Convert a collection of Either<A, B> into an Either<A[], B[]> array

Looking to convert an array of type Either<A, B>[] into Either<A[], B[]>

The goal here is to gather all the left-values (errors) if there is at least one, otherwise return all right answers.

This task may appear straightforward, but my current attempt seems overly complex:

const compress = <A, B>(arr: E.Either<A, B>[]): E.Either<A[], B[]> => 
  A.reduce(
    E.right([]),
    (acc: E.Either<A[], B[]>, v: E.Either<A, B>) => 
    E.match(
      (a: A) => E.match((aar: A[]) => E.left([...aar, a]), 
                        (bar: B[]) => E.left([a]))(acc),
      (b: B) => E.match((aar: A[]) => E.left(aar), 
                        (bar: B[]) => E.right([...bar, b]))(acc)
    )(v)
   )(arr);

There has to be a simpler way to accomplish this objective.

typescriptfunctional-programmingfp-ts

Answer №1

You can utilize the separate function to gather the lefts and rights in an array of Eithers. While this may suffice, you can achieve the precise type you desire by following:

import { separate } from 'fp-ts/Array';
import * as E from 'fp-ts/Either';
import { pipe } from 'fp-ts/lib/function';

declare const es: Array<E.Either<number, string>>;

const output: E.Either<string[], number[]> = pipe(
  es,
  separate,
  // Assuming you want the either to be on the left if any result was left.
  (s) => s.left.length > 0 ? E.left(s.left) : E.right(s.right),
);

Another option is sequenceArray, which serves as a shortcut when consolidating errors from various sources. However, it converts to a left as soon as a single left is found, so you won't receive all of the left values back at once (which could be a drawback if you need full visibility into all the lefts).

const output2: E.Either<string, readonly number[]> = pipe(
  es,
  E.sequenceArray,
);

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