确定两点是否彼此最接近的最快方法

Question

我的问题包括以下内容：我有两对角（在球坐标系中），它由两部分组成——一个方位角和一个余纬角。如果我们无限延长两个角（从而增加它们各自的半径），使一条长线指向这对角给定的方向，那么我的目标是确定

如果它们彼此相交或非常接近并且
它们相交的地方。

目前我试过几种方法：

最明显的方法是迭代比较每个半径，直到匹配或两者之间的距离足够小。（当我说比较每个半径时，我指的是将每个球面坐标转换为笛卡尔坐标，然后求出两者之间的欧氏距离）。然而，这个运行时间是 $O(n^{2})$，如果我试图扩展这个程序，它会非常慢
第二个最明显的方法是使用优化包来找到这个距离。不幸的是，我不能迭代地优化包，并且在一个实例之后优化算法重复相同的答案，这是没有用的。
最不明显的方法是直接计算（使用微积分）角度的精确半径。虽然这是一种快速的方法，但它并不是非常准确。

注意： 虽然交点总是在零原点 (0,0,0) 看起来很简单，但情况并非总是如此。有些点永远不会相交。

方法 (1) 的代码

def match1(azimuth_recon_1,colatitude_recon_1,azimuth_recon_2, colatitude_recon_2,centroid_1,centroid_2 ):


   # Constants: tolerance factor and extremely large distance
    tol = 3e-2
    prevDist = 99999999

    # Initialize a list of radii to loop through
    # Checking iteravely for a solution
    for r1 in list(np.arange(0,5,tol)):
        for r2 in list(np.arange(0,5,tol)):

            # Get the estimates
            estimate_1 = np.array(spher2cart(r1,azimuth_recon_1,colatitude_recon_1)) + np.array(centroid_1)
            estimate_2 = np.array(spher2cart(r2,azimuth_recon_2,colatitude_recon_2))+ np.array(centroid_2)

            # Calculate the euclidean distance between them
            dist = np.array(np.sqrt(np.einsum('i...,i...', (estimate_1 - estimate_2), (estimate_1 - estimate_2)))[:,np.newaxis])

        # Compare the distance to this tolerance
            if dist < tol: 
                if dist == 0:
                    return estimate_1, [], True
                else:
                    return estimate_1, estimate_2, False

            ## If the distance is too big break out of the loop
            if dist > prevDist:
                prevDist = 9999999
                break
            prevDist = dist

    return [], [], False

方法(3)的代码

def match2(azimuth_recon_1,colatitude_recon_1,azimuth_recon_2, colatitude_recon_2,centriod_1,centroid_2):

   # Set a Tolerance factor
    tol = 3e-2

    def calculate_radius_2(azimuth_1,colatitude_1,azimuth_2,colatitude_2):

        """Return radius 2 using both pairs of angles (azimuth and colatitude). Equation is provided in the document"""

        return 1/((1-(math.sin(azimuth_1)*math.sin(azimuth_2)*math.cos(colatitude_1-colatitude_2))
        +math.cos(azimuth_1)*math.cos(azimuth_2))**2)

    def calculate_radius_1(radius_2,azimuth_1,colatitude_1,azimuth_2,colatitude_2):

        """Returns radius 1 using both pairs of angles (azimuth and colatitude) and radius 2. 
    Equation provided in document"""

        return (radius_2)*((math.sin(azimuth_1)*math.sin(azimuth_2)*math.cos(colatitude_1-colatitude_2))
        +math.cos(azimuth_1)*math.cos(azimuth_2))

    # Compute radius 2
    radius_2 = calculate_radius_2(azimuth_recon_1,colatitude_recon_1,azimuth_recon_2,colatitude_recon_2)

    #Compute radius 1
    radius_1 = calculate_radius_1(radius_2,azimuth_recon_1,colatitude_recon_1,azimuth_recon_2,colatitude_recon_2)

    # Get the estimates
    estimate_1 = np.array(spher2cart(radius_1,azimuth_recon_1,colatitude_recon_1))+ np.array(centroid_1)
    estimate_2 = np.array(spher2cart(radius_2,azimuth_recon_2,colatitude_recon_2))+ np.array(centroid_2) 

    # Calculate the euclidean distance between them
    dist = np.array(np.sqrt(np.einsum('i...,i...', (estimate_1 - estimate_2), (estimate_1 - estimate_2)))[:,np.newaxis])

    # Compare the distance to this tolerance
    if dist < tol: 
        if dist == 0:
            return estimate_1, [], True
        else:
            return estimate_1, estimate_2, False
    else:
        return [], [], False

My question is two-fold:

Is there a faster and more accurate way to find the radii for both points?

If so, how do I do it?

编辑：我正在考虑只创建两个半径的两个 numpy 数组，然后通过 numpy 布尔逻辑比较它们。但是，我仍然会反复比较它们。有没有更快的方法来执行此比较？

Answer 1

在这种情况下使用 kd-tree。它会很容易地查找最小距离：

def match(azimuth_recon_1,colatitude_recon_1,azimuth_recon_2, colatitude_recon_2,centriod_1,centroid_2):

    cartesian_1 = np.array([np.cos(azimuth_recon_1)*np.sin(colatitude_recon_1),np.sin(azimuth_recon_1)*np.sin(colatitude_recon_1),np.cos(colatitude_recon_1)]) #[np.newaxis,:]
    cartesian_2 = np.array([np.cos(azimuth_recon_2)*np.sin(colatitude_recon_2),np.sin(azimuth_recon_2)*np.sin(colatitude_recon_2),np.cos(colatitude_recon_2)]) #[np.newaxis,:]

    # Re-center them via adding the centroid
    estimate_1 = r1*cartesian_1.T + np.array(centroid_1)[np.newaxis,:] 
    estimate_2 = r2*cartesian_2.T + np.array(centroid_2)[np.newaxis,:]

    # Add them to the output list
    n = estimate_1.shape[0]
    outputs_list_1.append(estimate_1)
    outputs_list_2.append(estimate_2)

    # Reshape them so that they are in proper format
    a = np.array(outputs_list_1).reshape(len(two_pair_mic_list)*n,3)
    b = np.array(outputs_list_2).reshape(len(two_pair_mic_list)*n,3)

    # Get the difference
    c = a - b

    # Put into a KDtree
    tree = spatial.KDTree(c)

    # Find the indices where the radius (distance between the points) is 3e-3 or less 
    indices = tree.query_ball_tree(3e-3)

这将输出距离为 3e-3 或更小的索引列表。现在您所要做的就是使用索引列表和估计列表来找到确切的点。好了，这将为您节省很多时间 space!

确定两点是否彼此最接近的最快方法

Fastest way to determine if two points are closest to one another

distance

python-3.x

numpy-einsum

numpy-ndarray