R：保存循环的结果

Question

我正在使用 R 编程语言。我正在学习循环以及如何存储循环的结果。例如，我编写了以下循环函数的代码（生成随机数据并适合不同的决策树）：

#load libraries    
    library(caret)
library(rpart)

#generate data

a = rnorm(1000, 10, 10)

b = rnorm(1000, 10, 5)

c = rnorm(1000, 5, 10)

group <- sample( LETTERS[1:2], 1000, replace=TRUE, prob=c(0.5,0.5) )
group_1 <- 1:1000

#put data into a frame
d = data.frame(a,b,c, group, group_1)

d$group = as.factor(d$group)

#place results in table
#final_table = matrix(1, nrow = 10, ncol=10)


e <- d



vec1 <- sample(200:300, 5)
vec2 <- sample(400:500,5)
vec3 <- sample(700:800,5)

for (i in seq_along(vec1)) { 
    for (j in seq_along(vec2)) {
        for (k in seq_along(vec3)) {
            # d <- e
            d$group_2 = as.integer(ifelse(d$group_1 < vec1[i] , 0, ifelse(d$group_1 >vec1[i]  & d$group_1 < vec2[j] , 1, ifelse(d$group_1 >vec2[j]  & d$group_1 < vec3[k] , 2,3))))
            
            
            
            
            d$group_2 = as.factor(d$group_2)
            
            fitControl <- trainControl(## 10-fold CV
                method = "repeatedcv",
                number = 2,
                ## repeated ten times
                repeats = 1)
            
            TreeFit <- train(group_2 ~ ., data = d[,-5],
                             method = "rpart",
                             trControl = fitControl)
            
            pred <- predict(
                TreeFit,
                d[,-5])
            
            con <- confusionMatrix(
                d$group_2,
                pred) 
            
            #update results into table
            #final_table[i,j] = con$overall[1]
            acc=c(vec1[i],vec2[j],vec3[k],con$overall[1])
            print(acc)
            
            
        }
    }
}

我有兴趣将“acc”的结果保存到表格（或数据框）中。我可以打印“acc”的所有值，但是当我正式查看“acc”的结果时：只显示最后一行。

我的问题：是否可以获取整个打印输出（即“acc”）并将其存储到表格中？

谢谢

Answer 1

您在此处发布的非常好的示例。要启动您的数据框，我们可以添加：

#place results in table
final_table = data.frame(vec1=double(),vec2=double(),vec3=double(),Accuracy=double())

我们可以将acc 的输出存储到其中：

#update results into table
acc=c(vec1[i],vec2[j],vec3[k],con$overall[1])
final_table<-rbind(final_table,acc)

Answer 2

您可以使用expand.grid 创建vec1、vec2 和vec3 的所有可能组合，并在数据框中的每次迭代中保存con$overall[1]。

library(caret)
library(rpart)

#generate data

a = rnorm(1000, 10, 10)
b = rnorm(1000, 10, 5)
c = rnorm(1000, 5, 10)
group <- sample( LETTERS[1:2], 1000, replace=TRUE, prob=c(0.5,0.5))
group_1 <- 1:1000

#put data into a frame
d = data.frame(a,b,c, group, group_1)
d$group = as.factor(d$group)

e <- d
vec1 <- sample(200:300, 5)
vec2 <- sample(400:500,5)
vec3 <- sample(700:800,5)
z <- 0
df <- expand.grid(vec1, vec2, vec3)
df$Accuracy <- NA

for (i in seq_along(vec1)) { 
  for (j in seq_along(vec2)) {
    for (k in seq_along(vec3)) {
      # d <- e
      d$group_2 = as.integer(ifelse(d$group_1 < vec1[i] , 0, ifelse(d$group_1 >vec1[i]  & d$group_1 < vec2[j] , 1, ifelse(d$group_1 >vec2[j]  & d$group_1 < vec3[k] , 2,3))))
      
      d$group_2 = as.factor(d$group_2)
      
      fitControl <- trainControl(## 10-fold CV
        method = "repeatedcv",
        number = 2,
        ## repeated ten times
        repeats = 1)
      
      TreeFit <- train(group_2 ~ ., data = d[,-5],
                       method = "rpart",
                       trControl = fitControl)
      
      pred <- predict(
        TreeFit,
        d[,-5])
      
      con <- confusionMatrix(
        d$group_2,
        pred) 
      
      #update results into table
      #final_table[i,j] = con$overall[1]
      z <- z + 1
      df$Accuracy[z] <- con$overall[1]
    }
  }
}

head(df)

#  Var1 Var2 Var3 Accuracy
#1  300  492  767    0.299
#2  202  492  767    0.299
#3  232  492  767    0.299
#4  293  492  767    0.376
#5  231  492  767    0.299
#6  300  435  767    0.331

Answer 3

还有一种替代方法，它将每次迭代的结果保存为列表的元素，然后将结果组合起来。通过在循环开始之前分配列表，我们可以避免代价高昂的growing in a loop。此外，这种方法在循环顺序发生变化的情况下也很稳健。

#load libraries    
library(caret)
library(rpart)

#generate data

a = rnorm(1000, 10, 10)

b = rnorm(1000, 10, 5)

c = rnorm(1000, 5, 10)

group <- sample( LETTERS[1:2], 1000, replace=TRUE, prob=c(0.5,0.5) )
group_1 <- 1:1000

#put data into a frame
d = data.frame(a,b,c, group, group_1)

d$group = as.factor(d$group)

#place results in table
#final_table = matrix(1, nrow = 10, ncol=10)

e <- d

vec1 <- sample(200:300,2)
vec2 <- sample(400:500,2)
vec3 <- sample(700:800,2)

result_list <- vector("list", length(vec1)*length(vec2)*length(vec3))
result_count <- 0L

for (i in seq_along(vec1)) { 
  for (j in seq_along(vec2)) {
    for (k in seq_along(vec3)) {
      # d <- e
      d$group_2 = as.integer(ifelse(d$group_1 < vec1[i] , 0, ifelse(d$group_1 >vec1[i]  & d$group_1 < vec2[j] , 1, ifelse(d$group_1 >vec2[j]  & d$group_1 < vec3[k] , 2,3))))
      d$group_2 = as.factor(d$group_2)
      
      fitControl <- trainControl(## 10-fold CV
        method = "repeatedcv",
        number = 2,
        ## repeated ten times
        repeats = 1)
      
      TreeFit <- train(group_2 ~ ., data = d[,-5],
                       method = "rpart",
                       trControl = fitControl)
      
      pred <- predict(
        TreeFit,
        d[,-5])
      
      con <- confusionMatrix(
        d$group_2,
        pred) 
      
      #update results into table
      #final_table[i,j] = con$overall[1]
      acc=c(vec1=vec1[i],vec2=vec2[j],vec3=vec3[k],con$overall[1])
      result_count <- result_count + 1L
      result_list[[result_count]] <- acc
      print(acc)
    }
  }
}

(final_table <- do.call(rbind, result_list))