我需要你的帮助,
在Java 8中使用收集器分组我需要将这样的列表分组
ValueObject {id=1, value=2.0}
ValueObject {id=2, value=2.0}
ValueObject {id=3, value=2.0}
ValueObject {id=4, value=3.0}
ValueObject {id=5, value=3.0}
ValueObject {id=6, value=4.0}
ValueObject {id=7, value=4.0}
ValueObject {id=8, value=4.0}
ValueObject {id=9, value=4.0}
ValueObject {id=10, value=4.0}
在另一个像这样的
GroupedObject {from=1, to=3, value=2.0}
GroupedObject {from=4, to=5, value=3.0}
GroupedObject {from=6, to=10, value=4.0}
这些是我正在使用的对象的定义
public class ValueObject {
private int id;
private double value;
public String getId() {
return id;
}
public float getValue() {
return value;
}
public void setValue(float value) {
this.value = value;
}
}
public class GroupedObject {
private int from;
private int to;
private double value;
public int getFrom() {
return from;
}
public void setFrom(int from) {
this.from = from;
}
public int getTo() {
return to;
}
public void setTo(int to) {
this.to = to;
}
public double getValue() {
return value;
}
public void setValue(double value) {
this.value = value;
}
}
这就是我以编程方式进行的方式。
public class Service {
public List<GroupedObject> groupToRange(List<ValueObject> list) {
List<GroupedObject> filtered = new ArrayList<>();
if (list.size() > 0) {
ValueObject current = list.get(0);
GroupedObject dto = new GroupedObject();
dto.setValue(current.getValue());
dto.setFrom(current.getId());
for (int i = 0; i < list.size(); i++) {
ValueObject vo = list.get(i);
if (vo.getValue() != current.getValue()) {
dto.setTo(current.getId());
filtered.add(dto);
dto = new GroupedObject();
dto.setValue(vo.getValue());
dto.setFrom(vo.getId());
current = vo;
} else {
current = vo;
}
if (i == list.size() - 1) {
dto.setTo(vo.getId());
filtered.add(dto);
}
}
}
return filtered;
}
}
这是单元测试
public class ServiceTest {
Service service = new Service();
@Test
public void testgGoupToRange() {
List entryList = new ArrayList<>();
entryList.add(new ValueObject(1, 2.0));
entryList.add(new ValueObject(2, 2.0));
entryList.add(new ValueObject(3, 2.0));
entryList.add(new ValueObject(4, 3.0));
entryList.add(new ValueObject(5, 3.0));
entryList.add(new ValueObject(6, 4.0));
entryList.add(new ValueObject(7, 4.0));
entryList.add(new ValueObject(8, 4.0));
entryList.add(new ValueObject(9, 4.0));
entryList.add(new ValueObject(10, 4.0));
List responseList = service.groupToRange(entryList);
responseList.forEach(obj-> System.out.println(obj.toString()));
assertNotNull(responseList);
assertEquals(3, responseList.size());
}
}
我找不到使用java 8和收藏家
的方法答案 0 :(得分:2)
这就是我提出的
List<ValueObject> values = Arrays.asList(new ValueObject(1, 2.0),
new ValueObject(2, 2.0),
new ValueObject(3, 3.0),
new ValueObject(4, 4.0),
new ValueObject(5, 4.0),
new ValueObject(6, 4.0));
Map<Double, IntSummaryStatistics> groupedValues = values.stream()
.collect(Collectors.groupingBy(ValueObject::getValue,
Collectors.summarizingInt(ValueObject::getId)));
List<GroupedObject> groupedObjects = groupedValues.entrySet()
.stream()
.map(groupedValue -> new GroupedObject(groupedValue.getValue().getMin(),
groupedValue.getValue().getMax(),
groupedValue.getKey()))
.collect(Collectors.toList());
System.out.println(groupedObjects);
我非常有信心有办法避免中间人Map<Double, IntSummaryStatistics
,但我还没有想出来。如果我这样做会更新。
编辑:请参阅@Tunaki的答案,获得1次答案。
这是输出
[GroupedObject{from=4, to=6, value=4.0}, GroupedObject{from=1, to=2, value=2.0}, GroupedObject{from=3, to=3, value=3.0}]