For hcat, what shape would you want the output to have?
With your example input, the result of vcat is straightforward: v = [[1],[2,3],[4,5]] vcat(v...) # -> [1, 2, 3, 4, 5] But for hcat, what output would you want? hcat(v...) throws a DimensionMismatch error stating that vectors must have the same length. //T On Friday, November 27, 2015 at 1:39:34 PM UTC+1, Cedric St-Jean wrote: foldl would work, but it's going to create a ton of temporary arrays. > > None of the proposed efficient solutions work with hcat... I suppose if > splatting is a problem I should allocate and fill in the array myself. > > On Friday, November 27, 2015 at 6:39:07 AM UTC-5, Glen O wrote: >> >> Any chance that foldl(vcat,arr_of_arr) will do the job? >> >> On Sunday, 22 November 2015 23:04:26 UTC+10, Cedric St-Jean wrote: >>> >>> I have a big vector of vectors. Is there any way to vcat/hcat them >>> without splatting? >>> >>> arr_of_arr = Vector[[1],[2,3],[4,5]] >>> vcat(arr_of_arr...) >>> >>> I'm asking because splatting big arrays is a performance issue (and IIRC >>> it blows the stack at some point). >>> >>
