Re: [R] lm and time series.

[rkevinburton]

That is the thing. As a new comer to 'R' I don't understand how to write a 
formula when all I have is a time series. I don't know how to express the 
independent and dependent variables in a formula when the object is a time 
series. So please just solve this simple example and I will extrapolate from 
there.

Say the units of the time series is days and the value at each point is the 
response. If I wanted to fit a straiight line through the following time series:

y <- 4:7
t <- ts(y)

So this is saying to me something like 4 units were sold on the first day, 5 on 
the second, 6 on the third, and 7 on the fourth.

So given the time series t I want to find the slope and inercept:

y = m*x + b

with x in days and the respoinse would be the number of units sold. I need to 
find 'm' and 'b'. If all I have is t (the time series above) then what would be 
the formula, and for that matter the arguments to lm to give the desired result?

fit <- lm(???)

Thank you.

Kevin

 stephen sefick <[EMAIL PROTECTED]> wrote: 
> So you want time as the independent variable?  Let's say that the
> units of y in your first example were seconds- couldn't you just use a
> regular lm and say that the units were seconds, minutes, or what ever?
>  I am probably out of my league here, but I am just not understanding
> what it is that you want.  a time series is just a series of data
> points indexed by time.  Arima maybe, or some other cool times series
> modeling approach- wavelet, spectral density- for frequency domain
> type things...  What are you trying to accomplish?
> 
> On Fri, Sep 5, 2008 at 5:47 PM,  <[EMAIL PROTECTED]> wrote:
> > I want to fit a function to time series. If I had:
> >
> > x <- 1:4
> > y <- 1:4
> >
> > lm(y~x)
> >
> > This would fit a simple line to the four points. But if it is represented 
> > as a time series
> >
> > x <- 1:4
> > t <- ts(x)
> >
> > lm()
> >
> > So I have a time series in the object t. How do I write a formula for lm? 
> > What do I put in the formula for x and y when I only have t (the time 
> > series).
> >
> > Kevin
> >
> >  stephen sefick <[EMAIL PROTECTED]> wrote:
> >> what do you want to do?
> >>
> >> On Fri, Sep 5, 2008 at 3:22 PM,  <[EMAIL PROTECTED]> wrote:
> >> > I am sorry but I looked at ?lm and could not see any guidance on 
> >> > writting a formula. If I have two arrays or a data set then I know how 
> >> > to do that (y ~ x) but for a time series I am not sure how to write y or 
> >> > x.
> >> >
> >> > Thank you.
> >> >
> >> > Kevin
> >> >
> >> >  Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> >> >> The Time Series section in ?lm should be self explanatory.   If you are 
> >> >> using
> >> >> diff's and lag's then look at the dyn package.
> >> >>
> >> >> On Fri, Sep 5, 2008 at 12:25 PM,  <[EMAIL PROTECTED]> wrote:
> >> >> > I did a ?lm and it said basically to be careful when using lm and a 
> >> >> > time series. But my question is probably more to do with my 
> >> >> > inexperience that anything. If I have a time series object 'ti' how 
> >> >> > do I write the formula? The response is the value at any particular 
> >> >> > time and the time is basically the index of the time series. But I 
> >> >> > don't know how to put that into a formula.
> >> >> >
> >> >> > Thank you.
> >> >> >
> >> >> > Kevin
> >> >
> >> > __
> >> > 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.
> >> >
> >>
> >>
> >>
> >> --
> >> Stephen Sefick
> >> Research Scientist
> >> Southeastern Natural Sciences Academy
> >>
> >> Let's not spend our time and resources thinking about things that are
> >> so little or so large that all they really do for us is puff us up and
> >> make us feel like gods. We are mammals, and have not exhausted the
> >> annoying little problems of being mammals.
> >>
> >>   -K. Mullis
> >
> >
> 
> 
> 
> -- 
> Stephen Sefick
> Research Scientist
> Southeastern Natural Sciences Academy
> 
> Let's not spend our time and resources thinking about things that are
> so little or so large that all they really do for us is puff us up and
> make us feel like gods. We are mammals, and have not exhausted the
> annoying little problems of being mammals.
> 
>   -K. Mullis

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Re: [R] Using R to generate reports

[Berwin A Turlach]

G'day all,

On Fri, 5 Sep 2008 18:10:37 -0400
"Jorge Ivan Velez" <[EMAIL PROTECTED]> wrote:

> Dear Bill,
> See http://www.statistik.lmu.de/~leisch/Sweave/

The following R News article might also be of interest:

Sven Garbade and Peter Burgard. Using R/Sweave in everyday
clinical practice. R News, 6(2):26-31, May 2006.

If one is new to LaTeX, the following articles might be of interest too:

Max Kuhn. Sweave and the open document format - the odfWeave
package. R News, 6(4):2-8, October 2006.

Gregor Gorjanc. Using sweave with lyx. R News, 8(1):2-9, May
2008.

Finally, check out:

http://bioinformatics.oxfordjournals.org/cgi/content/full/24/2/276

Cheers,

Berwin

=== Full address =
Berwin A TurlachTel.: +65 6515 4416 (secr)
Dept of Statistics and Applied Probability+65 6515 6650 (self)
Faculty of Science  FAX : +65 6872 3919   
National University of Singapore
6 Science Drive 2, Blk S16, Level 7  e-mail: [EMAIL PROTECTED]
Singapore 117546http://www.stat.nus.edu.sg/~statba

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Re: [R] Package for Tidying-up R-code

[Matthias Kohl]

http://cran.r-project.org/doc/manuals/R-exts.html#Tidying-and-profiling-R-code

hth,
Matthias

Don MacQueen wrote:
I don't know what kind of tidying you want, but ESS within emacs is a 
possibility.


-Don

At 6:32 PM +0900 9/5/08, Gundala Viswanath wrote:

Dear all,

Is there any such package?

I am thinking of something equivalent to Perl::Tidy.
For example with VI editor one can visual highlight a portion
of Perlcode and then issue the command:

!perltidy

then the code will be automatically arranged.

- Gundala Viswanath
Jakarta - Indonesia

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--
Dr. Matthias Kohl
www.stamats.de

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Re: [R] annotating objects in workspace

[Gabor Grothendieck]

See ?comment

On Fri, Sep 5, 2008 at 11:43 PM, Alexy Khrabrov <[EMAIL PROTECTED]> wrote:
> Is there a way to associate descriptions with the objects in the workspace,
> and later retrieve them to know what the object was created for?
>
> Thanks,
> Alexy
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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[R] annotating objects in workspace

[Alexy Khrabrov]

Is there a way to associate descriptions with the objects in the  
workspace, and later retrieve them to know what the object was created  
for?


Thanks,
Alexy

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Re: [R] xls to csv conversion via WinXP's context menu?

[Jorge Ivan Velez]

Dear Mark,
I think you don't need to convert your xls files into csv files to read them
using R. There are several ways to read them such as
the RODBC package and ?read.xls in gdata. Now, if you really want to do the
conversion, try ?xls2csv in gdata as well.

HTH,

Jorge



On Fri, Sep 5, 2008 at 6:45 PM, Mark Na <[EMAIL PROTECTED]> wrote:

> Frequently I need to convert a .xls to a .csv (for import into R) and I do
> this by opening the file in excel and saving it as a csv. I would rather do
> this using WinXP's context menu (right click the xls, choose "convert to
> csv") but I don't know of a utility that does this. Any ideas? Thanks, Mark
>
> __
> 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.
>

[[alternative HTML version deleted]]

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Re: [R] xls to csv conversion via WinXP's context menu?

[Gabor Grothendieck]

See ?xls2csv in the gdata package.

On Fri, Sep 5, 2008 at 6:45 PM, Mark Na <[EMAIL PROTECTED]> wrote:
> Frequently I need to convert a .xls to a .csv (for import into R) and I do
> this by opening the file in excel and saving it as a csv. I would rather do
> this using WinXP's context menu (right click the xls, choose "convert to
> csv") but I don't know of a utility that does this. Any ideas? Thanks, Mark

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Re: [R] Area of density

[Rolf Turner]

On 6/09/2008, at 8:32 AM, pragmatic wrote:



Hello!
Please, anybody help me.

Can I calculate area of density was created by:

D <- density(x)

In other words I want to know area under curve 'plot(D)'
and It's good to calculate area before and after 0 separately.


(1) You could use splinefun()

foo <- splinefun(D$x,D$y)
integrate(foo,min(D$x),0)
integrate(foo,0,max(D$x))

or you could use simp() from

http://finzi.psych.upenn.edu/R/Rhelp02a/archive/134918.html

with(D,simp(y=y[x<0],x=x[x<0]))
with(D,simp(y=y[x>0],x=x[x>0]))

You'll get different answers, neither of which is ``correct''.

cheers,

Rolf Turner

##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}

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[R] xls to csv conversion via WinXP's context menu?

[Mark Na]

Frequently I need to convert a .xls to a .csv (for import into R) and I 
do this by opening the file in excel and saving it as a csv. I would 
rather do this using WinXP's context menu (right click the xls, choose 
"convert to csv") but I don't know of a utility that does this. Any 
ideas? Thanks, Mark


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Re: [R] Area of density

[Duncan Murdoch]

On 05/09/2008 4:32 PM, pragmatic wrote:

Hello!
Please, anybody help me.

Can I calculate area of density was created by:

D <- density(x)

In other words I want to know area under curve 'plot(D)'


That should be 1.


and It's good to calculate area before and after 0 separately.


That's harder, but a good approximation should be sum(x < 0)/length(x) 
and sum(x > 0)/length(x).


Duncan Murdoch

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Re: [R] lm and time series.

[stephen sefick]

So you want time as the independent variable?  Let's say that the
units of y in your first example were seconds- couldn't you just use a
regular lm and say that the units were seconds, minutes, or what ever?
 I am probably out of my league here, but I am just not understanding
what it is that you want.  a time series is just a series of data
points indexed by time.  Arima maybe, or some other cool times series
modeling approach- wavelet, spectral density- for frequency domain
type things...  What are you trying to accomplish?

On Fri, Sep 5, 2008 at 5:47 PM,  <[EMAIL PROTECTED]> wrote:
> I want to fit a function to time series. If I had:
>
> x <- 1:4
> y <- 1:4
>
> lm(y~x)
>
> This would fit a simple line to the four points. But if it is represented as 
> a time series
>
> x <- 1:4
> t <- ts(x)
>
> lm()
>
> So I have a time series in the object t. How do I write a formula for lm? 
> What do I put in the formula for x and y when I only have t (the time series).
>
> Kevin
>
>  stephen sefick <[EMAIL PROTECTED]> wrote:
>> what do you want to do?
>>
>> On Fri, Sep 5, 2008 at 3:22 PM,  <[EMAIL PROTECTED]> wrote:
>> > I am sorry but I looked at ?lm and could not see any guidance on writting 
>> > a formula. If I have two arrays or a data set then I know how to do that 
>> > (y ~ x) but for a time series I am not sure how to write y or x.
>> >
>> > Thank you.
>> >
>> > Kevin
>> >
>> >  Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
>> >> The Time Series section in ?lm should be self explanatory.   If you are 
>> >> using
>> >> diff's and lag's then look at the dyn package.
>> >>
>> >> On Fri, Sep 5, 2008 at 12:25 PM,  <[EMAIL PROTECTED]> wrote:
>> >> > I did a ?lm and it said basically to be careful when using lm and a 
>> >> > time series. But my question is probably more to do with my 
>> >> > inexperience that anything. If I have a time series object 'ti' how do 
>> >> > I write the formula? The response is the value at any particular time 
>> >> > and the time is basically the index of the time series. But I don't 
>> >> > know how to put that into a formula.
>> >> >
>> >> > Thank you.
>> >> >
>> >> > Kevin
>> >
>> > __
>> > 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.
>> >
>>
>>
>>
>> --
>> Stephen Sefick
>> Research Scientist
>> Southeastern Natural Sciences Academy
>>
>> Let's not spend our time and resources thinking about things that are
>> so little or so large that all they really do for us is puff us up and
>> make us feel like gods. We are mammals, and have not exhausted the
>> annoying little problems of being mammals.
>>
>>   -K. Mullis
>
>



-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.

-K. Mullis

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Re: [R] Using R to generate reports

[Jorge Ivan Velez]

Dear Bill,
See http://www.statistik.lmu.de/~leisch/Sweave/


HTH,


Jorge


On Fri, Sep 5, 2008 at 5:40 PM, Bill Cunliffe <[EMAIL PROTECTED]> wrote:

> Hi all,
>
>
>
> I would like to generate nicely formatted reports on a daily basis.  The
> reports would be a combination of charts and tables, generated in an
> environment such as R.
>
>
>
> Ideally, this would be an automated process.  Does anyone know if this is
> feasible, and how I could achieve well formatted reports?  One idea would
> be
> to have R scripts process the data/generate the charts and then use Latex
> to
> create the report, referencing the charts produced by R.
>
>
>
> I've not used Latex before so I'm trying to get a feel for how to best
> proceed.
>
>
>
> Thanks for your input,
>
>
>
> Bill.
>
>
>[[alternative HTML version deleted]]
>
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>

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Re: [R] Multiple Correspondence Analysis

[Jorge Ivan Velez]

Dear Bill,
See http://finzi.psych.upenn.edu/R/library/ade4/html/dudi.acm.html

HTH,


Jorge



On Fri, Sep 5, 2008 at 5:00 PM, Bill Vorias <[EMAIL PROTECTED]>wrote:

> Is there a way to get the coordinates from a plot of an MCA object?
>
>
>[[alternative HTML version deleted]]
>
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>

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Re: [R] library/function that estimates parameters of well known distributions from empirical data?

[Ted Byers]

Thanks Ben

That was the one I'd remembered but couldn't find.

Mark Leeds also told me about DistributionFits(fBasics), which I hadn't
seen.  There seems to be only a little overlap between the two.

Could I trouble you to expand on AIC (esp. what the function name and
package is to apply it to the output from these two functions)?  I just read
the help provided for each and neither mentions AIC.

Thanks again Ben

Ted


Ben Bolker wrote:
> 
> Ted Byers  gmail.com> writes:
> 
>> 
>> 
>> I found this a few months ago, but for the life of me I can't remember
>> what
>> the function or package was, and I have had no luck finding it this week.
>> 
>> I have found, again, the functions for working with distributions like
>> Cauchy, F, normal, &c., and ks.test, but I have not found the functions
>> for
>> estimating the distribution parameters given a vector of values.
>> 
> 
>   Look at the fitdistr function in the MASS package.  Consider
> AIC comparisons for ranking the fits to these non-nested models.
> 
>   good luck
>Ben Bolker
> 
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> 
> 

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[R] Multiple Correspondence Analysis

[Bill Vorias]

Is there a way to get the coordinates from a plot of an MCA object?


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Re: [R] upgraded maps database for italy?

[Ray Brownrigg]

The "italy" (and "france") databases only exists in the maps package because somebody 
supplied me with the data in an appropriate format - latitude/longitude pairs separately 
for each polygon (provinces in this case).  If you have the appropriate data for what you 
want i.e. updated provinces and regioni, then I could generate a new database (no promises 
as to how quickly this will happen though).  You *could* do it yourself if you read all 
the references contain in the maps documentation, but I wouldn't recommend learning how to 
do this unless you intend doing it frequently.


However as Greg Snow comments, maptools may well be a better way to go and uses a more 
generally available data format.  The maps package, which traces its roots back to 'New' S 
(the Blue book), is rather showing its age, but there is no reason to delete it while some 
people still find it useful.


Regards,
Ray Brownrigg

livio spam wrote:

Dear UseRs,
I'm using the library maps. I'm drawing maps of Italy. The map available for
this library were prepared around *1989*:
"This italy database comes from the NUTS III (Tertiary Administrative Units
of the European Community) database of the United Nations Environment
Programme (UNEP) GRID-Geneva data sets. These were prepared around *1989*"
[cited: help("italy")].
The details of this map is by "province", which is not maybe optimal, but is
OK.
However, after 1989 many new province are born. So for example the new
provincia of Rimini has born and the sardinan Island has doubled them.

Some of you do know how to get upgraded maps of italy? I think it would
useful for all Italian mapers.

The second question is as follow: Italy is subdivided in "regioni". Each
"regione" is a set of many province; so, if I want to plot map region-based
index, how can I omit the borders of prince belonging to the same "regione"?
so, how plot maps to a higher unit level?

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[R] Articles about comparision between R and others softwares

[ricardo13]

Hi

Do you know some articles, papers, something than tell about comparision
between R and others softwares statisticals.

Thank You

Ricardo
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[R] Area of density

[pragmatic]

Hello!
Please, anybody help me.

Can I calculate area of density was created by:
> D <- density(x)
In other words I want to know area under curve 'plot(D)'
and It's good to calculate area before and after 0 separately.

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Re: [R] lm and time series.

[rkevinburton]

I want to fit a function to time series. If I had:

x <- 1:4
y <- 1:4

lm(y~x)

This would fit a simple line to the four points. But if it is represented as a 
time series

x <- 1:4
t <- ts(x)

lm()

So I have a time series in the object t. How do I write a formula for lm? What 
do I put in the formula for x and y when I only have t (the time series). 

Kevin

 stephen sefick <[EMAIL PROTECTED]> wrote: 
> what do you want to do?
> 
> On Fri, Sep 5, 2008 at 3:22 PM,  <[EMAIL PROTECTED]> wrote:
> > I am sorry but I looked at ?lm and could not see any guidance on writting a 
> > formula. If I have two arrays or a data set then I know how to do that (y ~ 
> > x) but for a time series I am not sure how to write y or x.
> >
> > Thank you.
> >
> > Kevin
> >
> >  Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> >> The Time Series section in ?lm should be self explanatory.   If you are 
> >> using
> >> diff's and lag's then look at the dyn package.
> >>
> >> On Fri, Sep 5, 2008 at 12:25 PM,  <[EMAIL PROTECTED]> wrote:
> >> > I did a ?lm and it said basically to be careful when using lm and a time 
> >> > series. But my question is probably more to do with my inexperience that 
> >> > anything. If I have a time series object 'ti' how do I write the 
> >> > formula? The response is the value at any particular time and the time 
> >> > is basically the index of the time series. But I don't know how to put 
> >> > that into a formula.
> >> >
> >> > Thank you.
> >> >
> >> > Kevin
> >
> > __
> > 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.
> >
> 
> 
> 
> -- 
> Stephen Sefick
> Research Scientist
> Southeastern Natural Sciences Academy
> 
> Let's not spend our time and resources thinking about things that are
> so little or so large that all they really do for us is puff us up and
> make us feel like gods. We are mammals, and have not exhausted the
> annoying little problems of being mammals.
> 
>   -K. Mullis

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[R] Recall: Using R to generate reports

[Bill Cunliffe]

Bill Cunliffe would like to recall the message, "Using R to generate
reports".
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[R] Using R to generate reports

[Bill Cunliffe]

Hi all,

 

I would like to generate nicely formatted reports on a daily basis.  The
reports would be a combination of charts and tables, generated in an
environment such as R.

 

Ideally, this would be an automated process.  Does anyone know if this is
feasible, and how I could achieve well formatted reports?  One idea would be
to have R scripts process the data/generate the charts and then use Latex to
create the report, referencing the charts produced by R.

 

I've not used Latex before so I'm trying to get a feel for how to best
proceed.

 

Thanks for your input,

 

Bill.


[[alternative HTML version deleted]]

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Re: [R] Package for Tidying-up R-code

[Don MacQueen]

I don't know what kind of tidying you want, but ESS within emacs is a 
possibility.


-Don

At 6:32 PM +0900 9/5/08, Gundala Viswanath wrote:

Dear all,

Is there any such package?

I am thinking of something equivalent to Perl::Tidy.
For example with VI editor one can visual highlight a portion
of Perlcode and then issue the command:

!perltidy

then the code will be automatically arranged.

- Gundala Viswanath
Jakarta - Indonesia

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--
--
Don MacQueen
Environmental Protection Department
Lawrence Livermore National Laboratory
Livermore, CA, USA
925-423-1062

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Re: [R] lm and time series.

[stephen sefick]

what do you want to do?

On Fri, Sep 5, 2008 at 3:22 PM,  <[EMAIL PROTECTED]> wrote:
> I am sorry but I looked at ?lm and could not see any guidance on writting a 
> formula. If I have two arrays or a data set then I know how to do that (y ~ 
> x) but for a time series I am not sure how to write y or x.
>
> Thank you.
>
> Kevin
>
>  Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
>> The Time Series section in ?lm should be self explanatory.   If you are using
>> diff's and lag's then look at the dyn package.
>>
>> On Fri, Sep 5, 2008 at 12:25 PM,  <[EMAIL PROTECTED]> wrote:
>> > I did a ?lm and it said basically to be careful when using lm and a time 
>> > series. But my question is probably more to do with my inexperience that 
>> > anything. If I have a time series object 'ti' how do I write the formula? 
>> > The response is the value at any particular time and the time is basically 
>> > the index of the time series. But I don't know how to put that into a 
>> > formula.
>> >
>> > Thank you.
>> >
>> > Kevin
>
> __
> 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.
>



-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.

-K. Mullis

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Re: [R] dealing with NAs in time series

[stephen sefick]

look at zoo
na.approx
this will interpolate in a couple of ways

On Fri, Sep 5, 2008 at 11:28 AM, Alexy Khrabrov <[EMAIL PROTECTED]> wrote:
> Certain timeseries I have had outliers, which I removed by assigning NA to
> their positions.  Now acf() refuses to go to work.  What's the right way to
> remove outliers from ts objects, and what are teh standard ways to
> interpolate NAs in them?
>
> Cheers,
> Alexy
>
> __
> 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.
>



-- 
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy

Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.

-K. Mullis

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Re: [R] comparing two files

[Jorge Ivan Velez]

Hi there,
See ?"%in%"  Here is an example:

> x=sample(1:10)
> x
 [1]  2  1 10  5  4  7  3  6  8  9
> y=sample(4:20)
> y
 [1] 17 16 18 10  8 19  4  5 11 13 12 14 20  6  7  9 15
> x[x%in%y]  # "x" values present in "y"
[1] 10  5  4  7  6  8  9

HTH,


Jorge



On Fri, Sep 5, 2008 at 2:57 PM, Rajasekaramya <[EMAIL PROTECTED]>wrote:

>
> Hi there
>
> I have two object on is a vector T and the other is  dataframe C.
> vector T has more no of rows when comapred with a dataframe Ccolumn[,2].I
> want to find the the missing rows in dataframe C.That is those values that
> are not matchig  in dataframe C[,2]
>
> Kindly give me suggestion on how to go about it.
>
> Ramya
>
> --
> View this message in context:
> http://www.nabble.com/comparing-two-files-tp19337486p19337486.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> 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.
>

[[alternative HTML version deleted]]

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Re: [R] comparing two files

[jim holtman]

?match

> T.v <- sample(1:20)
> C.df <- data.frame(a=1:10, b= sample(1:20, 10))
> T.v
 [1] 10 12  9  4 14 11 19  2 18  5 16  6  7 17  8 20  1 13  3 15
> C.df
a  b
1   1 10
2   2 17
3   3  8
4   4  5
5   5  2
6   6 16
7   7 19
8   8  7
9   9 18
10 10 14
> # find the missing values
> T.v[is.na(match(T.v, C.df[,2]))]
 [1] 12  9  4 11  6 20  1 13  3 15
>


On Fri, Sep 5, 2008 at 2:57 PM, Rajasekaramya <[EMAIL PROTECTED]> wrote:
>
> Hi there
>
> I have two object on is a vector T and the other is  dataframe C.
> vector T has more no of rows when comapred with a dataframe Ccolumn[,2].I
> want to find the the missing rows in dataframe C.That is those values that
> are not matchig  in dataframe C[,2]
>
> Kindly give me suggestion on how to go about it.
>
> Ramya
>
> --
> View this message in context: 
> http://www.nabble.com/comparing-two-files-tp19337486p19337486.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> 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.
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] comparing two files

[John Kane]

?which

which(C[,2]!= T)  # untested


--- On Fri, 9/5/08, Rajasekaramya <[EMAIL PROTECTED]> wrote:

> From: Rajasekaramya <[EMAIL PROTECTED]>
> Subject: [R]  comparing two files
> To: r-help@r-project.org
> Received: Friday, September 5, 2008, 2:57 PM
> Hi there
> 
> I have two object on is a vector T and the other is 
> dataframe C.
> vector T has more no of rows when comapred with a dataframe
> Ccolumn[,2].I
> want to find the the missing rows in dataframe C.That is
> those values that
> are not matchig  in dataframe C[,2]
> 
> Kindly give me suggestion on how to go about it.
> 
> Ramya
> 
> -- 
> View this message in context:
> http://www.nabble.com/comparing-two-files-tp19337486p19337486.html
> Sent from the R help mailing list archive at Nabble.com.
> 
> __
> 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.


  __
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Re: [R] lm and time series.

[rkevinburton]

I am sorry but I looked at ?lm and could not see any guidance on writting a 
formula. If I have two arrays or a data set then I know how to do that (y ~ x) 
but for a time series I am not sure how to write y or x.

Thank you.

Kevin

 Gabor Grothendieck <[EMAIL PROTECTED]> wrote: 
> The Time Series section in ?lm should be self explanatory.   If you are using
> diff's and lag's then look at the dyn package.
> 
> On Fri, Sep 5, 2008 at 12:25 PM,  <[EMAIL PROTECTED]> wrote:
> > I did a ?lm and it said basically to be careful when using lm and a time 
> > series. But my question is probably more to do with my inexperience that 
> > anything. If I have a time series object 'ti' how do I write the formula? 
> > The response is the value at any particular time and the time is basically 
> > the index of the time series. But I don't know how to put that into a 
> > formula.
> >
> > Thank you.
> >
> > Kevin

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Re: [R] boxplot including null info from dataframe, not with SQLite dataframe

[Ben Bolker]

Coey Minear  securecomputing.com> writes:

> 
> I have been trying to use R to gather some information from parsed log
> files (as part of examining some performance issues).  I parsed the
> log files and put the data into an SQLite database, and then used
> RSQLite to load the data into R.  The fields of interest are
> controller, action and total_time: controller and action have string
> values; total_time has a decimal value.
> 
> I first did the following box plot to find the problem controllers.
>   boxplot(total_time ~ controller, all_data)
> 
> Having identified one controller of interest (let's say
> "BadController"), I then wanted to then focus on the actions
> associated with that controller.  So I did this:
>   boxplot(total_time ~ action, subset(all_data, controller == 
> "BadController"))
> 
> This gave me a plot I was expecting: just the actions which are
> associated with "BadController".  However, I'd done this work on a
> FreeBSD system, and then I wanted to print it, and the easiest means
> seemed to re-plot using R on Windows.  So I wrote the data to a file,
> moved it to Windows and loaded it up there.  
> 
> On FreeBSD: 
>   write.table(all_data, "datafile.R")
> 
> On Windows:
>   all_data <- read.table("datafile.R")
> 

  I'm guessing that you want

bad <- subset(all_data,controller=="BadController")
bad$action <- factor(bad$action)
boxplot(total_time ~ action)

   Subsetting doesn't drop factor levels that don't
occur, which is an unfortunate design decision ...


   Ben Bolker

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Re: [R] library/function that estimates parameters of well known distributions from empirical data?

[Ben Bolker]

Ted Byers  gmail.com> writes:

> 
> 
> I found this a few months ago, but for the life of me I can't remember what
> the function or package was, and I have had no luck finding it this week.
> 
> I have found, again, the functions for working with distributions like
> Cauchy, F, normal, &c., and ks.test, but I have not found the functions for
> estimating the distribution parameters given a vector of values.
> 

  Look at the fitdistr function in the MASS package.  Consider
AIC comparisons for ranking the fits to these non-nested models.

  good luck
   Ben Bolker

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[R] comparing two files

[Rajasekaramya]

Hi there

I have two object on is a vector T and the other is  dataframe C.
vector T has more no of rows when comapred with a dataframe Ccolumn[,2].I
want to find the the missing rows in dataframe C.That is those values that
are not matchig  in dataframe C[,2]

Kindly give me suggestion on how to go about it.

Ramya

-- 
View this message in context: 
http://www.nabble.com/comparing-two-files-tp19337486p19337486.html
Sent from the R help mailing list archive at Nabble.com.

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and provide commented, minimal, self-contained, reproducible code.

[R] boxplot including null info from dataframe, not with SQLite dataframe

[Coey Minear]

I have been trying to use R to gather some information from parsed log
files (as part of examining some performance issues).  I parsed the
log files and put the data into an SQLite database, and then used
RSQLite to load the data into R.  The fields of interest are
controller, action and total_time: controller and action have string
values; total_time has a decimal value.

I first did the following box plot to find the problem controllers.
  boxplot(total_time ~ controller, all_data)

Having identified one controller of interest (let's say
"BadController"), I then wanted to then focus on the actions
associated with that controller.  So I did this:
  boxplot(total_time ~ action, subset(all_data, controller == "BadController"))

This gave me a plot I was expecting: just the actions which are
associated with "BadController".  However, I'd done this work on a
FreeBSD system, and then I wanted to print it, and the easiest means
seemed to re-plot using R on Windows.  So I wrote the data to a file,
moved it to Windows and loaded it up there.  

On FreeBSD: 
  write.table(all_data, "datafile.R")

On Windows:
  all_data <- read.table("datafile.R")

However, on Windows, when I get to the boxplot of the subset of data,
I'm seeing every action that's part of all_data, not just the ones
that are associated with "BadController".  Eventually, I found that
this wasn't a Windows vs. FreeBSD issue, because if I reload the data
from the file on FreeBSD, I start to see the same behavior.

Based on a few bug reports I found, it would seem that this is working
as designed.  So my question is: 
What do I do to get it to truly ignore the actions with no data?  I've
tried adding "drop=FALSE" to the subset call; that hasn't worked.  I
also tried specifically adding "na.action=NULL" to the boxplot call,
with no change.

I'm also curious what's different between a data frame loaded from
SQLite versus a data frame loaded from a file.

In the mean time, I'll either try to install RSQLite on Windows or get
postscript working on FreeBSD.  (My quick attempt with postscript in R
on FreeBSD was not drawing the bounding box nor the axes.)

Thanks for any help.



-- 
Coey Minear

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Re: [R] asscii2netcdf]

[Philip Twumasi-Ankrah]

Hi Marshall,

This link provides very good directions and examples you can use.

http://www.image.ucar.edu/GSP/Software/Netcdf/

A Smile costs Nothing  
 But Rewards Everything

Happiness is not perfected until it is shared
  -Jane Porter  
  


--- On Fri, 9/5/08, Marshall Mdoka <[EMAIL PROTECTED]> wrote:
From: Marshall Mdoka <[EMAIL PROTECTED]>
Subject: Re: [R] asscii2netcdf]
To: r-help@r-project.org
Date: Friday, September 5, 2008, 1:15 PM

Dear All,

A friend of mine seeks help in converting the attached ascii to netcdf. 
Any assistance will be sincerely appreciated.

Regards,

Marshall
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[[alternative HTML version deleted]]

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Re: [R] upgraded maps database for italy?

[Greg Snow]

If I remember correctly, the maps package only has a few dedicated maps that 
works with it.  The maptools package has similar (but not exactly the same) 
functionality and can work with shapefiles that are more commonly available.  
Maptools also integrates nicely with the sp and related packages that give many 
more options for working with the maps.

For the 2nd question, I would suggest plotting the map first using the 
province, then plot a second map over the top using the information on the 
regioni using a different color/line thickness for the boarders and no fill (so 
that the province show through).  This requires having 2 different map files, 
one with province and one with regioni information (there may be a better way 
with overlaying information, but this method works without having to think 
about it too much).

