有监督的kNN近邻算法:
(1)计算已知类别数据集中的点与当前点之间的距离
(2)按照距离递增次序排序
(3)选取与当前点距离最小的k个点
(4)确定前k个点所在类别的出现频率
(5)返回前k个点出现频率最高的类别作为当前点的预测分类
#数据样例
1 2:a
1 3:a1000 10000:d
#版本0:纯python
"kNN"
from math import sqrt
from collections import Counter
distance=lambda a,b:sqrt(sum(map(lambda ai,bi:pow(ai-bi,2),a,b))) if len(a)==len(b) else "Error0:data length match fail"
distance2=lambda a,b:distance([int(i) for i in a.split()],[int(i) for i in b.split()]) # for strings
#print(distance2('1 2 4 7 8','2 5 5 6 110'))
readData=lambda file:{line.split(':')[0]:line.strip().split(':')[1] for line in open(file)}
#print(readData())
def judgeSpot(fileIn='test0.txt',x='1 2',num=5):
distanceDict,data={},readData(fileIn)
for k in data:
distanceDict[str(distance2(x,k))]=data[k]
# sortDistance=sorted(distanceDict.items(),key=lambda x:float(x[0]))[:num]
# kindDict=[item[1] for item in sortDistance]
return sorted(dict(Counter(item[1] for item in sorted(distanceDict.items(),key=lambda x:float(x[0]))[:num])).items(),key=lambda x:x[1],reverse=True)[0][0]
#print(judgeSpot('1000 10000','test0.txt'),)
def judgeSpot2(dataIn,x='1 2',num=5):
distanceDict,data={},dataIn
for k in data:
distanceDict[str(distance2(x,k))]=data[k]
# sortDistance=sorted(distanceDict.items(),key=lambda x:float(x[0]))[:num]
# kindDict=[item[1] for item in sortDistance]
return sorted(dict(Counter(item[1] for item in sorted(distanceDict.items(),key=lambda x:float(x[0]))[:num])).items(),key=lambda x:x[1],reverse=True)[0][0]
print(judgeSpot('test0.txt','1000 10000'),)
#Rate of Right
def rateRight(fileIn='test0.txt',num=5):
countRight,data=0,readData(fileIn)
for k in data:
if judgeSpot2(data,k,num)==data[k]:
countRight+=1
return countRight/float(len(open(fileIn).readlines()))
print(rateRight())
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原文地址:http://blog.csdn.net/awsxsa/article/details/45955871