Convert Dict to DataFrame in Julia - dictionary

Question

Suppose I have a Dict defined as follows:
x = Dict{AbstractString,Array{Integer,1}}("A" => [1,2,3], "B" => [4,5,6])
I want to convert this to a DataFrame object (from the DataFrames module). Constructing a DataFrame has a similar syntax to constructing a dictionary. For example, the above dictionary could be manually constructed as a data frame as follows:
DataFrame(A = [1,2,3], B = [4,5,6])
I haven't found a direct way to get from a dictionary to a data frame but I figured one could exploit the syntactic similarity and write a macro to do this. The following doesn't work at all but it illustrates the approach I had in mind:
macro dict_to_df(x)
typeof(eval(x)) <: Dict || throw(ArgumentError("Expected Dict"))
return quote
DataFrame(
for k in keys(eval(x))
#eval ($k) = $(eval(x)[$k])
end
)
end
end
I also tried writing this as a function, which does work when all dictionary values have the same length:
function dict_to_df(x::Dict)
s = "DataFrame("
for k in keys(x)
v = x[k]
if typeof(v) <: AbstractString
v = string('"', v, '"')
end
s *= "$(k) = $(v),"
end
s = chop(s) * ")"
return eval(parse(s))
end
Is there a better, faster, or more idiomatic approach to this?

Answer 1

Another method could be
DataFrame(Any[values(x)...],Symbol[map(symbol,keys(x))...])
It was a bit tricky to get the types in order to access the right constructor. To get a list of the constructors for DataFrames I used methods(DataFrame).
The DataFrame(a=[1,2,3]) way of creating a DataFrame uses keyword arguments. To use splatting (...) for keyword arguments the keys need to be symbols. In the example x has strings, but these can be converted to symbols. In code, this is:
DataFrame(;[Symbol(k)=>v for (k,v) in x]...)
Finally, things would be cleaner if x had originally been with symbols. Then the code would go:
x = Dict{Symbol,Array{Integer,1}}(:A => [1,2,3], :B => [4,5,6])
df = DataFrame(;x...)