我在R中运行一个潜在类分析,并使用熵函数。我想了解为什么在输出中,它为较低的Nclasses生成一个结果,然后为更高的nclasses生成一个NaN。
我是一个软件初学者!
这里供参考的是输出和代码:
> entropy<-function (p) sum(-p*log(p))
> error_prior <- entropy(France_2class$P) # Class proportions
> error_post <- mean(apply(France_2class$posterior, 1, entropy))
> R2_entropy <- (error_prior - error_post) / error_prior
> R2_entropy
[1] 0.8121263
>
> entropy<-function (p) sum(-p*log(p))
> error_prior <- entropy(France_3class$P) # Class proportions
> error_post <- mean(apply(France_3class$posterior, 1, entropy))
> R2_entropy <- (error_prior - error_post) / error_prior
> R2_entropy
[1] 0.8139903
>
> entropy<-function (p) sum(-p*log(p))
> error_prior <- entropy(France_4class$P) # Class proportions
> error_post <- mean(apply(France_4class$posterior, 1, entropy))
> R2_entropy <- (error_prior - error_post) / error_prior
> R2_entropy
[1] NaN
>
> entropy<-function (p) sum(-p*log(p))
> error_prior <- entropy(France_5class$P) # Class proportions
> error_post <- mean(apply(France_5class$posterior, 1, entropy))
> R2_entropy <- (error_prior - error_post) / error_prior
> R2_entropy
[1] NaN
>
> entropy<-function (p) sum(-p*log(p))
> error_prior <- entropy(France_6class$P) # Class proportions
> error_post <- mean(apply(France_6class$posterior, 1, entropy))
> R2_entropy <- (error_prior - error_post) / error_prior
> R2_entropy
[1] NaN有人能帮忙吗?谢谢
发布于 2020-04-19 10:52:19
我想问题来自于entropy的定义。更准确地说,如果0包含在p中,那么您将获得NaN,例如,
> entropy(p1)
[1] 1.279854
> entropy(p2)
[1] NaN
> entropy(p3)
[1] 0.5004024要修复它,可以将na.omit添加到entropy函数中,如下所示
entropy<-function(p) sum(na.omit(-p*log(p)))然后你就可以看到
> entropy(p1)
[1] 1.279854
> entropy(p2)
[1] 0.5004024
> entropy(p3)
[1] 0.5004024数据
p1 <- c(0.1,0.2,0.3,0.4)
p2 <- c(0,0.2,0.8)
p3 <- c(0.2,0.8)https://stackoverflow.com/questions/61302401
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