我正在尝试对 R 中的数据集进行分区，2/3 用于训练，1/3 用于测试。我有一个分类变量和七个数值变量。每个观测值都被分类为 A、B、C 或 D。
为简单起见，假设分类变量 cl 是前 100 个观测值的 A，观测值 101 到 200 的 B，C 到 300，D 到 400。我试图得到一个分区A、B、C 和 D 中每一个的 2/3 的观察值（而不是简单地获得整个数据集的 2/3 的观察值，因为每个分类的数量可能不相等）。
当我尝试从数据子集（例如 sample(subset(data, cl=='A'))）中采样时，会重新排序列而不是行。
总而言之，我的目标是从 A、B、C 和 D 中的每一个中获得 67 个随机观察作为我的训练数据，并将 A、B、C 和 D 中每个的剩余 33 个观察作为测试数据存储.我发现了一个与我非常相似的问题，但它没有考虑多个变量。
【问题讨论】：
标签： r random partitioning
实际上有一个很好的包 caret 用于处理机器学习问题，它包含一个函数 createDataPartition() 几乎可以从每个级别采样 2/3提供的因素：
#2/3rds for training
library(caret)
inTrain = createDataPartition(df$yourFactor, p = 2/3, list = FALSE)
dfTrain=df[inTrain,]
dfTest=df[-inTrain,]
【讨论】：
这可能会更长，但我认为它更直观，可以在基础 R 中完成；)
# create the data frame you've described
x <-
data.frame(
cl =
c(
rep( 'A' , 100 ) ,
rep( 'B' , 100 ) ,
rep( 'C' , 100 ) ,
rep( 'D' , 100 )
) ,
othernum1 = rnorm( 400 ) ,
othernum2 = rnorm( 400 ) ,
othernum3 = rnorm( 400 ) ,
othernum4 = rnorm( 400 ) ,
othernum5 = rnorm( 400 ) ,
othernum6 = rnorm( 400 ) ,
othernum7 = rnorm( 400 )
)
# sample 67 training rows within classification groups
training.rows <-
tapply(
# numeric vector containing the numbers
# 1 to nrow( x )
1:nrow( x ) ,
# break the sample function out by
# the classification variable
x$cl ,
# use the sample function within
# each classification variable group
sample ,
# send the size = 67 parameter
# through to the sample() function
size = 67
)
# convert your list back to a numeric vector
tr <- unlist( training.rows )
# split your original data frame into two:
# all the records sampled as training rows
training.df <- x[ tr , ]
# all other records (NOT sampled as training rows)
testing.df <- x[ -tr , ]
【讨论】：
以下将在您的 data.frame 中添加一个带有值 "train" 或 "test" 的 set 列：
library(plyr)
df <- ddply(df, "cl", transform, set = sample(c("train", "test"), length(cl),
replace = TRUE, prob = c(2, 1)))
您可以使用基本的ave 函数获得类似的东西，但我发现ddply 对于这种特殊用途非常干净（可读）。
然后您可以使用subset 函数拆分数据：
train.data <- subset(df, set == "train")
test.data <- subset(df, set == "test")
跟进：要将每个组精确地分成 2/3 和 1/3 大小，您可以使用：
df <- ddply(df, "cl", transform,
set = sample(c(rep("train", round(2/3 * length(cl)),
rep("test", round(1/3 * length(cl)))))
【讨论】：
在构建我自己的函数以使用多个分层因素进行交叉验证的数据分区时遇到了这个问题。您可以通过将数据分成 3 个（或 N 个）大小相等的部分来构建此类数据集，同时将每个层内的观察值平均划分为这些部分，然后选择三分之一作为测试集，然后将其余部分组合为训练集。我会处理 R 中的诸如 list 元素。
这是我使用支持多个分层因素的基本包构建的一个函数，表示为您希望作为分层的字段的列号或列名（mtcars 数据集示例）。我认为它在功能上与 ddply 非常相似，除了您还可以使用列号并且生成的子集在 list 中给出：
# Function that partitions data into a number of equally (or almost-equally) sized bins that do not overlap, and returns the data bins as a list
# Useful for cross validation
partition_data <- function(
# Data frame to partition (default example: mtcars data, assuming rows correspond to observations)
dat = mtcars,
# Number of equally sized bins to partition to (default here: 2 bins)
bins = 2,
# Stratification element, homogeneous subpopulations according to a column that should be subsampled,
# Observations within a substrata are divided equally to the partitioned bins
stratum = NA
){
# Total number of observations
nobs <- dim(dat)[1]
# Allocation vector, to be used for randomly distributing the samples to the bins
loc <- rep(1:bins, times=ceiling(nobs/bins))[1:nobs]
# If the dataset is stratified, each subpopulation is distributed equally to the bins, otherwise the whole population is the "subpopulation"
if(missing(stratum)){
pops <- list(sample(1:dim(dat)[1]))
}else{
uniqs <- na.omit(as.matrix(unique(dat[,stratum])))
pops <- list()
for(i in 1:nrow(uniqs)){
# If some of the stratified fields include NA-values, these will not be included in the sampling
w <- apply(as.matrix(dat[,stratum]), MARGIN=1, FUN=function(x) all(x==uniqs[i,]))