Hope this helps,

--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111



> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of livio spam
> Sent: Friday, September 05, 2008 12:24 PM
> To: r-help@r-project.org
> Subject: [R] upgraded maps database for italy?
>
> Dear UseRs,
> I'm using the library maps. I'm drawing maps of Italy. The
> map available for this library were prepared around *1989*:
> "This italy database comes from the NUTS III (Tertiary
> Administrative Units of the European Community) database of
> the United Nations Environment Programme (UNEP) GRID-Geneva
> data sets. These were prepared around *1989*"
> [cited: help("italy")].
> The details of this map is by "province", which is not maybe
> optimal, but is OK.
> However, after 1989 many new province are born. So for
> example the new provincia of Rimini has born and the sardinan
> Island has doubled them.
>
> Some of you do know how to get upgraded maps of italy? I
> think it would useful for all Italian mapers.
>
> The second question is as follow: Italy is subdivided in
> "regioni". Each "regione" is a set of many province; so, if I
> want to plot map region-based index, how can I omit the
> borders of prince belonging to the same "regione"?
> so, how plot maps to a higher unit level?
>
> [[alternative HTML version deleted]]
>
> __
> 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.
>

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and provide commented, minimal, self-contained, reproducible code.

[R] upgraded maps database for italy?

[livio spam]

Dear UseRs,
I'm using the library maps. I'm drawing maps of Italy. The map available for
this library were prepared around *1989*:
"This italy database comes from the NUTS III (Tertiary Administrative Units
of the European Community) database of the United Nations Environment
Programme (UNEP) GRID-Geneva data sets. These were prepared around *1989*"
[cited: help("italy")].
The details of this map is by "province", which is not maybe optimal, but is
OK.
However, after 1989 many new province are born. So for example the new
provincia of Rimini has born and the sardinan Island has doubled them.

Some of you do know how to get upgraded maps of italy? I think it would
useful for all Italian mapers.

The second question is as follow: Italy is subdivided in "regioni". Each
"regione" is a set of many province; so, if I want to plot map region-based
index, how can I omit the borders of prince belonging to the same "regione"?
so, how plot maps to a higher unit level?

[[alternative HTML version deleted]]

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Re: [R] casting help please

[davidr]

I think I just remembered that in a melted data.frame the values have to
be called 'value' and tried renaming my third column so. Now it works
(without add.missing).

Thanks for this powerful piece of software!
-- David
(I was trying to convert a set of data in 'long' format to 'wide'
format. Basically I have a pretty long price history for many tickers
basically stacked up as indicated in the toy example and want them
reshaped with each ticker in a column with dates in the first column.
Maybe cast was not the best tool? But it works now.)

-Original Message-
From: hadley wickham [mailto:[EMAIL PROTECTED] 
Sent: Friday, September 05, 2008 1:10 PM
To: David Reiner <[EMAIL PROTECTED]>
Cc: r-help@r-project.org
Subject: [SPAM] - Re: [R] casting help please - Found word(s) list error
in the Text body

Hi David,

Does getting rid of the missings help?

prices <- prices[complete.cases(prices), ]

Also, you shouldn't need  add.missing=TRUE - what are you trying to do?

Hadley

On Fri, Sep 5, 2008 at 12:12 PM,  <[EMAIL PROTECTED]> wrote:
> I have a data.frame which I believe is melted already and am having
> trouble casting it to 'wide' format.
>
> It looks something like
>
>
>
>> (x <- data.frame(ticker=c(rep("A",5),rep("B",6)), date=c(1:5, 1:6),
> value=c(NA,100*exp(rnorm(10,0,.1)
>
>> cast(x, date ~ ticker) # this does what I want with toy data
>
>
>
> But when I use my real data frame
>
>
>
>> str(prices)
>
> 'data.frame':   308188 obs. of  3 variables:
>
>  $ Ticker: chr  "ticker1" " ticker1" " ticker1" " ticker1" ...
>
>  $ Date:Class 'Date'  num [1:308188] 12296 12297 12298 12299 12300
> ...
>
>  $ Price   : num  NA NA NA NA NA NA NA NA NA NA ...
>
>
>
> I get
>
>
>
>> prices.wide <- cast(prices, Date ~ Ticker, add.missing=TRUE)
>
> Error in data.frame(data[, c(variables), drop = FALSE], result =
> data$value) :
>
>  arguments imply differing number of rows: 308188, 0
>
>
>
> (I tried various other arguments to cast - all gave the same error
> message.)
>
> It is a fact that the various tickers have data for different date
> ranges in the data frame and there are lots of NA's,
>
> but the toy example above has different date ranges for the two
tickers
> and an NA, so I don't know what else to look for in my data,
>
> or what args to cast might make it work.
>
>
>
> Any insights or direction would be much appreciated.
>
>
>
> David L. Reiner, PhD
>
> Head Quant
>
> Rho Trading Securities, LLC
>
>
>
>
>[[alternative HTML version deleted]]
>
> __
> 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.
>



-- 
http://had.co.nz/

__
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Re: [R] casting help please

[hadley wickham]

Hi David,

Does getting rid of the missings help?

prices <- prices[complete.cases(prices), ]

Also, you shouldn't need  add.missing=TRUE - what are you trying to do?

Hadley

On Fri, Sep 5, 2008 at 12:12 PM,  <[EMAIL PROTECTED]> wrote:
> I have a data.frame which I believe is melted already and am having
> trouble casting it to 'wide' format.
>
> It looks something like
>
>
>
>> (x <- data.frame(ticker=c(rep("A",5),rep("B",6)), date=c(1:5, 1:6),
> value=c(NA,100*exp(rnorm(10,0,.1)
>
>> cast(x, date ~ ticker) # this does what I want with toy data
>
>
>
> But when I use my real data frame
>
>
>
>> str(prices)
>
> 'data.frame':   308188 obs. of  3 variables:
>
>  $ Ticker: chr  "ticker1" " ticker1" " ticker1" " ticker1" ...
>
>  $ Date:Class 'Date'  num [1:308188] 12296 12297 12298 12299 12300
> ...
>
>  $ Price   : num  NA NA NA NA NA NA NA NA NA NA ...
>
>
>
> I get
>
>
>
>> prices.wide <- cast(prices, Date ~ Ticker, add.missing=TRUE)
>
> Error in data.frame(data[, c(variables), drop = FALSE], result =
> data$value) :
>
>  arguments imply differing number of rows: 308188, 0
>
>
>
> (I tried various other arguments to cast - all gave the same error
> message.)
>
> It is a fact that the various tickers have data for different date
> ranges in the data frame and there are lots of NA's,
>
> but the toy example above has different date ranges for the two tickers
> and an NA, so I don't know what else to look for in my data,
>
> or what args to cast might make it work.
>
>
>
> Any insights or direction would be much appreciated.
>
>
>
> David L. Reiner, PhD
>
> Head Quant
>
> Rho Trading Securities, LLC
>
>
>
>
>[[alternative HTML version deleted]]
>
> __
> 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.
>



-- 
http://had.co.nz/

__
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Re: [R] Question about col.names in write.csv

[Julian Burgos]

Hi Luz,
No entiendo bien tu pregunta.  Querés grabar una tabla con nombres en 
las columnas y tambien en la primera fila?  Si es asi, tenés que 
asignarle los nombres a la tabla antes de grabar.  Por ejemplo:


mi.tabla=matrix(runif(30),ncol=3)

colnames(mi.tabla)=c("A","B","C")
rownames(mi.tabla)=c("D",rep("",9))

write.csv(mi.tabla,file="mi.archivo.csv")

Cuando asignás nombres a las filas o columnas, el vector tiene que tener 
la misma longitud que el numero de filas o columnas.  Entonces, para 
darle un nombre solamente a la primera fila, hice un vector con el 
nombre y nueve espacios en blanco (para completar las diez filas que 
tiene mi tabla).


Al grabar la tabla usando write.csv, el comportamiento "default" es 
guardar los nombres de las columnas y filas.



Saludos,

Julian

Luz Milena Zea Fernandez wrote:

Dear support, I don't ignore col.names in write.csv. I want to write names for 
the firts row. How can I do?

Thanks in advance

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[R] casting help please

[davidr]

I have a data.frame which I believe is melted already and am having
trouble casting it to 'wide' format.

It looks something like

 

> (x <- data.frame(ticker=c(rep("A",5),rep("B",6)), date=c(1:5, 1:6),
value=c(NA,100*exp(rnorm(10,0,.1)

> cast(x, date ~ ticker) # this does what I want with toy data

 

But when I use my real data frame

 

> str(prices)

'data.frame':   308188 obs. of  3 variables:

 $ Ticker: chr  "ticker1" " ticker1" " ticker1" " ticker1" ...

 $ Date:Class 'Date'  num [1:308188] 12296 12297 12298 12299 12300
...

 $ Price   : num  NA NA NA NA NA NA NA NA NA NA ...

 

I get

 

> prices.wide <- cast(prices, Date ~ Ticker, add.missing=TRUE)

Error in data.frame(data[, c(variables), drop = FALSE], result =
data$value) : 

  arguments imply differing number of rows: 308188, 0

 

(I tried various other arguments to cast - all gave the same error
message.)

It is a fact that the various tickers have data for different date
ranges in the data frame and there are lots of NA's,

but the toy example above has different date ranges for the two tickers
and an NA, so I don't know what else to look for in my data,

or what args to cast might make it work.

 

Any insights or direction would be much appreciated.

 

David L. Reiner, PhD

Head Quant

Rho Trading Securities, LLC

 


[[alternative HTML version deleted]]

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[R] Adding a legend to R graph device with several plots (no to individual plots!)

[Monica Pisica]

Hi Nelson,

I don't know if you got your answer or not but here it is my low-tech solution 
for what is worth. I like to do a layout as I want my graphs to be and add some 
space where I "plot" an invisible graph with dummy data and add the legend 
there. In this case I decided to put the legend on middle top of the layout. 

Because you used lapply to do your graphs it seems that I have to change the 
numbers in the p.names parameter because somehow it establishes the order of 
your graphs as well. I usually use the simple for loop because you never have 
loads of graphs in one page, so it is safe enough. 

I hope this helps a little.

Monica

# code starts

l <- layout(matrix(c(rep(1, 2), seq(2,5)), 3,2, byrow = T), c(3,3,3), 
c(1,3,3,3))
layout.show(l)
par(mar = c(1,1,1,1))

plot(seq(1,10), seq(1,10), type = "n", axes = F)
smartlegend(x= "center", y="top", c("1995","2006"), lty=2:1)
par(mar = c(4,4,3,2))

p.names <- c("2","3","4","5")### change p.names definition
lapply(p.names, function(x) {
d1 <- density(rnorm(100,0,1))
d2 <- density(rnorm(100,.3,1.2))
plot(range(d1$x, d2$x), range(d1$y, d2$y), type = "n", xlab = "", ylab = "") 
## change xlab, ylab in plot
lines(d1, col = "black", lty = 2)
lines(d2, col = "black", lty = 1)
})
mtext("Learning R the hard way", side = 1, line=-1.5, outer=TRUE)

 code ends

--
Message: 16
Date: Thu, 4 Sep 2008 09:35:42 -0400
From: "Nelson Villoria" 
Subject: [R] Adding a legend to R graph device with several plots (no
to individual plots!)
To: r-help@r-project.org
Message-ID:

Content-Type: text/plain; charset=ISO-8859-1

Dear Users, I already posted this question: it either went unnoticed,
or it is to basic (if this is so, please sent me a hint).
I would like to know if there is a way to add a common
legend to an arrangement of plots. In the example below, I get four
plots in my device. each one has a density for 1995 and one for 2006.
I have found that using legend or smartlegend I can add a legend to
each plot, but I am looking for something in the spirit of mtext. That
is, putting the legend anywhere I want on the device. In my situation
I have:

par(mfrow=c(2,2), ann = FALSE)
p.names <- c("1","2","3","4")
lapply(p.names, function(x) {
d1 <- density(rnorm(100,0,1))
d2 <- density(rnorm(100,.3,1.2))
plot(range(d1$x, d2$x), range(d1$y, d2$y), type = "n", xlab = NULL,
ylab = NULL)
lines(d1, col = "black", lty = 2)
lines(d2, col = "black", lty = 1)
})
mtext("Learning R the hard way", side = 1, line=-1.5, outer=TRUE)
smartlegend(x= "center", y="top", c("1995","2006"), lty=2:1)

the last line puts the legend on the lower-right plot. I'd like to put
it in the middle - or top of the device. Any suggestion?

Thanks!

Nelson




_


50F681DAD532637!5295.entry?ocid=TXT_TAGLM_WL_domore_092008
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Re: [R] Plot by column

[Henrique Dallazuanna]

You can try something like about this:

r[upper.tri(r)] <- NA
matplot(t(r), lty = 1, type = 'p', pch = 1, ylim = c(0.4, 1.2), col =
'black')


On Fri, Sep 5, 2008 at 12:27 PM, Zhang Yanwei - Princeton-MRAm <
[EMAIL PROTECTED]> wrote:

> Dear list,
>
>  I have the following matrix. How can I make the following plot?
>  1. The x-axis has index 1:7, and the first column is plotted against index
> 1, second against 2, and so on.
>  2. I want the points from the left upper conner including the antidiagonal
> to be plotted with col=2, and the rest with col=3
>  [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]
> [1,] 0.589 0.857 0.923 0.944 0.954 0.963 0.980
> [2,] 0.470 0.763 0.877 0.900 0.911 0.957 0.974
> [3,] 0.486 0.863 0.905 0.960 0.968 0.998 1.015
> [4,] 0.653 0.888 0.943 0.952 0.962 0.992 1.009
> [5,] 0.664 0.774 0.958 0.986 0.996 1.027 1.045
> [6,] 0.546 0.910 1.011 1.041 1.051 1.083 1.102
> [7,] 0.407 0.600 0.667 0.686 0.693 0.715 0.727
>
>
> The following the my code to do the above with the matrix called "r". How
> can I simply the code?
>
>
> plot(r[,1],type="n",ylim=c(0.4,1.2))
> for (i in 1:7)
>  {points(rep(i,8-i),r[,i][1:(8-i)],col=2) }
> for (i in 2:7)
>  {points(rep(i,(i-1)),r[,i][(9-i):7],col=3)}
>
> Thanks
>
>
> Sincerely,
> Yanwei Zhang
> Department of Actuarial Research and Modeling
> Munich Re America
> 
>
>[[alternative HTML version deleted]]
>
> __
> 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.
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

[[alternative HTML version deleted]]

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Re: [R] Lowest k values of list

[Greg Snow]

In addition to the other responses, you may want to look at the 'partial' 
argument to sort.  For large vectors it may speed things up to not sort 
everything if all you care about is the top and bottom few.

Try:

> tmp <- rnorm(100)
> plot( sort(tmp, partial=c(1,2,99,100) ) )

Hope this helps,

--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111



> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Markus Mühlbacher
> Sent: Friday, September 05, 2008 9:26 AM
> To: r-help@r-project.org
> Subject: [R] Lowest k values of list
>
> Hi @ all,
>
> how do I get the largest or lowest k values of list? It must
> similar to the min() / max() function, but I just don't get it.
>
> Best wishes,
> Markus
>
>
>
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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Re: [R] lm and time series.

[Gabor Grothendieck]

The Time Series section in ?lm should be self explanatory.   If you are using
diff's and lag's then look at the dyn package.

On Fri, Sep 5, 2008 at 12:25 PM,  <[EMAIL PROTECTED]> wrote:
> I did a ?lm and it said basically to be careful when using lm and a time 
> series. But my question is probably more to do with my inexperience that 
> anything. If I have a time series object 'ti' how do I write the formula? The 
> response is the value at any particular time and the time is basically the 
> index of the time series. But I don't know how to put that into a formula.
>
> Thank you.
>
> Kevin

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Re: [R] binary order combinations

[Dimitri Liakhovitski]

Henrik,
this is amazing! wow!
Thank you so much!
Dimitri

On 9/5/08, Henrik Bengtsson <[EMAIL PROTECTED]> wrote:
> names <- sprintf("V%d", 1:4);
> n <- length(names);
> stopifnot(n <= 32); # Theoretical upper limit
> x <- matrix(intToBits(1:(2^n-1)), ncol=2^n-1);
> x <- x[1:n,,drop=FALSE];
> keys <- apply(x, MARGIN=2, FUN=function(z) paste(names[as.logical(z)],
> collapse=":"));
> print(keys);
>
>  [1] "V1"  "V2"  "V1:V2"   "V3"
>  [5] "V1:V3"   "V2:V3"   "V1:V2:V3""V4"
>  [9] "V1:V4"   "V2:V4"   "V1:V2:V4""V3:V4"
> [13] "V1:V3:V4""V2:V3:V4""V1:V2:V3:V4"
>
> /H
>
> On Fri, Sep 5, 2008 at 8:23 AM, Dimitri Liakhovitski <[EMAIL PROTECTED]> 
> wrote:
> > I am not sure it can do it. Besides, I ran a test of combos from quantreg:
> >
> > library(quantreg)
> > H<-1:3
> > test.combos<-lapply(1:3,function(x)
> > {combn(H,x)
> > })
> >
> > Every time I tried it crashed my R...
> >
> > :(
> >
> > Dimitri
> >
> > On 9/5/08, roger koenker <[EMAIL PROTECTED]> wrote:
> >> Does ?combos in the quantreg package do what you want?
> >>
> >>
> >> url:www.econ.uiuc.edu/~rogerRoger Koenker
> >> email[EMAIL PROTECTED]Department of Economics
> >> vox: 217-333-4558University of Illinois
> >> fax:   217-244-6678Champaign, IL 61820
> >>
> >>
> >>
> >>
> >> On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:
> >>
> >> >
> >> > Dear all!
> >> >
> >> > I have a vector of names
> >> > names<-("V1", "V2", "V3",., "V15")
> >> >
> >> > I could create all possible combinations of these names (of all
> >> > lengths) using R:
> >> >
> >> > combos<-lapply(1:15,function(x)
> >> > {combn(names,x)
> >> > })
> >> >
> >> > I get a list with all possible combinations of elements of 'names'
> >> > that looks like this (just the very beginning of it):
> >> >
> >> > [[1]] - the first element contains all combinations of 1 name
> >> >[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
> >> [,14]
> >> > [1,] "V1" "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12" "V13"
> >> "V14"
> >> >[,15]
> >> > [1,] "V15"
> >> >
> >> > [[2]] - the second element contains all possible combinations of 2 names
> >> >[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]  [,10] [,11] [,12] [,13]
> >> > [1,] "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1"  "V1"  "V1"  "V1"  "V1"
> >> > [2,] "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12" "V13" 
> >> > "V14"
> >> > .
> >> > .
> >> > .
> >> > etc.
> >> >
> >> > My question is: Is there any way to re-arrange all sub-elements of the
> >> > above list (i.e., all possible combinations of names such as V1,
> >> > V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
> >> > according to this system:
> >> > V1=1
> >> > V2=2
> >> > V3=4
> >> > V4=8
> >> > V5=16, etc
> >> >
> >> > So, I'd like those combinations to be arranged in a vector in the
> >> > following order:
> >> > 1. V1 (because V1=1)
> >> > 2. V2 (because V2=2)
> >> > 3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
> >> > 4. V3 (because V3=4)
> >> > 5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
> >> > 6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
> >> > 7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
> >> > 8. V4 (because V4=8)
> >> > etc.
> >> >
> >> > Is it at all possible?
> >> > Or maybe there is a way to create the name combinations in such an
> >> > order in the first place?
> >> >
> >> > Thank you very much!
> >> > Dimitri Liakhovitski
> >> >
> >> > __
> >> > 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.
> >> >
> >>
> >>
> >
> > __
> > 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.
> >
>

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[R] lm and time series.

[rkevinburton]

I did a ?lm and it said basically to be careful when using lm and a time 
series. But my question is probably more to do with my inexperience that 
anything. If I have a time series object 'ti' how do I write the formula? The 
response is the value at any particular time and the time is basically the 
index of the time series. But I don't know how to put that into a formula.

Thank you.

Kevin

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Re: [R] how to draw the legend about color from 3d picture

[Greg Snow]

Look at the color.legend function in the plotrix package to see if it does what 
you want.  You may want to use the layout function to split the graphics device 
into 2 sections, one for the persp plot and the other to hold the legend.

Hope this helps,

--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111



> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of yk
> Sent: Friday, September 05, 2008 6:07 AM
> To: r-help@r-project.org
> Subject: [R] how to draw the legend about color from 3d picture
>
> I have drawed a picture with persp, it's 3d map with
> different color, indicate different altitude. In gnuplot, the
> corresponding command 'splot' will generate a picture beside
> to indicate the relationship between color and altitude. But
> in R, how to draw it? I have read the manual of legend, but
> they are all about how to draw a legend with colored text,
> not a continuous varing color with corresponding number.
>
> __
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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, minimal, self-contained, reproducible code.
>

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] binary order combinations

[roger koenker]

You need to read the help file ?combos, and then use it to do indexing
of your objects, it only knows how to construct the integer combinations
given  a pair (n,p).

url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax:   217-244-6678Champaign, IL 61820



On Sep 5, 2008, at 10:23 AM, Dimitri Liakhovitski wrote:

I am not sure it can do it. Besides, I ran a test of combos from  
quantreg:


library(quantreg)
H<-1:3
test.combos<-lapply(1:3,function(x)
{combn(H,x)
})

Every time I tried it crashed my R...

:(

Dimitri

On 9/5/08, roger koenker <[EMAIL PROTECTED]> wrote:

Does ?combos in the quantreg package do what you want?


url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax:   217-244-6678Champaign, IL 61820




On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:



Dear all!

I have a vector of names
names<-("V1", "V2", "V3",., "V15")

I could create all possible combinations of these names (of all
lengths) using R:

combos<-lapply(1:15,function(x)
{combn(names,x)
})

I get a list with all possible combinations of elements of 'names'
that looks like this (just the very beginning of it):

[[1]] - the first element contains all combinations of 1 name
  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [, 
13]

[,14]
[1,] "V1" "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11"  
"V12" "V13"

"V14"

  [,15]
[1,] "V15"

[[2]] - the second element contains all possible combinations of 2  
names
  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]  [,10] [,11] [,12]  
[,13]
[1,] "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1"  "V1"  "V1"   
"V1"  "V1"
[2,] "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12"  
"V13" "V14"

.
.
.
etc.

My question is: Is there any way to re-arrange all sub-elements of  
the

above list (i.e., all possible combinations of names such as V1,
V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
according to this system:
V1=1
V2=2
V3=4
V4=8
V5=16, etc

So, I'd like those combinations to be arranged in a vector in the
following order:
1. V1 (because V1=1)
2. V2 (because V2=2)
3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
4. V3 (because V3=4)
5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
8. V4 (because V4=8)
etc.

Is it at all possible?
Or maybe there is a way to create the name combinations in such an
order in the first place?

Thank you very much!
Dimitri Liakhovitski

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Re: [R] Lowest k values of list

[Jorge Ivan Velez]

Dear Markus,
I think that I missed something in my previous reply. Here is another
approach using the first k values of each list:

# Data set
set.seed(123)
mylist=list(a=rnorm(10),b=rpois(15,10),f=rnorm(30,12,3))

# Function to report the first k values
k.values=function(x,k){
res= x[order(x)][1:k]
names(res)=paste('value_',1:k,sep="")
res
}

# Example: first 3 values
res=lapply(mylist,k.values,k=3)
do.call(rbind,res)
 value_1value_2value_3
a -1.265061 -0.6868529 -0.5604756
b  3.00  4.000  6.000
f  7.353742  8.2038109  8.5855892


HTH,

Jorge




On Fri, Sep 5, 2008 at 11:26 AM, Markus Mühlbacher <[EMAIL PROTECTED]>wrote:

> Hi @ all,
>
> how do I get the largest or lowest k values of list? It must similar to the
> min() / max() function, but I just don't get it.
>
> Best wishes,
> Markus
>
>
>
>
> __
> 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.
>

[[alternative HTML version deleted]]

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Re: [R] binary order combinations

[Dimitri Liakhovitski]

Sorry,

I misread it first. I tried:

library(quantreg)
 test.combos<-lapply(1:15,function(x)
 {combos(15,x)
 })

It gives me a list with an order that is somewhat different from
before, but I am not sure it helps me much:

[[1]]
 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,]1234567891011121314
 [,15]
[1,]15

[[2]]
 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,]111111111 1 1 1 1 1
[2,]2   15   14   13   12   11   1098 7 6 5 4 3
 [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [,26]
[1,] 2 2 2 2 2 2 2 2 2 2 2 2
[2,] 3151413121110 9 8 7 6 5


Dimitri

On 9/5/08, roger koenker <[EMAIL PROTECTED]> wrote:
> You need to read the help file ?combos, and then use it to do indexing
> of your objects, it only knows how to construct the integer combinations
> given  a pair (n,p).
>
> url:www.econ.uiuc.edu/~rogerRoger Koenker
> email[EMAIL PROTECTED]Department of Economics
> vox: 217-333-4558University of Illinois
> fax:   217-244-6678Champaign, IL 61820
>
>
>
> On Sep 5, 2008, at 10:23 AM, Dimitri Liakhovitski wrote:
>
> > I am not sure it can do it. Besides, I ran a test of combos from quantreg:
> >
> > library(quantreg)
> > H<-1:3
> > test.combos<-lapply(1:3,function(x)
> > {combn(H,x)
> > })
> >
> > Every time I tried it crashed my R...
> >
> > :(
> >
> > Dimitri
> >
> > On 9/5/08, roger koenker <[EMAIL PROTECTED]> wrote:
> >
> > > Does ?combos in the quantreg package do what you want?
> > >
> > >
> > > url:www.econ.uiuc.edu/~rogerRoger Koenker
> > > email[EMAIL PROTECTED]Department of Economics
> > > vox: 217-333-4558University of Illinois
> > > fax:   217-244-6678Champaign, IL 61820
> > >
> > >
> > >
> > >
> > > On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:
> > >
> > >
> > > >
> > > > Dear all!
> > > >
> > > > I have a vector of names
> > > > names<-("V1", "V2", "V3",., "V15")
> > > >
> > > > I could create all possible combinations of these names (of all
> > > > lengths) using R:
> > > >
> > > > combos<-lapply(1:15,function(x)
> > > > {combn(names,x)
> > > > })
> > > >
> > > > I get a list with all possible combinations of elements of 'names'
> > > > that looks like this (just the very beginning of it):
> > > >
> > > > [[1]] - the first element contains all combinations of 1 name
> > > >  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
> > > >
> > > [,14]
> > >
> > > > [1,] "V1" "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12"
> "V13"
> > > >
> > > "V14"
> > >
> > > >  [,15]
> > > > [1,] "V15"
> > > >
> > > > [[2]] - the second element contains all possible combinations of 2
> names
> > > >  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]  [,10] [,11] [,12] [,13]
> > > > [1,] "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1"  "V1"  "V1"  "V1"
> "V1"
> > > > [2,] "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12" "V13"
> "V14"
> > > > .
> > > > .
> > > > .
> > > > etc.
> > > >
> > > > My question is: Is there any way to re-arrange all sub-elements of the
> > > > above list (i.e., all possible combinations of names such as V1,
> > > > V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
> > > > according to this system:
> > > > V1=1
> > > > V2=2
> > > > V3=4
> > > > V4=8
> > > > V5=16, etc
> > > >
> > > > So, I'd like those combinations to be arranged in a vector in the
> > > > following order:
> > > > 1. V1 (because V1=1)
> > > > 2. V2 (because V2=2)
> > > > 3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
> > > > 4. V3 (because V3=4)
> > > > 5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
> > > > 6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
> > > > 7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
> > > > 8. V4 (because V4=8)
> > > > etc.
> > > >
> > > > Is it at all possible?
> > > > Or maybe there is a way to create the name combinations in such an
> > > > order in the first place?
> > > >
> > > > Thank you very much!
> > > > Dimitri Liakhovitski
> > > >
> > > > __
> > > > 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.
> > > >
> > > >
> > >
> > >
> > >
> >
> > __
> > 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, mini

Re: [R] Lowest k values of list

[jim holtman]

Do you mean a vector? If so, head/tail will work for you:

> x <- runif(10)
> x
 [1] 0.26550866 0.37212390 0.57285336 0.90820779 0.20168193 0.89838968
0.94467527
 [8] 0.66079779 0.62911404 0.06178627
> # lowest 5
> head(sort(x), 5)
[1] 0.06178627 0.20168193 0.26550866 0.37212390 0.57285336
> # highest 5
> tail(sort(x), 5)
[1] 0.6291140 0.6607978 0.8983897 0.9082078 0.9446753
>


On Fri, Sep 5, 2008 at 11:26 AM, Markus Mühlbacher <[EMAIL PROTECTED]> wrote:
> Hi @ all,
>
> how do I get the largest or lowest k values of list? It must similar to the 
> min() / max() function, but I just don't get it.
>
> Best wishes,
> Markus
>
>
>
>
> __
> 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.
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] binary order combinations

[Henrik Bengtsson]

names <- sprintf("V%d", 1:4);
n <- length(names);
stopifnot(n <= 32); # Theoretical upper limit
x <- matrix(intToBits(1:(2^n-1)), ncol=2^n-1);
x <- x[1:n,,drop=FALSE];
keys <- apply(x, MARGIN=2, FUN=function(z) paste(names[as.logical(z)],
collapse=":"));
print(keys);

 [1] "V1"  "V2"  "V1:V2"   "V3"
 [5] "V1:V3"   "V2:V3"   "V1:V2:V3""V4"
 [9] "V1:V4"   "V2:V4"   "V1:V2:V4""V3:V4"
[13] "V1:V3:V4""V2:V3:V4""V1:V2:V3:V4"

/H

On Fri, Sep 5, 2008 at 8:23 AM, Dimitri Liakhovitski <[EMAIL PROTECTED]> wrote:
> I am not sure it can do it. Besides, I ran a test of combos from quantreg:
>
> library(quantreg)
> H<-1:3
> test.combos<-lapply(1:3,function(x)
> {combn(H,x)
> })
>
> Every time I tried it crashed my R...
>
> :(
>
> Dimitri
>
> On 9/5/08, roger koenker <[EMAIL PROTECTED]> wrote:
>> Does ?combos in the quantreg package do what you want?
>>
>>
>> url:www.econ.uiuc.edu/~rogerRoger Koenker
>> email[EMAIL PROTECTED]Department of Economics
>> vox: 217-333-4558University of Illinois
>> fax:   217-244-6678Champaign, IL 61820
>>
>>
>>
>>
>> On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:
>>
>> >
>> > Dear all!
>> >
>> > I have a vector of names
>> > names<-("V1", "V2", "V3",., "V15")
>> >
>> > I could create all possible combinations of these names (of all
>> > lengths) using R:
>> >
>> > combos<-lapply(1:15,function(x)
>> > {combn(names,x)
>> > })
>> >
>> > I get a list with all possible combinations of elements of 'names'
>> > that looks like this (just the very beginning of it):
>> >
>> > [[1]] - the first element contains all combinations of 1 name
>> >[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
>> [,14]
>> > [1,] "V1" "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12" "V13"
>> "V14"
>> >[,15]
>> > [1,] "V15"
>> >
>> > [[2]] - the second element contains all possible combinations of 2 names
>> >[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]  [,10] [,11] [,12] [,13]
>> > [1,] "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1"  "V1"  "V1"  "V1"  "V1"
>> > [2,] "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12" "V13" "V14"
>> > .
>> > .
>> > .
>> > etc.
>> >
>> > My question is: Is there any way to re-arrange all sub-elements of the
>> > above list (i.e., all possible combinations of names such as V1,
>> > V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
>> > according to this system:
>> > V1=1
>> > V2=2
>> > V3=4
>> > V4=8
>> > V5=16, etc
>> >
>> > So, I'd like those combinations to be arranged in a vector in the
>> > following order:
>> > 1. V1 (because V1=1)
>> > 2. V2 (because V2=2)
>> > 3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
>> > 4. V3 (because V3=4)
>> > 5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
>> > 6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
>> > 7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
>> > 8. V4 (because V4=8)
>> > etc.
>> >
>> > Is it at all possible?
>> > Or maybe there is a way to create the name combinations in such an
>> > order in the first place?
>> >
>> > Thank you very much!
>> > Dimitri Liakhovitski
>> >
>> > __
>> > 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.
>> >
>>
>>
>
> __
> 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.
>

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Re: [R] Lowest k values of list

[David Winsemius]

On Sep 5, 2008, at 11:26 AM, Markus Mühlbacher wrote:


Hi @ all,

how do I get the largest or lowest k values of list? It must similar  
to the min() / max() function, but I just don't get it.



Might depend on how you define "high" and "low". Consider these  
experiments:


> z <- list(1,4,"5",6,,2,3)
> head(sort(unlist(z)),n=2)
[1] "1" "2"
> tail(sort(unlist(z)),n=2)
[1] "5" "6"


> z <- list(1,2,3,4,"5",6,"a")
> tail(sort(unlist(z)),n=2)
[1] "6" "a"
> head(sort(unlist(z)),n=2)
[1] "1" "2"



--
David Winsemius

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Re: [R] Lowest k values of list

[Jorge Ivan Velez]

Dear Markus,
Try this:

# Data set
set.seed(123)
mylist=list(a=rnorm(10),b=rpois(15,10),f=rnorm(30,12,3))
mylist


# min and max
temp=lapply(mylist,function(x){
 res=c(min(x),max(x))
 names(res)=c('Min.','Max.')
 res
}
)

do.call(rbind,temp)
Min.  Max.
a -1.265061  1.715065
b  3.00 15.00
f  7.353742 18.506868


HTH,

Jorge



On Fri, Sep 5, 2008 at 11:26 AM, Markus Mühlbacher <[EMAIL PROTECTED]>wrote:

> Hi @ all,
>
> how do I get the largest or lowest k values of list? It must similar to the
> min() / max() function, but I just don't get it.
>
> Best wishes,
> Markus
>
>
>
>
> __
> 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.
>

[[alternative HTML version deleted]]

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Re: [R] Time Series (attribute)

[Gabor Grothendieck]

The zoo package can handle series with times of any class that
is totally ordered and provides certain methods.  That
includes all common time classes and allows for you to define
new time classes too.  See

library(zoo)
?zoo

and the three vignettes (i.e. pdf documents) that come with it.

On Fri, Sep 5, 2008 at 11:22 AM, Oliver Bandel
<[EMAIL PROTECTED]> wrote:
> Hello,
>
> what kind of advantages does R's time series offer?
>
> Would it be possible to use data with units?
>
> For example data from a real time series with certain
> sampling rates - could they be used so,
> that the sample-times are measured in millisceonds or
> microseconds, instead of an index of the positions?
>
> If so, this would be much easier to handle than just
> working with indices of a vector.
>
> If the above mentioned way would not be possible, for what
> is this timeseries-attribute (?) useful?
>
>
> Ciao,
>   Oliver
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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[R] dealing with NAs in time series

[Alexy Khrabrov]

Certain timeseries I have had outliers, which I removed by assigning  
NA to their positions.  Now acf() refuses to go to work.  What's the  
right way to remove outliers from ts objects, and what are teh  
standard ways to interpolate NAs in them?


Cheers,
Alexy

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Re: [R] Confidence Intervals on Hazard Plots

[David Winsemius]

On Sep 5, 2008, at 9:20 AM, Alan Cox wrote:

Hello,  Is it possible to create confidence intervals for hazard  
rates?  I'm creating two muhaz objects:


haz1 <- muhaz(NumDaysCustomer[cRV=="true"],status[cRV=="true"])
haz2 <- muhaz(NumDaysCustomer[cRV=="false"],status[cRV=="false"])

and plotting them.

There are many, many more observations in the cohort cRV=="false"  
than =="true".  And, there are many more observations with lifetimes  
in the middle of the range than at the ends.  I suspect that this is  
a common phenomenon.When I plot the two hazard rate curves, haz1  
looks very different than haz2, but I'd like to see if it's  
different enough.  How can I go about creating confidence intervals  
and plotting them?


?muhaz.object  says that var.min and bias.min could be extracted from  
haz1 and haz2, since you have not changed the default for bw.method.


The default haz.est vector is twice as long as the default var.min  
vector, so you will need to take care to only plot the haz.est values  
only for the the event times, or perhaps you could specify that  
n.est.grid be the same as n.min.grid at the time of creating your  
fits. Not sure if the second method might have any undesirable side- 
effects, however.


--
David Winsemius
Heritage Laboratories

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Re: [R] Upgrade to R 2.7.2 from Debian Repository

[Dirk Eddelbuettel]

Tom,

(That's a suitable question for r-sig-debian)

On 5 September 2008 at 03:08, Tom La Bone wrote:
| I am running R 2.7.1 on an eeepc with the standard Xandros Linux. I
| install/upgrade R binaries using synaptic from the
| http://cran.r-project.org/bin/linux/debian/etch-cran/ repository. R 2.7.2
| does not appear to be available from this repository. Will it be available
| soon or is there another repository I should be using?

The volunteer who is providing these backports for CRAN indicated a few days
ago that he is being slowed down due to network issues.  So sorry, right now
you either have to wait, fine someone else to build the .deb files on etch
for you, or do it yourself.  We do have the Ubuntu backports in place so if
you were contemplating switching your Xandros install with the Ubuntu/EEE
setup you'd be covered...  

Dirk

-- 
Three out of two people have difficulties with fractions.

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[R] Plot by column

[Zhang Yanwei - Princeton-MRAm]

Dear list,

 I have the following matrix. How can I make the following plot?
 1. The x-axis has index 1:7, and the first column is plotted against index 1, 
second against 2, and so on.
 2. I want the points from the left upper conner including the antidiagonal to 
be plotted with col=2, and the rest with col=3
  [,1]  [,2]  [,3]  [,4]  [,5]  [,6]  [,7]
[1,] 0.589 0.857 0.923 0.944 0.954 0.963 0.980
[2,] 0.470 0.763 0.877 0.900 0.911 0.957 0.974
[3,] 0.486 0.863 0.905 0.960 0.968 0.998 1.015
[4,] 0.653 0.888 0.943 0.952 0.962 0.992 1.009
[5,] 0.664 0.774 0.958 0.986 0.996 1.027 1.045
[6,] 0.546 0.910 1.011 1.041 1.051 1.083 1.102
[7,] 0.407 0.600 0.667 0.686 0.693 0.715 0.727


The following the my code to do the above with the matrix called "r". How can I 
simply the code?


plot(r[,1],type="n",ylim=c(0.4,1.2))
for (i in 1:7)
 {points(rep(i,8-i),r[,i][1:(8-i)],col=2) }
for (i in 2:7)
 {points(rep(i,(i-1)),r[,i][(9-i):7],col=3)}

Thanks


Sincerely,
Yanwei Zhang
Department of Actuarial Research and Modeling
Munich Re America


[[alternative HTML version deleted]]

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[R] Lowest k values of list

[Markus Mühlbacher]

Hi @ all,

how do I get the largest or lowest k values of list? It must similar to the 
min() / max() function, but I just don't get it. 

Best wishes,
Markus




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Re: [R] binary order combinations

[Dimitri Liakhovitski]

I am not sure it can do it. Besides, I ran a test of combos from quantreg:

library(quantreg)
H<-1:3
test.combos<-lapply(1:3,function(x)
{combn(H,x)
})

Every time I tried it crashed my R...

:(

Dimitri

On 9/5/08, roger koenker <[EMAIL PROTECTED]> wrote:
> Does ?combos in the quantreg package do what you want?
>
>
> url:www.econ.uiuc.edu/~rogerRoger Koenker
> email[EMAIL PROTECTED]Department of Economics
> vox: 217-333-4558University of Illinois
> fax:   217-244-6678Champaign, IL 61820
>
>
>
>
> On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:
>
> >
> > Dear all!
> >
> > I have a vector of names
> > names<-("V1", "V2", "V3",., "V15")
> >
> > I could create all possible combinations of these names (of all
> > lengths) using R:
> >
> > combos<-lapply(1:15,function(x)
> > {combn(names,x)
> > })
> >
> > I get a list with all possible combinations of elements of 'names'
> > that looks like this (just the very beginning of it):
> >
> > [[1]] - the first element contains all combinations of 1 name
> >[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
> [,14]
> > [1,] "V1" "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12" "V13"
> "V14"
> >[,15]
> > [1,] "V15"
> >
> > [[2]] - the second element contains all possible combinations of 2 names
> >[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]  [,10] [,11] [,12] [,13]
> > [1,] "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1"  "V1"  "V1"  "V1"  "V1"
> > [2,] "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12" "V13" "V14"
> > .
> > .
> > .
> > etc.
> >
> > My question is: Is there any way to re-arrange all sub-elements of the
> > above list (i.e., all possible combinations of names such as V1,
> > V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
> > according to this system:
> > V1=1
> > V2=2
> > V3=4
> > V4=8
> > V5=16, etc
> >
> > So, I'd like those combinations to be arranged in a vector in the
> > following order:
> > 1. V1 (because V1=1)
> > 2. V2 (because V2=2)
> > 3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
> > 4. V3 (because V3=4)
> > 5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
> > 6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
> > 7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
> > 8. V4 (because V4=8)
> > etc.
> >
> > Is it at all possible?
> > Or maybe there is a way to create the name combinations in such an
> > order in the first place?
> >
> > Thank you very much!
> > Dimitri Liakhovitski
> >
> > __
> > 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.
> >
>
>

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[R] Time Series (attribute)

[Oliver Bandel]

Hello,

what kind of advantages does R's time series offer?

Would it be possible to use data with units?

For example data from a real time series with certain
sampling rates - could they be used so,
that the sample-times are measured in millisceonds or
microseconds, instead of an index of the positions?

If so, this would be much easier to handle than just
working with indices of a vector.

If the above mentioned way would not be possible, for what
is this timeseries-attribute (?) useful?


Ciao,
   Oliver

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Re: [R] controlling lattice plot ticks with relation="free"

[Richard . Cotton]

> How do you persuade lattice to draw tick marks on both the left and 
right 
> side of the y-axis, when relation="free" in the scales component?
> 
> #Ticks appear on both sides
> histogram(~height|voice.part, data=singer)
> 
> ##Ticks only on left
> histogram(~height|voice.part, data=singer, 
> scales=list(y=list(relation="free")))

I had an off list reply that was very helpful, suggesting the use of 
panel.axis.   For posterity, here's my own answer to my problem, using 
yscale.component. It's not a particularly natural solution, and it 
requires manually increasing the horizontal spacing between plots (with 
the between agrument), but it works.

myyscale.component <- function(...)
{
  ans <- yscale.components.default(...)
  ans$right <- ans$left
  ans$right$labels$labels <- NULL
  ans
} 

histogram(~height|voice.part, data=singer, 
   scales=list(y=list(relation="free")),
   yscale.component=myyscale.component,
   between=list(x=.5))

Regards,
Richie.

Mathematical Sciences Unit
HSL



ATTENTION:

This message contains privileged and confidential inform...{{dropped:20}}

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Re: [R] binary order combinations

[roger koenker]

Does ?combos in the quantreg package do what you want?


url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax:   217-244-6678Champaign, IL 61820



On Sep 5, 2008, at 9:58 AM, Dimitri Liakhovitski wrote:


Dear all!

I have a vector of names
names<-("V1", "V2", "V3",., "V15")

I could create all possible combinations of these names (of all
lengths) using R:

combos<-lapply(1:15,function(x)
{combn(names,x)
})

I get a list with all possible combinations of elements of 'names'
that looks like this (just the very beginning of it):

[[1]] - the first element contains all combinations of 1 name
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [, 
13] [,14]
[1,] "V1" "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12"  
"V13" "V14"

[,15]
[1,] "V15"

[[2]] - the second element contains all possible combinations of 2  
names
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]  [,10] [,11] [,12]  
[,13]
[1,] "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1"  "V1"  "V1"  "V1"   
"V1"
[2,] "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12" "V13"  
"V14"

.
.
.
etc.

My question is: Is there any way to re-arrange all sub-elements of the
above list (i.e., all possible combinations of names such as V1,
V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
according to this system:
V1=1
V2=2
V3=4
V4=8
V5=16, etc

So, I'd like those combinations to be arranged in a vector in the
following order:
1. V1 (because V1=1)
2. V2 (because V2=2)
3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
4. V3 (because V3=4)
5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
8. V4 (because V4=8)
etc.

Is it at all possible?
Or maybe there is a way to create the name combinations in such an
order in the first place?

Thank you very much!
Dimitri Liakhovitski

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[R] binary order combinations

[Dimitri Liakhovitski]

Dear all!

I have a vector of names
names<-("V1", "V2", "V3",., "V15")

I could create all possible combinations of these names (of all
lengths) using R:

combos<-lapply(1:15,function(x)
{combn(names,x)
})

I get a list with all possible combinations of elements of 'names'
that looks like this (just the very beginning of it):

[[1]] - the first element contains all combinations of 1 name
 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[1,] "V1" "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12" "V13" "V14"
 [,15]
[1,] "V15"

[[2]] - the second element contains all possible combinations of 2 names
 [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]  [,10] [,11] [,12] [,13]
[1,] "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1" "V1"  "V1"  "V1"  "V1"  "V1"
[2,] "V2" "V3" "V4" "V5" "V6" "V7" "V8" "V9" "V10" "V11" "V12" "V13" "V14"
.
.
.
etc.

My question is: Is there any way to re-arrange all sub-elements of the
above list (i.e., all possible combinations of names such as V1,
V1:V3, V1:V2:V4:V5) in a binary system order. More specifically,
according to this system:
V1=1
V2=2
V3=4
V4=8
V5=16, etc

So, I'd like those combinations to be arranged in a vector in the
following order:
1. V1 (because V1=1)
2. V2 (because V2=2)
3. V1:V2 (because V1=1 and V2=2 so that 1+2=3)
4. V3 (because V3=4)
5. V1:V3 (because V1=1 and V3=4 so that 1+4=5)
6. V2:V3 (because V2=2 and V3=4 so that 2+4=6)
7. V1:V2:V3 (because V1=1 and V2=2 and V3=4 so that 1+2+4=7)
8. V4 (because V4=8)
etc.

Is it at all possible?
Or maybe there is a way to create the name combinations in such an
order in the first place?

Thank you very much!
Dimitri Liakhovitski

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Re: [R] ggplot2: Changes to grobs not saved to file output

[hadley wickham]

Unfortunately it's not particularly easy in the current version.  In
the next version, you can do:

p <- qplot(wt, mpg, data=mtcars, colour=cyl)
# Get the plot grob
grob <- ggplotGrob(p)
# Modify it place
grob <- geditGrob(grob, gPath("strip","label"), gp=gpar(fontface="bold"))

# Draw it
pdf(...)
grid.newpage()
grid.draw(grob)
dev.off()

I think in the current version you can do

qplot(wt, mpg, data=mtcars, colour=cyl)
grob <- grid.grab()

and then follow the remaining steps.

Regards,

Hadley

On Fri, Aug 29, 2008 at 8:58 AM, btcruiser <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> Maybe I missed something - most likely .:-(
>
> I create a gplot and then makes some changes to the plot using grid graphics
> functions. These changes show up on the display OK, but when I save using
> ggsave() the grid changes do not show up. How do I save the plot with these
> changes?
>
> Thanks in advance.
> --
> View this message in context: 
> http://www.nabble.com/ggplot2%3A-Changes-to-grobs-not-saved-to-file-output-tp19220492p19220492.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> 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.
>



-- 
http://had.co.nz/

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Re: [R] Density estimates in modelling framework

[hadley wickham]

Thanks for all your replies.  The suggestions I got were:

  * gss
  * logspline
  * locfit

I ended up using locfit because the interface was just right for my
needs, and it was faster than the alternatives.

