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r tidyr

Answers (4)

July 21, 2026 Score: 2 Rep: 79,216 Quality: Medium Completeness: 80%

The solutions below are 2 orders of magnitude faster and need much less memory.

original  
        bindrows(dftoadd)
    }
  ) |> 
    bindrows() |> 
    select(-z) |> 
    dropna()
}
rui1  dropna()
rui2  na.omit()
bench::mark(original(), rui1(), rui2(), col(), check = FALSE)
#> # A tibble: 4 × 6
#>   expression      min   median itr/sec memalloc gc/sec
#>               
#> 1 original()   14.5ms   15.3ms      62.1    1.99MB     9.55
#> 2 rui1()      314.9µs  353.8µs    2590.    47.66KB     8.43
#> 3 rui2()      201.2µs    238µs    3712.    26.06KB    10.4 
library(microbenchmark)
microbenchmark(original(), rui1(), rui2()) |> print(order = "median")
#> Unit: microseconds
#>        expr     min       lq      mean   median       uq     max neval cld
#>      rui2()   224.5   275.40   325.794   306.30   346.75   683.5   100   b
#>      rui1()   334.9   363.35   465.796   398.15   462.80  3173.3   100   b
#>  original() 13254.1 14440.80 16645.104 14879.60 17207.40 66733.8   100  a

Created on 2026-07-21 with reprex v2.1.1

When run outside reprex I get much less memory use, rui1 only 216B and rui2 0B.

I also tried timing with bigger data and the results were comparable to the results above. For instance, with a data set with 5000+ rows.

exbackup
July 21, 2026 Score: 2 Rep: 13,440 Quality: Medium Completeness: 60%

You can unnest and preserve the row_id. Then set ord = 1 for the unnested, and ord = 0 for the normal columns, then order by both. For 500k rows, this takes ~80 ms.

library(tibble)
ex = tibble(
  x = c("A", "B", "C", "D", "E"),
  y = 1:5,
  z = list(tibble(x = "A2", y = 1.5), NA, NA, tibble(x = "D2", y =  4.5), NA)
)
ex2
July 21, 2026 Score: 2 Rep: 106,421 Quality: Low Completeness: 40%

You can try

id
July 21, 2026 Score: 2 Rep: 7,578 Quality: Low Completeness: 50%

using {data.table}:

library(data.table)
ex  setkey(id)
datatablesolution 
        ## rowbind them all:
        rbindlist() |>
        ## order by row id (tibbles coming last):
        [order(id), -c("id")]
}

somewhat slower than Rui Barradas' solutions:

+ Unit: nanoseconds
               expr min lq  mean median uq max neval
               rui1  14 15 19.12     16 16 373   100
               rui2  15 16 19.83     17 19 274   100
 datatable_solution  14 15 21.42     16 16 592   100