ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all(): Silhouette performance algorithm

Question

我在 python 中实现了 kmeans 算法，并试图针对不同的 k 值计算集群的 silhouette 性能。以下是一小部分数据集的几个变量。

def avgdist(pt, clust):
    dists = []
    for elem in clust:
        dists.append(np.linalg.norm(pt-elem))
    return np.mean(dists)

def silhouette(data, clusts):
    s = []
    print("data-")
    print(data)
    for i in range(len(clusts)):
        for j in range(len(clusts[i])):
            clusts[i][j] = clusts[i][j].tolist()
    print("Clusters")
    print(clusts)
    for elem in data:
        a = []
        b = []
        print(elem)
        for clust in clusts:
            print(clust)
            if elem in clust: #Error in this line
                b.append(avgdist(elem, clust))
            else:
                a.append(avgdist(elem, clust))

        s.append((min(b)-min(a)/(max(min(b), min(a)))))
    return np.mean(s)

得到的终端输出如下-

data-
[[  0.   0.   5.]
 [  0.   0.   0.]
 [  0.   0.   0.]
 [  0.   0.   7.]
 [  0.   0.   0.]
 [  0.   0.  12.]
 [  0.   0.   0.]
 [  0.   0.   7.]
 [  0.   0.   9.]
 [  0.   0.  11.]]
Clusters
[[array([ 0.,  0.,  5.]), array([ 0.,  0.,  0.]), array([ 0.,  0.,  0.]), array([ 0.,  0.,  0.]), array([ 0.,  0.,  0.])], [array([ 0.,  0.,  7.]), array([  0.,   0.,  12.]), array([ 0.,  0.,  7.]), array([ 0.,  0.,  9.]), array([  0.,   0.,  11.])]]
[ 0.  0.  5.]
[[0.0, 0.0, 5.0], [0.0, 0.0, 0.0], [0.0, 0.0, 0.0], [0.0, 0.0, 0.0], [0.0, 0.0, 0.0]]

这个是随着注释行的错误得到的-

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

请帮忙，因为我不确定该错误在我的上下文中意味着什么。类似的问题让我对错误性质有了一些了解，但我认为这里不适用。

Edit - 我通过将错误行更改为-

解决了这个问题

.....
if elem.tolist() in clust: #Error in this line
    .....

Answer 1

你的问题是，如果列表列表 (clust) 包含另一个列表 (elem)，你会尝试在有问题的行中进行评估，这会导致 list/array 的 Truth/False 值，因为评估是按元素进行的：有问题的代码行将沿行评估

   if [True, False, ...]: #<- error here
       code

这将产生问题中的错误

不是保存列表的列表，convert/pack您的数据和集群元素到元组列表中，此评估将起作用。

Answer 2

假设你有

import numpy as np
data = np.array([[  0.,   0.,   5.],
                 [  0.,   0.,   0.],
                 [  0.,   0.,   0.],
                 [  0.,   0.,   7.],
                 [  0.,   0.,   0.],
                 [  0.,   0.,  12.],
                 [  0.,   0.,   0.],
                 [  0.,   0.,   7.],
                 [  0.,   0.,   9.],
                 [  0.,   0.,  11.]])

clusts = [[np.array([ 0.,  0.,  5.]), np.array([ 0.,  0.,  0.]), np.array([ 0.,  0.,  0.]), np.array([ 0.,  0.,  0.]), np.array([ 0.,  0.,  0.])], [np.array([ 0.,  0.,  7.]), np.array([  0.,   0.,  12.]), np.array([ 0.,  0.,  7.]), np.array([ 0.,  0.,  9.]), np.array([  0.,   0.,  11.])]]

更换呢

[...]
if elem in clust: #Error in this line
    [...]

来自

[...]
if any([compa.all() for compa in elem == clust]):
    [...]

其中 测试此类数组列表中是否存在一个 numpy 数组。

在Python3.6

下测试

总结

def silhouette(data, clusts):
    s = []
    print("data-")
    print(data)
    for i in range(len(clusts)):
        for j in range(len(clusts[i])):
            clusts[i][j] = clusts[i][j].tolist()
    print("Clusters")
    print(clusts)
    for elem in data:
        a = []
        b = []
        print(elem)
        for clust in clusts:
            print(clust)
            condition = any([compa.all() for compa in elem == clust])
            print(condition)
            if condition: #No error anymore in this line
                b.append(avgdist(elem, clust))
            else:
                a.append(avgdist(elem, clust))

        s.append((min(b)-min(a)/(max(min(b), min(a)))))
    return np.mean(s)

将打印（仅报告一个比较）

[...]
[ 0.  0.  5.]
[[0.0, 0.0, 5.0], [0.0, 0.0, 0.0], [0.0, 0.0, 0.0], [0.0, 0.0, 0.0], [0.0, 0.0, 0.0]]
True
[[0.0, 0.0, 7.0], [0.0, 0.0, 12.0], [0.0, 0.0, 7.0], [0.0, 0.0, 9.0], [0.0, 0.0, 11.0]]
False

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all(): Silhouette performance algorithm

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all(): Silhouette performance algorithm

python

numpy

k-means

python-3.x

silhouette