Convert struct fields and their values to DataFrames

New to Julia
1

Hi, I want to convert following struct to DataFrames:

struct MyStruct
x::Vector{Float64}
y::Vector{Float64}
end

I have written following function to convert MyStruct fields and their values to DataFrame:

function struct_to_dataframe(s::MyStruct)
# Create an empty data frame
df = DataFrame()
fields = fieldnames(s)
# Get the values for each field
values = [getfield(s,field) for field in fields]
push!(df, (field=fields, value=values))
return df
end

#Create instance
s = MyStruct(rand(10),rand(10))
#Convert to DF
df = struct_to_dataframe(s)

But the outcome is something like this:

1×2 DataFrame
Row │ field value
│ Tuple… Array…
─────┼─────────────────────────────────────────────
1 │ (:x, :y) [[0.976849, 0.653888, 0.956756, …

Whereas I want each field and its value in column form. How can I do that? Thanks!!!

2

Maybe you had a mistake at fieldnames. It require the struct inself, not instance. Use typeof function to get the name of class(structure).

julia> struct MyStruct
           x::Vector{Float64}
           y::Vector{Float64}
           end

julia> function struct_to_dataframe(s)
           fields = fieldnames(typeof(s))
           still_vector = [getfield(s,field) for field in fields]
           return DataFrame(hcat(still_vector...), collect(fields))
       end
struct_to_dataframe (generic function with 1 method)

julia> s = MyStruct(rand(10),rand(10));

julia> struct_to_dataframe(s)
10×2 DataFrame
 Row │ x         y
     │ Float64   Float64
─────┼──────────────────────
   1 │ 0.286858  0.140575
   2 │ 0.253701  0.821031
   3 │ 0.767813  0.474758
   4 │ 0.669886  0.74336
   5 │ 0.463448  0.956407
   6 │ 0.692433  0.715941
   7 │ 0.520875  0.447658
   8 │ 0.783861  0.00811924
   9 │ 0.928598  0.885618
  10 │ 0.681593  0.271318
3

@rmsmsgood thank you so much! Related to my question, I also want to ask if I set one of the field of MyStruct as an integer then I get a dimension mismatch error. How to take care of that?

4

Setting one of the filed of YourStruct as an integer, not an integer vector? You mean, this?

struct MyStruct
    x::Vector{Float64}
    y::Vector{Float64}
    k::Int64
end

I guess that should be a dimension error because x,y,k have different length and hcat makes them one matrix. If x,y has length 4, then the matrix has the form below

x[1] y[1] k[1]
x[2] y[2] 😡
x[3] y[3] 😡
x[4] y[4] 😡

so maybe hcat raise dimension mismatch error since :rage: doesn’t exist. You can take care of this in many way.

  1. Make integer to integer vector. repeat function maybe fit for you. eg) k \mapsto [k, k, k, \cdots , k].
  2. Just ignore that. In the for loop, you could check the length of each field or type, so if that is not an array then skip that.
5

If you mean something like this, the DataFrame constructor can handle automatic broadcasting of scalar elements


struct MS
    x::Vector{Float64}
    y::Vector{Float64}
    k::Int64
end
s = MS(rand(10),rand(10), rand(1:10))
vals=getfield.([s],1:fieldcount(typeof(s)))
using DataFrames
DataFrame(;zip(propertynames(s),vals)...)