我有一个与Resample相同的问题，即通过集群进行替换，即我想进行集群引导。使用rbindlist(lapply(resampled_ids, function(resampled_id) df[df$id == resampled_id,]))解决这个问题的最佳方法是有效的，但由于我有一个大的数据集，所以这个重新采样步骤相当缓慢。我的问题是，有可能加快速度吗？
使用sequence进行索引。用较大的data.frame:演示
df <- data.frame(id = rep.int(1:1e2, sample(100:200, 1e2, replace = TRUE))[1:1e4], X = rnorm(1e4))
resampled_ids <- sample(unique(df$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$X[idx])
针对rbindlist解决方案的基准测试：
library(data.table)
library(microbenchmark)
microbenchmark(rbindlist = rbindlist(lapply(resampled_ids, function(x) df[df$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$X[idx])})
#> Unit: microseconds
#> expr min lq mean median uq max neval
#> rbindlist 9480.4 9921.95 11470.567 10431.05 12555.35 31178.2 100
#> sequence 406.7 444.55 564.873 498.10 545.70 2818.4 100
注意，从索引向量创建新的data.frame比对原始data.frame进行行索引快得多。如果使用data.table，差异就不那么明显了，但令人惊讶的是，rbindlist解决方案变得更慢：
microbenchmark(rbindlist = rbindlist(lapply(resampled_ids, function(x) df[df$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$X[idx])})
#> Unit: microseconds
#> expr min lq mean median uq max neval
#> rbindlist 14877.4 15878.30 17181.572 16348.50 18527.6 22520.9 100
#> sequence1 795.0 1016.80 1187.266 1101.95 1326.7 2566.5 100
#> sequence2 386.4 441.75 556.226 473.70 500.9 3373.6 100
更新
回复jay.sf:的评论
lens <- tabulate(df$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])
CCD_ 9对应于CCD_。
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]), ]