对我来说,在回归情况下用随机森林预测的程序是非常重要的,因为我得到了不同的结果然后内置程序:
set.seed(1)
wyn <- randomForest(y = iris[, 1], x = iris[, -1])
(wyn) #Mean of squared residuals: 0.1341068
response=predict(wyn, iris[, -1]) # is it ok ? below to lines gives different output
mean((iris[,1]-wyn$predicted)^2) #returns 0.1341068
mean((iris[,1]-response)^2) # returns 0.07259335
plot(wyn$predicted,response) # almost but not identical
我在上面使用predict()
时出了什么问题?
答案 0 :(得分:1)
#returns out-of-bag error
mean((iris[,1]-wyn$predicted)^2) #returns 0.1341068
#returns error calculated on whole dataset
mean((iris[,1]-response)^2) # returns 0.07259335