Question

我想比较两个arraylist内容。

我以这种方式存储对象。

代表Arraylist 1：

Employee e1=new Employee();

e1.setID("1");
e1.setID("2");

ArrayList<Employee>list1 = new ArrayList<Employee>(); 

if(e1!=null){
    list1.add(e1);
}

代表Arraylist 2：

Employee e2=new Employee();

e2.setID("1");
e2.setID("2");
e2.setID("4");

ArrayList<Employee>list2 = new ArrayList<Employee>(); 

if(e2!=null){
    list2.add(e2);
}

现在我试图以这种方式比较上面的arraylist内容

ArrayList<Employee>unmatchedList = new ArrayList<Employee>();   

for (Employee l1 : list1){                               
    if(!list2.contains(l1.getID())){    
        System.out.println("list2 does not contains this ID"+l1.getID());   
        Employee e3=new Employee();          
        e3.setID(l1.getID());

        if(unmatchedList==null){            
        unmatchedList=new ArrayList<Employee>();            
        unmatchedList.add(e3);          
        }

        if(unmatchedList!=null){                            
            unmatchedList.add(e3);              
        }                       
    }
}

但我没有将正确的unmatchedList内容仅作为“4”。我得到unmatchedList为“1”和“2”这是错误的。那么如何才能在“unmatchedList”中获得不匹配的内容

Answer 1

如果您的班级Employee定义如下：

public class Employee {

private int id;

public Employee(int id){
    this.id = id;
}

public int getId() {
    return id;
}

@Override
public String toString() {
    return "Id : " + this.id;
}

@Override
public boolean equals(Object obj) {
    return (obj instanceof Employee) && this.id == ((Employee)obj).getId();
}

在

方法中，您可以检索不匹配的内容，如下所示：

 public static void main( String[] args )
{
   List<Employee> l1 = new ArrayList<Employee>();
   l1.add(new Employee(1));
   l1.add(new Employee(2));
   l1.add(new Employee(3));
   l1.add(new Employee(4));
   l1.add(new Employee(5));


   List<Employee> l2 = new ArrayList<Employee>();
   l2.add(new Employee(4));
   l2.add(new Employee(5));


   l1.removeAll(l2);
   System.out.println(l1);

}

这将打印：[Id : 1, Id : 2, Id : 3]

请注意，对于此项工作，您必须覆盖equals方法。

Answer 2

问题是这段代码： -

if(!list2.contains(l1.getID())){

您的list2是ArrayList<Employee>，您正在检查对l1.getId()的遏制。因此，您的if条件将始终为真。

您应该覆盖equals课程中的hashCode和Employee方法，并使用： -

if(!list2.contains(l1))

用于检查list2是否包含员工l1。

为什么在将id添加到列表之前将其设置为Employee 3次。它不会添加所有三个id，而只会添加一个e2.setID("1"); e2.setID("2"); e2.setID("4"); list.add(e2); // Will only add one Employee with id = 4，其中最后一个值设置为id。你需要纠正它： -

{{1}}

Answer 3

向您的员工添加equals方法，根据他们的ID对其进行比较，制作list2的副本并致电removeAll：

List<Employee> list1 = ...;
List<Employee> list2 = ...;
List<Employee> unmatchedList = new ArrayList<Employee>(list2);
unmatchedList.removeAll(list1);

有关完整示例，请参阅@ Dimitri的回答。

Answer 4

我认为你应该在你的场景中使用Set。可能这个例子对你有帮助。

Employee班

package com.yourcomp;

/**
 * <br>
 * <div style="width:600px;text-align:justify;">
 *
 * TODO: Class comment.
 *
 * </div>
 * <br>
 *
 */
public class Employee {

    private int id;

    /**
     * Constructor for Employee. <tt></tt>
     */
    public Employee() {
        this(-1);
    }

    /**
     * Constructor for Employee. <tt></tt>
     */
    public Employee(int id) {
        this.id = id;
    }

    /**
     * Gets the id.
     * 
     * @return <tt> the id.</tt>
     */
    public int getId() {
        return id;
    }

    /**
     * Sets the id.
     *
     * @param id <tt> the id to set.</tt>
     */
    public void setId(int id) {
        this.id = id;
    }

    @Override
    public boolean equals(Object obj) {
        return this.id == ((Employee)obj).id;
    }

    /* (non-Javadoc)
     * @see java.lang.Object#toString()
     */
    @Override
    public String toString() {
        return "Employee [id=" + id + "]";
    }

    /* (non-Javadoc)
     * @see java.lang.Object#hashCode()
     */
    @Override
    public int hashCode() {
        // TODO Auto-generated method stub
        return new Integer(id).hashCode();
    }



}

TestEmployee类：

package com.yourcomp;

import java.util.HashSet;
import java.util.Set;

/**
 * <br>
 * <div style="width:600px;text-align:justify;">
 *
 * TODO: Class comment.
 *
 * </div>
 * <br>
 * 
 */
public class TestEmployee {

    public static void main(String[] args) {
        // creating the first set with employees having ID's 1 to 5
        Set<Employee> set1 = new HashSet<Employee>();
        for(int i=1;i<5;i++) 
            set1.add(new Employee(i));

        System.out.println(set1); // printing it

        // creating the first set with employees having ID's 3 to 8.
        // note that now you have two employees with same ID, 3 & 4
        Set<Employee> set2 = new HashSet<Employee>();
        for(int i=3;i<8;i++)
            set2.add(new Employee(i));

        System.out.println(set2);// printing the second set

        // creates a final set to contain all elements from above two sets without duplicates
        Set<Employee> finalSet = new HashSet<Employee>();
        for(Employee employee:set1)
            finalSet.add(employee); // adds first set content


        for(Employee employee:set2)
            finalSet.add(employee); // adds second set content. If any duplicates found, it will be overwritten

        System.out.println(finalSet); // prints the final set



    }
}

和ouput

[Employee [id=1], Employee [id=2], Employee [id=3], Employee [id=4]]
[Employee [id=3], Employee [id=4], Employee [id=5], Employee [id=6], Employee [id=7]]
[Employee [id=1], Employee [id=2], Employee [id=3], Employee [id=4], Employee [id=5], Employee [id=6], Employee [id=7]]

Answer 5

我还建议使用Set来做差异，如果你可以使用Sets，那么我会使用Guava-Sets的差异方法： http://google-collections.googlecode.com/svn/trunk/javadoc/com/google/common/collect/Sets.html#difference%28java.util.Set,%20java.util.Set%29

比较两个arraylist内容并将不匹配的内容存储在另一个arraylist中

5 个答案: