Question

我想计算（x，y，z）坐标的欧几里得距离矩阵。我找到了计算（x，y）坐标的欧几里得距离矩阵的方法。

def euclidian_dist(x,y):
    diff = np.expand_dims(x, axis=1) - np.expand_dims(y, axis=0)
    distance_matrix = np.sqrt(np.sum(diff ** 2, axis=-1))
    return distance_matrix

if __name__=="__main__":
    #[[x1,y1],[x2,y2],...,[xn,yn]]
    p = np.array([[0.0, 0.0],[1., 0],[1, 1],[0, 1]])
    q = np.array([[0., 0],[1., 0],[0.3, 0.4],[0.7, 0.6]])
    euclidian_dist(p,q)

ans: array([[0.        , 1.        , 0.5       , 0.92195445],
            [1.        , 0.        , 0.80622577, 0.67082039],
            [1.41421356, 1.        , 0.92195445, 0.5       ],
            [1.        , 1.41421356, 0.67082039, 0.80622577]])

在下一步中，我想使用x，y，z坐标（[[x1，y1，z1]，[x2，y2，z2]，...，[xn，yn，zn]]）< / p>

并想要获得欧氏距离矩阵（形状：n✕n）。

您能否告诉我您是否知道解决方案？

Answer 1

对3D点使用相同的代码。效果很好：

def euclidian_dist(x,y):
    diff = np.expand_dims(x, axis=1) - np.expand_dims(y, axis=0)
    distance_matrix = np.sqrt(np.sum(diff ** 2, axis=-1))
    return distance_matrix

if __name__=="__main__":
    #[[x1,y1],[x2,y2],...,[xn,yn]]
    p = np.array([[0.0, 0.0, 0.0],[1.,1., 0],[1, 1, 1],[0, 0, 1]])
    q = np.array([[0., 0, 0],[1.,1., 0],[0.3,0.3, 0.4],[0.7, 0.6, 0.6]])
    print(euclidian_dist(p,q))

输出：

[[0.         1.41421356 0.58309519 1.1       ]
 [1.41421356 0.         1.06770783 0.78102497]
 [1.73205081 1.         1.15758369 0.64031242]
 [1.         1.73205081 0.73484692 1.00498756]]

numpy广播的计算方法距离矩阵

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