generated from LufsX/Challenge-kNN
This commit is contained in:
+16
-16
@@ -42,6 +42,8 @@ name = "黎世浩"
|
|||||||
|
|
||||||
# 这里写自己的代码实现
|
# 这里写自己的代码实现
|
||||||
|
|
||||||
|
import numpy as np
|
||||||
|
|
||||||
def classify0(inX, dataSet, labels, k=3):
|
def classify0(inX, dataSet, labels, k=3):
|
||||||
"""
|
"""
|
||||||
KNN 分类算法,使用马氏距离。
|
KNN 分类算法,使用马氏距离。
|
||||||
@@ -55,27 +57,25 @@ def classify0(inX, dataSet, labels, k=3):
|
|||||||
Returns:
|
Returns:
|
||||||
- int: 预测的分类标签。
|
- int: 预测的分类标签。
|
||||||
"""
|
"""
|
||||||
# 计算协方差矩阵的逆和均值
|
# 计算训练数据集的协方差矩阵和均值
|
||||||
inv_cov = np.linalg.inv(np.cov(dataSet.T)) # 计算协方差矩阵的逆
|
covMat = np.cov(dataSet, rowvar=False)
|
||||||
meanVals = np.mean(dataSet, axis=0) # 计算均值
|
meanVec = np.mean(dataSet, axis=0)
|
||||||
|
|
||||||
# 计算输入样本与训练集中每个样本的马氏距离
|
# 计算待分类样本与训练集中每个样本的马氏距离
|
||||||
diffMat = dataSet - meanVals
|
invCovMat = np.linalg.inv(covMat)
|
||||||
mahalDist = np.array([np.sqrt((inX - meanVals).dot(inv_cov).dot((inX - meanVals).T)) * np.sqrt(np.sum((diff - meanVals).dot(inv_cov).dot(diff - meanVals.T)) / (diffMat.shape[0] - 1)) for diff in diffMat])
|
diffVecs = dataSet - meanVec
|
||||||
|
dist = np.sqrt(np.sum(inX - meanVec, axis=1).dot(invCovMat).dot((inX - meanVec).T))
|
||||||
|
|
||||||
# 确保 k 不大于训练集中的样本数量
|
# 获取距离最近的 k 个样本的索引
|
||||||
k = min(k, len(mahalDist))
|
kNearestIndices = np.argsort(dist)[:k]
|
||||||
|
|
||||||
# 对距离进行排序,选择距离最小的 k 个样本
|
# 统计这些邻居的标签
|
||||||
sortedDistIndices = np.argsort(mahalDist)
|
|
||||||
classCount = {}
|
classCount = {}
|
||||||
|
for i in kNearestIndices:
|
||||||
|
voteLabel = labels[i]
|
||||||
|
classCount[voteLabel] = classCount.get(voteLabel, 0) + 1
|
||||||
|
|
||||||
# 遍历距离最小的 k 个样本,统计每个类别的出现次数
|
# 返回出现次数最多的标签
|
||||||
for i in range(k):
|
|
||||||
voteIlabel = labels[sortedDistIndices[i]]
|
|
||||||
classCount[voteIlabel] = classCount.get(voteIlabel, 0) + 1
|
|
||||||
|
|
||||||
# 选择出现次数最多的类别作为预测标签
|
|
||||||
sortedClassCount = sorted(classCount.items(), key=lambda x: x[1], reverse=True)
|
sortedClassCount = sorted(classCount.items(), key=lambda x: x[1], reverse=True)
|
||||||
return sortedClassCount[0][0]
|
return sortedClassCount[0][0]
|
||||||
|
|
||||||
|
|||||||
Reference in New Issue
Block a user