Reputation: 42110
I am trying to understand foldM
of Foldable
in cats
trying a simple example: suppose I need to sum up the numbers in a list while the running sum is positive and break when it is not.
val sumUp: (Int, Int) => Option[Int] = (x, y) => {
println(s"x = $x, y = $y")
val sum = x + y
if (sum > 0) Some(sum) else None
}
scala> val xs = List(1, 2, 3, -2, -5, 1, 2, 3)
xs: List[Int] = List(1, 2, 3, -2, -5, 1, 2, 3)
scala> Foldable[Stream].foldM(xs.toStream, 0)(sumUp)
x = 0, y = 1
x = 1, y = 2
x = 3, y = 3
x = 6, y = -2
x = 4, y = -5
res27: Option[Int] = None
Now I need to write new function sumUp2
to get the input stream tail, which starts where the running sum becomes <= 0 and the foldM
breaks. For example, I need to get something like this:
scala> val tail = Foldable[Stream].foldM(xs.toStream, 0)(sumUp2)
tail: Stream[Int] = Stream(-5, ?)
scala>tail.toList
res28: List[Int] = List(-5, 1, 2, 3)
How to write sumUp2
?
Upvotes: 2
Views: 1288
Reputation: 42110
I wrote sumUp2
to return Either[Int, (Int, Int)]
: left is the number of visited elements and the right is a pair of the number of visited elements and running sum.
type IntOr[A] = Either[Int, A]
val sumUp2: ((Int, Int), Int) => IntOr[(Int, Int)] = (pair, y) => {
val (size, x) = pair
val sum = x + y
println(s"sum = $sum, y = $y")
if (sum > 0) (size + 1, sum).asRight else size.asLeft
}
We know that foldM
stops when sumUp2
returns Left
so sumUp2
won't be invoked for all elements:
scala> val r = Foldable[Stream].foldM(xs.toStream, (0, 0))(sumUp2)
sum = 1, y = 1
sum = 3, y = 2
sum = 6, y = 3
sum = 4, y = -2
sum = -1, y = -5
r: IntOr[(Int, Int)] = Left(4)
Given r: Either[Int, (Int, Int)]
we can get the tail:
scala> r match { case Right(_) => Nil; case Left(n) => xs.drop(n) }
res63: List[Int] = List(-5, 1, 2, 3)
The solution seems to work fine but doesn't look nice to me. How would you improve it ?
Upvotes: 1
Reputation: 56537
What you can do is accumulate two values (in a tuple): the running sum up until it turns negative or zero; and the tail, which starts to accumulate values then.
val sumUp2: ((Int, List[Int]), Int) => Id[(Int, List[Int])] = (x, y) => {
val sum = if (x._1 < 0) x._1 else x._1 + y
if (sum > 0) (sum, x._2) else (-1, x._2 ++ List(y))
}
Then, you can get the tail from the second element in the tuple:
val xs = List(1, 2, 3, -2, -5, 1, 2, 3)
val res = Foldable[Stream].foldM(xs.toStream, (0, List[Int]()))(sumUp2)
println(res._2)
Fiddle here.
Upvotes: 3