You can use a type guard[1] as an argument to Array.filter, but the function has to be explicitly typed as such.

I don't know why the type isn't narrowed in Array.filter like it is in if statements without this weird workaround.

  const array: (number | string)[] = [];
  
  const mixedArray = array.filter(value => typeof value === 'string');
  // mixedArray: (number | string)[]
  const arrayOfString = array.filter((value): value is string => typeof value === 'string');
  // arrayOfString: string[]

This example in Typescript playground: https://www.typescriptlang.org/play?#code/MYewdgzgLgBAhgJwXA...

[1]: https://www.typescriptlang.org/docs/handbook/advanced-types....

Oh so there is an overload!

filter<U extends T>(pred: (a: T) => a is U): U[];

Additionally, getting TS better at inferring type guards is an open issue (literally): https://github.com/microsoft/TypeScript/issues/38390