python - How do I iterate over an ndarray without using for/while loops?

Question

For two given 1-d arrays or lists I can calculate the squared Euclidean distance via the function

import numpy as np

def npdistance(x1, x2):

    return sum((np.array(x1)-np.array(x2))**2)

Now for a given vector v and 2d-array X I would like to find the shortest squared Euclidean distance of any vector contained in X to the vector u without iterating over the elements of X with for/while loops. My attempt is

def npnearest(u, X):
    L=npdistance(u,X)
    return min(L)

which does not give me what I want. For example

 npnearest(np.array([1,1,1]), np.array([[1,1,1],[2,3,4]]))

would give me 16 instead of 0. How can I do it?

标签： pythonnumpynumpy-ndarray

Answer 1

在 numpy 的情况下，更喜欢np.sumand np.min，而不是 Python buildinssum和min.

我们可以适应npdistance2D numpy 向量：

def npdistance(x1, x2):
    return np.sum((np.array(x1)-np.array(x2))**2, axis=1)

考虑矩阵x2：

x2 = np.array([[1,1,1],[2,3,4]])

矩阵x2有两个轴：

• 第零个是向量号：x2[0]isnp.array([1, 1, 1])和x2[1]is np.array([2, 3, 4]),
• 第一个轴用于向量维度：x2[1][1]是3（第一个向量的第二个元素）。

我们执行求和以获得每个向量axis=1的距离。

• 没有np.sum axis=1它会返回标量，
• 使用 buildinsum给出所有向量的总和 (ala axis=0)。

npnearest在这种情况下工作正常。

def npnearest(u, X):
    L=npdistance(u,X)
    return min(L)

npnearest(np.array([1,1,1]), np.array([[1,1,1],[2,3,4]]))

给出 0。