Generating multiple random (x, y) coordinates, exc

问题:

I want to generate a bunch (x, y) coordinates from 0 to 2500 that excludes points that are within 200 of each other without recursion.

Right now I have it check through a list of all previous values to see if any are far enough from all the others. This is really inefficient and if I need to generate a large number of points it takes forever.

So how would I go about doing this?

回答1:

This is a variant on Hank Ditton's suggestion that should be more efficient time- and memory-wise, especially if you're selecting relatively few points out of all possible points. The idea is that, whenever a new point is generated, everything within 200 units of it is added to a set of points to exclude, against which all freshly-generated points are checked.

import random

radius = 200
rangeX = (0, 2500)
rangeY = (0, 2500)
qty = 100  # or however many points you want

# Generate a set of all points within 200 of the origin, to be used as offsets later
# There's probably a more efficient way to do this.
deltas = set()
for x in range(-radius, radius+1):
    for y in range(-radius, radius+1):
        if x*x + y*y <= radius*radius:
            deltas.add((x,y))

randPoints = []
excluded = set()
i = 0
while i<qty:
    x = random.randrange(*rangeX)
    y = random.randrange(*rangeY)
    if (x,y) in excluded: continue
    randPoints.append((x,y))
    i += 1
    excluded.update((x+dx, y+dy) for (dx,dy) in deltas)
print randPoints

回答2:

I would overgenerate the points, target_N < input_N, and filter them using a KDTree. For example:

import numpy as np
from scipy.spatial import KDTree
N   = 20
pts = 2500*np.random.random((N,2))

tree = KDTree(pts)
print tree.sparse_distance_matrix(tree, 200)

Would give me points that are "close" to each other. From here it should be simple to apply any filter:

  (11, 0)   60.843426339
  (0, 11)   60.843426339
  (1, 3)    177.853472309
  (3, 1)    177.853472309

回答3:

Some options:

• Use your algorithm but implement it with a kd-tree that would speed up nearest neighbours look-up
• Build a regular grid over the [0, 2500]^2 square and 'shake' all points randomly with a bi-dimensional normal distribution centered on each intersection in the grid
• Draw a larger number of random points then apply a k-means algorithm and only keep the centroids. They will be far away from one another and the algorithm, though iterative, could converge more quickly than your algorithm.

回答4:

This has been answered, but it's very tangentially related to my work so I took a stab at it. I implemented the algorithm described in this note which I found linked from this blog post. Unfortunately it's not faster than the other proposed methods, but I'm sure there are optimizations to be made.

import numpy as np
import matplotlib.pyplot as plt

def lonely(p,X,r):
    m = X.shape[1]
    x0,y0 = p
    x = y = np.arange(-r,r)
    x = x + x0
    y = y + y0

    u,v = np.meshgrid(x,y)

    u[u < 0] = 0
    u[u >= m] = m-1
    v[v < 0] = 0
    v[v >= m] = m-1

    return not np.any(X[u[:],v[:]] > 0)

def generate_samples(m=2500,r=200,k=30):
    # m = extent of sample domain
    # r = minimum distance between points
    # k = samples before rejection
    active_list = []

    # step 0 - initialize n-d background grid
    X = np.ones((m,m))*-1

    # step 1 - select initial sample
    x0,y0 = np.random.randint(0,m), np.random.randint(0,m)
    active_list.append((x0,y0))
    X[active_list[0]] = 1

    # step 2 - iterate over active list
    while active_list:
        i = np.random.randint(0,len(active_list))
        rad = np.random.rand(k)*r+r
        theta = np.random.rand(k)*2*np.pi

        # get a list of random candidates within [r,2r] from the active point
        candidates = np.round((rad*np.cos(theta)+active_list[i][0], rad*np.sin(theta)+active_list[i][1])).astype(np.int32).T

        # trim the list based on boundaries of the array
        candidates = [(x,y) for x,y in candidates if x >= 0 and y >= 0 and x < m and y < m]

        for p in candidates:
            if X[p] < 0 and lonely(p,X,r):
                X[p] = 1
                active_list.append(p)
                break
        else:
            del active_list[i]

    return X

X = generate_samples(2500, 200, 10)
s = np.where(X>0)
plt.plot(s[0],s[1],'.')

And the results: