我试图划分，我在R，2/3用于训练和1/3用于测试的数据集。 我有一个分类变量，和七个数值变量。 每个观测被分类为A，B，C或D.
为简单起见，假设该分类变量，CL，是一种用于第100个观测，B为观察101到200，C 300为止，和d直到400我试图让有2/3分区的每个A，B，C，和d的观测（而不是简单地对获得的整个数据集，因为它可能会不具有每个分类的等量的意见的2/3）。
当尝试从数据，例如子集采样sample(subset(data, cl=='A'))中，列被重新排序，而不是行。
总之，我的目标是让每个A，B，C，和d为我的训练数据67个随机观察，并存储剩余33个观察每个A，B，C，和d作为测试数据。 我发现我的一个非常类似的问题，但它并没有在多个变量因素。
这可能会更长，但我认为它更直观，可以在基础R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$cl ,
# use the sample function within
# each classification variable group
sample ,
# send the size = 67 parameter
# through to the sample() function
size = 67
)
# convert your list back to a numeric vector
tr <- unlist( training.rows )
# split your original data frame into two:
# all the records sampled as training rows
training.df <- x[ tr , ]
# all other records (NOT sampled as training rows)
testing.df <- x[ -tr , ]
实际上有一个用于处理机器学习问题的一个很好的包插入符号 ，它包含一个函数createDataPartition（），它几乎做到这一点取样从供给因素的每个级别2 /三分之二：
#2/3rds for training
library(caret)
inTrain = createDataPartition(df$yourFactor, p = 2/3, list = FALSE)
dfTrain=df[inTrain,]
dfTest=df[-inTrain,]
下面将增加一个set列与价值观"train"或"test"你的data.frame：
library(plyr)
df <- ddply(df, "cl", transform, set = sample(c("train", "test"), length(cl),
replace = TRUE, prob = c(2, 1)))
你可以使用基本类似的ave功能，但我发现ddply很干净（读取）这个特殊用法。
然后，您可以使用您的数据拆分subset功能：
train.data <- subset(df, set == "train")
test.data <- subset(df, set == "test")
追问：每个组拆分成完全相同2/3和1/3的大小，你可以使用：
df <- ddply(df, "cl", transform,
set = sample(c(rep("train", round(2/3 * length(cl)),
rep("test", round(1/3 * length(cl)))))
碰到了这个问题，而构建自己的功能与多种因素分层交叉验证的数据分区。 你可以通过将数据分成3（或N）个大小相等的部分，而将观测值的每个阶层内平均的部分，然后选择三分之一作为测试集，然后结合其余作为训练集构建这样的数据集。 我会处理，如R. 列表元素
这里是我使用基础包支持多个分层因素中，表示为的是要具有与地层（mtcars数据集的例子）的列数或场的列名内置的功能。 我认为这是在功能，而类似ddply，不同之处在于，你也可以使用列数和所产生的子集列表中给出：
# 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; 在这个假设的例子人愿意为“VS”到所有的垃圾箱中平等地表示和“AM”的因素：
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")]))
在讨论的是，与安东尼的回答data.frame这种特定的数据集：
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
