• 决策树题目(是否适合打网球)简单学习-手算


    题目:

    大致结果手绘图:

    详细计算过程:

     一级筛选:

    一级筛选
    1.天气
    from math import*
    sum = -(9/14) * log(9/14,2) - (5/14) * log(5/14,2)
    a =  5/14*(-(2/5) * log(2/5,2) - (3/5) * log(3/5,2))#晴
    b =  4/14*(-1 * log(1,2))#阴
    c =  5/14*(-(3/5) * log(3/5,2) - (2/5) * log(2/5,2))#雨
    sum1 = a + b + c
    print(sum - sum1)
    结果
    #0.2467498197744391
    
    2.温度
    from math import*
    sum = -(9/14) * log(9/14,2) - (5/14) * log(5/14,2)
    a =  4/14*(-(1/2) * log(1/2,2) - (1/2) * log(1/2,2))#热
    b =  6/14*(-(2/3) * log(2/3,2) - (2/3) * log(2/3,2))#温
    c =  4/14*(-(3/4) * log(3/4,2) - (1/4) * log(1/4,2))#凉爽
    sum1 = a + b + c
    print(sum - sum1)
    结果
    #0.08851363698450365
    
    3.湿度
    from math import*
    sum = -(9/14) * log(9/14,2) - (5/14) * log(5/14,2)
    a =  7/14*(-(3/7) * log(3/7,2) - (4/7) * log(4/7,2))#高
    b =  7/14*(-(6/7) * log(6/7,2) - (1/7) * log(1/7,2))#中
    
    sum1 = a + b 
    print(sum - sum1)
    结果
    #0.15183550136234136
    
    4.是否有风
    from math import*
    sum = -(9/14) * log(9/14,2) - (5/14) * log(5/14,2)
    a =  6/14*(-(3/6) * log(3/6,2) - (3/6) * log(3/6,2))#是
    b =  8/14*(-(6/8) * log(6/8,2) - (2/8) * log(2/8,2))#否
    
    sum1 = a + b 
    print(sum - sum1)
    结果
    #0.04812703040826927
    
    所以一级筛选选择天气

    二级筛选:

    2.1晴
    2.1.1温度
    from math import*
    sum = -(2/5) * log(2/5,2) - (3/5) * log(3/5,2)
    a =  2/5*(-(1/2) * log(1/2,2) - (1/2) * log(1/2,2))#温
    #热和凉爽都为0
    sum1 = a
    print(sum - sum1)
    结果
    #0.5709505944546686
    
    2.1.2湿度
    由于中和高都为0,所以信息增益最大
    
    2.1.3是否有风
    from math import*
    sum = -(2/5) * log(2/5,2) - (3/5) * log(3/5,2)
    a =  2/5*(-(1/2) * log(1/2,2) - (1/2) * log(1/2,2))#是
    b =  3/5*(-(1/3) * log(1/3,2) - (2/3) * log(2/3,2))#否
    sum1 = a + b
    print(sum - sum1)
    结果
    #0.01997309402197489
    
    所以在二级筛选中晴天时选择湿度
    
    2.2阴————全部适合
    
    2.3雨
    2.3.1温度
    from math import*
    sum = -(2/5) * log(2/5,2) - (3/5) * log(3/5,2)
    a =  2/5*(-(1/2) * log(1/2,2) - (1/2) * log(1/2,2))#凉爽
    b =  3/5*(-(1/3) * log(1/3,2) - (2/3) * log(2/3,2))#温
    #热为0
    sum1 = a + b
    print(sum - sum1)
    结果
    #0.01997309402197489
    
    2.3.2湿度
    from math import*
    sum = -(2/5) * log(2/5,2) - (3/5) * log(3/5,2)
    a =  2/5*(-(1/2) * log(1/2,2) - (1/2) * log(1/2,2))#高
    b =  3/5*(-(1/3) * log(1/3,2) - (2/3) * log(2/3,2))#中
    sum1 = a + b
    print(sum - sum1)
    结果
    #0.01997309402197489
    
    2.3.3是否有风
    由于有风和无风都为0,所以信息增益最大
    
    所以在二级筛选中晴天时选择是否有风

    关于手算有不懂的的地方可以问我~

    (争取早点弄出程序,手算费脑子。。。)

  • 相关阅读:
    jQuery基础学习5——JavaScript方法获取页面中的元素
    jQuery基础学习4——jQuery容错性
    jQuery基础学习3——jQuery库冲突
    jQuery基础学习2——DOM和jQuery对象
    Windows Server 2003 R2 64位简体中文版下载
    jQuery基础学习1
    关闭Linux里边的selinux
    linux下安装apache详解
    Autolayout学习(1)-了解Autoreszing
    新浪微博客户端(64)-下拉放大
  • 原文地址:https://www.cnblogs.com/zhying99/p/10582166.html
Copyright © 2020-2023  润新知