[R] regression methods for circular(?) data.

[(Ted Harding)]

On 26-Sep-05 Witold Eryk Wolski wrote:
> Hi,
> 
> I do not know the intercept and slope.
> And you have to know them in order to do something like:
> ix<-(y < 0.9*(x-50)/200
> 
> I am right?
> 
> cheers

Although I really knew them from the way you generated the data,
I "pretended" I did not know them.

Read below: "If you know the modulus (in your case 1.0)" -- I did
assume that this was known, i.e. that the data "wrap round" to 0
above 1.0. Also: "the constants 0.9/200, -50 being chosen to give
a good separation on the graph" -- I plotted the data, and saw that
the "wrapped" data were well separated, and that 0.9*(x-50)/200
was an adequate discriminant function. This was estimated purely by
eye, by looking at the graph, to find some line that went between
the two groups of data; no attempt was made to calculate anything
precisely. Apart from assuming that the modulus was 1.0, and that
the well-separated data at the bottom right of the graph were
"wrapped round" data, no other information was used by me!

So the question remains: If you can assume that the modulus is 1.0,
and that the wrapped-round data will be well separated, then all
is simple. All you need to do is to "unwrap" the "wrapped" data
by adding 1.0, having first identified them by virtue of their
obvious separation. Then you can estimate the slope by using 'lm'.

But:-- if you, Witold, can not assume these two things for your
real data, what can we assume in considering your question?
Is the modulus unknown, for instance? Is the scatter so large that
the groups are not well separated? Might we have "twice-wrapped"
data (i.e. original y > 2)? 

In short, do your real data look like the data you sent us, and
are they wrapped at 1.0? or what?

With thanks, and best wishes,
Ted.

> (Ted Harding) wrote:
>> On 26-Sep-05 nwew wrote:
>> 
>>>Dear R-users,
>>>
>>>I have the following data
>>>
>>>x <- runif(300,min=1,max=230)
>>>
>>>y <- x*0.005 + 0.2
>>>y <- y+rnorm(100,mean=0,sd=0.1)
>>>y <- y%%1 #  <------- modulo operation
>>>plot(x,y)
>>>
>>>and would like to recapture the slope (0.005) and intercept(0.2).
>>>I wonder if there are any clever algorithms to do this. I was
>>>looking at the function lm.cirucalar. Is this the method to use?
>>>If, which of the references is best too look at?
>>>
>>>Eryk
>> 
>> 
>> Hi Eryk,
>> 
>> If you know the modulus (in your case 1.0) and you get data that
>> look like the result of your "plot(x,y)", then I wouldn't mess
>> about.
>> 
>> I would simply do something like
>> 
>> y1<-y
>> ix <- ix<-(y < 0.9*(x-50)/200)
>> y1[ix] <- y1[ix]+1.0
>> lm(y1~x)
>> 
>> (the constants 0.9/200, -50 being chosen to give a good separation
>> on the graph).
>> 
>> On the other hand, if there are good reasons why this very simple
>> approach is not suitable, then if we knew what they were a more
>> helpful reply would be easier to formulate!
>> 
>> Best wishes,
>> Ted.
>> 
>> 
>> --------------------------------------------------------------------
>> E-Mail: (Ted Harding) <Ted.Harding at nessie.mcc.ac.uk>
>> Fax-to-email: +44 (0)870 094 0861
>> Date: 26-Sep-05                                       Time: 15:56:48
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E-Mail: (Ted Harding) <Ted.Harding at nessie.mcc.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 26-Sep-05                                       Time: 18:08:28
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