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
软件包的knn3
玩具示例。似乎最后一个电话返回了预测概率的列表。虽然s
的预测概率为1的列表明预测的物种为s
,但在其他一些行中,物种c
的预测概率为0.2,而物种v
的概率为0.8。在这种情况下,预测结果是什么?我猜这是v
的物种,因为它的预测概率更高?
是否有一个函数调用可以快速评估knn
模型的准确性?
首先,保存您的预测:
fit=knn3Train(train, test, cl, k = 5, prob = TRUE)
然后,您需要一个混乱矩阵:
cm = as.matrix(table(Actual = cl, Predicted = fit))
现在您可以计算准确性:
sum(diag(cm))/length(cl)
或其他任何其他性能测量值:https://en.wikipedia.org/wiki/precision_and_recall