Question

我正在尝试了解如何使用Java 8 Streams API。

例如，我有这两个类：

basepath = "pathtodir/DataFiles/"
for filename in os.listdir(basepath):
    if filename.endswith(".log"): 
        print(os.path.join("./DataFiles", filename))

        with open(basepath + filename) as openfile:    
            for line in openfile:
            ........

我有十个用户的public class User { private String name; public String getName() { return name; } public void setName(String name) { this.name = name; } } public class UserWithAge { private String name; public String getName() { return name; } public void setName(String name) { this.name = name; } private int age; public int getAge() { return age; } public void setAge(int age) { this.age = age; } }，我希望将其转换为具有相同名称且年龄恒定的十个用户的List<User>（例如，27）。如何使用Java 8 Streams API（没有循环，不修改上述类）？

Answer 1

您可以使用流的map()功能将列表中的每个User实例转换为UserWithAge实例。

List<User> userList = ... // your list

List<UserWithAge> usersWithAgeList = userList.stream()
        .map(user -> {
                // create UserWithAge instance and copy user name
                UserWithAge userWithAge = new UserWithAge();
                userWithAge.setName(user.getName());
                userWithAge.setAge(27);
                return userWithAge;
         })
         .collect(Collectors.toList()); // return the UserWithAge's as a list

Answer 2

虽然你可以做到这一点，但你不应该这样做。

List<UserWithAge> userWithAgeList = new ArrayList<UserWithAge>();

userList.stream().forEach(user -> {
                UserWithAge userWithAge = new UserWithAge();
                userWithAge.setName(user.getName());
                userWithAge.setAge(27);
                userWithAgeList.add(userWithAge);
});

Answer 3

public class ListIteratorExp {

    public static void main(String[] args) {
        List<Person> list = new ArrayList<>();
        Person p1 = new Person();
        p1.setName("foo");

        Person p2 = new Person();
        p2.setName("bee");

        list.add(p1);
        list.add(p2);

        list.stream().forEach(p -> {
            String name = p.getName();
            System.out.println(name);
        });
    }

}

class Person{
    private String name;

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

}

output:-
vishal
thakur

列表迭代＆amp;使用Java 8 Streams API设置值

3 个答案: