R:如何将数据框分成训练,验证和测试集?

时间:2016-03-17 18:34:42

标签: r machine-learning training-data

我正在使用R来进行机器学习。遵循标准的机器学习方法,我想将我的数据随机分成训练,验证和测试数据集。我如何在R?中做到这一点?

我知道有一些关于如何分成2个数据集的相关问题(例如这个post),但是如何对3个分割数据集进行分析并不明显。顺便说一句,正确的方法是使用3个数据集(包括验证集来调整超参数)。

7 个答案:

答案 0 :(得分:13)

这两个组的链接方法(使用floor)自然不会扩展到三个。我会做

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)

检查结果:

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 

set.seed(1)之前运行,结果看起来像

$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

可以像res$testres[["test"]]一样访问Data.frame。

cut是基于共享进行分区的标准工具。

答案 1 :(得分:6)

遵循此post中显示的方法,这里使用R代码将数据帧划分为三个新的数据帧,以进行测试,验证和测试。这三个子集不重叠。

# 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, ]

答案 2 :(得分:4)

其中一些似乎过于复杂,这里使用样本将任何数据集拆分为3个甚至任意数量的集合都是一种简单的方法。

# 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,]

如果您更愿意使用可重复使用的代码:

# 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]

答案 3 :(得分:0)

这是一个60,20,20分割的解决方案,也确保没有重叠。然而,适应分裂是一个麻烦。如果有人能帮助我,我很感激

   # 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

答案 4 :(得分:0)

Caret 还支持使用函数 createDataPartition

进行数据拆分

如果您的结果 y 是不平衡因素(是 >>> 否或否 >>> 是) 随机抽样发生在每个类内,并应保留数据的整体类分布。

示例:

library(caret)
set.seed(123)
table(iris$Species=="setosa")
## 
## FALSE  TRUE 
##   100    50

注意我们的结果是不平衡的

分裂

trainIndex <- createDataPartition(iris$Species=="setosa", p = .8, 
                                  list = FALSE, 
                                  times = 1)
train = iris[ trainIndex,]
test = iris[-trainIndex,]

验证

table(train$Species == "setosa")

## 
## FALSE  TRUE 
##    80    40
table(test$Species == "setosa")
## 
## FALSE  TRUE 
##    20    10

注意我们保留了整体的类分布

答案 5 :(得分:-1)

我认为我的方法是最简单的方法:

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,它将数据分成70%的训练数据和其余的(idxNotTrain)。 然后,其余部分再次分为验证数据集(33%,总数据的10%)和其余部分(测试数据,66%,总数据的20%)。

答案 6 :(得分:-2)

如果这样可行,请告诉我。只是简化版

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, ]