Question

我在一个2200x1700的空间中有（x，y）坐标的73,000个点，我想将所有（x，y）点归为一组，这些点落入半径为35个单位的圆中。为此，我必须为时间复杂度为O（73000x73000）的循环运行嵌套，因为我正在计算每个点之间的距离，并对计算出的距离小于35的点进行分组。

我需要为此优化的解决方案！

temp_cluster=cluster
for key,value in cluster.items():
    pointa=np.array((key[0],key[1]))
    for key2,value2 in temp_cluster.items():
        pointb=np.array((key2[0],key2[1]))
        distance=np.linalg.norm(pointa-pointb)
        if(distance<=35):
            cluster[(key[0],key[1])]=cluster[(key[0],key[1])]+1## here we are counting how many points lies within a radius of 35 for that particular point

Answer 1

您可以做的一种优化是根据点的x坐标对点进行排序。然后，对于每个点，您只需要考虑在该点的x差之内具有35坐标的点。通过二进制搜索，您可以找到该条带内的点（即，找到点(x1, y1)这样的点abs(x-x1) <= 35)。大致步骤是

sort(points) according to their x coordinates
for each point (x, y):
    count = 0
    start_index = starting index of points with x-x1 >= 35 (find using binary search)
    end_index = end index of point with x1-x<=35 (find using binary search)
    for each point(x1, y1) in (start_index, end_index):
        distance = calculate_distance(x, y, x1, y1)
        if distance <= 35:
            count = count + 1
    cluster[(x,y)] = count

如何在较大空间内将点分组

1 个答案: