1

我想有效地从 a 中按组随机抽样data.table，但应该可以为每组抽样不同的比例。

如果我想从每个组中抽取分数，我可以从这个问题和相关sampling_fraction答案中得到启发，做一些类似的事情：

DT = data.table(a = sample(1:2), b = sample(1:1000,20))

group_sampler <- function(data, group_col, sample_fraction){
  # this function samples sample_fraction <0,1> from each group in the data.table
  # inputs:
  #   data - data.table
  #   group_col - column(s) used to group by
  #   sample_fraction - a value between 0 and 1 indicating what % of each group should be sampled
  data[,.SD[sample(.N, ceiling(.N*sample_fraction))],by = eval(group_col)]
}

# what % of data should be sampled
sampling_fraction = 0.5

# perform the sampling
sampled_dt <- group_sampler(DT, 'a', sampling_fraction)

但是，如果我想从第 1 组中抽取 10%，从第 2 组中抽取 50%，该怎么办？

4

2 回答 2

4

您可以使用.GRPbut 来确保匹配正确的组。您可能希望将其定义group_col为因子变量。

group_sampler <- function(data, group_col, sample_fractions) {
  # this function samples sample_fraction <0,1> from each group in the data.table
  # inputs:
  #   data - data.table
  #   group_col - column(s) used to group by
  #   sample_fraction - a value between 0 and 1 indicating what % of each group should be sampled
  stopifnot(length(sample_fractions) == uniqueN(data[[group_col]]))
  data[, .SD[sample(.N, ceiling(.N*sample_fractions[.GRP]))], keyby = group_col]
}

针对 chinsoon12 的评论进行编辑：

拥有函数的最后一行会更安全（而不是依赖于正确的顺序）：

data[, .SD[sample(.N, ceiling(.N*sample_fractions[[unlist(.BY)]]))], keyby = group_col]

然后你sample_fractions作为一个命名向量传递：

group_sampler(DT, 'a', sample_fractions= c(x = 0.1, y = 0.9))
于 2019-10-15T13:33:38.217 回答
0

这是一个使用查找表的选项（因此不依赖于向量或组的顺序）。

library(data.table)
DT = data.table(group = sample(1:2), val = sample(1:1000,20))

sample_props <- data.table(group = 1:2, prop = c(.1,.5))

group_sampler <- function(data, group_col, sample_props){
  # this function samples sample_fraction <0,1> from each group in the data.table
  # inputs:
  #   data - data.table with data
  #   group_col - column(s) used to group by (must be in both data.tables)
  #   sample_props - data.table with sample proportions
  ret <- merge(DT, sample_props, by = group_col)
  ret <- ret[,.SD[sample(.N, ceiling(.N*prop))], eval(group_col)]
  return(ret[,prop := NULL][])
}

# perform the sampling
group_sampler(DT, 'group', sample_props)
#>    group val
#> 1:     1 721
#> 2:     2 542
#> 3:     2 680
#> 4:     2 613
#> 5:     2 170
#> 6:     2 175

由reprex 包（v0.3.0）于 2019 年 10 月 15 日创建

于 2019-10-15T13:38:50.210 回答