How do I access a Type's constructors as functions in Julia

Question

This would be useful for object construction by chaining. For example, say I would like to create a DataFrame by piping a Dict to it. As in,

merge(dict1, dict2) |> DataFrame

But DataFrame here returns the type DataFrame rather than the constructor I need. How can I access the constructor? I can see the signatures with methods(DataFrame) but can't access the actual function.

Isaiah Norton · Accepted Answer · 2013-10-05 21:17:38Z

1

This doesn't work for any type because the |> (pipe) method does not exist for the signature (Any, DataType).

I haven't tried with DataFrame, but the following trivial example works:

type Foo
  x::Int
end

|>(a::Any, T::DataType) = T(a)

test = 1 |> Foo

edited Oct 5, 2013 at 21:17

answered Oct 5, 2013 at 0:50

Isaiah Norton

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2 Comments

John Myles White Over a year ago

To elaborate on Isaiah's answer, when you write a |> b, you are really calling, |>(a, b). In the example you give, this means you would be calling |>(a::Dict, b::Type{DataFrame}).

Sean Mackesey Over a year ago

@Isaiah Thanks Isaiah. I understand that in the example I gave it accesses the type. I am wondering though if it is possible to access the constructor of the type to store in a variable. Then pipe could be used without a new method, though the pipe was really just being used as an example of a possible use case. Something like: `cnst = Foo.constructors[1]; 1 |> cnst.

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