How to keep union type inside other type instead of destructure it

Issue

I have this type

type Test = {
  checked: boolean;
  selection: 'female' | 'male'
}

And, I would like to transform it into another type such as

type Test = {
  checked: Record<string, boolean>;
  selection: Record<string, 'female' | 'male'>
}

Then, I wrote this transform

type NestedTest<T> = T extends object
  ? { [P in keyof T]: NestedTest<T[P]> }
  : Record<string, T>;

But, when I call it like

type Test2 = NestedTest<Test>;

Then I got the unexpected results below

type Test2 = {
    checked: Record<string, false> | Record<string, true>;
    selection: Record<string, "female"> | Record<string, "male">;
}

As you can see, typescript auto splits these union types into a union of Record type instead of only one.

I don't why it is like that. If you can, please tell me what I misunderstood? and how to fix it.

Here is my playground

Thank you so much!

Solution

You basically need to stop the unions from distributing to seperate Records. See Distributive Conditional Types in the TypeScript docs.

type NestedTest<T> = [T] extends [object]
  ? { [P in keyof T]: NestedTest<T[P]> }
  : Record<string, T>;

If we surround T and object with square brackets, the elements of the union will not be distributed to different Records.

Playground

Answered By - Tobias S.