CODEWITHSUNDEEP
lambdajava-8
Luckylukee
Luckylukee

Reputation: 595

How to create a Function2 instance using lambdas in Java?

I am trying to implement a lambda Function2 in Java:

JavaPairRDD<String, Integer> reducedCounts = counts.reduceByKey(new Function2<Integer, Integer, Integer>() {
    @Override
    public Integer call(final Integer value0, final Integer value1) {
        return Integer.valueOf(value0.intValue() + value1.intValue());
    }
});

The above is a sample, but how to change it to a lamb8 format? I wanna define as a separate function

Function2<...> f2 = LAMBDA;
counts.reduceByKey(f2);

Upvotes: 1

Views: 2193

Answers (2)

Grzegorz Piwowarek
Grzegorz Piwowarek

Reputation: 13793

You can go for a method reference which is the easiest option in this case:

Function2<Integer, Integer, Integer> f2 = Integer::sum;
counts.reduceByKey(f2)

which is equal to:

Function2<Integer, Integer, Integer> f2 = (i1, i1) -> i1 + i2;
counts.reduceByKey(f2)

Also, such simplification is possible because you perform a lot of unnecessary boxing/unboxing in order to calculate the sum of two integers.

Upvotes: 3

Misha
Misha

Reputation: 28143

Since Function2 is a functional interface, you can implement it with a simple lambda expression:

Function2<Integer,Integer,Integer> plus = (x,y)->Integer.valueOf(x.intValue()+y.intValue());

If your logic is more than a simple expression, you can use a block:

Function2<Integer, Integer, Integer> plus = (x, y) -> {
   // do stuff
   return Integer.valueOf(x.intValue()+y.intValue());
};

In case of a simple integer addition, you can take advantage of auto-unboxing and write this more simply as:

Function2<Integer, Integer, Integer> plus = (x, y) -> x + y;

You can also take advantage of sum method in Integer class to use a method reference:

Function2<Integer, Integer, Integer> plus = Integer::sum;

Upvotes: 2

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