[SOLVED] Faster "Resample with replacement by cluster"

Issue

I have the same question as Resample with replacement by cluster, i.e. I want to do cluster bootstrapping. The best answer’s approach to that question using rbindlist(lapply(resampled_ids, function(resampled_id) df[df$id == resampled_id,])) works, but because I have a big dataset, this resampling step is rather slow. My question is, is it possible to speed this up?

Solution

Use sequence to index. Demonstrated with a larger data.frame:

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])

Benchmarking against the rbindlist solution:

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

Note that creating a new data.frame from indexed vectors is much faster than row-indexing the original data.frame. The difference is much less pronounced if a data.table is used, but, surprisingly, the rbindlist solution becomes even slower:

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

Update

To address the comment from jay.sf:

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 corresponds to the index of resampled_ids.

Answered By – jblood94

Answer Checked By – Candace Johnson (BugsFixing Volunteer)

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