Convert List[Either[A, B]] to Either[List[A], List[B]]

Question

How to convert List[Either[String, Int]] to Either[List[String], List[Int]] using a method similar to cats sequence? For example, xs.sequence in the following code

import cats.implicits._
val xs: List[Either[String, Int]] = List(Left("error1"), Left("error2"))
xs.sequence

returns Left(error1) instead of required Left(List(error1, error2)).

KevinWrights' answer suggests

val lefts = xs collect {case Left(x) => x }
def rights = xs collect {case Right(x) => x}
if(lefts.isEmpty) Right(rights) else Left(lefts)

which does return Left(List(error1, error2)), however does cats provide out-of-the-box sequencing which would collect all the lefts?

Answer 1

Another variation on the same theme (similar to this answer), all imports included:

import scala.util.Either
import cats.data.Validated
import cats.syntax.traverse._
import cats.instances.list._

def collectErrors[A, B](xs: List[Either[A, B]]): Either[List[A], List[B]] = {
  xs.traverse(x => Validated.fromEither(x.left.map(List(_)))).toEither
}

If you additionally import cats.syntax.either._, then the toValidated becomes available, so you can also write:

xs.traverse(_.left.map(List(_)).toValidated).toEither

and if you additionally replace the left.map by bimap(..., identity), you end up with @DmytroMitin's wonderfully concise solution.