https://stackoverflow.com/questions/17408880
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RussianIn general, all 6 fold functions apply a binary operator to each element of a collection. The result of each step is passed on to the next step (as input to one of the binary operator's two arguments). This way we can cumulate a result.
reduceLeft and reduceRight cumulate a single result.
foldLeft and foldRight cumulate a single result using a start value.
scanLeft and scanRight cumulate a collection of intermediate cumulative results using a start value.
Accumulate
From LEFT and forwards...
With a collection of elements abc and a binary operator add we can explore what the different fold functions do when going forwards from the LEFT element of the collection (from A to C):
val abc = List("A", "B", "C")
def add(res: String, x: String) = {
println(s"op: $res + $x = ${res + x}")
res + x
}
abc.reduceLeft(add)
// op: A + B = AB
// op: AB + C = ABC // accumulates value AB in *first* operator arg `res`
// res: String = ABC
abc.foldLeft("z")(add) // with start value "z"
// op: z + A = zA // initial extra operation
// op: zA + B = zAB
// op: zAB + C = zABC
// res: String = zABC
abc.scanLeft("z")(add)
// op: z + A = zA // same operations as foldLeft above...
// op: zA + B = zAB
// op: zAB + C = zABC
// res: List[String] = List(z, zA, zAB, zABC) // maps intermediate results
From RIGHT and backwards...
If we start with the RIGHT element and go backwards (from C to A) we'll notice that now the second argument to our binary operator accumulates the result (the operator is the same, we just switched the argument names to make their roles clear):
def add(x: String, res: String) = {
println(s"op: $x + $res = ${x + res}")
x + res
}
abc.reduceRight(add)
// op: B + C = BC
// op: A + BC = ABC // accumulates value BC in *second* operator arg `res`
// res: String = ABC
abc.foldRight("z")(add)
// op: C + z = Cz
// op: B + Cz = BCz
// op: A + BCz = ABCz
// res: String = ABCz
abc.scanRight("z")(add)
// op: C + z = Cz
// op: B + Cz = BCz
// op: A + BCz = ABCz
// res: List[String] = List(ABCz, BCz, Cz, z)
.
De-cumulate
From LEFT and forwards...
If instead we were to de-cumulate some result by subtraction starting from the LEFT element of a collection, we would cumulate the result through the first argument res of our binary operator minus:
val xs = List(1, 2, 3, 4)
def minus(res: Int, x: Int) = {
println(s"op: $res - $x = ${res - x}")
res - x
}
xs.reduceLeft(minus)
// op: 1 - 2 = -1
// op: -1 - 3 = -4 // de-cumulates value -1 in *first* operator arg `res`
// op: -4 - 4 = -8
// res: Int = -8
xs.foldLeft(0)(minus)
// op: 0 - 1 = -1
// op: -1 - 2 = -3
// op: -3 - 3 = -6
// op: -6 - 4 = -10
// res: Int = -10
xs.scanLeft(0)(minus)
// op: 0 - 1 = -1
// op: -1 - 2 = -3
// op: -3 - 3 = -6
// op: -6 - 4 = -10
// res: List[Int] = List(0, -1, -3, -6, -10)
From RIGHT and backwards...
But look out for the xRight variations now! Remember that the (de-)cumulated value in the xRight variations is passed to the second parameter res of our binary operator minus:
def minus(x: Int, res: Int) = {
println(s"op: $x - $res = ${x - res}")
x - res
}
xs.reduceRight(minus)
// op: 3 - 4 = -1
// op: 2 - -1 = 3 // de-cumulates value -1 in *second* operator arg `res`
// op: 1 - 3 = -2
// res: Int = -2
xs.foldRight(0)(minus)
// op: 4 - 0 = 4
// op: 3 - 4 = -1
// op: 2 - -1 = 3
// op: 1 - 3 = -2
// res: Int = -2
xs.scanRight(0)(minus)
// op: 4 - 0 = 4
// op: 3 - 4 = -1
// op: 2 - -1 = 3
// op: 1 - 3 = -2
// res: List[Int] = List(-2, 3, -1, 4, 0)
The last List(-2, 3, -1, 4, 0) is maybe not what you would intuitively expect!
As you see, you can check what your foldX is doing by simply running a scanX instead and debug the cumulated result at each step.
Bottom line
- Cumulate a result with
reduceLeftorreduceRight. - Cumulate a result with
foldLeftorfoldRightif you have a start value. Cumulate a collection of intermediate results with
scanLeftorscanRight.Use a xLeft variation if you want to go forwards through the collection.
- Use a xRight variation if you want to go backwards through the collection.
