Preallocate a data frame of known size in Julia
julia
Solution
To preallocate a data frame, you must pre-allocate its columns. You can create three columns full of `missing` values by simply doing `[fill(missing, 10000) for _ in 1:3]`, but that doesn't actually allocate anything at all because those vectors can only hold one value — `missing` — and thus they can't be changed to hold other values later. One way to do this is by using to `Vector` constructors that can hold either `Missing` or `Float64`:
julia> DataFrame([Vector{Union{Missing, Float64}}(missing, 10000) for _ in 1:3], [:a, :b, :c])
10000×3 DataFrame
Row │ a b c
│ Float64? Float64? Float64?
───────┼──────────────────────────────
1 │ missing missing missing
2 │ missing missing missing
⋮ │ ⋮ ⋮ ⋮
10000 │ missing missing missing
9997 rows omitted
Note that rather than `Real`, this is using the concrete `Float64` — this will have significantly better performance.
(this answer was edited to reflect DataFrames v1.0 syntax)
Problem
When I'm running simulations, I like to initialize a big, empty array and fill it up as the simulation iterates through to the end. I do this with something like `res = Array(Real,(n_iterations,n_parameters))`. However, it would be nice to have named columns, which I think means using a DataFrame. Yet when I try to do something like `res_df = convert(DataFrame,res)` it throws an error. I would like a more concise approach than doing something like `res_df = DataFrame(a=Array(Real,N),b=Array(Real,N),c=Array(Real,N),....)` as suggested by the answers to: julia create an empty dataframe and append rows to it