R：您如何计算KNN的预测准确性？

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
  ...

I'm using the toy example for knn3 from the caret package. It seems like the last call returns a list of predicted probabilities. While the columns where the predicted probability is 1 for s suggests that the predicted species is s , there are some other rows where the predicted probability of species c is 0.2, and 0.8 for species v . In that case, what is the predicted outcome? I'm guessing it's species v since its predicted probability is higher?

Is there a function call that can quickly assess the accuracy of knn model fit here?

Answer 1

First, save your predictions:

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

Then, you need a confusion matrix:

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

Now you can calculate accuracy:

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

Or any number of other performance measurements: https://en.wikipedia.org/wiki/Precision_and_recall