Question

我有一个ArrayList，其中有一些HashMap<String, String>。所以，我想在地图中比较相同的值。当我找到相同的值时，我想保留一张它们的地图。例如，考虑第二个地图和第五个地图（在arraylist中）具有相同的值。我想保留第二张地图并从arraylist中删除第五张。我尝试用迭代器，但我不能这样做。这似乎很复杂。你能举个例子吗？

这是我的最后一次尝试：

private HashMap<String, String> mapValues = new HashMap<String, String>();
private HashMap<String, String> mapValues2 = new HashMap<String,String>(); 
private HashMap<Integer, String> mval = new HashMap<Integer, String>();

//i take the ArrayList with the maps for comparison private
ArrayList<HashMap<String, String>> check(ArrayList<HashMap<String, String>> list) {           

 //a new ArrayList. It will have the maps(HashMap<key, value>) with no same values.
 ArrayList<HashMap<String, String>> listFinal = new ArrayList<HashMap<String, String();

    for (int i = 0; i < list.size(); i++) {
        mapValues = list.get(i);
        mval.put(i, mapValues.get("value"));
    }

    for (int i = 0; i < mval.size(); i++) {
        HashMap<String, String> newMapValues = new HashMap<String, String>();
        mapValues2 = list.get(i);
        String iVal = mapValues2.get("value");
        newMapValues = list.get(i);
        int flag = -1;
        int remove = -1;

        for (int j = i+1; j < mval.size()-1; j++) {
            String jVal = mval.get(j);
            if (val.compareTo(jVal) == 0) {
                flag = i;
                remove = j;
            }
        }
        if (flag == -1) {
            listFinal.add(newMapValues );
        } else if (flag != -1) {
            listFinal.remove(remove);
        }   
    }
}

Answer 1

List<Map<String, String>> mapList = new ArrayList<Map<String, String>>(); //Assuming string-string pairs for simplicity...
//... filling up list and maps...
Set<String> valueSet = new HashSet<String>();
for(Iterator<Map<String, String> mapIt = mapList.iterator(); mapIt.hasNext();) {
    final Map<String, String> map = mapIt.next();
    boolean hasDuplicate = false;
    for(final String mapValue : map.values()) {
        if(valueSet.contains(mapValue)
            hasDuplicate = true;
    }
    if(hasDuplicate)
        mapIt.remove();
    valueSet.addAll(map.values());
}

希望有人能够证明这一点，因为我没有在IDE中输入它而我还没有喝咖啡。

编辑：好的，以前的版本错了，因为地狱。请改为检查。

编辑2：刚刚意识到这也行不通。它可能会删除地图3，因为它具有地图2的重叠值，但地图2被删除，因为地图1的某些其他重重值。结果：仅保留地图1并删除地图2和3但是地图3没有地图1的欺骗。这比我想象的要复杂一点。最好喝咖啡......

Answer 2

只是大声思考，但我的方法是这样的：

创建一个Set，您可以在其中存储已在地图中找到的值。

每次在列表的新位置获取Map时，检查Set中是否存在Map的元素，如果是，则从ArrayList中删除Map（它是重复的），如果没有，将Map的值添加到Set and Carry on。

确保使用Iterator的remove方法从ArrayList中删除Map！

Answer 3

创建Set<HashMap<String,String>>并将list的每个成员添加到其中。问题解决了！

如果你绝对需要一个ArrayList而不是Set，你可以从ArrayList创建一个新的Set，但无论如何都是这样的：让Java做为你工作。与标准库相比，您不太可能在集合处理方面做得更好。

Answer 4

将Map键与Arraylist值进行比较

public static void main(String[] args) {

        Iterator<Entry<String, CustomerContactVO>> it = getVO().entrySet().iterator();

