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machine-learning - 从安德鲁·摩尔的教程中获取信息

转载 作者:行者123 更新时间:2023-11-30 09:38:41 25 4
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我现在正在自学机器学习。

一个关于信息增益的简单问题。如何计算图片数据的信息增益?我不明白。

有人能解释一下如何从第一行得到 0.992385 吗?

非常感谢! enter image description here

最佳答案

设 g(x) = -x * log(x)/log(2)

总人数为48842。

H(财富,关系)= g[52/48842] + g[111/48842] + g[309/48842] + g[1093/48842] + g[1238/48842] + g[1276/48842] + g[1454/48842] + g[4816/48842] + g[7470/48842] + g[8846/48842] + g[10870/48842] + g[11307/48842] = 2.7835

H(财富)= g(贫穷/总计)+ g(富裕/总计)= g[0.239282] + g[0.760718] = 0.793844

H(关系) = g(丈夫/总计) + g(非家庭成员/总计) + ... = g[0.0308341] + g[0.0477253] + g[0.10493] + g[0.155215] + g[0.257627] + g[0.403669] = 2.15508

H(财富 | 关系) = H(财富, 关系) - H(关系) = 2.7835 - 2.15508 = 0.628421

IG = H(财富) - H(财富|关系) = H(财富) + H(关系) - H(财富,关系) = 0.165423

这里是用 Mathematica 编写的源代码。如果您觉得需要查看其他语言的源代码,请在下面用您的首选语言发表评论。如果我有时间,我会把它写下来。 - 干杯,汉斯

Mathematica 中的源代码

(* =============================================== ====== *)

m = {{10870, 8846}, {11307, 1276}, {1454, 52}, {7470, 111}, 
{4816, 309}, {1238, 1093}};

iTot = Total[ Flatten[m]];

h[x_] := -x * Log[2, x];

fHAll = Sum[ h[ m[[i, j]]/ iTot ], {i, 6}, {j, 2}] // N;

fHWealth = h[ Total[ m[[All, 1]]]/iTot] + h[ Total[ m[[All, 2]]]/iTot] // N ;

fHRelation = Sum[
h[ Total[ m[[i ]]]/iTot] , {i, Length[m]}] // N;

fWealthGivenRelation = fHAll - fHRelation;

Print[" H(relation, wealth) = ", fHAll];
Print[" H(relation) = ", fHRelation];
Print[" H(wealth) = ", fHWealth];
Print[" H(wealth | relation) = ", fWealthGivenRelation];
Print[" IG = MI = ", fHWealth - fWealthGivenRelation, " = ",
fHWealth + fHRelation - fHAll];

(* ===================输出====================*)

 H(relation, wealth) = 2.7835
H(relation) = 2.15508
H(wealth) = 0.793844
H(wealth | relation) = 0.628421
IG = MI = 0.165423 = 0.165423

哎呀,我没有回答你的主要问题。这就是答案。

H(财富|关系=丈夫)= g(10870/(10870 + 8846)) + g(8846/(10870 + 8846))= 0.992385

关于machine-learning - 从安德鲁·摩尔的教程中获取信息,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12719091/

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