给定 R^p 中的 2 个多维点，如何在通过这两个点的线上投影 N 个个体？

Question

I have 2 points in R^p, and i want to project N individuals on the line passing through those two points. To be exact, i want the coordinates of every N individuals on the 1D space created by this line.

The operations will be repeated multiple times on different points, so an automatic method is needed. I was wondering if it exists a package to find the equation of the line. The projection should be easy to implement.

Edit (Vishwas comment) :

Given two points y1 = (1,4) & y2 = (5,6) do a projection ( https://en.wikipedia.org/wiki/Projection_(mathematics) ) of {x1,...,xN} in R2 onto the line passing through y1 and y2.

Expected output: the coordinates of each projected points {x1,...,xN} in the original space.

Answer 1

Your projection of point xi is zi = y1 + ti × (y2 - y1) for some suitable ti. The remaining difference xi - zi is orthogonal to the line, so the dot product (xi - zi) • (y2 - y1) = 0. Put in the definition of zi and you get an equation that is linear in ti. You can solve it for ti, then use that to compute zi.

If you want to do this operation for many points, you could come up with a precomputation that expresses this whole operation eg as a matrix multiplication. As the line you project to dues not necessarily pass through the origin, you need an assume transformation, which can be expressed as an (p+1) × (p+1) matrix acting on homogeneous coordinates.

Let's switch nomenclature to single letter symbols. Let your line be defined by points a and b. We project point x to z = a + t (b - a). So you get 0 = (x - a - t (b - a)) • (b - a) = (x - a) • (b - a) - t (b - a) • (b - a) which you can solve for t = ((x - a) • (b - a)) / ((b - a) • (b - a)).

Here is a bit of Python code which you can use to compare against your implementation:

def project(p, a, b, xs):
  ab = [b[i] - a[i] for i in range(p)]
  abab = sum(i*i for i in ab)
  zs = []
  for x in xs:
    t = sum((x[i] - a[i])*ab[i] for i in range(p)) / abab
    z = [a[i] + t*ab[i] for i in range(p)]
    zs.append(z)
  return zs