Java 8 partition data from a list to two separate list

Question

I have deriving two separate groups from a list, error data and error free data:

List<ProductHolder> errorOnes = holderList.stream()
                                                      .filter(holder-> (holder.getRecord().isX() || holder.getRecord().isY()))
                                                      .collect(Collectors.toList());

List<ProductHolder> nonErrorOnes = holderList.stream()
                                                      .filter(holder-> (!holder.getRecord().isX() && !holder.getRecord().isY()))
                                                      .collect(Collectors.toList());

The above has multiple calls to .stream().

Is there a way I can partition the above into separate lists using single call to .stream() ? Is there a performance gain in doing so?

how to then access these data? I presume I need to hold use a Map?

Eran · Accepted Answer · 2020-10-28T13:44:06.493

3

~~First of all, your two conditions do not partition the original List. i.e. the second condition is not the negation of the first.~~

If you want a true partitioning, you can use Collectors.partitioningBy:

Map<Boolean, List<ProductHolder>> partition = 
    holderList.stream()
              .collect(Collectors.partitioningBy(holder-> (holder.getRecord().isX() || holder.getRecord().isY())));

The first partition (that satisfies the condition) will be obtained via partition.get(true), and the second via partition.get(false).

edited Oct 28 '20 at 13:44

answered Oct 28 '20 at 13:33

Eran

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updated the condition. would the answer still apply ? – M06H Oct 28 '20 at 13:40
@M06H Yes (except for the first sentence that addressed the old version). See [Collectors.partitioningBy](https://docs.oracle.com/en/java/javase/15/docs/api/java.base/java/util/stream/Collectors.html#partitioningBy(java.util.function.Predicate)) – Hulk Oct 28 '20 at 13:41
@M06H yes, now `partition.get(true)` would contain the elements of `errorOnes` and `partition.get(false)` the elements of `nonErrorOnes`. – Eran Oct 28 '20 at 13:42

Menelaos · Answer 2 · 2020-10-28T14:00:15.547

Intro

You can use Java 8 Collectors partitioningBy. So you could split your list into a map of lists and then get the two lists from there.

Short Solution

Map<Boolean, List<ProductHolder>> groups =
            holderList.stream().collect(Collectors.groupingBy(s ->!s.getRecord().isX() && !s.getRecord().isY()));
    
    List<ProductHolder> resultOne = groups.get(true);
    List<ProductHolder> resultTwo = groups.get(false);

Full Example

@Data @Builder
    static
    class ProductHolder{
        PRecord record;
    }

    @Data @Builder
    static class PRecord{
        boolean x;  boolean y;
    }


  List<ProductHolder> holderList = new ArrayList<>() {{
                add(ProductHolder.builder().record(PRecord.builder().x(true).y(false).build()).build());
                add(ProductHolder.builder().record(PRecord.builder().x(true).y(true).build()).build());
                add(ProductHolder.builder().record(PRecord.builder().x(false).y(false).build()).build());
                add(ProductHolder.builder().record(PRecord.builder().x(false).y(false).build()).build());
                add(ProductHolder.builder().record(PRecord.builder().x(false).y(false).build()).build());
            }};

Map<Boolean, List<ProductHolder>> groups =
                holderList.stream().collect(Collectors.groupingBy(s ->!s.getRecord().isX() && !s.getRecord().isY()));

        List<ProductHolder> resultOne = groups.get(true);
        List<ProductHolder> resultTwo = groups.get(false);

References

https://www.baeldung.com/java-list-split

Java 8 partition data from a list to two separate list

2 Answers2

Intro

Short Solution

Full Example

References