math Kernel Library) support and thus with no
multithreading.
I would suggest doing a quick research on how to re-compile R with MKL
support and maybe you'll be good to go.
On Fri, Mar 14, 2014 at 9:45 PM, Russell Bainer
R 2.11.1 is really old, but that's not likely to be your problem. I'd
guess that file n1.R doesn't contain what you think it contains. If you
run the source() command from within R, does that work?
Duncan Murdoch
On 14-03-15 3:18 AM, Ashis Deb wrote:
I had this code like :--
using System.
On 14-03-14 11:03 PM, Mike Miller wrote:
On Fri, 14 Mar 2014, Duncan Murdoch wrote:
On 14-03-14 8:59 PM, Mike Miller wrote:
What I'm using:
R version 3.0.1 (2013-05-16) -- "Good Sport"
Copyright (C) 2013 The R Foundation for Statistical Computing
Platform: x86_64-unknown-linux-gnu (64-bit)
Sent from my iPhone
__
R-help@r-project.org mailing list
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, minimal, self-contained, reproducible code.
Hi All,
I've run across an odd phenomenon and I am wondering if someone might be
able to provide insight as to what is going on. I'm running some R code
that was provided by a collaborator, who is not a very experienced R
programmer (e.g., the code is functional but not very efficient). When I
run
Dear Sir,
How I can obtain the predicted values of Y variable with fitting smooth spline
regressions (in GAMs) using R software?
in my research, temperature (X variable) is as smoother and fish catch values
(Y variable) is as response
variable.
I would be grateful if you could help me.
Attaching RData files is unfortunately not supported.
Try reading [1] and using dput to share your reproducible data.
[1]
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
---
Jeff Newmill
I had this code like :--
using System.Text;
using RDotNet;
using RDotNet.NativeLibrary;
using Stats;
namespace ConsoleApplication1
{
class Program
{
static void Main(string[] args)
{
string rhome =
System.Environment.GetEnvironmentVariable("R_HOME");
On 15 Mar 2014, at 04:03 , Mike Miller wrote:
> On Fri, 14 Mar 2014, Duncan Murdoch wrote:
>
>> On 14-03-14 8:59 PM, Mike Miller wrote:
>>> What I'm using:
>>> R version 3.0.1 (2013-05-16) -- "Good Sport"
>>> Copyright (C) 2013 The R Foundation for Statistical Computing
>>> Platform: x86_64-unk
I don't have access to a computer that will run R at the moment but if I were
tackling this, I would first look at the code for summary.survfit and adapt the
code that extracts the median ( the 50th percentile ).
--
David
Sent from my iPhone
> On Mar 15, 2014, at 1:43 PM, Lucy Leigh wrote:
>
Post output from:
dput(data2)
My guess is the 3rd column is a factor vector.
--
David
Sent from my iPhone
> On Mar 15, 2014, at 12:03 PM, Mike Miller wrote:
>
>> On Fri, 14 Mar 2014, Duncan Murdoch wrote:
>>
>>> On 14-03-14 8:59 PM, Mike Miller wrote:
>>> What I'm using:
>>> R version 3.0.1
This may be of interest:
http://stackoverflow.com/questions/20724872/median-from-survfit-object-and-textconnection
Sent from my iPhone
> On Mar 15, 2014, at 1:43 PM, Lucy Leigh wrote:
>
> Hello,
>
> I am using the function survfit in the 'survival' package. Calling the
> function produces th
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