k-近邻算法实例

Posted 2020-10-17 jihite

1. 简单例子

步骤

1.1 计算已知点和被求点的距离

1.2 按距离递增排序

1.3 求出距离最近的前k个点的类别最大值作为目标分类

from numpy import *
import operator

def createDateSet():
    group = array([[1.0,1.1], [1.0,1.0], [0,0], [0,0.1]])
    labels = [‘A‘, ‘A‘, ‘B‘, ‘B‘]
    return group, labels


def classify0(inX, dataSet, labels, k):
    dataSetSize = dataSet.shape[0]
    diffMat = tile(inX, (dataSetSize,1)) - dataSet
    sqDiffMat = diffMat ** 2
    sqDistances = sqDiffMat.sum(axis=1)
    distance = sqDistances ** 0.5
    sortDistIndices = distance.argsort()
    classCount = {}
    for i in range(k):
        voteIlable = labels[sortDistIndices[i]]
        classCount[voteIlable] = classCount.get(voteIlable, 0) + 1
    sortedClassCount = sorted(classCount.iteritems(),
                              key = operator.itemgetter(1), reverse=True)
    return sortedClassCount[0][0]

if __name__ == "__main__":
    group, labels = createDateSet()
    inX = [1.1, 0.2]
    k = 3
    aimClass = classify0(inX, group, labels, k)
    print aimClass

语法解析

a. shape()得到矩阵的各个维度的长度

b. tile，举例

>>> a
[1, 2]
>>> tile(a, 2)
array([1, 2, 1, 2])
>>> tile(a, (2,2))
array([[1, 2, 1, 2],
       [1, 2, 1, 2]])
>>> tile(a, (3, 2,2))
array([[[1, 2, 1, 2],
        [1, 2, 1, 2]],

       [[1, 2, 1, 2],
        [1, 2, 1, 2]],

       [[1, 2, 1, 2],
        [1, 2, 1, 2]]])

c. sortDistIndices = distance.argsort() 得到排序后的名次，越大名次越大

d. sortedClassCount = sorted(classCount.iteritems(), key = operator.itemgetter(1), reverse=True) 对字典的值进行逆序（降序）排序