[Scala MOOC I] Lec5: Lists

5.1 - More Functions on Lists

already known methods:

xs.head
xs.tail

sublist and ele access:

  • xs.length
  • xs.last
  • xs.init: all elementh except last element
  • xs.take(n): sublist of first n elements
  • xs.drop(n): the rest of list after taking first n elements
  • xs(n): = xs.apply(n) , element at index n

More methods:

  • concatenation: xs ++ ys (::: is legacy usage)
  • xs.reverse
  • xs.updated(n,x): return a same list, except xs(n)=x (Note: Lists are immutable, so cannot modify)
  • xs.indexOf(x): index or -1
  • xs.contains(x): same as xs.indexOf(x)>=0

complexity: head, tail: simple to implement ⇒ complexity of last?

def last[T](xs: List[T]): T = xs match{
    case List() => throw new Error("last of Nil")
    case List(x) => x
    case y::ys => last(ys)
}

⇒ complexity of last = O(n)

implement init:

def init[T](xs: List[T]): T = xs match{
    case List() => throw new Error("init of Nil")
    case List(x) => List()
    case y::ys => y::init(ys)
}

⇒ complexity = O(length of xs)

implement concatenation :::/++ (NB: ::: is right-associative, xs:::ys = ys. :::(xs). )

def concat[T](xs: List[T], ys: List[T]) = xs match{
    case List() => ys
    case z::zs => x:concat(xs, ys)
}

⇒ complexity of concat = O(length of xs)

implement of reverse

def reverse[T](xs: List[T]) = xs match{
    case List() => List()
    case y::ys => reverse(ys) ++ y
}

⇒ complexity of reverse: every call contains a concat, thus complexity=O(n2)

exercice: remove nth element:

def removeAt[T](n:Int, xs: List[T]): List[T] = 
(xs take n ) ++ (xs drop n+1)

5.2 - Pairs and Tuples

example: sort list faster than insertion sort → merge sort.

sort 2 sublist, merge them ⇒ list is sorted

def msort(xs: List[Int]): List[Int] = {
    val n = xs.length/2
    if(n==0) xs
    else{
        def merge(xs: List[Int], ys: List[Int]) = ... // see below
        val (fst, snd) = x splitAt n // splitAt returns 2 sublists
        merge(msort(fst), msort(snd))
    }
}

merge:

def merge(xs: List[Int], ys: List[Int]): List[Int] = xs match {
    case Nil => ys
    case x::zs => match ys{
        case Nil => xs
        case y::ws => {
            if(x<y) x::merge(zs, ys)
            else y::merge(xs, ws)
        }
    }
}

Pair/Tuple

written as (x, y) in scala. pair can be used as patterns : similar for tuples.

val pair = ("a", 2)
val (label, value) = pair

tuple implementation:

⇒ can use _1 _2 to access elements

exercice: rewrite the merge function using a pattern matching over pairs:

def merge(xs: List[Int], ys: List[Int]): List[Int] =
(xs, ys) match {
    case (Nil, ys) => ys
    case (xs, Nil) => xs
    case (x:zs, y:ws) => 
        if(x<y) x::merge(zs, ys)
        else y::merge(xs, ws)
}

5.3 - Implicit Parameters

pb: how to apply msort to list of other element types.

using type parameters ? msort[T]⇒ the compare operator is not always defined !

pass the lt function as a parameter:

def msort[T](xs: List[T])(lt: (T,T)=>Boolean) = ...

another option: scala.math.Ordering[T]

impor math.Ordering
def msort[T](xs: List[T])(ord: Ordering) = ...// use ord.lt(x,y)
msort(nums)(Ordering.Int)

pb: pass each time the function parameter is cumbersome... ⇒ use implicite parameters

def msort[T](xs: List[T])(implicite ord: Ordering) = ...// use ord.lt(x,y)

⇒ the function calls can ignore the implicite parameter, the compiler will figure it out.

5.4 - Higher-Order List Functions

functions over list have similar pattern:

  • transform each element
  • retrive elements that satisfy some cretirion
  • combing elements using an operator

map

apply an operation to every elements.

abstract class List[T]{
    def map[U](f: T=>U): List[U] = this match {
        case Nil => this
        case x:xs => f(x)::xs.map(f)
    }
}

filtering

def filter(p: T=>Boolean): List[T] = this match {
    case Nil => this
    case x:xs => f(p(x)) x::xs.filter(p) else xs.filter(p)
}

other methods that extracts sublist:

exercice: implement a function pack:

def pack[T](xs: List[T]): List[List[T]] = xs match {
    case Nil => Nil
    case x::ys => {
        val (head, tail) = xs span (c => c==x)
        head :: pack(tail)
    }
}

exercice2: implement a function encode:

def encode[T](xs: List[T]): List[(T, Int)] = xs match {
    case Nil => Nil
    case x::ys => {
        val (head, tail) = xs span (c => c==x)
        (x, head.length) :: pack(tail)
    }
}

another version: use the pack:

def encode[T](xs: List[T]): List[(T, Int)] = 
    pack(xs) map (l => (l.head, l.length))

5.5 - Reduction of Lists

fold/reduce: combine elements using an operator.

reduceLeft

(can apply only to non-empty lists) inserts a binary operator between adj elements:

ex.

def sum(xs: List[Int]) = (0::xs) reduceLeft ( (x,y)=> x+y)
def prod(xs: List[Int]) = (1::xs) reduceLeft ( (x,y)=> x*y)

write shorter function values using underscore _: every _ represents a new parameter

def sum(xs: List[Int]) = (0::xs) reduceLeft ( _+_ )
def prod(xs: List[Int]) = (1::xs) reduceLeft ( _*_ )

foldLeft

foldLeft is like reduceLeft, but can apply on Nil, and takes an accumulator z => returns z when calling on Nil.

def sum(xs: List[Int]) = (xs foldleft 0) ( _+_ )
def prod(xs: List[Int]) = (xs foldleft 1) ( _*_ )

foldRight/reduceRight

dual functions to foldLeft and reduceLeft, but produce a tree leaned to right

if the operation is associative and communitive, foldLeft and foldRight should give same results. Other times need to think.

ex. concat

if apply foldLeft ⇒ type error, because the :: operator will be applied to 2 elements of type T.

5.6 - Reasoning About Concat

proof of programs

structural induction

pb: prove some properties of concat:

类似数学归纳法:

ex. prove (xs ++ ys) ++ zs = xs ++ (ys ++ zs): induction on xs

def concat[T](xs: List[T], ys: List[T]) = xs match{
    case Nil => ys
    case z::zs => x:concat(xs, ys)
}
  • base case: xs=Nil

(Nil ++ ys ) ++ zs = Nil ++ (ys ++ zs)

  • induction step: x::xs

5.7 - A Larger Equational Proof on Lists

pb: want to prove that xs.reverse.revese == xs

  • base case: Nil.reverse.revers = Nil
  • induction step

pb: cannot advance ⇒ generalize the argument.

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