Java 将对象从一个列表添加到另一种类型的列表的 Lambda 表达式
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原文地址: http://stackoverflow.com/questions/39042897/
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Lambda expression to add objects from one list to another type of list
提问by Débora
There is a List<MyObject>and it's objects are required to create object that will be added to another List with different elements : List<OtherObject>.
有一个List<MyObject>并且它的对象需要创建将添加到具有不同元素的另一个列表的对象：List<OtherObject>。
This is how I am doing,
这就是我的做法，
List<MyObject> myList = returnsList();
List<OtherObj> emptyList = new ArrayList();
for(MyObject obj: myList) {
OtherObj oo = new OtherObj();
oo.setUserName(obj.getName());
oo.setUserAge(obj.getMaxAge());
emptyList.add(oo);
}
I'm looking for a lamdbaexpression to do the exact same thing.
我正在寻找一个lamdba表达式来做完全相同的事情。
采纳答案by ByeBye
If you define constructor OtherObj(String name, Integer maxAge)you can do it this java8 style:
如果你定义构造函数，OtherObj(String name, Integer maxAge)你可以这样做 java8 风格：
myList.stream()
.map(obj -> new OtherObj(obj.getName(), obj.getMaxAge()))
.collect(Collectors.toList());
This will map all objects in list myListto OtherObjand collect it to new Listcontaining these objects.
这会将列表中的所有对象映射myList到OtherObj并将其收集到List包含这些对象的新对象。
回答by CoderCroc
You can create a constructor in OtherObjectwhich uses MyObjectattributes,
您可以创建一个OtherObject使用MyObject属性的构造函数，
public OtherObject(MyObject myObj) {
this.username = myObj.getName();
this.userAge = myObj.getAge();
}
and you can do following to create OtherObjects from MyObjects,
您可以执行以下操作以OtherObject从MyObjects创建s，
myObjs.stream().map(OtherObject::new).collect(Collectors.toList());
回答by kots_14
I see that this is quite old post. However, this is my take on this based on the previous answers. The only modification in my answer is usage of .collect(ArrayList::new, ArrayList::add,ArrayList:addAll).
我看到这是一个很老的帖子。但是，这是我根据以前的答案对此的看法。我的答案中唯一的修改是使用.collect(ArrayList::new, ArrayList::add,ArrayList:addAll).
Sample code :
示例代码：
List<OtherObj> emptyList = myList.stream()
.map(obj -> {
OtherObj oo = new OtherObj();
oo.setUserName(obj.getName());
oo.setUserAge(obj.getMaxAge());
return oo; })
.collect(ArrayList::new, ArrayList::add,ArrayList::addAll);