pops[[i]] <- sample(which(w))
}
}
indices <- vector(length=nobs)
# Assign the group indices according to permutated samples within each subpopulation
indices[unlist(pops)] <- loc
# Assign observations to separate locations in a list
partitioned_data <- lapply(unique(indices), FUN=function(x) dat[x==indices,])
# Return the result
partitioned_data
}
它是如何工作的示例；在这个假设的例子中，人们希望因子“vs”和“am”在所有的 bin 中得到同样的表示：
set.seed(1)
# Stratified sampling, so that combinations of binary covariates vs = {0,1} & am = {0,1} appear equally over the randomized bins of data
pt <- partition_data(mtcars, stratum=c("vs", "am"), bins=3)
# Instances are distributed equally
lapply(pt, FUN=function(x) table(x[,c("vs","am")]))
#> lapply(pt, FUN=function(x) table(x[,c("vs","am")]))
#[[1]]
# am
#vs 0 1
# 0 4 2
# 1 3 2
#
#[[2]]
# am
#vs 0 1
# 0 4 2
# 1 2 3
#
#[[3]]
# am
#vs 0 1
# 0 4 2
# 1 2 2
# 10 or 11 samples (=rows) per partition of data (data had 11 columns)
lapply(pt, FUN=dim)
# Training set containing 2/3 of the stratified samples
# Constructed by dropping out the first third of samples
train <- do.call("rbind", pt[-1])
# Test set containing the remaining 1/3
test <- pt[[1]]
# 21 samples in training dataset
print(dim(train))
# 11 samples in testing dataset
print(dim(test))
> print(train)
mpg cyl disp hp drat wt qsec vs am gear carb
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 4 Drive 21.4 6 258.0 110 3.08 3.215 19.44 1 0 3 1
Merc 450SE 16.4 8 275.8 180 3.07 4.070 17.40 0 0 3 3
Cadillac Fleetwood 10.4 8 472.0 205 2.93 5.250 17.98 0 0 3 4
Fiat 128 32.4 4 78.7 66 4.08 2.200 19.47 1 1 4 1
Toyota Corona 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1
Dodge Challenger 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2
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
Volvo 142E 21.4 4 121.0 109 4.11 2.780 18.60 1 1 4 2
Hornet Sportabout 18.7 8 360.0 175 3.15 3.440 17.02 0 0 3 2
Duster 360 14.3 8 360.0 245 3.21 3.570 15.84 0 0 3 4
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 450SLC 15.2 8 275.8 180 3.07 3.780 18.00 0 0 3 3
Honda Civic 30.4 4 75.7 52 4.93 1.615 18.52 1 1 4 2
Pontiac Firebird 19.2 8 400.0 175 3.08 3.845 17.05 0 0 3 2
Porsche 914-2 26.0 4 120.3 91 4.43 2.140 16.70 0 1 5 2
Lotus Europa 30.4 4 95.1 113 3.77 1.513 16.90 1 1 5 2
Ferrari Dino 19.7 6 145.0 175 3.62 2.770 15.50 0 1 5 6
> print(test)
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
Valiant 18.1 6 225.0 105 2.76 3.460 20.22 1 0 3 1
Merc 240D 24.4 4 146.7 62 3.69 3.190 20.00 1 0 4 2
Merc 280C 17.8 6 167.6 123 3.92 3.440 18.90 1 0 4 4
Merc 450SL 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3
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
Toyota Corolla 33.9 4 71.1 65 4.22 1.835 19.90 1 1 4 1
AMC Javelin 15.2 8 304.0 150 3.15 3.435 17.30 0 0 3 2
Fiat X1-9 27.3 4 79.0 66 4.08 1.935 18.90 1 1 4 1
Maserati Bora 15.0 8 301.0 335 3.54 3.570 14.60 0 1 5 8
# Example of sampling without stratification; the binary covariates 'vs' and 'am' are probably not distributed equally over the bins
lapply(pt2 <- partition_data(mtcars, bins=3), FUN=function(x) table(x[,c("vs","am")]))
# Stratified according to a single covariate (cylinders)
lapply(pt3 <- partition_data(mtcars, stratum="cyl", bins=3), FUN=function(x) table(x[,c("cyl")]))
在讨论过的这个特定数据集中，使用 Anthony 回答中的 data.frame：
xpt <- partition_data(x, stratum="cl", bins=3)
# Same as:
#xpt <- partition_data(x, stratum=1, bins=3)
train_xpt <- do.call("rbind", xpt[-1])
test_xpt <- xpt[[1]]
#> summary(train_xpt[,"cl"])
# A B C D
#67 66 67 67
#> summary(test_xpt[,"cl"])
# A B C D
#33 34 33 33
【讨论】：