定义k-1个簇质心——SKlearn KMeans答案

【问题标题】：Define k-1 cluster centroids -- SKlearn KMeans定义k-1个簇质心——SKlearn KMeans
【发布时间】：2021-03-03 09:34:33
【问题描述】：

我正在对部分标记的数据集执行二元分类。我对它的 1 有一个可靠的估计，但对它的 0 没有。

来自 sklearn KMeans 文档：

init : {‘k-means++’, ‘random’ or an ndarray}
Method for initialization, defaults to ‘k-means++’:   
If an ndarray is passed, it should be of shape (n_clusters, n_features) and gives the initial centers.

我想传递一个 ndarray，但我只有 1 个可靠的质心，而不是 2 个。

有没有办法最大化第 K-1 个质心和第 K 个质心之间的熵？或者，有没有办法手动初始化K-1个质心，剩下的使用K++？

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相关问题：

This 试图定义具有 n-1 个特征的 K 个质心。（我想定义具有 n 个特征的 k-1 个质心）。

Here is a description of what I want，但它被其中一位开发人员解释为错误，并且“易于实现[能够]”

【问题讨论】：

标签： python scikit-learn k-means

【解决方案1】：

我有理由相信这可以按预期工作，但如果您发现错误，请纠正我。（由geeks for geeks拼凑而成）：


import sys

def distance(p1, p2): 
    return np.sum((p1 - p2)**2)


def find_remaining_centroid(data, known_centroids, k = 1): 
    ''' 
    initialized the centroids for K-means++ 
    inputs: 
        data - Numpy array containing the feature space
        known_centroid - Numpy array containing the location of one or multiple known centroids
        k - remaining centroids to be found
    '''
    n_points = data.shape[0]

    # Initialize centroids list
    if known_centroids.ndim > 1:
        centroids = [cent for cent in known_centroids]
    
    else:
        centroids = [np.array(known_centroids)]

    # Perform casting if necessary
    if isinstance(data, pd.DataFrame):
        data = np.array(data)
        
    # Add a randomly selected data point to the list  
    centroids.append(data[np.random.randint( 
            n_points), :])
    
    # Compute remaining k-1 centroids
    for c_id in range(k - 1):
        ## initialize a list to store distances of data 
        ## points from nearest centroid 
        dist = np.empty(n_points)

        for i in range(n_points):
            point = data[i, :] 
            d = sys.maxsize 

            ## compute distance of 'point' from each of the previously 
            ## selected centroid and store the minimum distance 
            for j in range(len(centroids)): 
                temp_dist = distance(point, centroids[j]) 
                d = min(d, temp_dist) 

            dist[i] = d

        ## select data point with maximum distance as our next centroid 
        next_centroid = data[np.argmax(dist), :] 
        centroids.append(next_centroid) 

        # Reinitialize distance array for next centroid
        dist = np.empty(n_points)
    

    
    return centroids[-k:]

它的用法：

# For finding a third centroid:
third_centroid = find_remaining_centroid(X_train, np.array([presence_seed, absence_seed]), k = 1)

# For finding the second centroid:
second_centroid = find_remaining_centroid(X_train, presence_seed, k = 1)

presence_seed和absence_seed是已知的质心位置。

【讨论】：