ItF*_*eak 11 python numpy scikit-learn data-science
我有一个点列表,它们是 kmeans 算法的惯性值。
为了确定最佳集群数量,我需要找到这条曲线开始变平的点。
数据示例
以下是我的值列表的创建和填充方式:
sum_squared_dist = []
K = range(1,50)
for k in K:
km = KMeans(n_clusters=k, random_state=0)
km = km.fit(normalized_modeling_data)
sum_squared_dist.append(km.inertia_)
print(sum_squared_dist)
Run Code Online (Sandbox Code Playgroud)
我怎样才能找到一个点,这条曲线的节距增加(曲线正在下降,所以一阶导数为负)?
我的方法
derivates = []
for i in range(len(sum_squared_dist)):
derivates.append(sum_squared_dist[i] - sum_squared_dist[i-1])
Run Code Online (Sandbox Code Playgroud)
我想使用肘部方法找到任何给定数据的最佳聚类数。有人可以帮助我如何找到惯性值列表开始变平的点吗?
编辑数据点
:
[7342.1301373073857, 6881.7109460930769, 6531.1657905495022,
6356.2255554679778, 6209.8382535595829, 6094.9052166741121,
5980.0191582610196, 5880.1869867848218, 5779.8957906367368,
5691.1879324562778, 5617.5153566271356, 5532.2613232619951,
5467.352265375117, 5395.4493783888756, 5345.3459908298091,
5290.6769823693812, 5243.5271656371888, 5207.2501206569532,
5164.9617535255456]
Run Code Online (Sandbox Code Playgroud)
Kev*_*vin 28
我研究了一个以Kneedle 算法为模型的 Python 包。它发现x=5曲线开始变平的点。文档和论文更详细地讨论了选择拐点的算法。
y = [7342.1301373073857, 6881.7109460930769, 6531.1657905495022,
6356.2255554679778, 6209.8382535595829, 6094.9052166741121,
5980.0191582610196, 5880.1869867848218, 5779.8957906367368,
5691.1879324562778, 5617.5153566271356, 5532.2613232619951,
5467.352265375117, 5395.4493783888756, 5345.3459908298091,
5290.6769823693812, 5243.5271656371888, 5207.2501206569532,
5164.9617535255456]
x = range(1, len(y)+1)
from kneed import KneeLocator
kn = KneeLocator(x, y, curve='convex', direction='decreasing')
print(kn.knee)
5
import matplotlib.pyplot as plt
plt.xlabel('number of clusters k')
plt.ylabel('Sum of squared distances')
plt.plot(x, y, 'bx-')
plt.vlines(kn.knee, plt.ylim()[0], plt.ylim()[1], linestyles='dashed')
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
16433 次 |
| 最近记录: |