Question

我有一个MyObjects列表... MyObject {int id，String name}。现在我想将列表拆分为具有相同“id”值的子列表，任何人都可以建议这样做的有效方法。

Answer 1

// create the thing to store the sub lists
Map<Integer, List<MyObject>> subs = new HashMap<Integer, List<MyObject>>();

// iterate through your objects
for(MyObject o : list){

    // fetch the list for this object's id
    List<MyObject> temp = subs.get(o.getId());

    if(temp == null){
        // if the list is null we haven't seen an
        // object with this id before, so create 
        // a new list
        temp = new ArrayList<MyObject>();

        // and add it to the map
        subs.put(o.getId(), temp);
    }

    // whether we got the list from the map
    // or made a new one we need to add our
    // object.
    temp.add(o);
}

Answer 2

使用Guava：

ListMultimap<Integer, MyObject> myObjectsById = Multimaps.index(myObjects,
    new Function<MyObject, Integer>() {
      public Integer apply(MyObject myObject) {
        return myObject.id;
      }
    });

Answer 3

如果您使用的是JDK 1.8，则可以使用优雅的解决方案，如：

Map<Integer, List<MyObject>> myObjectsPerId =
    myObjects.stream().collect(Collectors.groupingBy(MyObject::getId));

Answer 4

循环浏览元素，检查其id值，并将其放在Hashtable中，并以id为键。这就是O（N），这和你将获得的效率一样高。

Answer 5

使用JDK 1.8：

List<MyObject> objects= new ArrayList();
Map<Integer, List<MyObject>> obejctMap = new HashMap();
objects.stream().map(MyObject::getId).distinct().forEach(id -> obejctMap .put(id,
              objects.stream().filter(object -> id.equals(object.getId())).collect(Collectors.toList())));

Answer 6

ArrayList<MyObject> list=new ArrayList<MyObject>();
//fill Objects..
HashMap<Integer,ArrayList<MyObject>> hash=new HashMap<Integer,ArrayList<MyObject>>();
for(MyObject elem:list)//iterate the list
{
ArrayList<MyObject> tmp=null; //temporary variable 
if((tmp=hash.get(elem.getId()))==null) // check if id already present in map
 {
  tmp=new ArrayList<MyObject>();   
  hash.put(elem.getId(),tmp); //if not put a new array list
 }
names.add(elem); //if present add the name to arraylist
}

基于对象属性将java集合拆分为子集合

6 个答案: