137 lines
4.6 KiB
Markdown
137 lines
4.6 KiB
Markdown
# 在 Python 中 K-Means 从零开始聚类[算法讲解]
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> 原文：<https://www.askpython.com/python/examples/k-means-clustering-from-scratch>
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K-Means 是一种非常流行的聚类技术。K-means 聚类是另一类非监督学习算法，用于找出给定数据集中的数据聚类。
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在本文中，我们将使用 [Numpy 模块](https://www.askpython.com/python-modules/numpy/python-numpy-module)从头开始实现 K-Means 聚类算法。
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## K-means 聚类算法的 5 个步骤
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**第一步。随机选取 k 个数据点作为我们的初始质心。**
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**第二步。**用 k 个质心找出训练集中每个数据点之间的距离(我们的目的是欧几里德距离)。
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**第三步。**现在根据找到的距离将每个数据点分配到最近的质心。
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**第四步。**通过取每个聚类组中的点的平均值来更新质心位置。
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**第五步。重复步骤 2 到 4，直到我们的质心不变。**
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我们可以使用像肘方法这样的方法来选择 K(聚类数)的最佳值。
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## 实现 K 均值聚类算法
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现在让我们用代码实现上面的步骤。导入 numpy 模块，然后浏览这里的其余代码，以了解 K-Means 聚类是如何在代码中实现的。
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```py
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#Importing required modules
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import numpy as np
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from scipy.spatial.distance import cdist
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#Function to implement steps given in previous section
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def kmeans(x,k, no_of_iterations):
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idx = np.random.choice(len(x), k, replace=False)
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#Randomly choosing Centroids
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centroids = x[idx, :] #Step 1
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#finding the distance between centroids and all the data points
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distances = cdist(x, centroids ,'euclidean') #Step 2
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#Centroid with the minimum Distance
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points = np.array([np.argmin(i) for i in distances]) #Step 3
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#Repeating the above steps for a defined number of iterations
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#Step 4
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for _ in range(no_of_iterations):
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centroids = []
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for idx in range(k):
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#Updating Centroids by taking mean of Cluster it belongs to
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temp_cent = x[points==idx].mean(axis=0)
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centroids.append(temp_cent)
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centroids = np.vstack(centroids) #Updated Centroids
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distances = cdist(x, centroids ,'euclidean')
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points = np.array([np.argmin(i) for i in distances])
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return points
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```
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上面函数为我们训练集中的每个数据点返回一个聚类标签数组。
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## 测试 K 均值聚类
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我们将使用 digits 数据集(内置在 sklearn 模块中)来测试我们的功能。可以参考[这篇](https://www.askpython.com/python/examples/plot-k-means-clusters-python)文章，了解更多绘制 K-Means 聚类的方法。
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```py
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#Loading the required modules
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import numpy as np
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from scipy.spatial.distance import cdist
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from sklearn.datasets import load_digits
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from sklearn.decomposition import PCA
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from sklearn.cluster import KMeans
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import matplotlib.pyplot as plt
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#Defining our function
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def kmeans(x,k, no_of_iterations):
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idx = np.random.choice(len(x), k, replace=False)
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#Randomly choosing Centroids
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centroids = x[idx, :] #Step 1
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#finding the distance between centroids and all the data points
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distances = cdist(x, centroids ,'euclidean') #Step 2
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#Centroid with the minimum Distance
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points = np.array([np.argmin(i) for i in distances]) #Step 3
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#Repeating the above steps for a defined number of iterations
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#Step 4
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for _ in range(no_of_iterations):
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centroids = []
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for idx in range(k):
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#Updating Centroids by taking mean of Cluster it belongs to
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temp_cent = x[points==idx].mean(axis=0)
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centroids.append(temp_cent)
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centroids = np.vstack(centroids) #Updated Centroids
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distances = cdist(x, centroids ,'euclidean')
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points = np.array([np.argmin(i) for i in distances])
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return points
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#Load Data
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data = load_digits().data
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pca = PCA(2)
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#Transform the data
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df = pca.fit_transform(data)
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#Applying our function
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label = kmeans(df,10,1000)
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#Visualize the results
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u_labels = np.unique(label)
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for i in u_labels:
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plt.scatter(df[label == i , 0] , df[label == i , 1] , label = i)
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plt.legend()
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plt.show()
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```
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Plotting K Means Clusters
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输出结果看起来很有希望。我们的实现是可行的。
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## 结论
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在本文中，我们使用 Python 从头开始创建了一个 K-Means 聚类算法。我们还讲述了制作 K-Means 算法的步骤，最后在 Digits 数据集上测试了我们的实现。你可以在维基百科的[页面上阅读 K-means 聚类算法的理论](https://en.wikipedia.org/wiki/K-means_clustering)
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