Fold with previous element - scala

Question

Given val as: Seq[Int] = ...
A lot of times I need to apply an operation to two consecutive elements, e.g.
By the way I don't like
for (i <- 1 until as.size) {
// do something with as(i) and as(i - 1)
}
Or by another
as.tail.foldLeft((0, as.head)) { (acc, e) =>
// do something with acc._2 and e
// and try to not forget returning (_, e)
}
How do I writer better code for this scenario?

Answer 1

You could zip the sequence as with its own tail:
for ((prev, curr) <- as zip as.tail) {
// do something with `prev` and `curr`
}
Or you could use sliding:
for (window <- as.sliding(2)) {
val prev = window(0)
val curr = window(1)
// do something with `prev` and `curr`
}

Answer 2

Here's one way to supply the head of your sequence to every subsequent element.
val sq:Seq[Int] = Seq(. . .)
sq.headOption.fold(sq){hd =>
sq.tail.map(/*map() or fold() with the hd value*/)
}
Note that this is safe for collections of 1 or zero elements.

Answer 3

Can make your own fold which supports previous element.
Safe with 1 or zero element collections.
def foldLeftWithPrevious[A, B](as: Seq[A], accumulator: B)(f: (B, A, A) => B): B = {
#scala.annotation.tailrec
def foldLeftInner(list2: Seq[A], previous: A, accumulator: B, f: (B, A, A) => B): B = {
if (list2.isEmpty) accumulator
else foldLeftInner(list2.tail, list2.head, f(accumulator, previous, list2.head), f)
}
if (as.length <= 1) accumulator
else foldLeftInner(as.tail, as.head, accumulator, f)
}
Feel free to test it with this snippet.
val foldLeftTest = Seq(1)
foldLeftWithPrevious(foldLeftTest, 0)((accum, previous, current) => {
println("accum = " + accum)
println("previous = " + previous)
println("current = " + current)
println("accum will be... " + accum + " + " + previous + " + " + current)
println("which is... " + (accum + previous + current))
accum + previous + current
})