416
votes

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();
12
"but the syntax does not feel right" What makes you think that? This is the usual idiom. Maybe you want to use mapToLong to avoid overflows, depending on the values your map can have. - Alexis C.
@JBNizet I find i -> i very clear, personally. Well, yes you need to know that the value will be automatically unboxed, but it's true since Java 5... - Alexis C.
@AlexisC. it's understandable because it's passed to mapToInt(), and because I'm an experienced developer. But i -> i, without context, looks like a noop. Integer::intValue is more verbose, but makes the unboxing operation explicit. - JB Nizet
@JBNizet People that calls the method foo(int i) do not write foo(myInteger.intValue()); each time they call it (or at least I expect not!!). I agree with you that Integer::intValue is more explicit but I think the same applies here. People should just learn it once and then you're done :-). It's not like if it was some magic obfuscation. - Alexis C.
@JB Nizet: well, i -> i looks like a no-op and conceptionally, it is a no-op. Sure, under the hood Integer.intValue() gets called, but even deeper under the hood, that methods gets inlined to become exactly the no-op that it looks like in the source code. Integer::intValue has the bonus point of not creating a synthetic method in the byte code but it’s not what should drive your decision of how to organize your source code. - Holger

12 Answers

557
votes

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();
168
votes

I suggest 2 more options:

integers.values().stream().mapToInt(Integer::intValue).sum();
integers.values().stream().collect(Collectors.summingInt(Integer::intValue));

The second one uses Collectors.summingInt() collector, there is also a summingLong() collector which you would use with mapToLong.


And a third option: Java 8 introduces a very effective LongAdder accumulator designed to speed-up summarizing in parallel streams and multi-thread environments. Here, here's an example use:

LongAdder a = new LongAdder();
map.values().parallelStream().forEach(a::add);
sum = a.intValue();
91
votes

From the docs

Reduction operations A reduction operation (also called a fold) takes a sequence of input elements and combines them into a single summary result by repeated application of a combining operation, such as finding the sum or maximum of a set of numbers, or accumulating elements into a list. The streams classes have multiple forms of general reduction operations, called reduce() and collect(), as well as multiple specialized reduction forms such as sum(), max(), or count().

Of course, such operations can be readily implemented as simple sequential loops, as in:

int sum = 0;
for (int x : numbers) {
   sum += x;
}

However, there are good reasons to prefer a reduce operation over a mutative accumulation such as the above. Not only is a reduction "more abstract" -- it operates on the stream as a whole rather than individual elements -- but a properly constructed reduce operation is inherently parallelizable, so long as the function(s) used to process the elements are associative and stateless. For example, given a stream of numbers for which we want to find the sum, we can write:

int sum = numbers.stream().reduce(0, (x,y) -> x+y);

or:

int sum = numbers.stream().reduce(0, Integer::sum);

These reduction operations can run safely in parallel with almost no modification:

int sum = numbers.parallelStream().reduce(0, Integer::sum);

So, for a map you would use:

integers.values().stream().mapToInt(i -> i).reduce(0, (x,y) -> x+y);

Or:

integers.values().stream().reduce(0, Integer::sum);
29
votes

You can use reduce method:

long sum = result.stream().map(e -> e.getCreditAmount()).reduce(0L, (x, y) -> x + y);

or

long sum = result.stream().map(e -> e.getCreditAmount()).reduce(0L, Integer::sum);
18
votes

You can use reduce() to sum a list of integers.

int sum = integers.values().stream().reduce(0, Integer::sum);
11
votes

You can use collect method to add list of integers.

List<Integer> list = Arrays.asList(2, 4, 5, 6);
int sum = list.stream().collect(Collectors.summingInt(Integer::intValue));
6
votes

This would be the shortest way to sum up int type array (for long array LongStream, for double array DoubleStream and so forth). Not all the primitive integer or floating point types have the Stream implementation though.

IntStream.of(integers).sum();
6
votes

I have declared a list of Integers.

ArrayList<Integer> numberList = new ArrayList<Integer>(Arrays.asList(1, 2, 3, 4, 5));

You can try using these different ways below.

Using mapToInt

int sum = numberList.stream().mapToInt(Integer::intValue).sum();

Using summarizingInt

int sum = numberList.stream().collect(Collectors.summarizingInt(Integer::intValue)).getSum();

Using reduce

int sum = numberList.stream().reduce(Integer::sum).get().intValue();
6
votes

May this help those who have objects on the list.

If you have a list of objects and wanted to sum specific fields of this object use the below.

List<ResultSom> somList = MyUtil.getResultSom();
BigDecimal result= somList.stream().map(ResultSom::getNetto).reduce(
                                             BigDecimal.ZERO, BigDecimal::add);
0
votes

Unfortunately looks like the Stream API only returns normal streams from, say, List<Integer>#stream(). Guess they're pretty much forced to because of how generics work.

These normal Streams are of generic objects so don't have specialized methods like sum() etc. so you have to use the weird re-stream "looks like a no-op" conversion by default to get to those methods... .mapToInt(i -> i).

Another option is using "Eclipse Collections" which are like an expanded java Stream API

IntLists.immutable.ofAll(integers.values()).sum();

-1
votes
class Pojo{
    int num;

    public Pojo(int num) {
        super();
        this.num = num;
    }

    public int getNum() {
        return num;
    }

    public void setNum(int num) {
        this.num = num;
    }
}

List<Pojo> list = new ArrayList<Pojo>();
            list.add(new Pojo(1));
            list.add(new Pojo(5));
            list.add(new Pojo(3));
            list.add(new Pojo(4));
            list.add(new Pojo(5));

            int totalSum = list.stream().mapToInt(pojo -> pojo.getNum()).sum();
            System.out.println(totalSum);
-1
votes

Most of the aspects are covered. But there could be a requirement to find the aggregation of other data types apart from Integer, Long(for which specialized stream support is already present). For e.g. stram with BigInteger For such a type we can use reduce operation like

list.stream().reduce((bigInteger1, bigInteger2) -> bigInteger1.add(bigInteger2))