Finding Max with Lambda Expression in Java

Question

This is my code

    List ints = Stream.of(1,2,4,3,5).collect(Collectors.toList());
    Integer maxInt = ints.stream()
                                 


        

Answer 1

This function (note -> is for closures and not to be confused with => which is for comparison)

i -> i

just means you need to compare the entire object as it is. i.e. if I have an i you need to compare i

A less trivial example might be

max(Comparator.comparing(i -> -i))

which will give you the minimum or

max(Comparator.comparing(i -> Math.abs(100-i))

gives you a value which is farthest from 100.

max(Comparator.comparing(i -> i.toString()))

which will give you the maximum comparing as a String i.e. "9" > "10" as a string.

Answer 2

Comparator.comparing expects a function which maps the source object to the value which actually gets compared - in your case, as you don't want to pre-process the value to be compared, i is simply mapped to itself.

Answer 3

The method Comparator.comparing(…) is intended to create a Comparator which uses an order based on a property of the objects to compare. When using the lambda expression i -> i, which is a short writing for (int i) -> { return i; } here, as a property provider function, the resulting Comparator will compare the values itself. This works when the objects to compare have a natural order as Integer has.

So

Stream.of(1,2,4,3,5).max(Comparator.comparing(i -> i))
.ifPresent(maxInt->System.out.println("Maximum number in the set is " + maxInt));

does the same as

Stream.of(1,2,4,3,5).max(Comparator.naturalOrder())
.ifPresent(maxInt->System.out.println("Maximum number in the set is " + maxInt));

though the latter is more efficient as it is implemented as singleton for all types which have a natural order (and implement Comparable).

The reason why max requires a Comparator at all, is because you are using the generic class Stream which might contain arbitrary objects.

This allows, e.g. to use it like streamOfPoints.max(Comparator.comparing(p->p.x)) to find the point with the largest x value while Point itself does not have a natural order. Or do something like streamOfPersons.sorted(Comparator.comparing(Person::getAge)).

When using the specialized IntStream you can use the natural order directly which is likely to be more efficient:

IntStream.of(1,2,4,3,5).max()
.ifPresent(maxInt->System.out.println("Maximum number in the set is " + maxInt));

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To illustrate the difference between “natural order” and a property based order:

Stream.of("a","bb","aaa","z","b").max(Comparator.naturalOrder())
.ifPresent(max->System.out.println("Maximum string in the set is " + max));

this will print

Maximum string in the set is z

as the natural order of Strings is the lexicographical order where z is greater than b which is greater than a

On the other hand

Stream.of("a","bb","aaa","z","b").max(Comparator.comparing(s->s.length()))
.ifPresent(max->System.out.println("Maximum string in the set is " + max));

will print

Maximum string in the set is aaa

as aaa has the maximum length of all Strings in the stream. This is the intended use case for Comparator.comparing which can be made even more readable when using method references, i.e. Comparator.comparing(String::length) which almost speaks for itself…