python - nltk k-means 聚类或纯 python 的 k-means

Question

虽然我看到了很多与此相关的问题，但我并没有真正得到答案，可能是因为我是使用 nltk 集群的新手。我确实需要对聚类新手进行基本解释，特别是关于 NLTK K 均值聚类的向量表示以及如何使用它。我有一个单词列表，例如 [cat、dog、kitten、puppy 等] 和另外两个单词列表，例如 [carnivore、herbivore、pet 等] 和 [mammal、domestic etc]。我希望能够使用第一个作为均值或质心，基于第一个单词列表对最后两个单词列表进行聚类。我试过了，我收到了这样的断言错误:

clusterer = cluster.KMeansClusterer(2, euclidean_distance, initial_means=means)
  File "C:\Python27\lib\site-packages\nltk\cluster\kmeans.py", line 64, in __init__
    assert not initial_means or len(initial_means) == num_means

AND
    print clusterer.cluster(vectors, True)
  File "C:\Python27\lib\site-packages\nltk\cluster\util.py", line 55, in cluster
    self.cluster_vectorspace(vectors, trace)
  File "C:\Python27\lib\site-packages\nltk\cluster\kmeans.py", line 82, in cluster_vectorspace
    self._cluster_vectorspace(vectors, trace)
  File "C:\Python27\lib\site-packages\nltk\cluster\kmeans.py", line 113, in _cluster_vectorspace
    index = self.classify_vectorspace(vector)
  File "C:\Python27\lib\site-packages\nltk\cluster\kmeans.py", line 137, in classify_vectorspace
    dist = self._distance(vector, mean)
  File "C:\Python27\lib\site-packages\nltk\cluster\util.py", line 118, in euclidean_distance
    diff = u - v
TypeError: unsupported operand type(s) for -: 'numpy.ndarray' and 'numpy.ndarray'

我认为向量表示中有一些我的意思。向量表示的基本示例和示例代码将受到高度赞赏。任何使用 nltk 或纯 python 的解决方案都将受到赞赏。预先感谢您的友好回复

Answer 1

如果我正确理解你的问题，这样的事情应该有效。 kmeans的难点在于找到聚类中心，如果你已经找到了中心或者知道你想要什么中心，你可以:为每个点找到到每个聚类中心的距离，并将该点分配给最近的聚类中心。

(顺便说一句 sklearn 是一个非常适合集群和机器学习的软件包。)

在您的示例中，它应该如下所示:

Levenstein

# levenstein function is not my implementation; I copied it from the 
# link above 
def levenshtein(s1, s2):
    if len(s1) < len(s2):
        return levenshtein(s2, s1)

    # len(s1) >= len(s2)
    if len(s2) == 0:
        return len(s1)

    previous_row = xrange(len(s2) + 1)
    for i, c1 in enumerate(s1):
        current_row = [i + 1]
        for j, c2 in enumerate(s2):
            insertions = previous_row[j + 1] + 1 # j+1 instead of j since previous_row and current_row are one character longer
            deletions = current_row[j] + 1       # than s2
            substitutions = previous_row[j] + (c1 != c2)
            current_row.append(min(insertions, deletions, substitutions))
        previous_row = current_row

    return previous_row[-1]

def get_closest_lev(cluster_center_words, my_word):
    closest_center = None
    smallest_distance = float('inf')
    for word in cluster_center_words:
        ld = levenshtein(word, my_word)
        if ld < smallest_distance:
            smallest_distance = ld
            closest_center = word
    return closest_center

def get_clusters(cluster_center_words, other_words):
    cluster_dict = {}
    for word in cluster_center_words:
        cluster_dict[word] = []
    for my_word in other_words:
        closest_center = get_closest_lev(cluster_center_words, my_word)
        cluster_dict[closest_center].append(my_word)
    return cluster_dict

示例:

cluster_center_words = ['dog', 'cat']
other_words = ['dogg', 'kat', 'frog', 'car']

结果:

>>> get_clusters(cluster_center_words, other_words)
{'dog': ['dogg', 'frog'], 'cat': ['kat', 'car']}