获得每个点之间的距离，并找到曲线自身的位置

Question

I am programming a random generated spline curve by first generating control points and then interpolate with spicy.splev. Here is an example.

Points are given like this:

np.array =[[  1.00000000e+01  -4.65000000e+02]
           [  1.78319153e+01  -4.60252579e+02]
          ...]

I now want to get the distance of every point with every other of the spline to see if at one point the spline comes too close to itself which includes self-collision.

Before and after every point there should be an interval where points are ignored as these are always the closest points to each point:

def collision(splinePoints, interval):
    length = len(splinePoints)    
    mylist = []
    i = -1
    for item in splinePoints:
        i += 1
        first = item
        lowerLimit = i - interval
        uperLimit = i + interval
        if lowerLimit >= 0:
            for item in splinePoints[:lowerLimit]:
                mylist.append(first)
                mylist.append(item)
        if uperLimit <= length:
            for item in splinePoints[uperLimit:]:
                mylist.append(first)
                mylist.append(item)
    return np.amin(lengthOfLines(np.array(mylist)))

Lengths of lines is checked with this:

def lengthOfLines(points):
    return np.sqrt(np.sum(np.diff(points.T)**2, axis=0))

It somehow works, but not always. I am also struggling with debugging as the generated data is big and hard to read check or compare. Any idea how to do it better?

Answer 1

All pairwise distances are obtained with pdist method of scipy.spatial package. It returns a flat array of distances, with redundancies eliminated. The utility function squareform unpacks them to a symmetric square matrix, which is often more convenient.

You also want to find the nearest point that is not directly before or after the given point on the curve. In the example below, I penalize the distances between neighbors (within 20 index values) by setting those distances to infinity. Then argmin find the nearest point for everyone, and I visualize it by drawing a red line to that nearest point.

import numpy as np
from scipy.spatial.distance import pdist, squareform

t = np.linspace(0, 10, 50)
points = np.stack(((t+5)*np.cos(t), (t+5)*np.sin(t)), axis=-1)  # for example

distances = squareform(pdist(points))    # distance matrix
i, j = np.meshgrid(np.arange(t.size), np.arange(t.size))
distances[np.abs(i-j) <= 20] = np.inf    # don't count neighbors
nearest = np.argmin(distances, axis=0)   # nearest to each

plt.plot(points[:, 0], points[:, 1])
for k in range(len(t)):
  npoint = points[nearest[k]]
  plt.plot([points[k, 0], npoint[0]], [points[k, 1], npoint[1]], 'r')
plt.axes().set_aspect('equal', 'datalim')
plt.show()