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模型评估【分类问题】

慎壹 2022-03-30 阅读 94

模型评估

文章目录

符号含义
y y y真实值【一个向量】
y ^ \hat y y^预测值【一个向量】

分类评估(Metrics for Classification)

准确度(Accuracy)

Eg:

y y y = (0,0,1,0,1)

y ^ \hat y y^ = (1,0,1,0,1)
A c c u a r c y = s u m ( y = = y ^ ) / l e n ( y ) = s u m ( ( 0 , 0 , 1 , 0 , 1 ) = = ( 1 , 0 , 1 , 0 , 1 ) ) / 5 = 4 / 5 \begin{aligned} Accuarcy &= sum(y == \hat y)/len(y) \\ &= sum((0,0,1,0,1) == (1,0,1,0,1))/5\\ &= 4/5 \end{aligned} Accuarcy=sum(y==y^)/len(y)=sum((0,0,1,0,1)==(1,0,1,0,1))/5=4/5


精度(Precision)

当有时候数据正负类分布不平衡,此时得到的准确度就有一点偏离模型真正能够达到的准确度,因此需要其他指标来权衡利弊。

Eg:

y y y = (0,0,1,0,1)

y ^ \hat y y^ = (1,0,1,0,1)

预测正确率
P r e c i s i o n = s u m ( y = = 1   a n d   y ^ = = 1 ) / s u m ( y ^ = = 1 ) = s u m ( ( 0 , 0 , 1 , 0 , 1 ) = = 1   a n d   ( 1 , 0 , 1 , 0 , 1 ) = = 1 ) / s u m ( ( 1 , 0 , 1 , 0 , 1 ) = = 1 ) = 2 / 3 \begin{aligned} Precision &= sum(y==1\ and\ \hat y==1)/sum(\hat y==1) \\ &= sum((0,0,1,0,1)==1\ and\ (1,0,1,0,1)==1)/ sum((1,0,1,0,1)==1)\\ &= 2/3 \end{aligned} Precision=sum(y==1 and y^==1)/sum(y^==1)=sum((0,0,1,0,1)==1 and (1,0,1,0,1)==1)/sum((1,0,1,0,1)==1)=2/3


召回率(Recall)

Eg:

y y y = (0,0,1,0,1)

y ^ \hat y y^ = (1,0,1,0,1)

预测正确率
R e c a l l = s u m ( y = = 1   a n d   y ^ = = 1 ) / s u m ( y = = 1 ) = s u m ( ( 0 , 0 , 1 , 0 , 1 ) = = 1   a n d   ( 1 , 0 , 1 , 0 , 1 ) = = 1 ) / s u m ( ( 0 , 0 , 1 , 0 , 1 ) = = 1 ) = 2 / 2 \begin{aligned} Recall &= sum(y==1\ and\ \hat y==1)/sum( y==1) \\ &= sum((0,0,1,0,1)==1\ and\ (1,0,1,0,1)==1)/ sum((0,0,1,0,1)==1)\\ &= 2/2 \end{aligned} Recall=sum(y==1 and y^==1)/sum(y==1)=sum((0,0,1,0,1)==1 and (1,0,1,0,1)==1)/sum((0,0,1,0,1)==1)=2/2


F1

F 1 = 2 ∗ P ∗ R / ( P + R ) F1 = 2*P*R/(P+R) F1=2PR/(P+R)

Eg:

在以上的例子中计算F1
F 1 = 2 ∗ P ∗ R / ( P + R ) = ( 2 ∗ 2 / 3 ∗ 2 / 2 ) / ( 2 / 3 + 2 / 2 ) = 4 / 5 \begin{aligned} F1 &= 2*P*R/(P+R) \\ &= (2*2/3*2/2)/(2/3+2/2)\\ &= 4/5 \end{aligned} F1=2PR/(P+R)=(22/32/2)/(2/3+2/2)=4/5

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