我在从 ArrayList 中删除重复对象时遇到问题。我将 XML 解析为我所说的 IssueFeed 对象。这包括症状、问题、解决方案。

我的大多数对象都是独一无二的，没有共同的症状、问题、解决方案，但有些对象具有相同的症状但有不同的问题。

我试图完成几件事。

1. 捕获与重复 Arraylist 具有相同症状的对象
2. 从主列表中删除重复项，至少保留 1 个具有该症状的项进行显示。
3. 当用户点击我们知道有重复的项目时，在我的 ListView /适配器中设置重复数据 Arraylist。

我采取的步骤。

1. 我已尝试对对象进行排序并且我能够捕获重复项，但不确定如何从主列表中删除除一个以外的所有对象。
2. 2 在列表之间循环并查找不是自身的对象和 symptom = symptom，然后删除并更新我的重复数组和主数组。

一些代码

IssueFeed - 对象

public IssueFeed(String symptom, String problem, String solution) {
    this.symptom = symptom;
    this.problem = problem;
    this.solution = solution;
}
public String getSymptom() {
    return symptom;
}
public String getProblem() {
    return problem;
}
public String getSolution() {
    return solution;
}

我的 ArrayList<IssueFeed>的

duplicateDatalist = new ArrayList<IssueFeed>(); // list of objects thats share a symptom

list_of_non_dupes = new ArrayList<IssueFeed>(); // list of only objects with unique symptom

mIssueList = mIssueParser.parseLocally(params[0]); // returns ArrayList<IssueFeed> of all objects

我可以通过以下方式获得副本 sort下面的代码。

Collections.sort(mIssueList, new Comparator<IssueFeed>(){
            public int compare(IssueFeed s1, IssueFeed s2) {
                if (s1.getSymptom().matches(s2.getSymptom())) {
                    if (!duplicateDatalist.contains(s1)) {
                        duplicateDatalist.add(s1);
                        System.out.print("Dupe s1 added" + " " + s1.getSymptom() + ", " + s1.getProblem() + "\n");
                    }
                    if (!duplicateDatalist.contains(s2)) {
                        duplicateDatalist.add(s2);
                        System.out.print("Dupe s2 added" + " " + s2.getSymptom() + ", " + s2.getProblem() + "\n");
                    }
                }
                return s1.getSymptom().compareToIgnoreCase(s2.getSymptom());
            }
        });

现在我需要创建新的非欺骗列表，这段代码只添加了所有对象。 :/

for (int j = 0; j < mIssueList.size(); j++) {
            IssueFeed obj = mIssueList.get(j);

            for (int i = 0; i < mIssueList.size(); i++) {
                IssueFeed obj_two = mIssueList.get(j);

                if (obj.getSymptom().matches(obj_two.getSymptom())) {
                    if (!list_non_dupes.contains(obj_two)) {
                        list_non_dupes.add(obj_two);
                    }
                    break;
                } else {
                    if (!list_non_dupes.contains(obj_two)) {
                        list_non_dupes.add(obj_two);
                    }
                }
            }
        }

最佳答案

如果您可以修改 IssueFeed 对象，请考虑覆盖 equals() 和 hashCode() 方法并使用集合查找重复项.例如

import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedHashSet;
import java.util.List;
import java.util.Set;

class IssueFeed {
    private String symptom;
    private String problem;
    private String solution;

    public IssueFeed(String symptom, String problem, String solution) {
        this.symptom = symptom;
        this.problem = problem;
        this.solution = solution;
    }
    public String getSymptom() {
        return symptom;
    }
    public String getProblem() {
        return problem;
    }
    public String getSolution() {
        return solution;
    }
    @Override
    public int hashCode() {
        final int prime = 31;
        int result = 1;
        result = prime * result + ((symptom == null) ? 0 : symptom.hashCode());
        return result;
    }
    @Override
    public boolean equals(Object obj) {
        if (this == obj)
            return true;
        if (obj == null)
            return false;
        if (getClass() != obj.getClass())
            return false;
        IssueFeed other = (IssueFeed) obj;
        if (symptom == null) {
            if (other.symptom != null)
                return false;
        } else if (!symptom.equals(other.symptom))
            return false;
        return true;
    }
    @Override
    public String toString() {
        return "IssueFeed [symptom=" + symptom + ", problem=" + problem
                + ", solution=" + solution + "]";
    }
}

public class Sample {

    public static void main(String[] args) {
        List<IssueFeed> mainList = new ArrayList<IssueFeed>(
                Arrays.asList(new IssueFeed[] {
                        new IssueFeed("sym1", "p1", "s1"),
                        new IssueFeed("sym2", "p2", "s2"),
                        new IssueFeed("sym3", "p3", "s3"),
                        new IssueFeed("sym1", "p1", "s1") }));
        System.out.println("Initial List : " + mainList);
        Set<IssueFeed> list_of_non_dupes = new LinkedHashSet<IssueFeed>();
        List<IssueFeed> duplicateDatalist = new ArrayList<IssueFeed>(); 
        for(IssueFeed feed : mainList){
            if(!list_of_non_dupes.add(feed)) {
                duplicateDatalist.add(feed);
            }
        }
        mainList = new ArrayList<IssueFeed>(list_of_non_dupes); // Remove the duplicate items from the main list, leaving at least 1 item with that symptom to be display
        list_of_non_dupes.removeAll(duplicateDatalist); // list of only objects with unique symptom
        System.out.println("Fina main list : " + mainList);
        System.out.println("Unique symptom" + list_of_non_dupes);
        System.out.println("Duplicate symptom" + duplicateDatalist);
    }
}

关于java - 从 ArrayList 中捕获重复项，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/25044726/