参考资料：

https://baike.baidu.com/item/%E7%9F%A9%E9%98%B5/18069?fr=aladdin

http://blog.csdn.net/c406495762/article/details/75172850

基础概念：

#0取行数，1取得列数

import numpy as np

dataSetSize = dataSet.shape[0]

 

 

diffMat = np.tile(inX, (dataSetSize, 1)) - dataSet

tile第二个参数为元组，元组第一个参数代表复制的个数，纵(list的个数)， 第二个参数为横向（单个list中的元素复制的份数）

例如：

>>> a=[0,1,2] >>> b=tile(a,9)>>> barray([0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1,       2, 0, 1, 2])>>> >>> b=tile(a,(9,2))>>> barray([[0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2]])>>> b=tile(a,(9,1))>>> barray([[0, 1, 2],       [0, 1, 2],       [0, 1, 2],       [0, 1, 2],       [0, 1, 2],       [0, 1, 2],       [0, 1, 2],       [0, 1, 2],       [0, 1, 2]])>>> b=tile(a,(9,12))>>> barray([[0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0,        1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0,        1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0,        1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0,        1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0,        1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0,        1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0,        1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0,        1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2],       [0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0,        1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2]])

 

矩阵减法

>>> aarray([[101,  20],       [101,  20],       [101,  20],       [101,  20]])>>> barray([[  1, 101],       [  5,  89],       [108,   5],       [115,   8]])>>> a-barray([[100, -81],       [ 96, -69],       [ -7,  15],       [-14,  12]])

 

#二维特征相减后平方

sqDiffMat = diffMat**2

>>> carray([[100, -81],       [ 96, -69],       [ -7,  15],       [-14,  12]])>>> c**2array([[10000,  6561],       [ 9216,  4761],       [   49,   225],       [  196,   144]])

#sum()所有元素相加，sum(0)列相加，sum(1)行相加

sqDistances = sqDiffMat.sum(axis=1)

>>> d=c**2>>> d.sum(axis=1)array([16561, 13977,   274,   340]) >>> d.sum(axis=0)array([19461, 11691])

#开方，计算出距离

distances = sqDistances**0.5

>>> f=e**0.5>>> farray([ 128.68954892,  118.22436297,   16.55294536,   18.43908891])

sortedDistIndices = distances.argsort()

#定一个记录类别次数的字典 ,按大小排序，打印index

>>> farray([ 128.68954892,  118.22436297,   16.55294536,   18.43908891])>>> g=f.argsort()>>> garray([2, 3, 1, 0])

取出前k个元素的类别

>>> for i  in range(3):...     print(i) ...     labels[sortedDistIndices[i]]... 0'动作片'1'动作片'2'爱情片'

 

>>> for i  in range(3):...     voteIlabel = labels[sortedDistIndices[i]]...     classCount.get(voteIlabel,0) ... 000>>> for i  in range(3):...     voteIlabel = labels[sortedDistIndices[i]]...     classCount.get(voteIlabel,0) + 1... 111

#key=operator.itemgetter(1)根据字典的值进行排序    #key=operator.itemgetter(0)根据字典的键进行排序    #reverse降序排序字典    sortedClassCount = sorted(classCount.items(),key=operator.itemgetter(1),reverse=True)    print('#######',sortedClassCount)

classCount~~~~~~ {}classCount~~~~~~ {
   '动作片': 2, '爱情片': 1}####### [('动作片', 2), ('爱情片', 1)]动作片

 

第一段代码总算跑通了，虽然到现在为止还不理解什么事人工智能，哈哈，