        List<CustomerOutPut> customerOutPutsList = new ArrayList<CustomerOutPut>();

        while(it.hasNext()){
            Entry<String, CustomerContactVO> ent = it.next();
            String contAcctIDKey = ent.getKey();
            String email = ent.getValue().getEmailID();
            CustomerOutPut customerOutPut = new CustomerOutPut();
            customerOutPut.setContactAcctIDVo(contAcctIDKey);
            customerOutPut.setEmailIDVo(email);

            for (CustomerPreferenceVO customerPreferenceVO : perfVo()) {
                if(customerPreferenceVO.getContactAcctID()!=null && customerPreferenceVO.getContactAcctID().equals(contAcctIDKey)){
                    customerOutPut.setContactAcctIDRef(customerPreferenceVO.getContactAcctID());
                    customerOutPut.setMktIndRef(customerPreferenceVO.getMktInd());
                    customerOutPut.setPrefIndRef(customerPreferenceVO.getPrefInd());
                }
            }

            customerOutPutsList.add(customerOutPut);
        }

        for (CustomerOutPut customerOutPut : customerOutPutsList) {
            System.out.println(customerOutPut.toString());
        }
    }

Answer 5

package com.test.examples;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;

import com.test.vo.CustomerContactVO;
import com.test.vo.CustomerOutPut;
import com.test.vo.CustomerPreferenceVO;

public class TestExampleOne {

    public static Map<String, CustomerContactVO> getVO(){

        Map<String, CustomerContactVO> contactVOMap = new HashMap<String, CustomerContactVO>();
        CustomerContactVO v = new CustomerContactVO();
        v.setContactAcctID("60011151");
        v.setEmailID("raj@gmail.com");

        CustomerContactVO v1 = new CustomerContactVO();
        v1.setContactAcctID("60011152");
        v1.setEmailID("raj1@gmail.com");

        CustomerContactVO v2 = new CustomerContactVO();
        v2.setContactAcctID("60011153");
        v2.setEmailID("raj2@gmail.com");

        CustomerContactVO v3 = new CustomerContactVO();
        v3.setContactAcctID("60011154");
        v3.setEmailID("raj3@gmail.com");

        CustomerContactVO v4 = new CustomerContactVO();
        v4.setContactAcctID("60011155");
        v4.setEmailID("raj4@gmail.com");

        contactVOMap.put("60011151", v);
        contactVOMap.put("60011152", v1);
        contactVOMap.put("60011153", v2);
        contactVOMap.put("60011154", v3);
        contactVOMap.put("60011155", v4);

        return contactVOMap;
    }

    public static List<CustomerPreferenceVO> perfVo(){
        CustomerPreferenceVO prefVo = new CustomerPreferenceVO();
        prefVo.setContactAcctID("60011151");
        prefVo.setMktInd("500");
        prefVo.setPrefInd("Y");


        CustomerPreferenceVO prefVo1 = new CustomerPreferenceVO();
        prefVo1.setContactAcctID("60011153");
        prefVo1.setMktInd("302");
        prefVo1.setPrefInd("N");

        CustomerPreferenceVO prefVo2 = new CustomerPreferenceVO();
        prefVo2.setContactAcctID("60011154");
        prefVo2.setMktInd("302");
        prefVo2.setPrefInd("Y");

        List<CustomerPreferenceVO> list = new ArrayList<CustomerPreferenceVO>();
        list.add(prefVo);
        list.add(prefVo1);
        list.add(prefVo2);

        return list;
    }

    public static void main(String[] args) {

        Iterator<Entry<String, CustomerContactVO>> it = getVO().entrySet().iterator();
        List<CustomerOutPut> customerOutPutsList = new ArrayList<CustomerOutPut>();

        while(it.hasNext()){

            Entry<String, CustomerContactVO> ent = it.next();
            String contAcctIDKey = ent.getKey();
            String email = ent.getValue().getEmailID();
            CustomerOutPut customerOutPut = new CustomerOutPut();
            customerOutPut.setContactAcctIDVo(contAcctIDKey);
            customerOutPut.setEmailIDVo(email);

            for (CustomerPreferenceVO customerPreferenceVO : perfVo()) {

                if(customerPreferenceVO.getContactAcctID()!=null && 
                        customerPreferenceVO.getContactAcctID().equals(contAcctIDKey)){

                    customerOutPut.setContactAcctIDRef(customerPreferenceVO.getContactAcctID());
                    customerOutPut.setMktIndRef(customerPreferenceVO.getMktInd());
                    customerOutPut.setPrefIndRef(customerPreferenceVO.getPrefInd());

                }
            }

            customerOutPutsList.add(customerOutPut);
        }

        for (CustomerOutPut customerOutPut : customerOutPutsList) {
            System.out.println(customerOutPut.toString());
        }
    }

}

比较ArrayList中的HashMap值

5 个答案: