python實現(xiàn)K最近鄰算法

#! /usr/bin/env python3
# -*- coding: utf-8 -*-
# fileName ： KNNdistance.py
# author : zoujiameng@aliyun.com.cn

import math

def getMaxLocate(target): # 查找target中最大值的locate
 maxValue = float("-inFinIty")
 for i in range(len(target)):
  if maxValue < target[i]:
   maxValue = target[i]
   flag = i
 return flag

def KDistance(K, dest, source):
 destlen = len(dest)
 source1len = len(source[1])
 sourcelen = len(source)
 KNN = []
 locate = source # 準(zhǔn)備從source中剔除N-K個最大值

 if destlen == source1len:
  for i in range(sourcelen):
   delta = 0
   for j in range(source1len):# 畢達哥拉斯公式
    delta += (dest[j] - source[i][j])*(dest[j] - source[i][j])
   KNN.append(math.sqrt(delta))

  for k in range(sourcelen, K, -1):
   flag = getMaxLocate(KNN)
   #print("%s 最大元素位置為%d" % (KNN, flag))
   KNN.remove(KNN[flag]);
   locate.remove(locate[flag])# 移除對應(yīng)位置的元素
   #print(locate)
  return locate # 返回最終K個最接近的元素
 else:
  return None

#! /usr/bin/env python3
# -*- coding: utf-8 -*-
# fileName ： KNNdistance.py
# author : zoujiameng@aliyun.com.cn

if __name__ == "__main__":

 history = {}
 history[5, 1, 0] = 300
 history[3, 1, 1] = 225
 history[1, 1, 0] = 75
 history[4, 0, 1] = 200
 history[4, 0, 0] = 150
 history[2, 0, 0] = 50

 dest = [4, 1, 0]
 source = []
 for i in history:
  source.append(i)
 print(source)

 from KNNdistance import KDistance
 K = 4
 locate = KDistance(K, dest, source)
 avg = 0
 for i in range(len(locate)):
  avg+=history[locate[i]]
 avg/=K
 print("回歸結(jié)果：今天應(yīng)該烤%d個面包" % round(avg))