TS 4.1 You can use as clauses in mapped types to filter out properties in one go: type Methods<T> = { [P in keyof T as T[P] extends Function ? P : never]: T[P] }; type A_Methods = Methods<A>; // { render: () => void; } When the type specified in an as clause resolves … Read more
As you may be aware, object types in TypeScript are matched via structural subtyping and are thus open and extendible. If Base is an object type and Sub extends Base, then a Sub is a kind of Base, and you should be allowed to use a Sub wherever a Base is required: const sub: Sub … Read more
Edit: Since Typescript 4.1 there is a way of doing this directly with Key Remapping, avoiding the Pick combinator. Please see the answer by Oleg where it’s introduced. type RemoveIndex<T> = { [ K in keyof T as string extends K ? never : number extends K ? never : K ] : T[K] }; … Read more
In Typescript 2.8 you can explicitly eliminate the modifier: type Foo2 = { [P in keyof Foo]-?: number; } Or use the Required type that is built into newer versions. If you are using an older version you can use this workaround: type Helper<T, TNames extends string> = { [P in TNames]: (T & { … Read more
This can be done with conditional types and indexed access types, like this: type KeysMatching<T, V> = {[K in keyof T]-?: T[K] extends V ? K : never}[keyof T]; and then you pull out the keys whose properties match string like this: const key: KeysMatching<Thing, string> = ‘other’; // ERROR! // ‘”other”‘ is not assignable … Read more