SCALA: Fold method with conditions - scala

Question

I am still learning the basics of Scala, therefore I am asking for your understanding. Is it any possible way to use fold method to print only names beginning with "A"
Object Scala {
val names: List[String] = List("Adam", "Mick", "Ann");
def main(args: Array[String]) {
println(names.foldLeft("my list of items starting with A: ")(_+_));
}
}
}

Answer 1

Have a look at the signature of foldLeft
def foldLeft[B](z: B)(op: (B, A) => B): B
where
z is the initial value
op is a function taking two arguments, namely accumulated result so far B, and the next element to be processed A
returns the accumulated result B
Now consider this concrete implementation
val names: List[String] = List("Adam", "Mick", "Ann")
val predicate: String => Boolean = str => str.startsWith("A")
names.foldLeft(List.empty[String]) { (accumulated: List[String], next: String) =>
if (predicate(next)) accumulated.prepended(next) else accumulated
}
here
z = List.empty[String]
op = (accumulated: List[String], next: String) => if (predicate(next)) accumulated.prepended(next) else accumulated
Usually we would write this inlined and rely on type inference so we do not have two write out full types all the time, so it becomes
names.foldLeft(List.empty[String]) { (acc, next) =>
if (next.startsWith("A")) next :: acc else acc
}
// val res1: List[String] = List(Ann, Adam)
On of the key ideas when working with List is to always prepend an element instead of append
names.foldLeft(List.empty[String]) { (accumulated: List[String], next: String) =>
if (predicate(next)) accumulated.appended(next) else accumulated
}
because prepending is much more efficient. However note how this makes the accumulated result in reverse order, so
List(Ann, Adam)
instead of perhaps required
List(Adam, Ann)
so often-times we perform one last traversal by calling reverse like so
names.foldLeft(List.empty[String]) { (acc, next) =>
if (next.startsWith("A")) next :: acc else acc
}.reverse
// val res1: List[String] = List(Adam, Ann)

Answer 2

The answer from #Mario Galic is a good one and should be accepted. (It's the polite thing to do).
Here's a slightly different way to filter for starts-with-A strings.
val names: List[String] = List("Adam", "Mick", "Ann")
println(names.foldLeft("my list of items starting with A: "){
case (acc, s"A$nme") => acc + s"A$nme "
case (acc, _ ) => acc
})
//output: "my list of items starting with A: Adam Ann"