更快 "Resample with replacement by cluster"

Question

我有与相同的问题，即我想进行集群引导。使用 rbindlist(lapply(resampled_ids, function(resampled_id) df[df$id == resampled_id,])) 解决该问题的最佳答案是有效的，但是因为我有一个大数据集，所以这个重采样步骤相当慢。我的问题是，是否可以加快速度？

Answer 1

f = data.frame( id=c(1,1,2,2,2,3,3), X = rnorm(7) )

试试这个：

ind_id <- split(seq_along(f$id), f$id)
samp_id <- sample(names(ind_id), replace = TRUE)
f[unlist(ind_id[samp_id]), ]

Answer 2

使用sequence索引。用更大的 data.frame 进行演示：

df <- data.frame(id = rep.int(1:1e2, sample(100:200, 1e2, replace = TRUE))[1:1e4], X = rnorm(1e4))
resampled_ids <- sample(unique(df$id), replace = TRUE)

idx <- sequence(tabulate(df$id)[resampled_ids], match(unique(df$id), df$id)[resampled_ids])
s <- data.frame(id = df$id[idx], X = df$X[idx])

针对 rbindlist 解决方案进行基准测试：

library(data.table)
library(microbenchmark)

microbenchmark(rbindlist = rbindlist(lapply(resampled_ids, function(x) df[df$id %in% x,])),
               sequence = {idx <- sequence(tabulate(df$id)[resampled_ids], match(unique(df$id), df$id)[resampled_ids])
                           data.frame(id = df$id[idx], X = df$X[idx])})
#> Unit: microseconds
#>       expr    min      lq      mean   median       uq     max neval
#>  rbindlist 9480.4 9921.95 11470.567 10431.05 12555.35 31178.2   100
#>   sequence  406.7  444.55   564.873   498.10   545.70  2818.4   100

请注意，从索引向量创建新的 data.frame 比 row-indexing 原始 data.frame 快得多。如果使用 data.table，差异就不那么明显了，但令人惊讶的是，rbindlist 解决方案变得更慢：

microbenchmark(rbindlist = rbindlist(lapply(resampled_ids, function(x) df[df$id %in% x,])),
               sequence1 = df[sequence(tabulate(df$id)[resampled_ids], match(unique(df$id), df$id)[resampled_ids]),],
               sequence2 = {idx <- sequence(tabulate(df$id)[resampled_ids], match(unique(df$id), df$id)[resampled_ids])
                            data.frame(id = df$id[idx], X = df$X[idx])})
#> Unit: microseconds
#>       expr    min     lq      mean   median       uq     max neval
#>  rbindlist 9431.9 9957.7 11101.545 10508.15 12395.25 15363.3   100
#>  sequence1 4284.5 4550.3  4866.891  4674.80  5009.90  8350.1   100
#>  sequence2  414.1  455.6   541.590   508.40   551.40  2881.1   100

    setDT(df)
    
microbenchmark(rbindlist = rbindlist(lapply(resampled_ids, function(x) df[df$id %in% x,])),
               sequence1 = df[sequence(tabulate(df$id)[resampled_ids], match(unique(df$id), df$id)[resampled_ids]),],
               sequence2 = {idx <- sequence(tabulate(df$id)[resampled_ids], match(unique(df$id), df$id)[resampled_ids])
                            data.table(id = df$id[idx], X = df$X[idx])})
#> Unit: microseconds
#>       expr     min       lq      mean   median      uq     max neval
#>  rbindlist 14877.4 15878.30 17181.572 16348.50 18527.6 22520.9   100
#>  sequence1   795.0  1016.80  1187.266  1101.95  1326.7  2566.5   100
#>  sequence2   386.4   441.75   556.226   473.70   500.9  3373.6   100

更新

解决来自 jay.sf 的评论：

lens <- tabulate(df$id)[resampled_ids]
idx <- sequence(lens, match(unique(df$id), df$id)[resampled_ids])
s <- data.frame(cluster = rep.int(seq_along(resampled_ids), lens), id = df$id[idx], X = df$X[idx])

cluster对应resampled_ids.

的索引

更快 "Resample with replacement by cluster"

Faster "Resample with replacement by cluster"

r

performance

bootstrapping

resampling