vtkSplineRepresentation 的 Actor ()

Question

我有几个 numpy 数组，我想比较它们并为给定数组找到最接近的数组。我可以使用https://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.distance.cdist.html计算这些数组之间的距离。但是，有没有办法从几个 numpy 数组中找到两个最接近的数组？

因为我得到了阵列print(arr.shape)给(300,)

Answer 1

要找到最接近的两个，您需要计算距离矩阵，然后找到该矩阵中的最小值以获得彼此最接近的坐标（使用矩阵您将获得坐标的索引）。from scipy.spatial import distanceimport numpy as np coords = np.array([  (35.0456, -85.2672),  (35.1174, -89.9711),  (35.9728, -83.9422),  (36.1667, -86.7833)])distances = distance.cdist(coords, coords, 'euclidean')# If you don't do that then the distance to self will always # be the min. distance in the matrix (always 0): np.fill_diagonal(distances, np.inf)min_index = (np.argmin(distances))closest = np.unravel_index(min_index, distances.shape)一旦closest定义了索引，您就可以获得关于最近坐标对的所有信息：print(f"The two closest are {closest}")print(f"They are at distance {distances[closest]}")print(f"Resp. coordinates {coords[closest[0]]} and {coords[closest[1]]}")输出：The two closest are (0, 2)They are at distance 1.6171965990565296Resp. coordinates [ 35.0456 -85.2672] and [ 35.9728-83.9422]最后，请注意所有这些输入也将起作用：coords = np.array([ [35.0456, -85.2672], [35.1174, -89.9711] ])arr1 = [35.0456, -85.2672]arr2 = [35.1174, -89.9711]coords = np.array([arr1, arr2])

Answer 2

编写一个距离函数，然后使用 itertools 计算列表对之间的距离如何？例如：a_1 = [0,3,4,5]a_2 = [4,7,8,9]a_3 = [12, 34, 44]from itertools import combinationsdef distance(list1, list2):    """Distance between two vectors."""    squares = [(p-q) ** 2 for p, q in zip(list1, list2)]    return sum(squares) ** .5distances = []for pair in combinations([a_1, a_2, a_3], 2):    distances.append(pair)    distances.append(distance(list(pair[0]), list(pair[1])))结果： [([0, 3, 4, 5], [4, 7, 8, 9]), 8.0, ([0, 3, 4, 5], [12, 34, 44]), 52.009614495783374, ([4, 7, 8, 9], [12, 34, 44]), 45.70557952810576]