Hadley

On Fri, Aug 29, 2008 at 2:03 PM, hadley wickham <[EMAIL PROTECTED]> wrote:
> Hi all,
>
> Do any packages implement density estimation in a modelling framework?
>  I want to be able to do something like:
>
> dmodel <- density(~ a + b, data = mydata)
> predict(dmodel, newdata)
>
> This isn't how sm or KernSmooth or base density estimation works.  Are
> there other packages that do density estimation?  Or is there some
> reason that this is a bad idea.
>
> Hadley
>
> --
> http://had.co.nz/
>



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Re: [R] Use of colour in plots

[Petr PIKAL]

Hi

[EMAIL PROTECTED] napsal dne 05.09.2008 16:24:35:

> Here is an example doing the same type of thing. 
> It should be easy enough to adapt. 
> 
> Good luck
> 
> ===
> x <- runif(100, 0, 1)
> y <- runif(100, 0, 1)
> z <- data.frame(x,y)
> 
>  plot(subset(z, z$y >=.5), col="red", ylim=c(min(z$y),
>  max(z$y)), pch=16) 
>  points(subset(z, z$y <=.49), col="blue", pch=16)
> ===

Or

third <- (z$y >=.5)+1
plot(z, col=third, pch=16)

Just tell to col a vector of colors with appropriate use of logical.

Or you can use col = as.numeric(some factor), which is quite convenient 
use of factors feature which is not desired in other cases.
See warning section of factor help page.

Regards

> 
> 
> --- On Fri, 9/5/08, Steve Murray <[EMAIL PROTECTED]> wrote:
> 
> > From: Steve Murray <[EMAIL PROTECTED]>
> > Subject: [R] Use of colour in plots
> > To: r-help@r-project.org
> > Received: Friday, September 5, 2008, 9:10 AM
> > Dear all,
> > 
> > I have 3 datasets all of which share the same longitude and
> > latitude values, which I'm looking to plot onto a
> > scattergraph. The third dataset has values which can only be
> > either '1' or '2'. So to incorporate all
> > three datasets onto two axes, I'm wondering if I can
> > plot dataset1 and dataset2 as normal, but then use colour to
> > determine whether these points are either values '1'
> > or '2' according to the third dataset.
> > 
> > If so, how would I go about doing this in R, and what
> > format would the command take?
> > 
> > Thanks for any help offered,
> > 
> > Steve
> > 
> > __
> > 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.
> 
> __
> 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.

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[R] using nls to fit a curve to data

[jpl]

Hi,

I am trying to fit a curve to data.  My command line is:

model10=nls(offspring~((A*c^k)/gamma(k))*((degdays-alpha)^(k-1))*exp(-c*(degdays-alpha)),
start=list(A=30,k=2,c=.018,alpha=131))

I get the error message:
Error in numericDeriv(form[[3]], names(ind), env) : 
Missing value or an infinity produced when evaluating the model
In addition: Warning message:
full precision was not achieved in 'gammafn' 


My starting values yield a curve very similar t to other models which I
fitted successfully.  So any suggestions?  I've tried different starting
values. 
-- 
View this message in context: 
http://www.nabble.com/using-nls-to-fit-a-curve-to-data-tp19332210p19332210.html
Sent from the R help mailing list archive at Nabble.com.

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[R] Example on Maximum likelihood estimation using R

[VincentLee]

Hi all,

I am very new to R too, but I read that R is powerful.

May I know given a set of data, are there any simple examples on using mle()
to estimate parameters of a lognormal and weibull distribution ?

Hope to hear from you soon.
Thank you

Vincent  




Ravi Varadhan wrote:
> 
> Hi,
> 
> 
> You should re-write your function `fr' as follows:
> 
> 
> fr <- function(par){
> a <- par[1]
> b <- par[2]
> p <- par[3]
> lambda <- par[4]
> 
> l <- 0.5*(lambda + b*p + (1-p)*(lambda-b))
> l^2 > lambda*b*p
> delta <- sqrt(abs(l^2 - b*p*lambda))
> mt <- a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
> logl <- sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
> return(-logl)
> }
> 
> However, I don't understand what the following fragment is doing in `fr':
> 
> l^2 > lambda*b*p
> 
> Can you clarfy that?
> 
> Ravi.
> 
> 
> ---
> 
> Ravi Varadhan, Ph.D.
> 
> Assistant Professor, The Center on Aging and Health
> 
> Division of Geriatric Medicine and Gerontology 
> 
> Johns Hopkins University
> 
> Ph: (410) 502-2619
> 
> Fax: (410) 614-9625
> 
> Email: [EMAIL PROTECTED]
> 
> Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html
> 
>  
> 
> 
> 
> 
> 
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> On
> Behalf Of toh
> Sent: Thursday, September 04, 2008 9:15 PM
> To: r-help@r-project.org
> Subject: Re: [R] Maximum likelihood estimation
> 
> 
> Yes I'm trying to optimize the parameters a, b, p and lambda where a > 0,
> b
>> 0 and 0 < p < 1. I attached the error message that I got when I run mle. 
> 
> 
>> t <- c(1:90)
>> y <-
>> c(5,10,15,20,26,34,36,43,47,49,80,84,108,157,171,183,191,200,204,211,2
>> 17,226,230,
> +
> 234,236,240,243,252,254,259,263,264,268,271,277,290,309,324,331,346,367,375,
> 381,
> +
> 401,411,414,417,425,430,431,433,435,437,444,446,446,448,451,453,460,463,463,
> 464,
> +
> 464,465,465,465,466,467,467,467,468,469,469,469,469,470,472,472,473,473,473,
> 473,
> + 473,473,473,475,475,475,475)
>> 
>> library(stats4)
>> fr <- function(a, b, p, lambda){
> + l <- 0.5*(lambda + b*p + (1-p)*(lambda-b))
> + l^2 > lambda*b*p
> + delta <- sqrt(abs(l^2 - b*p*lambda))
> + mt <- 
> + a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
> + logl <- sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
> + return(-logl)
> + }
>> 
>> mle(start=list(a=12,b=0.01,p=0.5,lambda=0.01),fr, method="L-BFGS-B",
> + lower = c(0.002, 0.002, 0.002, 0.002), upper = c(Inf, Inf, 0.999,
> Inf),control=list(fnscale=-1))
> Error in optim(start, f, method = method, hessian = TRUE, ...) : 
>   non-finite finite-difference value [3]
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> 
> Prof Brian Ripley wrote:
>> 
>>>From ?optim
>> 
>>fn: A function to be minimized (or maximized), with first
>>argument the vector of parameters over which minimization is
>>to take place.  It should return a scalar result.
>> 
>> I think you intended to optimize over c(a,b,p, lambda), so you need to 
>> specify them as a single vector.
>> 
>> You may be making life unnecessarily hard for yourself: see function 
>> mle() in package stats4.
>> 
>> Showing your code without a verbal description of what you are doing 
>> nor the error message you got is less helpful than we need.
>> 
>> On Wed, 3 Sep 2008, toh wrote:
>> 
>>>
>>> Hi R-experts,
>>> I'm new to R in mle. I tried to do the following but just couldn't 
>>> get it right. Hope someone can point out the mistakes. thanks a lot.
>>>
>>> t <- c(1:90)
>>> y <-
>>> c(5,10,15,20,26,34,36,43,47,49,80,84,108,157,171,183,191,200,204,211,
>>> 217,226,230,
>>>
>>> 234,236,240,243,252,254,259,263,264,268,271,277,290,309,324,331,346,3
>>> 67,375,381,
>>>
>>> 401,411,414,417,425,430,431,433,435,437,444,446,446,448,451,453,460,4
>>> 63,463,464,
>>>
>>>
> 464,465,465,465,466,467,467,467,468,469,469,469,469,470,472,472,473,473,473,
> 473,
>>> 473,473,473,475,475,475,475)
>>> fr <- function(a, b, p, lambda){
>>> l <- 0.5*(lambda + b*p + (1-p)*(lambda-b))
>>> l^2 > lambda*b*p
>>> delta <- sqrt(abs(l^2 - b*p*lambda))
>>> mt <- 
>>> a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
>>> logl <- sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
>>> return(-logl)
>>> }
>>> optim(c(15,0.01,0.5,0.01),fr, method="L-BFGS-B", lower = c(0.002, 
>>> 0.002, 0.002, 0.002), upper = c(Inf, Inf, 0.999,
>>> Inf),control=list(fnscale=-1))
>>>
>>> --
>>> View this message in context:
>>> http://www.nabble.com/Maximum-likelihood-estimation-tp19304249p193042
>>> 49.html Sent from the R help mailing list archive at Nabble.com.
>>>
>>> __
>>> 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-gu

Re: [R] Use of colour in plots

[John Kane]

Here is an example doing the same type of thing.  
It should be easy enough to adapt. 

Good luck

===
x <- runif(100, 0, 1)
y <- runif(100, 0, 1)
z <- data.frame(x,y)

 plot(subset(z, z$y >=.5), col="red", ylim=c(min(z$y),
 max(z$y)), pch=16) 
 points(subset(z, z$y <=.49), col="blue", pch=16)
===


--- On Fri, 9/5/08, Steve Murray <[EMAIL PROTECTED]> wrote:

> From: Steve Murray <[EMAIL PROTECTED]>
> Subject: [R] Use of colour in plots
> To: r-help@r-project.org
> Received: Friday, September 5, 2008, 9:10 AM
> Dear all,
> 
> I have 3 datasets all of which share the same longitude and
> latitude values, which I'm looking to plot onto a
> scattergraph. The third dataset has values which can only be
> either '1' or '2'. So to incorporate all
> three datasets onto two axes, I'm wondering if I can
> plot dataset1 and dataset2 as normal, but then use colour to
> determine whether these points are either values '1'
> or '2' according to the third dataset.
> 
> If so, how would I go about doing this in R, and what
> format would the command take?
> 
> Thanks for any help offered,
> 
> Steve
> 
> __
> 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.

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] (with subject)

[Dr Eberhard W Lisse]

Oh, YES, thank you!

This weekend I'll try and figure out how to plot these events on
a 24 hour scale, i.e I'll aggregate the SQL query on the time
But not on the date) to see how many of those fall outside of
normal working hours :-)-O

greetings, el

On 05 Sep 2008, at 00:59 , Gabor Grothendieck wrote:


I assume the problem is that you want the axis to have all 12
months but your data is much shorter.  Try this:

mos <- seq(as.Date("2008-01-01"), length = 12, by = "month")
plot(range(mos), range(rawData$y), type = "n", xaxt = "n")
lines(rawData$Date, rawData$y)
axis(1, mos, month.abb)


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[R] Question about col.names in write.csv

[Luz Milena Zea Fernandez]

Dear support, I don't ignore col.names in write.csv. I want to write names for 
the firts row. How can I do?

Thanks in advance

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Re: [R] Error: can not allocate vector of 117.3 Mb

[Tom La Bone]

To see some useful discussions of this problem in various settings search
R-Help for the phrase "cannot allocate vector of size".

Tom



ram basnet wrote:
> 
> Hi R users,
>  
> I am doing multiscale bootstrapping for clustering through pvclust
> package. I have large data set (observations 182 and variables 5546). When
> i tried to make bootstrapping, then i got message as "Bootstrap (r =
> 0.5)Error: cannot allocate vector of size 117.3 Mb".
> I am new R user and could not understand what is the problem and also
> don't know the easiet solution.
> If someone knows, pls help me.
> Thanks in advance.
>  
> Ram Kumar Basnet
> Graduate student.
> 
> 
>   
>   [[alternative HTML version deleted]]
> 
> 
> __
> 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.
> 
> 

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Re: [R] Maximum likelihood estimation

[Ravi Varadhan]

Hi,


You should re-write your function `fr' as follows:


fr <- function(par){
a <- par[1]
b <- par[2]
p <- par[3]
lambda <- par[4]

l <- 0.5*(lambda + b*p + (1-p)*(lambda-b))
l^2 > lambda*b*p
delta <- sqrt(abs(l^2 - b*p*lambda))
mt <- a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
logl <- sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
return(-logl)
}

However, I don't understand what the following fragment is doing in `fr':

l^2 > lambda*b*p

Can you clarfy that?

Ravi.


---

Ravi Varadhan, Ph.D.

Assistant Professor, The Center on Aging and Health

Division of Geriatric Medicine and Gerontology 

Johns Hopkins University

Ph: (410) 502-2619

Fax: (410) 614-9625

Email: [EMAIL PROTECTED]

Webpage:  http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html

 





-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of toh
Sent: Thursday, September 04, 2008 9:15 PM
To: r-help@r-project.org
Subject: Re: [R] Maximum likelihood estimation


Yes I'm trying to optimize the parameters a, b, p and lambda where a > 0, b
> 0 and 0 < p < 1. I attached the error message that I got when I run mle. 


> t <- c(1:90)
> y <-
> c(5,10,15,20,26,34,36,43,47,49,80,84,108,157,171,183,191,200,204,211,2
> 17,226,230,
+
234,236,240,243,252,254,259,263,264,268,271,277,290,309,324,331,346,367,375,
381,
+
401,411,414,417,425,430,431,433,435,437,444,446,446,448,451,453,460,463,463,
464,
+
464,465,465,465,466,467,467,467,468,469,469,469,469,470,472,472,473,473,473,
473,
+ 473,473,473,475,475,475,475)
> 
> library(stats4)
> fr <- function(a, b, p, lambda){
+ l <- 0.5*(lambda + b*p + (1-p)*(lambda-b))
+ l^2 > lambda*b*p
+ delta <- sqrt(abs(l^2 - b*p*lambda))
+ mt <- 
+ a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
+ logl <- sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
+ return(-logl)
+ }
> 
> mle(start=list(a=12,b=0.01,p=0.5,lambda=0.01),fr, method="L-BFGS-B",
+ lower = c(0.002, 0.002, 0.002, 0.002), upper = c(Inf, Inf, 0.999,
Inf),control=list(fnscale=-1))
Error in optim(start, f, method = method, hessian = TRUE, ...) : 
  non-finite finite-difference value [3]












Prof Brian Ripley wrote:
> 
>>From ?optim
> 
>fn: A function to be minimized (or maximized), with first
>argument the vector of parameters over which minimization is
>to take place.  It should return a scalar result.
> 
> I think you intended to optimize over c(a,b,p, lambda), so you need to 
> specify them as a single vector.
> 
> You may be making life unnecessarily hard for yourself: see function 
> mle() in package stats4.
> 
> Showing your code without a verbal description of what you are doing 
> nor the error message you got is less helpful than we need.
> 
> On Wed, 3 Sep 2008, toh wrote:
> 
>>
>> Hi R-experts,
>> I'm new to R in mle. I tried to do the following but just couldn't 
>> get it right. Hope someone can point out the mistakes. thanks a lot.
>>
>> t <- c(1:90)
>> y <-
>> c(5,10,15,20,26,34,36,43,47,49,80,84,108,157,171,183,191,200,204,211,
>> 217,226,230,
>>
>> 234,236,240,243,252,254,259,263,264,268,271,277,290,309,324,331,346,3
>> 67,375,381,
>>
>> 401,411,414,417,425,430,431,433,435,437,444,446,446,448,451,453,460,4
>> 63,463,464,
>>
>>
464,465,465,465,466,467,467,467,468,469,469,469,469,470,472,472,473,473,473,
473,
>>  473,473,473,475,475,475,475)
>> fr <- function(a, b, p, lambda){
>> l <- 0.5*(lambda + b*p + (1-p)*(lambda-b))
>> l^2 > lambda*b*p
>> delta <- sqrt(abs(l^2 - b*p*lambda))
>> mt <- 
>> a/p*(1-exp(-l*t)*cosh(delta*t)-(l-b*p)*exp(-t)*sinh(delta*t)/delta)
>> logl <- sum(diff(y)*log(diff(mt))-diff(mt)-lfactorial(diff(y)))
>> return(-logl)
>> }
>> optim(c(15,0.01,0.5,0.01),fr, method="L-BFGS-B", lower = c(0.002, 
>> 0.002, 0.002, 0.002), upper = c(Inf, Inf, 0.999,
>> Inf),control=list(fnscale=-1))
>>
>> --
>> View this message in context:
>> http://www.nabble.com/Maximum-likelihood-estimation-tp19304249p193042
>> 49.html Sent from the R help mailing list archive at Nabble.com.
>>
>> __
>> 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.
>>
> 
> -- 
> Brian D. Ripley,  [EMAIL PROTECTED]
> Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
> University of Oxford, Tel:  +44 1865 272861 (self)
> 1 South Parks Road, +44 1865 272866 (PA)
> Oxford OX1 3TG, UKFax:  +44 1865 272595
> 
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the po

[R] text processing for plots

[David Carslaw]

Hi R people,

I want to write some functions to automate the plotting of some expressions
that use some of the plotmath capabilities.

