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Re: fit plane to xz axis of data

  • To: mathgroup at smc.vnet.net
  • Subject: [mg89626] Re: fit plane to xz axis of data
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Sun, 15 Jun 2008 06:14:26 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <g30399$mes$1@smc.vnet.net>

will parr wrote:

[snip]

>> The following will find the best-fitting (i.e.,
>> least-squares) plane:
>>
>> m = Mean[data]; {u,w,v] =
>> = SingularValueDecomposition[#-m&/@data];
>>
>> v is a rotation matrix. Tr[w,List] (i.e., the
>> diagonal elements of w)
>> gives the norms of the projections of the points onto
>> the new axes.
>> They are in decreasing order, so
>>
>>   ({x,y,z}-m).v[[All,3]] == 0
>>
>> gives the equation which the best-fitting plane
>> satisfies.
>>
> 
> I'm really interested in this solution, but when I to use this method, I get an answer that I don't really understand, and cannot display (in the form of a plane).
> 
> using the original data in this message:
> 
> In[122]:= m = Mean[data]; {u, w, v} = 
>  SingularValueDecomposition[# - m & /@ data];
> 
> In[123]:= Tr[w, List]
> 
> Out[123]= {16.3822, 9.63356, 5.71769}
> 
> In[124]:= ({x, y, z} - m).v[[All, 3]] == 0
> 
> Out[124]= 
> 0.984797 (-9.86988 + x) + 0.134151 (0.269052+ y) + 
>   0.110358 (-1.44535 + z) == 0
> 
> I don't understand this output, it is not of the form:
> 
> ax + by + cz

[snip]

Solve the equation returned by ({x, y, z} - m).v[[All, 3]] == 0 for z. 
For instance,

     plane = z /. Solve[%, z][[1]] // Expand

returns

     89.1934- 8.92363 x - 1.2156 y

Regards,
- Jean-Marc


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