从数据帧分层随机抽样

Question

我有一个格式的数据框:

head(subset)
# ants  0 1 1 0 1 
# age   1 2 2 1 3
# lc    1 1 0 1 0

我需要根据年龄和lc创建带有随机样本的新数据框.例如,我想要30个年龄的样本:1和lc:1,30个样本来自年龄:1和lc:0等.

我确实看过随机抽样方法;

newdata <- function(subset, age, 30)

但这不是我想要的代码.

Answer 1

我建议使用stratified我的"splitstackshape"包或sample_n"dplyr"包:

## Sample data
set.seed(1)
n <- 1e4
d <- data.table(age = sample(1:5, n, T), 
                lc = rbinom(n, 1 , .5),
                ants = rbinom(n, 1, .7))
# table(d$age, d$lc)

对于stratified,您基本上指定数据集,分层列和表示每个组所需大小的整数或表示要返回的分数的小数(例如,.1表示每组的10%).

library(splitstackshape)
set.seed(1)
out <- stratified(d, c("age", "lc"), 30)
head(out)
#    age lc ants
# 1:   1  0    1
# 2:   1  0    0
# 3:   1  0    1
# 4:   1  0    1
# 5:   1  0    0
# 6:   1  0    1

table(out$age, out$lc)
#    
#      0  1
#   1 30 30
#   2 30 30
#   3 30 30
#   4 30 30
#   5 30 30

对于sample_n首先要创建一个分组表(使用group_by),然后指定想要观测次数.如果你想要比例采样,你应该使用sample_frac.

library(dplyr)
set.seed(1)
out2 <- d %>%
  group_by(age, lc) %>%
  sample_n(30)

# table(out2$age, out2$lc)

Answer 2

这是一些数据:

set.seed(1)
n <- 1e4
d <- data.frame(age = sample(1:5,n,TRUE), 
                lc = rbinom(n,1,.5),
                ants = rbinom(n,1,.7))

您需要一个拆分应用组合策略,您split可以d在其中使用data.frame(在此示例中),对每个子样本中的行/观察进行采样,然后将其组合在一起rbind.以下是它的工作原理:

sp <- split(d, list(d$age, d$lc))
samples <- lapply(sp, function(x) x[sample(1:nrow(x), 30, FALSE),])
out <- do.call(rbind, samples)

结果:

> str(out)
'data.frame':   300 obs. of  3 variables:
 $ age : int  1 1 1 1 1 1 1 1 1 1 ...
 $ lc  : int  0 0 0 0 0 0 0 0 0 0 ...
 $ ants: int  1 1 0 1 1 1 1 1 1 1 ...
> head(out)
         age lc ants
1.0.2242   1  0    1
1.0.4417   1  0    1
1.0.389    1  0    0
1.0.4578   1  0    1
1.0.8170   1  0    1
1.0.5606   1  0    1

Answer 3

请参阅strata包装采样中的功能.该函数选择分层简单随机抽样并作为结果给出样本.添加了额外的两列 - 包含概率(Prob)和分层指示符(Stratum).查看示例.

require(data.table)
require(sampling)

set.seed(1)
n <- 1e4
d <- data.table(age = sample(1:5, n, T), 
                lc = rbinom(n, 1 , .5),
                ants = rbinom(n, 1, .7))

# Sort
setkey(d, age, lc)

# Population size by strata
d[, .N, keyby = list(age, lc)]
#     age lc    N
#  1:   1  0 1010
#  2:   1  1 1002
#  3:   2  0  993
#  4:   2  1 1026
#  5:   3  0 1021
#  6:   3  1  982
#  7:   4  0  958
#  8:   4  1  940
#  9:   5  0 1012
# 10:   5  1 1056

# Select sample
set.seed(2)
s <- data.table(strata(d, c("age", "lc"), rep(30, 10), "srswor"))

# Sample size by strata
s[, .N, keyby = list(age, lc)]
#     age lc  N
#  1:   1  0 30
#  2:   1  1 30
#  3:   2  0 30
#  4:   2  1 30
#  5:   3  0 30
#  6:   3  1 30
#  7:   4  0 30
#  8:   4  1 30
#  9:   5  0 30
# 10:   5  1 30

Answer 4

这是一个单行使用data.table：

set.seed(1)
n <- 1e4
d <- data.table(age  = sample(1:5, n, T),
                lc   = rbinom(n,   1, .5),
                ants = rbinom(n,   1, .7))

out <- d[, .SD[sample(1:.N, 30)], by=.(age, lc)]

# Check
out[, table(age, lc)]
##    lc
## age  0  1
##   1 30 30
##   2 30 30
##   3 30 30
##   4 30 30
##   5 30 30