Selecting a specific comparator method when sorting

[Marc Petit-Huguenin]

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On 12/06/2012 11:30 AM, Christian Mallwitz wrote:
> Hi
> 
> I have been experimenting with some lambda sorting examples. Examples:
> 
> List<String> list = new ArrayList<>(); list.add("496"); list.add("6"); 
> list.add("8128"); list.add("28"); list.add("33550336");
> 
> list .stream()
> 
> .sorted(Comparators.comparing((IntFunction<String>) Integer::parseInt)) //
> (1) .sorted(Comparators.<String>comparing(Integer::parseInt)) // (2)
> 
> .forEach(x -> { System.out.println(x); });
> 
> I understand in case (1) syntactically speaking the (IntFunction<String>)
> part is just a type cast. What is the name and specification for the
> <String> bit in (2)? Could you point me to the relevant Java spec parts?
> 
> Additionally I want to share my attempt at a Schwartz Transformation - I
> have two versions: one version (A) requiring an additional class and one
> version (B) emitting a warning due to a type cast (which should be safe).
> Which one would you prefer? Any other feedback?
> 
> Call as in:
> 
> schwartz(list.stream(), Integer::parseInt).forEach(x -> { 
> System.out.println(x); });
> 
> A)
> 
> public static<T, R extends Comparable<? super R>> Stream<T> 
> schwartz(Stream<T> stream, Function<T, R> f) {
> 
> // class Pair - type of second element of pair must be Comparable final
> class Pair<F, S extends Comparable<? super S>> { public final F first; 
> public final S second; public Pair(F first, S second){ this.first = first; 
> this.second = second; } }
> 
> return stream .map(t -> new Pair<>(t, f.apply(t))) .sorted((p1, p2) ->
> p1.second.compareTo(p2.second)) .map(p -> p.first); }
> 
> B)
> 
> @SuppressWarnings("unchecked") public static<T, R extends Comparable<?
> super R>> Stream<T> schwartz(Stream<T> stream, Function<T, R> f) {
> 
> return stream .map(t -> new Object[]{t, f.apply(t)}) .sorted((o1, o2) ->
> ((Comparable) o1[1]).compareTo(o2[1])) .map(o -> ((T) o[0])); }
> 

If we still had MapStream:

public static<T, R extends Comparable<? super R>> Stream<T> schwartz(Stream<T>
stream, Mapper<T, R> f) {

  return stream.mapped(f)
    .swap()
    .sorted((p1, p2) -> p1.compareTo(p2))
    .values();
}

- -- 
Marc Petit-Huguenin
Email: marc at petit-huguenin.org
Blog: http://blog.marc.petit-huguenin.org
Profile: http://www.linkedin.com/in/petithug
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