开发者

Scala: “any” and “all” functions

开发者 https://www.devze.com 2023-03-14 04:44 出处:网络
my Haskell* is a bit rusty, so i can imagine that I’m missing the obvious: def any[A](s: Traversable[A], f: A => Boolean): Boolean = {

my Haskell* is a bit rusty, so i can imagine that I’m missing the obvious:

def any[A](s: Traversable[A], f: A => Boolean): Boolean = {
    s.foldLeft(false)((bool, elem) => bool || f(elem))
}

Does one of these properties apply to the it?

  1. predefined somewhere in the Scala libs
  2. circumstantial, and faster written as some one-liner
  3. wrong (I didn’t test it, sorry ;))

*actually SML,开发者_开发问答 but that’s 99% the same, but known by nobody under the sun.


  1. It's predefined and is called exists. And forall would be the "all" function you are looking for.

    scala> Vector(3, 4, 5).exists(_ % 2 == 0)
    res1: Boolean = true
    
    scala> Vector(3, 4, 5).forall(_ % 2 == 0)
    res2: Boolean = false
    
  2. You can make it more performant using a for loop with a break (from scala.util.control.Breaks). (See the standard library implementation of exists and forall.)

  3. It's correct.


Methods exist on the Traversable trait which are equivalent to any and all:

def all[A](xs: Traversable[A], p: A => Boolean): Boolean = xs forall p

def any[A](xs: Traversable[A], p: A => Boolean): Boolean = xs exists p


  1. No it isn't predifined with those names. You can use exists from Traversable package.
  2. The biggest disadvantage of your implementation is that will necessary consume all of your traversible, when, for any, if any is true, if could already give you your answer. The same goes for all. But one could easily implement this so that it doesn't evaluate the whole sequence. Another solution would be to implement a monad for this type of operation. Then you would call:

    a and b and c which is equivalent to a.and(b).and(c)

  3. It is correct.

BTW, another function that I find missing is a sum function.


How about exists:

scala> List(1,2,3).exists(_ > 2)
res12: Boolean = true

It's on Traversable.

0

精彩评论

暂无评论...
验证码 换一张
取 消