Define a list of optional keys for Typescript Record

Question

I want to type an object which can only have keys 'a', 'b' or 'c'.

So I can do it as follows:

Interface IList {
    a?: string;
    b?: string;
    c?: string;
}

They are all optional! Now I was wondering if this can be written with Record in just one line

type List = Record<'a' | 'b' | 'c', string>;

The only issue is that all keys need to be defined. So I ended up with

type List = Partial<Record<'a' | 'b' | 'c', string>>;

This works, but I can imagine there is a better way to do this without Partial. Is there an other way to make the keys optional inside Record ?

Answer 1

There is no way to specify the optionality of members of Record. They are required by definition

type Record<K extends keyof any, T> = {
    [P in K]: T; // Mapped properties are not optional, and it's not a homomorphic mapped type so it can't come from anywhere else.
};

You can define your own type if this is a common scenario for you:

type PartialRecord<K extends keyof any, T> = {
  [P in K]?: T;
};
type List =  PartialRecord<'a' | 'b' | 'c', string>

Or you can define PartialRecord using the predefined mapped types as well:

type PartialRecord<K extends keyof any, T> =  Partial<Record<K, T>>