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。每个观察结果分为A，B，C或D.
为了简单起见，我们假设分类变量cl对于前100次观测是A，对于观察101到200是C，C到300，D到400.我正在尝试获得一个分区A / B，C和D中每一个的观察结果的2/3（而不是简单地得到整个数据集的2/3的观察结果，因为它可能没有相同数量的每个分类）。
当我尝试从数据的子集（例如sample(subset(data, cl=='A'))）进行采样时，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中的每一个随机观察67个作为我的训练数据，并将A，B，C和D中的每一个的剩余33个观测值存储为测试数据。我发现了一个与我非常相似的问题，但它没有考虑到多个变量。
答案 0 :(得分：17)
实际上有一个很好的软件包插入符用于处理机器学习问题，它包含一个函数 createDataPartition（），几乎从每个级别抽取2 / 3rds提供的因素：
#2/3rds for training
library(caret)
inTrain = createDataPartition(df$yourFactor, p = 2/3, list = FALSE)
dfTrain=df[inTrain,]
dfTest=df[-inTrain,]
答案 1 :(得分：5)
这可能会更长，但我认为它更直观，可以在基础R中完成;）
# create the data frame you've described
x <-
data.frame(
cl =
c(
rep( 'A' , 100 ) ,
rep( 'B' , 100 ) ,
rep( 'C' , 100 ) ,
rep( 'D' , 100 )
) ,
othernum1 = rnorm( 400 ) ,
othernum2 = rnorm( 400 ) ,
othernum3 = rnorm( 400 ) ,
othernum4 = rnorm( 400 ) ,
othernum5 = rnorm( 400 ) ,
othernum6 = rnorm( 400 ) ,
othernum7 = rnorm( 400 )
)
# sample 67 training rows within classification groups
training.rows <-
tapply(
# numeric vector containing the numbers
# 1 to nrow( x )
1:nrow( x ) ,
# break the sample function out by
# the classification variable
x$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 , ]
答案 2 :(得分：4)
以下内容会为您的data.frame添加值为set或"train"的{{1}}列：
"test"您可以使用基本library(plyr)
df <- ddply(df, "cl", transform, set = sample(c("train", "test"), length(cl),
replace = TRUE, prob = c(2, 1)))
函数获得类似的东西，但我发现ave非常干净（可读）用于此特定用法。
然后，您可以使用ddply功能分割数据：
subset跟进：要将每个组精确地分成2/3和1/3大小，您可以使用：
train.data <- subset(df, set == "train")
test.data <- subset(df, set == "test")
答案 3 :(得分：1)
在构建我自己的函数时遇到了这个问题，该函数用于分割数据以进行交叉验证，并具有多个分层因子。您可以通过将数据划分为3个（或N个）相等大小的部分来构建此类数据集，同时将每个层次中的观察值等分为这些部分，然后选择三分之一作为测试集合，然后将休息结合为训练集合。我会在R。
中处理诸如 list 元素这是我使用支持多个分层因子的基础包构建的函数，表示为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，但您也可以使用列号，并且生成的子集在列表中给出：
# 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
}
它是如何工作的例子;在这个假设的例子中，人们希望因子'vs'和'am'在所有箱子中均等地表示：
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")]))
在这个讨论过的特定数据集中，使用了Anthony的答案中的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