Strip Separator Elements

Question

Is there a good way to use Java threads to alternate elements in a thread with a separator of the same type?

// Expected result in is list: [1, 0, 2, 0, 3] List<Integer> is = Stream.of(1, 2, 3).intersperse(0).collect(toList()); 

This is similar to the intersperse function in Haskell and other functional languages.

I saw many examples of how to join strings in a similar way, but did not find solutions for shared lists.

Answer 1

You can do this with flatMap, but after the last element you will get an extra separator:

 List<Integer> is = IntStream.of(1, 2, 3) .flatMap(i -> IntStream.of(i, 0)) .collect(toList()); 

Here's another way, without a trailing separator:

 List<Integer> is = IntStream.of(1, 2, 3) .flatMap(i -> IntStream.of(0, i)) .skip(1) .collect(toList()); 

This time we add a separator in front of each source element and get rid of the leading separator.

Answer 2

You can do a similar thing that Collectors.joining does with String with a little helper method:

 static <T> Collector<T,List<T>,List<T>> intersperse(T delim) { return Collector.of(ArrayList::new, (l,e)-> { if(!l.isEmpty()) l.add(delim); l.add(e); }, (l,l2)-> { if(!l.isEmpty()) l.add(delim); l.addAll(l2); return l; }); 

then you can use it similarly to Collectors.joining(delimiter) :

 List<Integer> l=Stream.of(1, 2, 3).collect(intersperse(0)); 

produces

[1, 0, 2, 0, 3]

Please note that this is thread safe because of the collect method that protects these operations.

---

If you do not want to list, but insert delimiters as an intermediate stream, you can do this as follows:

 Integer delimiter=0; Stream.of(1, 2, 3)/** or any other stream source */ .map(Stream::of) .reduce((x,y)->Stream.concat(x, Stream.concat(Stream.of(delimiter), y))) .orElse(Stream.empty()) /* arbitrary stream operations may follow */