如何将模型保存为带有循环的向量？

Question

所以我有这个任务，我必须创建 3 个不同的模型 (r)。 我可以毫无问题地单独完成它们。 但是，我想更进一步，创建一个函数，用 for 循环训练所有这些。 （我知道我可以创建一个每次训练 3 个模型的函数。我不是在寻找问题的其他解决方案，我想这样做（或以类似的方式），因为现在我有 3 个模型，但想象一下我想训练20！

我尝试创建一个列表来存储所有三个模型，但我一直有一些警告。

library(caret)
library(readr)
library(rstudioapi)
library(e1071)
library(dplyr)
library(rpart)

TrainingFunction <- function(method,formula,data,tune) {
 fitcontrol <-  trainControl(method = "repeatedcv", repeats = 4)

 if(method == "rf") {Model <- train(formula, data = data,method = method, trcontrol = fitcontrol , tunelenght = tune)}
 else if (method == "knn"){    
   preObj <- preProcess(data[, c(13,14,15)], method=c("center", "scale"))
   data <- predict(preObj, data)
   Model <- train(formula, data = data,method = method, trcontrol = fitcontrol , tunelenght = tune)  
 }
 else if (method == "svm"){Model <- svm(formula, data = data,cost=1000 , gamma = 0.001)}
   Model
 }

所以这是我创建的一个训练函数，它有效，但现在我想一次训练所有三个！

所以我试过这个：

methods <- c("rf","knn","svm") 
Models <- vector(mode = "list" , length = length(methods))
for(i in 1:length(methods))
{Models[i] <- TrainingFunction(methods[i],Volume~.,List$trainingSet,5)}

这是警告：

Warning messages:
1: In Models[i] <- TrainingFunction(methods[i], Volume ~ ., List$trainingSet,  :
  number of items to replace is not a multiple of replacement length
2: In Models[i] <- TrainingFunction(methods[i], Volume ~ ., List$trainingSet,  :
  number of items to replace is not a multiple of replacement length
3: In svm.default(x, y, scale = scale, ..., na.action = na.action) :
  Variable(s) ‘ProductType.GameConsole’ constant. Cannot scale data.
4: In Models[i] <- TrainingFunction(methods[i], Volume ~ ., List$trainingSet,  :
  number of items to replace is not a multiple of replacement length

当我做模型时，输出是这样的：

[[1]]
[1] "rf"

[[2]]
[1] "knn"

[[3]]
svm(formula = formula, data = data, cost = 1000, gamma = 0.001)

Answer 1

我认为问题来自这一行：

{Models[i] <- TrainingFunction(methods[i],Volume~.,List$trainingSet,5)}

如果要将模型分配到列表的第 i 个位置，则应使用双括号进行操作，如下所示：

{Models[[i]] <- TrainingFunction(methods[i],Volume~.,List$trainingSet,5)}

另一种选择是使用 lapply 而不是显式循环，这样你就可以完全避免这个问题：

train_from_method <- function(methods) {TrainingFunction(methods,Volume~.,List$trainingSet,5)}
Models <- lapply(species_vector, train_from_method)

Answer 2

考虑switch以避免许多if和else特别是如果扩展到 20 个模型。 然后使用lapply构建一个没有初始化或迭代赋值的列表：

TrainingFunction <- function(method, formula, data, tune) {
 fitcontrol <-  trainControl(method = "repeatedcv", repeats = 4)

 Model <- switch(method,
     "rf" = train(formula, data = data, method = method, 
                  trcontrol = fitcontrol, tunelength = tune)
     "knn" = {    
          preObj <- preProcess(data[,c(13,14,15)], 
                               method=c("center", "scale"))
          data <- predict(preObj, data)
          train(formula, data = data, method = method, 
                trcontrol = fitcontrol, tunelength = tune)  
          }
     "svm" = svm(formula, data = data, cost = 1000, gamma = 0.001)
 )
}

methods <- c("rf","knn","svm") 

Model_list <-lapply(methods, function(m)
    TrainingFunction(m, Volume~., List$trainingSet, 5))