20

I have a List of Integer say list1, and I want to get another list list2 which will contain the cumulative sum up until the current index from start. How can I do this using Stream API java 8 ?

List<Integer> list1 = new ArrayList<>();
list1.addAll(Arrays.asList(1, 2, 3, 4));
List<Integer> list2 = new ArrayList<>();
// initialization
list2.add(list1.get(0));
for(int i=1;i<list1.size();i++) {
// increment step
    list2.add(list2.get(i-1) + list1.get(i));
}

How can I change above imperative style code into declarative one ?

list2 should be [1, 3, 6, 10]
Stefan Zobel
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run_time_error
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    You'll notice from the answers that any solution using streams is going to be inefficient. Streams aren't really intended for this use case. (In general, streams aren't intended to replace all imperative code.) – Louis Wasserman Mar 20 '19 at 16:48
  • @LouisWasserman I agree. But in this case I do not care about efficiency (probably should have mentioned it in the question). I wanted to write the same thing with something other than above mentioned imperative style. I could not do it myself, as I was stuck because the solution is kind of mixture of map and reduce operation – run_time_error Mar 20 '19 at 17:17
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    Man, the lack of tuples really hurts here. – Alexander Mar 20 '19 at 22:19
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    An extremely similar question was asked yesterday: https://stackoverflow.com/questions/55230261/map-to-a-running-sum-in-java-8 – Jacob G. Mar 21 '19 at 00:41

5 Answers5

21

Streams are not suited for this kind of task, as there is state involved (the cumulative partial sum). Instead, you could use Arrays.parallelPrefix:

Integer[] arr = list1.toArray(Integer[]::new);

Arrays.parallelPrefix(arr, Integer::sum);

List<Integer> list2 = Arrays.asList(arr);

This first copies list1 to an array by using Collection.toArray, which is available since JDK 11. If you are not on Java 11 yet, you could replace the first line with the traditional toArray call:

Integer[] arr = list1.toArray(new Integer[0]);

This solution doesn't use streams, yet it's declarative, because Arrays.parallelPrefix receives the cumulative operation as an argument (Integer::sum in this case).

Time complexity is O(N), though there might be some non-minor constant costs involved associated with setting up the infrastructure needed for parallel processing. However, according to the docs:

Parallel prefix computation is usually more efficient than sequential loops for large arrays

So it seems it's worth giving this approach a try.

Also, it's worth mentioning that this approach works because Integer::sum is an associative operation. This is a requirement.

fps
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    Never heard of parallel prefix. thank you very much for teaching me a new concept. :). I will definitely give a try. – run_time_error Mar 20 '19 at 17:26
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    @run_time_error [Here's more reading](https://en.wikipedia.org/wiki/Prefix_sum) about the subject. – fps Mar 20 '19 at 18:26
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    Don’t make your life unnecessary hard. Use `list1.toArray(new Integer[0])`. As elaborated in [this great article](https://shipilev.net/blog/2016/arrays-wisdom-ancients/), using a zero size is even more efficient in the commonly used Hotspot JVM. That’s one of the reasons why [JDK 11’s new method](https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/util/Collection.html#toArray(java.util.function.IntFunction)) just does the same: “*The default implementation calls the generator function with zero and then passes the resulting array to `toArray(T[])`.*” – Holger Mar 21 '19 at 07:47
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    @run_time_error also worth reading: [How does newly introduced Arrays.parallelPrefix(…) in Java 8 work?](https://stackoverflow.com/q/52961981/2711488) – Holger Mar 21 '19 at 07:56
6

For every index: iterate from zero to that index, get each element, and get the sum
Box the ints to Integers
Collect to a list

IntStream.range(0, list1.size())
    .map(i -> IntStream.rangeClosed(0, i).map(list1::get).sum())
    .boxed()
    .collect(Collectors.toList());

You're adding every number together every time, rather than reusing the previous cumulative result, but streams do not lend themselves to looking at results from previous iterations.

You could write your own collector but at this point, honestly why are you even bothering with streams?

list1.stream()
    .collect(
        Collector.of(
            ArrayList::new,
            (a, b) -> a.add(a.isEmpty() ? b : b + a.get(a.size() - 1)),
            (a, b) -> { throw new UnsupportedOperationException(); }
        )
    );
Michael
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6

An O(n) (works only sequentially) solution would be the following, but I don't find it very elegant. I guess it is a matter of taste

AtomicInteger ai = new AtomicInteger();
List<Integer> collect = list1.stream()
                             .map(ai::addAndGet)
                             .collect(Collectors.toList());
System.out.println(collect); // [1, 3, 6, 10]
Yassin Hajaj
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    This solution only works as long as the stream is sequential. Parallelising it would break this completely. – Ben R. Mar 20 '19 at 17:30
  • @ben In parallel stream you couldn't have cumulative sum right? – Venkataraghavan Yanamandram Mar 21 '19 at 08:40
  • Ruslan's answer allows for parallel streaming. The only stateful operation is over the original list itself, which we can probably assume with relative safety is read-only. Each item in the list in his solution could query over the list independently of the other stream elements. – Ben R. Mar 21 '19 at 09:21
  • @VenkataraghavanYanamandram with other solutions yes, but they are `O(n^2)` – Yassin Hajaj Mar 21 '19 at 09:22
3

You can use sublist to sum up until the current index from start:

List<Integer> list = IntStream.range(0, list1.size())
        .mapToObj(i -> list1.subList(0, i + 1).stream().mapToInt(Integer::intValue).sum())
        .collect(Collectors.toList());
Ruslan
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2

You can just use Stream.collect() for that:

List<Integer> list1 = Arrays.asList(1, 2, 3, 4);
List<Integer> list2 = list1.stream()
        .collect(ArrayList::new, (sums, number) -> {
            if (sums.isEmpty()) {
                sums.add(number);
            } else {
                sums.add(sums.get(sums.size() - 1) + number);
            }
        }, (sums1, sums2) -> {
            if (!sums1.isEmpty()) {
                int sum = sums1.get(sums1.size() - 1);
                sums2.replaceAll(num -> sum + num);
            }
            sums1.addAll(sums2);
        });

This solution also works for parallel streams. Use list1.parallelStream() or list1.stream().parallel() instead of list1.stream().

The result in both cases is: [1, 3, 6, 10]

Samuel Philipp
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  • This should be the accepted answer as it is correct regardless whether the stream is parallel or not and O(n) in any case. – michid Feb 03 '22 at 09:50
  • Here's my take at making this stateless: https://stackoverflow.com/a/70969797/402428 – michid Feb 03 '22 at 10:45