Thanks a lot!
Precisely what I had in mind.
One last question (an extension of the previous one): can this be
extended to points in 3D? Once again, given the distance matrix, can I
reconstruct a set of coordinates (among many possible) for the points in
three-dimensional space?
Cheers
Lorenzo
On 12/15/2011 07:22 PM, Peter Langfelder wrote:
On Thu, Dec 15, 2011 at 10:08 AM, Lorenzo Isella
<lorenzo.ise...@gmail.com> wrote:
Dear All,
I am struggling with the following problem: I am given a NxN symmetric
matrix P ( P[i,i]=0, i=1...N and P[i,j]>0 for i!=j) which stands for the
relative distances of N points.
I would like use it to get the coordinates of the N points in a 2D plane. Of
course, the solution is not unique (given one solution, I can translate or
rotate all the points by the same amount and generate another solution), but
any correct solution will do for me.
Any idea about how I can achieve that? Is there any clustering package that
can help me?
Many thanks.
If your matrix really corresponds to distances of points (in 2
dimensions), you can try multidimensional scaling, function
cmdscale().
This little code illustrates that cmdscale recovers the 2-dimensional
points used to generate a distance matrix, up to a shift and rotation:
# Generate 10 random points in 2 dimensions
nPoints = 10;
nDim = 2;
set.seed(10);
points = matrix(runif(nPoints * nDim), nPoints, nDim);
# Their distance:
dst = dist(points)
# Classical multidimensional scaling
mds = cmdscale(dst);
# Distance of the points calculated by mds
dst2 = dist(mds);
# The two distances are equal
all.equal(as.vector(dst), as.vector(dst2))
HTH,
Peter
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