An example is a string supplied to a plot call such as:

plot(1, 1, ylab = "pm10 (ug/m3)")

This should actually appear like:

plot(1, 1, ylab = expression("PM"[10] * " (" * mu * "g m" ^-3 * ")"))

 i.e. pollutant name, units

What I would like to do is write a function that will automatically detect
certain strings, reformat them, and supply it as an argument to plot.  I
could just search for the whole string, but I would like something more
flexible because there will be many combinations of pollutant (pm10, nox,
no2...) and units (ug/m3, mg/m3, ng/m3...).

My question is this - is there a way to separately process the pollutant and
units and supply the correct overall expression to plot?

I'm not sure of the best strategy for this having read through previous
posts etc., and would appreciate your help!

Many thanks.

David Carslaw

-
Institute for Transport Studies
University of Leeds
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[R] Confidence Intervals on Hazard Plots

[Alan Cox]

Hello,  Is it possible to create confidence intervals for hazard rates?  I'm 
creating two muhaz objects:  

haz1 <- muhaz(NumDaysCustomer[cRV=="true"],status[cRV=="true"])
haz2 <- muhaz(NumDaysCustomer[cRV=="false"],status[cRV=="false"])

and plotting them.  

There are many, many more observations in the cohort cRV=="false" than 
=="true".  And, there are many more observations with lifetimes in the middle 
of the range than at the ends.  I suspect that this is a common phenomenon.
When I plot the two hazard rate curves, haz1 looks very different than haz2, 
but I'd like to see if it's different enough.  How can I go about creating 
confidence intervals and plotting them?

Thanks,  Alan 


-- 
Alan Cox 
Director, User Experience 
iContact, Corp. 
p 919.459.1038 f 919.287.2475 

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Re: [R] cluster a distance(analogue)-object using agnes(cluster)

[Gavin Simpson]

On Thu, 2008-09-04 at 11:28 +0200, Martin Maechler wrote:
> > "B" == Birgitle  <[EMAIL PROTECTED]>
> > on Tue, 2 Sep 2008 03:02:31 -0700 (PDT) writes:
> 
> B> I try to perform a clustering using an existing dissimilarity matrix 
> that I
> B> calculated using distance (analogue)
> B> I tried two different things. One of them worked and one not and I 
> don`t
> B> understand why.
> B> Here the code:
> 
> B> not working example
> 
> B> library(cluster)
> B> library(analogue)
> 
> B> iris2 <- as.data.frame(iris)
> 
> why that? After the above,  iris2  is identical() to iris !
> 
> B> str(iris2)
> B> 'data.frame':  150 obs. of  5 variables:
> B> $ Sepal.Length: num  5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
> B> $ Sepal.Width : num  3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
> B> $ Petal.Length: num  1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
> B> $ Petal.Width : num  0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
> B> $ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 
> 1 1
> B> 1 1 1 ...
> 
> B> Test.Gower <- distance(iris2, method ="mixed")
> 
> why not just 
> daisy(iris2, metric = "gower")
> 
> daisy() is in cluster which has been a recommended R package
> "forever".
> 
> So the solution (here, not in general!)
> is to stay with package 'cluster' and use
> daisy() before agnes().

And as the author of distance(), I agree *completely*, and say so in the
Environmetrics Task View.

distance() was written for a very specific task and that it does what
you want it to, Birgit, is side effect of the way I wrote distance().

Anyway, Birgit, you'll be glad to know that the example you included
works in the R-forge version of analogue (0.5-4 to be). Get it from:

https://r-forge.r-project.org/R/?group_id=69

or directly from within R

install.packages("analogue", repos="http://R-Forge.R-project.org";)

if you want to use distance(), but I'd be using the cluster package
tools myself for the problem you emailed about.

You also probably want to use as.dist() on the output from distance() to
store the matrix in a more compact form. Because of how I wanted to
calculate distances (between two data sets), the matrix output was
essential, but this output is storing (and computing) redundant data in
the case you cite (a single matrix).

All the best,

G

> 
> Regards,
> Martin Maechler, ETH Zurich  
>{same city!  feel free to phone me..}
> 
> B> Test.Gower.agnes<-agnes(Test.Gower, diss=T)
> B> Fehler in agnes(Test.Gower, diss = T) : 
> B> (list) Objekt kann nicht nach 'logical' umgewandelt werden
> B> Error in agnes(Test.Gower, diss=T).
> B> (list) object can`t be transformed to "logical"
> 
> B> working example only numerics used:
> 
> B> library(cluster)
> B> library(analogue)
> 
> B> irisPart<-subset(iris, select= Sepal.Length:Petal.Width)
> B> Dist.Gower <- distance(irisPart, method ="mixed")
> B> AgnesA <- agnes(Dist.Gower, method="average", diss=TRUE) 
> 
> B> Would be great if somebody could help me.
> B> The dataset that I would like to use for the clustering also contains
> B> factors.
> B> and gives me the same Error message as in the not working example.
> 
> B> Thanks in advance
> 
> B> B.
> 
> __
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-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
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 ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
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[R] Use of colour in plots

[Steve Murray]

Dear all,

I have 3 datasets all of which share the same longitude and latitude values, 
which I'm looking to plot onto a scattergraph. The third dataset has values 
which can only be either '1' or '2'. So to incorporate all three datasets onto 
two axes, I'm wondering if I can plot dataset1 and dataset2 as normal, but then 
use colour to determine whether these points are either values '1' or '2' 
according to the third dataset.

If so, how would I go about doing this in R, and what format would the command 
take?

Thanks for any help offered,

Steve

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] how to draw the legend about color from 3d picture

[Duncan Murdoch]

On 9/5/2008 8:06 AM, yk wrote:

I have drawed a picture with persp, it's 3d map with different color,
indicate different altitude. In gnuplot, the corresponding command
'splot' will generate a picture beside to indicate the relationship
between color and altitude. But in R, how to draw it? I have read the
manual of legend, but they are all about how to draw a legend with
colored text, not a continuous varing color with corresponding number.


I don't think there's an automatic way to do this (though probably some 
package provides one).


If you want to write your own, take a look at example(filled.contour). 
The legend it draws is done by this code:


plot.new()
plot.window(xlim = c(0, 1), ylim = range(levels), xaxs = "i",
yaxs = "i")
rect(0, levels[-length(levels)], 1, levels[-1], col = col)
if (missing(key.axes)) {
if (axes)
axis(4)
}
else key.axes
box()

(but it has done a lot of setup using layout() and par() before this).

Duncan Murdoch

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Re: [R] Plotting using 'if' statements

[Steve Murray]

Thanks again for a very quick reply! It worked really well too!



_
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Re: [R] Error: can not allocate vector of 117.3 Mb

[Uwe Ligges]

ram basnet wrote:

Hi R users,
 
I am doing multiscale bootstrapping for clustering through pvclust package. I have large data set (observations 182 and variables 5546). When i tried to make bootstrapping, then i got message as "Bootstrap (r = 0.5)Error: cannot allocate vector of size 117.3 Mb".

I am new R user and could not understand what is the problem and also don't 
know the easiet solution.
If someone knows, pls help me.


Please read ?Memory

Uwe Ligges



Thanks in advance.
 
Ram Kumar Basnet

Graduate student.


  
	[[alternative HTML version deleted]]






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[R] Error: can not allocate vector of 117.3 Mb

[ram basnet]

Hi R users,
 
I am doing multiscale bootstrapping for clustering through pvclust package. I 
have large data set (observations 182 and variables 5546). When i tried to make 
bootstrapping, then i got message as "Bootstrap (r = 0.5)Error: 
cannot allocate vector of size 117.3 Mb".
I am new R user and could not understand what is the problem and also don't 
know the easiet solution.
If someone knows, pls help me.
Thanks in advance.
 
Ram Kumar Basnet
Graduate student.


  
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Orthogonalization algorithms

[Williams, Robin]

Please ignore this, it appears I have found what I'm looking for in the
far package.   


Robin Williams 
Met Office summer intern - Health Forecasting 
[EMAIL PROTECTED] 
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Williams, Robin
Sent: Friday, September 05, 2008 1:18 PM
To: r-help@r-project.org
Subject: [R] Orthogonalization algorithms

Hi,
  I have eight vectors that I would like to orthogonalize preferably
using R. The vectors are of considerable length, however due to their
nature I know they satisfy the conditions needed to apply the
Gram-Schmidt algorithm. Before I embark on some R coding, I wanted to
check that there is no facility / function already around that computes
the orthogonalized set of vectors? I have performed an RSiteSearch etc
with no luck. 
  Anything that could help me along the way would be much appreciated. 
Thanks in advance,
Robin Williams
Met Office summer intern - Health Forecasting
[EMAIL PROTECTED] 
 

[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

[R] Orthogonalization algorithms

[Williams, Robin]

Hi, 
  I have eight vectors that I would like to orthogonalize preferably
using R. The vectors are of considerable length, however due to their
nature I know they satisfy the conditions needed to apply the
Gram-Schmidt algorithm. Before I embark on some R coding, I wanted to
check that there is no facility / function already around that computes
the orthogonalized set of vectors? I have performed an RSiteSearch etc
with no luck. 
  Anything that could help me along the way would be much appreciated. 
Thanks in advance,
Robin Williams
Met Office summer intern - Health Forecasting
[EMAIL PROTECTED] 
 

[[alternative HTML version deleted]]

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and provide commented, minimal, self-contained, reproducible code.

[R] how to draw the legend about color from 3d picture

[yk]

I have drawed a picture with persp, it's 3d map with different color,
indicate different altitude. In gnuplot, the corresponding command
'splot' will generate a picture beside to indicate the relationship
between color and altitude. But in R, how to draw it? I have read the
manual of legend, but they are all about how to draw a legend with
colored text, not a continuous varing color with corresponding number.

__
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Plotting using 'if' statements

[Henrique Dallazuanna]

Use this:

plot(January[c(3, 6)])

On Fri, Sep 5, 2008 at 8:52 AM, Steve Murray <[EMAIL PROTECTED]> wrote:

>
> Sorry! Hopefully just one more question!
>
> I'm now trying to plot columns 3 and 6. I've tried:
>
> > plot(January[3:6])
>
> ...but this plots columns 4 and 5 too!
>
>
> I've also tried:
>
> > plot(January[3],January[6])
> > plot(January[3,6])
> > plot(January[3]:January[6])
>
> But these don't produce the right results (instead, errors!)
>
> Again, I'm sure it's a just a small (but important) detail that I'm
> missing, and I'd be very grateful if anyone could help me out.
>
> Many thanks again,
>
> Steve
>
> 
> Date: Thu, 4 Sep 2008 13:52:03 -0300
> From: [EMAIL PROTECTED]
> To: [EMAIL PROTECTED]
> Subject: Re: [R] Plotting using 'if' statements
> CC: r-help@r-project.org
>
> Use '&':
>
> plot(January[January[,3]> 0 & January[,3] < 2, 3:4])
>
> On Thu, Sep 4, 2008 at 1:47 PM, Steve Murray  wrote:
>
>
>
> Ah that's great, thank you very much.
>
>
>
> As a follow-on, in the same format, how would I plot where column 3 is
> greater than 0 *but also less than 2*?
>
>
>
> Once again, any help is much appreciated.
>
>
>
>
>
> Thanks,
>
>
>
> Steve
>
>
>
> 
>
> Date: Thu, 4 Sep 2008 13:39:12 -0300
>
> From: [EMAIL PROTECTED]
>
> To: [EMAIL PROTECTED]
>
> Subject: Re: [R] Plotting using 'if' statements
>
> CC: r-help@r-project.org
>
>
>
> Try this:
>
>
>
> plot(January[January[,3]> 0, 3:4])
>
>
>
> On Thu, Sep 4, 2008 at 1:28 PM, Steve Murray  wrote:
>
>
>
>
>
>
>
>
>
>
>
> Dear all,
>
>
>
>
>
>
>
> I have a dataset of four columns, and I wish to plot (as a scatter graph)
> the values of the third column where the values are greater than zero, and
> the fourth column.
>
>
>
>
>
>
>
> I tried doing this via the plot command itself, but got into a bit of a
> mess (resulting in errors!). My dataframe is called 'January':
>
>
>
>
>
>
>
> > plot(January[3(>0):4])
>
>
>
> Error: unexpected '>' in "plot(January[3(>"
>
>
>
>
>
>
>
>
>
>
>
> After a few variations on this, I thought I'd try making a new object which
> includes all values from the third column of January>0 (to plot in a
> separate step) as follows:
>
>
>
>
>
>
>
> > JanFilter <- January[3]>0
>
>
>
>
>
>
>
> No error here. However, when I display the 'values' of JanFilter, it shows
> that instead of keeping the numerical values, the above operation simply
> displays the results of the logical test:
>
>
>
>
>
>
>
> > head(JanFilter)
>
>
>
>   Value
>
>
>
> [1,]   FALSE
>
>
>
> [2,]TRUE
>
>
>
> [3,]TRUE
>
>
>
> [4,]TRUE
>
>
>
> [5,]TRUE
>
>
>
> [6,]TRUE
>
>
>
>
>
>
>
>
>
>
>
> This is obviously no good for plotting the numerical values on axes!
>
>
>
>
>
>
>
> So my question is, how do I perform 'if' statements in order to filter out
> various parts of a dataset, for plotting on a graph.
>
>
>
>
>
>
>
> Many thanks,
>
>
>
>
>
>
>
> Steve
>
>
>
>
>
>
>
> __
>
>
>
> 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.
>
>
>
>
>
>
>
>
>
> --
>
> Henrique Dallazuanna
>
> Curitiba-Paraná-Brasil
>
> 25° 25' 40" S 49° 16' 22" O
>
>
>
>
>
>
>
> _
>
> Make a mini you and download it into Windows Live Messenger
>
> http://clk.atdmt.com/UKM/go/111354029/direct/01/
>
>
>
> --
> Henrique Dallazuanna
> Curitiba-Paraná-Brasil
>
> 25° 25' 40" S 49° 16' 22" O
>
>
>
> _
> Win New York holidays with Kellogg's & Live Search
> http://clk.atdmt.com/UKM/go/111354033/direct/01/




-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Plotting using 'if' statements

[Steve Murray]

Sorry! Hopefully just one more question!

I'm now trying to plot columns 3 and 6. I've tried:

> plot(January[3:6])

...but this plots columns 4 and 5 too!


I've also tried:

> plot(January[3],January[6])
> plot(January[3,6])
> plot(January[3]:January[6])

But these don't produce the right results (instead, errors!)

Again, I'm sure it's a just a small (but important) detail that I'm missing, 
and I'd be very grateful if anyone could help me out.

Many thanks again,

Steve


Date: Thu, 4 Sep 2008 13:52:03 -0300
From: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: Re: [R] Plotting using 'if' statements
CC: r-help@r-project.org

Use '&':

plot(January[January[,3]> 0 & January[,3] < 2, 3:4])

On Thu, Sep 4, 2008 at 1:47 PM, Steve Murray  wrote:



Ah that's great, thank you very much.



As a follow-on, in the same format, how would I plot where column 3 is greater 
than 0 *but also less than 2*?



Once again, any help is much appreciated.





Thanks,



Steve





Date: Thu, 4 Sep 2008 13:39:12 -0300

From: [EMAIL PROTECTED]

To: [EMAIL PROTECTED]

Subject: Re: [R] Plotting using 'if' statements

CC: r-help@r-project.org



Try this:



plot(January[January[,3]> 0, 3:4])



On Thu, Sep 4, 2008 at 1:28 PM, Steve Murray  wrote:











Dear all,







I have a dataset of four columns, and I wish to plot (as a scatter graph) the 
values of the third column where the values are greater than zero, and the 
fourth column.







