Java – How to Sum a List of Integers with Java Streams

javajava-8java-stream

I want to sum a list of Integers. It works as follows, but the syntax does not feel right. Could the code be optimized?

Map<String, Integer> integers;
integers.values().stream().mapToInt(i -> i).sum();

Best Answer

This will work, but the i -> i is doing some automatic unboxing which is why it "feels" strange. mapToInt converts the stream to an IntStream "of primitive int-valued elements". Either of the following will work and better explain what the compiler is doing under the hood with your original syntax:

integers.values().stream().mapToInt(i -> i.intValue()).sum();
integers.values().stream().mapToInt(Integer::intValue).sum();

Related Solutions

Java 8 Stream – Add Elements to List and Sum

What you want here is to collect your elements inside 2 collectors: the first one would collect into a list, and the second one would sum the price.

Since there are no such collectors in the Stream API itself, we can easily construct our own. Let's create a class ResultHolder that will hold the result of the Stream pipeline: this is the list of decimals and the sum.

class ResultHolder {
    List<BigDecimal> list = new ArrayList<>();
    BigDecimal sum = BigDecimal.ZERO;
}

Finally, we can use it with:

ResultHolder resultHolder = 
    listOfPricedObjects.stream()
            .map(PricedObject::getPrice)
            .collect(
                ResultHolder::new,
                (r, p) -> { r.list.add(p); r.sum = r.sum.add(p); },
                (r1, r2) -> { r1.list.addAll(r2.list); r1.sum = r1.sum.add(r2.sum); }
            );
System.out.println(resultHolder.list);
System.out.println(resultHolder.sum);

This will work in a parallel pipeline and will keep the initial order of the list, contrary to the other answers.

Best Answer

Related Solutions

Java 8 Stream – Add Elements to List and Sum

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