R：如何將數據幀分成訓練，驗證和測試集？

Question

我使用R鍵做機器學習。根據標準的機器學習方法，我想我的數據隨機分成培訓，驗證和測試數據集。我如何在R中做到這一點？

我知道如何分解成2個數據集（例如post）有一些相關的問題，但不清楚如何爲3個分割數據集執行此操作。順便說一句，正確的方法是用3個數據集（包括驗證組來調整你的超參數）。

Answer 1

遵循本post所示的方法，在這裏工作R代碼裏面的數據幀分成了測試，驗證和測試三個新dataframes。三個子集不重疊。

# Create random training, validation, and test sets 

# Set some input variables to define the splitting. 
# Input 1. The data frame that you want to split into training, validation, and test. 
df <- mtcars 

# Input 2. Set the fractions of the dataframe you want to split into training, 
# validation, and test. 
fractionTraining <- 0.60 
fractionValidation <- 0.20 
fractionTest  <- 0.20 

# Compute sample sizes. 
sampleSizeTraining <- floor(fractionTraining * nrow(df)) 
sampleSizeValidation <- floor(fractionValidation * nrow(df)) 
sampleSizeTest  <- floor(fractionTest  * nrow(df)) 

# Create the randomly-sampled indices for the dataframe. Use setdiff() to 
# avoid overlapping subsets of indices. 
indicesTraining <- sort(sample(seq_len(nrow(df)), size=sampleSizeTraining)) 
indicesNotTraining <- setdiff(seq_len(nrow(df)), indicesTraining) 
indicesValidation <- sort(sample(indicesNotTraining, size=sampleSizeValidation)) 
indicesTest  <- setdiff(indicesNotTraining, indicesValidation) 

# Finally, output the three dataframes for training, validation and test. 
dfTraining <- df[indicesTraining, ] 
dfValidation <- df[indicesValidation, ] 
dfTest  <- df[indicesTest, ] 

Answer 2

對於兩個組（使用floor）的這種鏈接方法沒有自然延伸到三。我會做

spec = c(train = .6, test = .2, validate = .2) 

g = sample(cut(
    seq(nrow(df)), 
    nrow(df)*cumsum(c(0,spec)), 
    labels = names(spec) 
)) 

res = split(df, g) 

檢查結果：

sapply(res, nrow)/nrow(df) 
# train  test validate 
# 0.59375 0.18750 0.21875 
# or... 
addmargins(prop.table(table(g))) 
# train  test validate  Sum 
# 0.59375 0.18750 0.21875 1.00000 

隨着set.seed(1)運行之前，結果看起來像

$train 
        mpg cyl disp hp drat wt qsec vs am gear carb 
Mazda RX4   21.0 6 160.0 110 3.90 2.620 16.46 0 1 4 4 
Mazda RX4 Wag  21.0 6 160.0 110 3.90 2.875 17.02 0 1 4 4 
Datsun 710  22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 
Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2 
Merc 240D   24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2 
Merc 230   22.8 4 140.8 95 3.92 3.150 22.90 1 0 4 2 
Merc 280   19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 
Merc 280C   17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4 
Merc 450SE  16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3 
Merc 450SL  17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 
Merc 450SLC  15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3 
Fiat 128   32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1 
Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1 
Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 
AMC Javelin  15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2 
Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2 
Fiat X1-9   27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1 
Porsche 914-2  26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2 
Volvo 142E  21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2 

$test 
        mpg cyl disp hp drat wt qsec vs am gear carb 
Valiant   18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1 
Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4 
Toyota Corona  21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 
Camaro Z28   13.3 8 350.0 245 3.73 3.840 15.41 0 0 3 4 
Ford Pantera L  15.8 8 351.0 264 4.22 3.170 14.50 0 1 5 4 
Ferrari Dino  19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6 

$validate 
        mpg cyl disp hp drat wt qsec vs am gear carb 
Hornet 4 Drive  21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1 
Duster 360   14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4 
Lincoln Continental 10.4 8 460.0 215 3.00 5.424 17.82 0 0 3 4 
Chrysler Imperial 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 
Honda Civic   30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2 
Lotus Europa  30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2 
Maserati Bora  15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8 

Data.frames可以像res$test或res[["test"]]訪問。

cut是基於股劃分的標準工具。

Answer 3

我覺得我的做法是最簡單的一個：

idxTrain <- sample(nrow(dat),as.integer(nrow(dat)*0.7)) 
idxNotTrain <- which(! 1:nrow(dat) %in% idxTrain) 
idxVal <- sample(idxNotTrain,as.integer(length(idxNotTrain)*0.333)) 
idxTest <- idxNotTrain[which(! idxNotTrain %in% idxVal)] 

首先，將數據分成70％的訓練數據，其餘（idxNotTrain）。然後，其餘部分再次被分成驗證數據集（33％，總數據的10％）和其餘部分（測試數據，66％，總數據的20％）。

Answer 4

其中一些看起來過於複雜，這裏有一個簡單的方法，使用樣本將任何數據集分成3個或甚至任意數量的集合。

# Simple into 3 sets. 
idx <- sample(seq(1, 3), size = nrow(iris), replace = TRUE, prob = c(.8, .2, .2)) 
train <- iris[idx == 1,] 
test <- iris[idx == 2,] 
cal <- iris[idx == 3,] 

如果你寧願重用代碼：

# Or a function to split into arbitrary number of sets 
test_split <- function(df, cuts, prob, ...) 
{ 
    idx <- sample(seq(1, cuts), size = nrow(df), replace = TRUE, prob = prob, ...) 
    z = list() 
    for (i in 1:cuts) 
    z[[i]] <- df[idx == i,] 
    z 
} 
z <- test_split(iris, 4, c(0.7, .1, .1, .1)) 

train <- z[1] 
test <- z[2] 
cal <- z[3] 
other <- z[4] 

Answer 5

這裏是一個解決方案可以與60，20，20分也確保沒有重疊。然而，適應分裂是個麻煩。如果有人可以幫助我，我很感激它

# Draw a random, stratified sample including p percent of the data  
    idx.train <- createDataPartition(y = known$return_customer, p = 0.8, list = FALSE) 
    train <- known[idx.train, ] # training set with p = 0.8 
    # test set with p = 0.2 (drop all observations with train indeces) 
    test <- known[-idx.train, ] 
    idx.validation <- createDataPartition(y = train$return_customer, p = 0.25, list = FALSE) # Draw a random, stratified sample of ratio p of the data 
    validation <- train[idx.validation, ] #validation set with p = 0.8*0.25 = 0.2 
    train60 <- train[-idx.validation, ] #final train set with p= 0.8*0.75 = 0.6 

Answer 6

讓我知道這是否會工作。只是一個簡化版本

sample_train<- sample(seq_len(nrow(mtcars)), size = floor(0.60*nrow(mtcars))) 
sample_valid<- sample(seq_len(nrow(mtcars)), size = floor(0.20*nrow(mtcars))) 
sample_test <- sample(seq_len(nrow(mtcars)), size = floor(0.20*nrow(mtcars))) 

train  <- mtcars[sample_train, ] 
validation<- mtcars[sample_valid, ] 
test  <- mtcars[sample_test, ]