I tried doing this via the plot command itself, but got into a bit of a mess 
(resulting in errors!). My dataframe is called 'January':







> plot(January[3(>0):4])



Error: unexpected '>' in "plot(January[3(>"











After a few variations on this, I thought I'd try making a new object which 
includes all values from the third column of January>0 (to plot in a separate 
step) as follows:







> JanFilter <- January[3]>0







No error here. However, when I display the 'values' of JanFilter, it shows that 
instead of keeping the numerical values, the above operation simply displays 
the results of the logical test:







> head(JanFilter)



   Value



[1,]   FALSE



[2,]TRUE



[3,]TRUE



[4,]TRUE



[5,]TRUE



[6,]TRUE











[[elided Hotmail spam]]







So my question is, how do I perform 'if' statements in order to filter out 
various parts of a dataset, for plotting on a graph.







Many thanks,







Steve







__



R-help@r-project.org mailing list







PLEASE do read the posting guide http://www.R-project.org/posting-guide.html



and provide commented, minimal, self-contained, reproducible code.









--

Henrique Dallazuanna

Curitiba-Paraná-Brasil

25° 25' 40" S 49° 16' 22" O







_







--
Henrique Dallazuanna
Curitiba-Paraná-Brasil

25° 25' 40" S 49° 16' 22" O



_
Win New York holidays with Kellogg’s & Live Search

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and provide commented, minimal, self-contained, reproducible code.

Re: [R] how to write a escape sequences in to a file

[Henrique Dallazuanna]

Try this:

cat(Str1, Str2, file = "out.txt", sep = "\n")

On Fri, Sep 5, 2008 at 2:21 AM, Kurapati, Ravichandra (Ravichandra) <
[EMAIL PROTECTED]> wrote:

>
>
> Hi
>
>
>
>   Str1<-"hai "
>
>   Str2<-"hru"
>
>
>
>  I want to write these 2 strings in a file  separated by
> newline.
>
>
>
>
>
>   How can I get this.
>
> Thanks
>
> K.Ravichandra
>
>
>
>
>
>
>[[alternative HTML version deleted]]
>
> __
> 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.
>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

[[alternative HTML version deleted]]

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Re: [R] Fitted probabilities in conditional logit regression

[Gavin Simpson]

On Sun, 2008-08-31 at 09:49 -0400, Min Chen wrote:
> Dear R-help,
> 
> I'm doing conditional logit regression for a discrete choice model.
> I want to know whether there's a way to get the fitted probabilities. In
> Stata, "predict" works for clogit, but it seems that in R "predict" does
> not.
> 
> Thank you very much!
> 
> Best wishes.

Can you explain/demonstrate how/why "predict" doesn't work for clogit? I
can get predictions from a clogit model:

## from ?clogit
mod <- clogit(case~spontaneous+induced+strata(stratum),data=infert)
predict(mod)
predict(mod, type = "expected")

all work for that example to give the "fitted" values.

See ?predict.coxph for details of the predict function. I'm not familiar
with either the code or the methods, but 'type = "expected"' seems to
give what you want.

HTH

G
-- 
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
 Dr. Gavin Simpson [t] +44 (0)20 7679 0522
 ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
 Gower Street, London  [w] http://www.ucl.ac.uk/~ucfagls/
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[R] controlling lattice plot ticks with relation="free"

[Richard . Cotton]

How do you persuade lattice to draw tick marks on both the left and right 
side of the y-axis, when relation="free" in the scales component?

#Ticks appear on both sides
histogram(~height|voice.part, data=singer)

##Ticks only on left
histogram(~height|voice.part, data=singer, 
scales=list(y=list(relation="free")))

Providing tck as a vector of length 2 doesn't seem to work, neither does 
setting alternating=3 (since the argument is ignored, and I don't really 
want the labels on both sides - just the ticks).

I'm sure I'm missing something obvious.

Regards,
Richie.

Mathematical Sciences Unit
HSL



ATTENTION:

This message contains privileged and confidential inform...{{dropped:20}}

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[R] Upgrade to R 2.7.2 from Debian Repository

[Tom La Bone]

I am running R 2.7.1 on an eeepc with the standard Xandros Linux. I
install/upgrade R binaries using synaptic from the
http://cran.r-project.org/bin/linux/debian/etch-cran/ repository. R 2.7.2
does not appear to be available from this repository. Will it be available
soon or is there another repository I should be using?

Tom

-- 
View this message in context: 
http://www.nabble.com/Upgrade-to-R-2.7.2-from-Debian-Repository-tp19328325p19328325.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Package for Tidying-up R-code

[Dimitris Rizopoulos]

check Section 3.1 of the `Writing R Extensions' manual.


I hope it helps.

Best,
Dimitris


Gundala Viswanath wrote:

Dear all,

Is there any such package?

I am thinking of something equivalent to Perl::Tidy.
For example with VI editor one can visual highlight a portion
of Perlcode and then issue the command:

!perltidy

then the code will be automatically arranged.

- Gundala Viswanath
Jakarta - Indonesia

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and provide commented, minimal, self-contained, reproducible code.



--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center

Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043399
Fax: +31/(0)10/7044657

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[R] Package for Tidying-up R-code

[Gundala Viswanath]

Dear all,

Is there any such package?

I am thinking of something equivalent to Perl::Tidy.
For example with VI editor one can visual highlight a portion
of Perlcode and then issue the command:

!perltidy

then the code will be automatically arranged.

- Gundala Viswanath
Jakarta - Indonesia

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[R] Add marks to a point pattern

[Vanessa Santos]

My name is Vanessa and I am from Spain.I have in Excel a list of children that 
go and don´t go to the dentist, and I want to add these categorical marks to 
the point pattern in package spatstat.How can I do it?
Thanks for your help.

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Re: [R] restricted bootstrap

[ONKELINX, Thierry]

Grant,

Have you considered a gls model instead of a lm model? In a gls model
one can model the correlation between the measures. So you won't need to
select a subset of your data. You can kind gls in the nlme package.

HTH,

Thierry




ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium 
tel. + 32 54/436 185
[EMAIL PROTECTED] 
www.inbo.be 

To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey

-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Namens Grant Gillis
Verzonden: donderdag 4 september 2008 14:57
Aan: r-help@r-project.org
Onderwerp: Re: [R] restricted bootstrap

Hello Professor Ripely,

Sorry for not being clear.  I posted after a long day of struggling.
Also
my toy distance matrix should have been symmetrical.

Simply put I have spatially autocorrelated data collected from many
points.
I would like to do a linear regression on these data.  To deal with the
autocrrelation I want to resample a subset of my data with replacement
but I
need to restrict subsets such that no two locations where data was
collected
are closer than Xm apart (further apart than the autocrrelation in the
data).

Thanks for having a look at this for me.  I will look up the hard-core
spatial point process.

Grant

2008/9/4 Prof Brian Ripley <[EMAIL PROTECTED]>

> I see nothing here to do with the 'bootstrap', which is sampling with
> replacement.
>
> Do you know what you mean exactly by 'randomly sample'?  In general
the way
> to so this is to sample randomly (uniformly, whatever) and reject
samples
> that do not meet your restriction.   For some restrictions there are
more
> efficient algorithms, but I don't understand yours.  (What are the
'rows'?
>  Do you want to sample rows in space or xy locations?  How come 'dist'
is
> not symmetric?)  For some restrictions, an MCMC sampling scheme is
needed,
> the hard-core spatial point process being a related example.
>
>
> On Wed, 3 Sep 2008, Grant Gillis wrote:
>
>  Hello List,
>>
>> I am not sure that I have the correct terminology here (restricted
>> bootstrap) which may be hampering my archive searches.  I have quite
a
>> large
>> spatially autocorrelated data set.  I have xy coordinates and the
>> corresponding pairwise distance matrix (metres) for each row.  I
would
>> like
>> to randomly sample some number of rows but restricting samples such
that
>> the
>> distance between them is larger than the threshold of
autocorrelation.  I
>> have been been unsuccessfully trying to link the 'sample' function to
>> values
>> in the distance matrix.
>>
>> My end goal is to randomly sample M thousand rows of data N thousand
times
>> calculating linear regression coefficients for each sample but am
stuck on
>> taking the initial sample. I believe I can figure out the rest.
>>
>>
>> Example Question
>>
>> I would like to radomly sample 3 rows further but withe the
restriction
>> that
>> they are greater than 100m apart
>>
>> example data:
>> main data:
>>
>> y<- c(1, 2, 9, 5, 6)
>> x<-c( 1, 3, 5, 7, 9)
>> z<-c(2, 4, 6, 8, 10)
>> a<-c(3, 9, 6, 4 ,4)
>>
>> maindata<-cbind(y, x, z, a)
>>
>>y x x a
>> [1,] 1 1 1 3
>> [2,] 2 3 3 9
>> [3,] 9 5 5 6
>> [4,] 5 7 7 4
>> [5,] 6 9 9 4
>>
>> distance matrix:
>> row1<-c(0, 123, 567, 89)
>> row2<-c(98, 0, 345, 543)
>> row3<-c(765, 90, 0, 987)
>> row4<-c(654, 8, 99, 0)
>>
>> dist<-rbind(row1, row2, row3, row4)
>>
>>[,1] [,2] [,3] [,4]
>> row10  123  567   89
>> row2   980  345  543
>> row3  765   900  987
>> row4  6548   990
>>
>> Thanks for all of the help in the past and now
>>
>> Cheers
>> Grant
>>
>>[[alternative HTML version deleted]]
>>
>> __
>> 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.
>>
>>
> --
> Brian D. Ripley,  [EMAIL PROTECTED]
> Professor of Applied Statistics,
http://www.stats.ox.ac.uk/~ripley/
> University of Oxford, Tel:  +44 1865 272861 (self)
> 1 South Parks Road, +44 1865 272866 (PA)
> Oxford OX1 3TG, UKFax:  +44 1865 272595
>

[[alternative HTML version deleted]]

[R] Odp: Help for par(mfrow)

[Petr PIKAL]

Hi
see from plot.boot help page

Side Effects
All screens are closed and cleared and a number of plots are produced on 
the current graphics device. Screens are closed but not cleared at 
termination of this function. 

You need to change a behaviour of this function what shall not be simple.

Or maybe you can produce set of graphs in one pdf file or in several png 
files and then to add them together by external means

Regards
Petr


[EMAIL PROTECTED] napsal dne 03.09.2008 12:03:03:

> Hi, R users
> I have a question with par(mfrow). I try to histograms and qqplots 
> form boot output for 5 statistics but par(mfrow=c(5,2)) or 
> par(mfrow=c(5,1)) does not work. R still display each figure 
> separately. What did I do wrong? (I check ?par)
> > c1.boot<-boot(c1data,c1.fun,R=999)
> > c1.boot
> 
> ORDINARY NONPARAMETRIC BOOTSTRAP
> Call:
> boot(data = c1data, statistic = c1.fun, R = 999)
> 
> Bootstrap Statistics :
> originalbiasstd. error
> t1*  0.30447696  0.0014228306  0.01026153
> t2*  0.05967183 -0.0014505285  0.01274977
> t3*  1.11852318 -0.0009339349  0.31586852
> t4*  4.12856813 -0.0310221125  0.39602210
> t5* -1.29958196  0.0222876889  0.67625299
> 
> > par(mfrow=c(5,2))
> > for (i in 1:5) {plot(c1.boot, index=i, col=i)}
> Many Thanks
> Chunhao
> 
> __
> 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.

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] reading and parsing gzipped files

[Prof Brian Ripley]

On Thu, 4 Sep 2008, Dmitriy Skvortsov wrote:


Hi all,I have  large  compressed text tab delimited  files,
I am trying to write efficient function to read them,
I am using   gzfile()  and readLines()

zz <- gzfile("exampl.txt.gz", "r")  # compressed file
system.time(temp1<-readLines(zz ))
close(zz)

which work fast, and create vector of strings.
The problem is to parse the result, if  I use strsplit  it takes longer then
decompress file manually , read it with scan and erase it.

Can anybody recommend  an efficient way of parsing large vector  ~200,000
entries


'parse'?  What is wrong with using read.delim (reading 'tab delimited 
files' is its job)?  It (and scan) work with gzfile connections, so there 
is no need to decompress manually.


See the 'R Data Import/Export Manual' for how to use read.delim efficiently.


Dmitriy

[[alternative HTML version deleted]]

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


PLEASE do.


--
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] how to write a escape sequences in to a file

[Prof Brian Ripley]

A newline is "\n".  So "hai \nhru".

On Fri, 5 Sep 2008, Kurapati, Ravichandra (Ravichandra) wrote:




Hi



  Str1<-"hai "

  Str2<-"hru"



 I want to write these 2 strings in a file  separated by
newline.





  How can I get this.

Thanks

K.Ravichandra






[[alternative HTML version deleted]]

__
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.



--
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

__
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.

Re: [R] Trouble with R CMD check: I can't seem to get dependencies right (maybe I'm using R_LIBS incorrectly?)

[Prof Brian Ripley]

'Writing R Extensions' does say

@quotation Note
@code{R CMD check} and @code{R CMD build} run @R{} with
@option{--vanilla}, so none of the user's startup files are read.  If
you need @env{R_LIBS} set (to find packages in a non-standard library)
you will need to set it in the environment.
@end quotation

~/.Renviron is one of those startup files (and so is ~/.Rprofile).

On Thu, 4 Sep 2008, Dylan Arena wrote:


Hi there,


I'm in the following directory:

~/Documents/Rstuff/diceFiles/dice_1.1

The directory "dice" is in this directory, with all the usual build
files (DESCRIPTION, NAMESPACE, etc).  I'm trying to run the following
command:

R CMD check dice

(where "dice" is the name of the package I'm checking), and I get the following:

* checking package dependencies ... ERROR
Packages required but not available:
 gtools

My DESCRIPTION file does (correctly) list gtools as a dependency, so
I'm happy this check is ensuring its availability.  What I can't
figure out is how to get the check command to see that gtools is in
fact installed.  I've created a file called .Renviron in my home
directory that has the following line in it:

R_LIBS=~/Documents/Rstuff/library

And in the directory specified by R_LIBS above is the gtools package.
(I've also got a file called .Rprofile in my home directory, with the
line .libPaths("~/Documents/Rstuff/library")
in it.)  My reading of the R-intro manual is that setting R_LIBS in
this way should allow the R CMD check command to search in the
directory specified by R_LIBS when looking for gtools.  I'm pretty
sure at this point, though, that my understanding is incorrect.

At this point I'm in over my depth.  I'm writing because I'm hopeful
that to someone on this list my problem is trivially solvable.  My
basic question is: what must I do to get R CMD check to find the
gtools package?


Please let me know any ideas you might have for me,
Dylan

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.



--
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] algorithm to create unique identifiers

[Prof Brian Ripley]

For a much simpler solution that does always work for numbers, see 
unique's methods for matrices and data frames.


On Thu, 4 Sep 2008, Henrik Bengtsson wrote:


On Thu, Sep 4, 2008 at 8:44 PM, Ralph S. <[EMAIL PROTECTED]> wrote:


Hi all,

I am trying to create a unique identifier for each row, combining 
numbers from three columns.


Do you know if there is a general formula to do this (or some manual 
where I can read about this)?


I figure I can use the numeric entries of the columns as "coordinates" 
and multiply them with different coefficients (different magnitudes) to 
get the unique ID - but it would be nice to read about such algorithms 
in general.


What are you numbers?  Are they in a fixed range?  Integers or reals?
If fixed range integers, it is easy.  Think regular numerical
representation, e.g. binary, octadecimal, decimal and hexadecimal.

For a more generic solution that works with any data types, see e.g.
MD5 [http://en.wikipedia.org/wiki/MD5].  It is not guaranteed to
generated unique codes, but it is extremely rare that two different
inputs gives the same MD5 code.  MD5 (and others) are implemented in
the 'digest' packages, e.g.


library(digest)
digest(list(a=1, b=list(1:10, c=letters)))

[1] "73e0ae066a97bfff7f79d41c65b55fde"

My $.02

/Henrik




Any links/input would be great -

Ralph


--
Brian D. Ripley,  [EMAIL PROTECTED]
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:  +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax:  +44 1865 272595

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.