CODEWITHSUNDEEP
dataframejulia
Shayan
Shayan

Reputation: 6385

How to convert a vector of vectors into a DataFrame in Julia, without for loop?

I created a Vector of Vectors, named all_arrays in Julia in this way for a specific purpose:

using DataFrames
using StatsBase

list_of_numbers = 1:17

all_arrays = [zeros(Float64, (17,)) for i in 1:1000]
round = 1
while round != 1001
    random_array = StatsBase.sample(1:17 , length(list_of_numbers))
    random_array = random_array/sum(random_array)

    if (0.0 in random_array) || (random_array in all_arrays)
        continue
    end

    all_arrays[round] = random_array
    round += 1
    println(round)
end

The dimension of all_arrays is:

julia> size(all_arrays)
(1000,)

Then I want to convert all_arrays into a DataFrame with 1000*17 dimensions (Note that each vector in the all_arrays is a (17,) shape Vector). I tried This way:

df = DataFrames.DataFrame(zeros(1000,17) , :auto)
for idx in 1:length(all_arrays)
    df[idx , :] = all_arrays[idx]
end

But I'm looking for a straightforward way for this instead of a for loop and a prebuilt DataFrame! Is there any?

Upvotes: 2

Views: 1296

Answers (1)

Bogumił Kamiński
Bogumił Kamiński

Reputation: 69949

If you want simple code use (the length of the code is the same as below, but I find it conceptually simpler):

DataFrame(mapreduce(permutedims, vcat, all_arrays), :auto)

For such small data as you described this should be efficient enough.

If you want something faster use:

DataFrame([getindex.(all_arrays, i) for i in 1:17], :auto, copycols=false)

Here is a benchmark:

julia> using BenchmarkTools

julia> @btime DataFrame(mapreduce(permutedims, vcat, $all_arrays), :auto);
  7.257 ms (3971 allocations: 65.22 MiB)

julia> @btime DataFrame([getindex.($all_arrays, i) for i in 1:17], :auto, copycols=false);
  41.000 μs (88 allocations: 140.66 KiB)

Upvotes: 3

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