可复制地将数据拆分为R中的培训和测试
在R中采样/分割数据的常用方法是使用可复制地将数据拆分为R中的培训和测试,r,cross-validation,sampling,reproducible-research,robustness,R,Cross Validation,Sampling,Reproducible Research,Robustness,在R中采样/分割数据的常用方法是使用sample,例如在行号上。例如: require(data.table) set.seed(1) population <- as.character(1e5:(1e6-1)) # some made up ID names N <- 1e4 # sample size sample1 <- data.table(id = sort(sample(population, N))) # randomly sample N ids
sample
,例如在行号上。例如:
require(data.table)
set.seed(1)
population <- as.character(1e5:(1e6-1)) # some made up ID names
N <- 1e4 # sample size
sample1 <- data.table(id = sort(sample(population, N))) # randomly sample N ids
test <- sample(N-1, N/2, replace = F)
test1 <- sample1[test, .(id)]
[1] 9999
然而,即使我们已经设定了种子,相同的行分割也会产生非常不同的测试集:
test2 <- sample2[test, .(id)]
nrow(test1)
[1] 2653
可以对特定的ID进行采样,但如果忽略或添加了观察结果,这将是不可靠的
有什么方法可以使拆分对数据的更改更具鲁棒性?也就是说,分配测试未更改的观察值,不分配丢弃的观察值,并重新分配新的观察值?使用哈希函数并在其最后一位的mod上采样:
md5_bit_mod <- function(x, m = 2L) {
# Inputs:
# x: a character vector of ids
# m: the modulo divisor (modify for split proportions other than 50:50)
# Output: remainders from dividing the first digit of the md5 hash of x by m
as.integer(as.hexmode(substr(openssl::md5(x), 1, 1)) %% m)
}
[1] 5057
nrow(test1a)
样本量不完全是5000,因为赋值是概率的,但由于大数定律,在大样本中不应该是问题
另见:和
nrow(merge(test1, test2))
md5_bit_mod <- function(x, m = 2L) {
# Inputs:
# x: a character vector of ids
# m: the modulo divisor (modify for split proportions other than 50:50)
# Output: remainders from dividing the first digit of the md5 hash of x by m
as.integer(as.hexmode(substr(openssl::md5(x), 1, 1)) %% m)
}
test1a <- sample1[md5_bit_mod(id) == 0L, .(id)]
test2a <- sample2[md5_bit_mod(id) == 0L, .(id)]
nrow(merge(test1a, test2a))
nrow(test1a)