Question

这适用于K-Means算法。这是作业，所以我不想用内置Kmeans功能。我有2个numpy数组。一个是质心。另一个是数据点。我试图找到每个质心到每个数据点的距离。我不知道如何将数组传递给我的函数以便打印。我想结束与质心一样多的距离数组。然后我可以比较阵列中的每个距离，选择最小的距离距离并将该点指定给其中一个群集。然后找出每个聚类的平均值数字成为我的新质心。

import numpy as np

centroids = np.array([[3,44],[5,15]])
dataPoints = np.array([[2,4],[17,4],[45,2],[45,7],[16,32],[32,14],[20,56],[68,33]])
def distance(a,b):
    for x in a: #for each point in centroids array
        for y in b:#for each point in the dataPoints array
            print np.sqrt((a[0] - b[0])**2 + (a[1] - b[1])**2)#print the distance

distance (randPoints, dataPoints)#call the function with the data

我得到的输出：

[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]
[ 12.04159458  41.48493703]

我在做什么，这显然是错的？我应该最终得到2个不同的阵列，每个阵列有8个距离。

Answer 1

import numpy as np

centroids = np.array([[3,44],[5,15]])
dataPoints = np.array([[2,4],[17,4],[45,2],[45,7],[16,32],[32,14],[20,56],[68,33]])

def size(vector):
    return np.sqrt(sum(x**2 for x in vector))

def distance(vector1, vector2):
    return size(vector1 - vector2)

def distances(array1, array2):
    return [[distance(vector1, vector2) for vector2 in array2] for vector1 in array1]

print(distances(centroids, dataPoints))

Answer 2

我厌倦了为1,2和3d数组进行距离计算的化身，所以我拼凑了一个模仿来自scipy的pdist和cdist的函数，但使用了很多人在这个网站上使用的einsum。至少在我的脑海里很容易理解，而且einsum用于其他目的。请考虑以下内容。如果你需要提取最接近的x值等，你可以使用然后使用排序（sort，argsort等）。希望你发现它有用

a = np.array([[1, 2], [3, 4], [5, 6]])
b = np.array([[6, 5], [4, 3], [2, 1]])

def e_dist(a, b, metric='euclidean'):
    """Distance calculation for 1D, 2D and 3D points using einsum
    : a, b   - list, tuple, array in 1,2 or 3D form
    : metric - euclidean ('e','eu'...), sqeuclidean ('s','sq'...), 
    :-----------------------------------------------------------------------
    """
    a = np.asarray(a)
    b = np.atleast_2d(b)
    a_dim = a.ndim
    b_dim = b.ndim
    if a_dim == 1:
        a = a.reshape(1, 1, a.shape[0])
    if a_dim >= 2:
        a = a.reshape(np.prod(a.shape[:-1]), 1, a.shape[-1])
    if b_dim > 2:
        b = b.reshape(np.prod(b.shape[:-1]), b.shape[-1])
    diff = a - b
    dist_arr = np.einsum('ijk,ijk->ij', diff, diff)
    if metric[:1] == 'e':
        dist_arr = np.sqrt(dist_arr)
    dist_arr = np.squeeze(dist_arr)
    return dist_arr

e_dist(a, b)
array([[ 5.8,  3.2,  1.4],
       [ 3.2,  1.4,  3.2],
       [ 1.4,  3.2,  5.8]])

e_dist(a[0], b)
array([ 5.8,  3.2,  1.4])

e_dist(a[:2], b)
array([[ 5.8,  3.2,  1.4],
       [ 3.2,  1.4,  3.2]])

如何找到不同numpy阵列中两点之间的距离？

2 个答案: