Yes. Nevermind tibbles, the [rep(NA, len),] construction only happens to work
because len will always be >= the number of rows in value[[1L]], witness
> (1:10)[rep(NA, 20)]
[1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> (1:20)[rep(NA, 10)]
[1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
> (1:20)[rep(NA_integer_, 10)]
[1] NA NA NA NA NA NA NA NA NA NA
> (1:10)[rep(NA_integer_, 20)]
[1] NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
-pd
> On 21 Nov 2020, at 16:55 , Mario Annau <mario.an...@gmail.com> wrote:
>
> Hello,
>
> using the `unsplit()` function with tibbles currently leads to the
> following error:
>
>> mtcars_tb <- as_tibble(mtcars, rownames = NULL)
>> s <- split(mtcars_tb, mtcars_tb$gear)
>> unsplit(s, mtcars_tb$gear)
> Error: Must subset rows with a valid subscript vector.
> ℹ Logical subscripts must match the size of the indexed input.
> x Input has size 15 but subscript `rep(NA, len)` has size 32.
> Run `rlang::last_error()` to see where the error occurred.
>
> Tibble seems to (rightly) complain, that a logical vector has been used for
> subsetting which does not have the same length as the data.frame (rows).
> Since `NA` is a logical value, the subset should be changed to
> `NA_integer_` in `unsplit()`:
>
>> unsplit
> function (value, f, drop = FALSE)
> {
> len <- length(if (is.list(f)) f[[1L]] else f)
> if (is.data.frame(value[[1L]])) {
> x <- value[[1L]][rep(*NA_integer_*, len), , drop = FALSE]
> rownames(x) <- unsplit(lapply(value, rownames), f, drop = drop)
> }
> else x <- value[[1L]][rep(NA, len)]
> split(x, f, drop = drop) <- value
> x
> }
>
> Cheers,
> Mario
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Office: A 4.23
Email: pd....@cbs.dk Priv: pda...@gmail.com
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