Using monoids to fold data structures 

LYAHFGG:

Because there are so many data structures that work nicely with folds, the Foldable type class was introduced. Much like Functor is for things that can be mapped over, Foldable is for things that can be folded up!

The equivalent in Cats is also called Foldable. Here’s the typeclass contract:

/**
 * Data structures that can be folded to a summary value.
 *
 * In the case of a collection (such as `List` or `Set`), these
 * methods will fold together (combine) the values contained in the
 * collection to produce a single result. Most collection types have
 * `foldLeft` methods, which will usually be used by the associationed
 * `Fold[_]` instance.
 *
 * Foldable[F] is implemented in terms of two basic methods:
 *
 *  - `foldLeft(fa, b)(f)` eagerly folds `fa` from left-to-right.
 *  - `foldLazy(fa, b)(f)` lazily folds `fa` from right-to-left.
 *
 * Beyond these it provides many other useful methods related to
 * folding over F[A] values.
 *
 * See: [[https://www.cs.nott.ac.uk/~gmh/fold.pdf A tutorial on the universality and expressiveness of fold]]
 */
@typeclass trait Foldable[F[_]] extends Serializable { self =>

  /**
   * Left associative fold on 'F' using the function 'f'.
   */
  def foldLeft[A, B](fa: F[A], b: B)(f: (B, A) => B): B

  /**
   * Right associative lazy fold on `F` using the folding function 'f'.
   *
   * This method evaluates `b` lazily (in some cases it will not be
   * needed), and returns a lazy value. We are using `A => Fold[B]` to
   * support laziness in a stack-safe way.
   *
   * For more detailed information about how this method works see the
   * documentation for `Fold[_]`.
   */
  def foldLazy[A, B](fa: F[A], lb: Lazy[B])(f: A => Fold[B]): Lazy[B] =
    Lazy(partialFold[A, B](fa)(f).complete(lb))

  /**
   * Low-level method that powers `foldLazy`.
   */
  def partialFold[A, B](fa: F[A])(f: A => Fold[B]): Fold[B]
  ....
}

We can use this as follows:

scala> import cats._, cats.data._, cats.implicits._
import cats._
import cats.data._
import cats.implicits._

scala> Foldable[List].foldLeft(List(1, 2, 3), 1) {_ * _}
res0: Int = 6

Foldable comes with some useful functions/operators, many of them taking advantage of the typeclasses. Let’s try the fold. Monoid[A] gives us empty and combine, so that’s enough information to fold over things.

  /**
   * Fold implemented using the given Monoid[A] instance.
   */
  def fold[A](fa: F[A])(implicit A: Monoid[A]): A =
    foldLeft(fa, A.empty) { (acc, a) =>
      A.combine(acc, a)
    }

Let’s try this out.

scala> Foldable[List].fold(List(1, 2, 3))(Monoid[Int])
res1: Int = 6

There’s a variant called foldMap that accepts a function.

  /**
   * Fold implemented by mapping `A` values into `B` and then
   * combining them using the given `Monoid[B]` instance.
   */
  def foldMap[A, B](fa: F[A])(f: A => B)(implicit B: Monoid[B]): B =
    foldLeft(fa, B.empty) { (b, a) =>
      B.combine(b, f(a))
    }

Since the standard collection library doesn’t implement foldMap, we can now use this as an operator.

scala> List(1, 2, 3).foldMap(identity)(Monoid[Int])
res2: Int = 6

Another useful thing is that we can use this to convert the values into newtype.

scala> :paste
// Entering paste mode (ctrl-D to finish)
class Conjunction(val unwrap: Boolean) extends AnyVal
object Conjunction {
  @inline def apply(b: Boolean): Conjunction = new Conjunction(b)
  implicit val conjunctionMonoid: Monoid[Conjunction] = new Monoid[Conjunction] {
    def combine(a1: Conjunction, a2: Conjunction): Conjunction =
      Conjunction(a1.unwrap && a2.unwrap)
    def empty: Conjunction = Conjunction(true)
  }
  implicit val conjunctionEq: Eq[Conjunction] = new Eq[Conjunction] {
    def eqv(a1: Conjunction, a2: Conjunction): Boolean =
      a1.unwrap == a2.unwrap
  }
}

// Exiting paste mode, now interpreting.

defined class Conjunction
defined object Conjunction

scala> val x = List(true, false, true) foldMap {Conjunction(_)}
x: Conjunction = Conjunction@4d5

scala> x.unwrap
res3: Boolean = false

This surely beats writing Conjunction(true) for each of them and connecting them with |+|.

We will pick it up from here later.