R：你如何计算 KNN 的预测精度？

Question

library(caret)
irisFit1 <- knn3(Species ~ ., iris)

irisFit2 <- knn3(as.matrix(iris[, -5]), iris[,5])

data(iris3)
train <- rbind(iris3[1:25,,1], iris3[1:25,,2], iris3[1:25,,3])
test <- rbind(iris3[26:50,,1], iris3[26:50,,2], iris3[26:50,,3])
cl <- factor(c(rep("s",25), rep("c",25), rep("v",25)))
> knn3Train(train, test, cl, k = 5, prob = TRUE)
 [1] "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "s" "c"
[27] "c" "v" "c" "c" "c" "c" "c" "v" "c" "c" "c" "c" "c" "c" "c" "c" "c" "c" "c" "c" "c" "c" "c" "c" "v" "c"
[53] "c" "v" "v" "v" "v" "v" "c" "v" "v" "v" "v" "c" "v" "v" "v" "v" "v" "v" "v" "v" "v" "v" "v"
attr(,"prob")
              c s         v
 [1,] 0.0000000 1 0.0000000
 [2,] 0.0000000 1 0.0000000
 [3,] 0.0000000 1 0.0000000
 [4,] 0.0000000 1 0.0000000
 [5,] 0.0000000 1 0.0000000
 [6,] 0.0000000 1 0.0000000
  ...

我正在使用 caret 包中 knn3 的玩具示例。好像最后一个调用 returns 预测概率的列表。虽然 s 的预测概率为 1 的列表明预测的物种是 s，但在其他一些行中，物种 c 的预测概率为 0.2，而物种的预测概率为 0.8 v。在那种情况下，预期的结果是什么？我猜它是物种 v 因为它的预测概率更高？

这里有没有可以快速评估knn模型拟合精度的函数调用？

Answer 1

首先，保存您的预测：

fit=knn3Train(train, test, cl, k = 5, prob = TRUE)

那么，你需要一个混淆矩阵：

cm = as.matrix(table(Actual = cl, Predicted = fit))

现在您可以计算准确率：

sum(diag(cm))/length(cl)

或任意数量的其他性能测量：https://en.wikipedia.org/wiki/Precision_and_recall

R：你如何计算 KNN 的预测精度？

R: how do you calculate prediction accuracy for KNN?

r

classification

prediction

knn