Question

你能帮我用ArrayList构建带有对象的Hashmap。

public class Store {

    private int storeId;
    private String storeName;
    private int deliveryDays;
    private List<String> servingZipCodes;
    private StoreAddress storeAddress;


    public Store() {
    }

    public Store(int storeId,String storeName,int deliveryDays,List<String> servingZipCodes,StoreAddress storeAddress){
        this.storeId = storeId;
        this.storeName=storeName;
        this.deliveryDays=deliveryDays;
        this.servingZipCodes=servingZipCodes;
        this.storeAddress = storeAddress;
    }

//getters and setters.
}

和StoreAddress类

public class StoreAddress {

    private String streetAddress1;
    private String streetAddress2;
    private String streetAddress3;
    private String city;
    private String state;
    private String postalCode;
    private String country;
    private String phone;

    public StoreAddress() {
    }

    public StoreAddress(String streetAddress1, String streetAddress2, String streetAddress3, String city, String state, String postalCode, String country, String phone) {
        this.streetAddress1 = streetAddress1;
        this.streetAddress2 = streetAddress2;
        this.streetAddress3 = streetAddress3;
        this.city = city;
        this.state = state;
        this.postalCode = postalCode;
        this.country = country;
        this.phone = phone;
    }

这是用于测试的Test类。

public class Test {

    public static void main(String args[]){

        List<Store> storeList=new ArrayList();
        StoreAddress storeAddress1 = new StoreAddress("1500 Polaris Pkwy",null,null,"Columbus","OH","43240","US","9165452345");
        StoreAddress storeAddress2 = new StoreAddress("160 Easton Town Center",null,null,"Columbus","OH","43240","US","9165452345");
        storeList.add(new Store(1,"Store 1",7,new ArrayList<String>(Arrays.asList("43240","43241","43242")),storeAddress1));
        storeList.add(new Store(2,"Store 2",7,new ArrayList<String>(Arrays.asList("43240","43082","43081")),storeAddress2));
        Map<String,List<Store>> zipCodeStoreList = null;
        storeList.forEach(store -> {
            List<String> servingZipCodes = store.getServingZipCodes();
            servingZipCodes.stream().filter(x -> x.matches(store.getStoreAddress().getPostalCode().substring(0,5))).map(x ->new HashMap<String, Object>(){{
                put(x, store);
            }});
        });

虽然它可以在Java 7中使用，但在java 8中寻找解决方案。

Key: 43240, value: Store1 , Store2
Key: 43241, value: null
Key: 43242, value: null
Key: 43082, value: null
Key: 43081, value: null

Answer 1

正如我所说，您应该首先考虑，而不是如何，然后问题很容易解决。然后你可以获得自描述代码，例如：

//             v--- union all zip codes from stores
Stream<String> union = storeList.stream().map(Store::getServingZipCodes)
                                         .flatMap(List::stream)
                                         .distinct();

//             v--- find out intersections between zip codes
Stream<String> intersection = union.filter(zip ->
        storeList.stream().map(Store::getServingZipCodes)
                          .allMatch(it -> it.contains(zip))
);

//                       v--- create a Map simply from intersections 
Map<String, List<Store>> result = intersection.collect(toMap(
        Function.identity(),
        unused -> storeList
));

输出

assert result.get("43240") == [store1, store2];
assert result.get("others") == null;

Answer 2

我首先使用给定的邮政编码过滤List Store的{{1}}。这个给定的邮政编码是变量名key。然后，我使用密钥将生成的过滤器放入Map变量zipCodeStoreList。

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;

/**
 *
 * @author blj0011
 */
public class Test
{

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args)
    {
        List<Store> storeList=new ArrayList();
        StoreAddress storeAddress1 = new StoreAddress("1500 Polaris Pkwy",null,null,"Columbus","OH","43240","US","9165452345");
        StoreAddress storeAddress2 = new StoreAddress("160 Easton Town Center",null,null,"Columbus","OH","43240","US","9165452345");

        storeList.add(new Store(1,"Store 1",7,new ArrayList(Arrays.asList("43240","43241","43242")),storeAddress1));
        storeList.add(new Store(2,"Store 2",7,new ArrayList(Arrays.asList("43240","43082","43081")),storeAddress2));
        Map<String,List<Store>> zipCodeStoreList = new HashMap();

        String key = "43241";//See if a Store associates with this zip code.
        //Add the key and the results from filtering the storeList based on the key.
        zipCodeStoreList.put(key, storeList.stream().filter(x -> x.getServingZipCodes().contains(key)).collect(Collectors.toList()));
        //Print results
        for(Map.Entry<String, List<Store>> entry : zipCodeStoreList.entrySet())
        {
            for(Store store : entry.getValue())
            {
                System.out.println("filter 1: " + entry.getKey() + " - " + store.getStoreName());
            }
        }        
    }

}

修改版本，其中列出了与特定邮政编码相关联的所有Stores。

删除

String key = "43081"; zipCodeStoreList.put(key, storeList.stream().filter(x -> x.getServingZipCodes().contains(key)).collect(Collectors.toList()));

替换为

//Add the zip code belong to at least one store plus a zip code that belongs to no store. List<String> allZipCodes = new ArrayList(Arrays.asList("43240","43082","43081", "43240","43241","43242", "55555")); for(String zipCode : allZipCodes) { zipCodeStoreList.put(zipCode, storeList.stream().filter(x -> x.getServingZipCodes().contains(zipCode)).collect(Collectors.toList())); }

使用lambda表达式列出到Hashmap

2 个答案:

输出