Question

如何根据Java 8中的“String”属性从对象列表中删除重复项？我只能找到Int或long属性的代码，但是当string是不同的情况并获得唯一列表时，无法找到字符串比较。

以下是我想要实现的计划。

public class Diggu {

    class Employee{
        private int id;
        private String name;
        public Employee(int id, String name){
            this.id = id;
            this.name = name;
        }

        public int getId() {
            return id;
        }

        public void setId(int id) {
            this.id = id;
        }

        public String getName() {
            return name;
        }

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



    }

    public void ra(){
        List<Employee> employee = Arrays.asList(new Employee(1, "John"), new Employee(3, "JOHN"), new Employee(2, "BOB"));
        System.out.println(""+ employee.size());
        List<Employee> unique = employee.stream()
                                .collect(collectingAndThen(toCollection(() -> new TreeSet<>(comparing(Employee::getName))),
                                                           ArrayList::new));
        System.out.println(""+ unique.size());
    }
    public static void main(String[] args) {
        new Diggu().ra();
    }
}

结果：

run:
3
3
BUILD SUCCESSFUL (total time: 2 seconds)

它应该是3和2

Answer 1

JOHN和Comparator是不同的字符串，使用不区分大小写的Comparator.comparing(Employee::getName, String.CASE_INSENSITIVE_ORDER)

// Given an array of elements (here just Ints):
let array = (0..<1000).map { _ in Int(arc4random_uniform(100)) }

// Sort it:
let sorted = array.sorted()

// Define an empty result (array of elements) which is a variable 
// and which gets modified in the subsequent reduce function:
var unique: [Int] = []

// A tailored reduce which depends on a sorted array and appends 
// to the result IFF that element is not the last in result:
let result = sorted.reduce(into: unique) { (result, element) in
    if let last = result.last, last == element {
    } else {
        result.append(element)
    }
}

根据Java 8

1 个答案: