どのようにこの3つの注文リストを1つの行に同じ名前のアイテムのマージを1つのリストにマージするのですか？

Question

私は顧客がショップから購入したアイテムを3つ注文します。 以下は注文の線のモデルです。以下

public class ShopOrderItem { 

private String name; 
private String itemcode; 
private int count; 
private double price; 
private String taxRate; 



public ShopOrderItem() { 
    super(); 
} 
public ShopOrderItem(String name, String itemcode, int count, double price, 
     String taxRate) { 
    super(); 
    this.name = name; 
    this.itemcode = itemcode; 
    this.count = count; 
    this.price = price; 
    this.taxRate = taxRate; 
} 
public String getName() { 
    return name; 
} 
public void setName(String name) { 
    this.name = name; 
} 
public String getItemcode() { 
    return itemcode; 
} 
public void setItemcode(String itemcode) { 
    this.itemcode = itemcode; 
} 
public int getCount() { 
    return count; 
} 
public void setCount(int count) { 
    this.count = count; 
} 
public double getPrice() { 
    return price; 
} 
public void setPrice(double price) { 
    this.price = price; 
} 
public String getTaxRate() { 
    return taxRate; 
} 
public void setTaxRate(String taxRate) { 
    this.taxRate = taxRate; 
} 

}

と3桁を取得する方法です。

public class testList { 

ShopOrderItem a = new ShopOrderItem("apple","001",2,(double)4.00,"0.17"); 
ShopOrderItem b = new ShopOrderItem("banana","002",1,(double)2.00,"0.17"); 
ShopOrderItem c = new ShopOrderItem("light","003",1,(double)30.00,"0.17"); 
ShopOrderItem d = new ShopOrderItem("apple","001",5,(double)4.00,"0.17"); 
ShopOrderItem e = new ShopOrderItem("light","003",10,(double)30.00,"0.17"); 
ShopOrderItem f = new ShopOrderItem("apple","001",1,(double)4.00,"0.17"); 
ShopOrderItem g = new ShopOrderItem("pen","004",3,(double)3.00,"0.17"); 

List<ShopOrderItem> orderA = new ArrayList<ShopOrderItem>(); 
orderA.add(a); 
orderA.add(b); 
orderA.add(c); 

List<ShopOrderItem> orderB = new ArrayList<ShopOrderItem>(); 
orderB.add(d); 
orderB.add(e); 

List<ShopOrderItem> orderC = new ArrayList<ShopOrderItem>(); 
orderC.add(f); 
orderC.add(g); 

}

私は1つのリストに、この3リストをマージしたいです。 同じ名前の場合は、カウントを追加します。 例えば、私の新しいリストは1行、名前はリンゴ、カウントは3オーダーのリンゴの合計です。 どのようにこれらの3つのリストをマージしますか？

Answer 1

使用法：groupByNameAndSumCount(orderA, orderB, orderC)

static List<ShopOrderItem> groupByNameAndSumCount(List<ShopOrderItem>... orders) { 
    List<ShopOrderItem> merged = new ArrayList<>(); 
    for (List<ShopOrderItem> order : orders) { 
     merged.addAll(order); 
    } 

    // group all by name as this is our invariant 
    HashMap<String, ShopOrderItem> grouped = new HashMap<>(); 
    for (ShopOrderItem item : merged) { 
     String name = item.getName(); 
     if (grouped.containsKey(name)) { 
      ShopOrderItem current = grouped.get(name); 
      // sum count 
      current.setCount(current.getCount() + item.getCount()); 
     } else { 
      grouped.put(name, item); 
     } 
    } 

    // reusing to reduce memory footprint 
    merged.clear(); 

    // prepare result 
    for (String key : grouped.keySet()) { 
     merged.add(grouped.get(key)); 
    } 
    return merged; 
} 

それは、先生がお役に立てば幸い！ Javaの8を使用して

Answer 2

3つのリストを1つのリストに追加し、並べ替えてマップに入れることができます。地図をリストに変換します。

List<ShopOrderItem> merged = new ArrayList<ShopOrderItem>(); 
    merged.addAll(orderA); 
    merged.addAll(orderB); 
    merged.addAll(orderC); 

    Collections.sort(merged, new Comparator<ShopOrderItem>() { 

     @Override 
     public int compare(ShopOrderItem o1, ShopOrderItem o2) { 
      return (Integer.parseInt(o1.getItemcode()) - Integer.parseInt(o2.getItemcode())); 

     } 
    }); 
    HashMap<String,Integer> map = new HashMap<String,Integer>(); 
    for (ShopOrderItem shopOrderItem : merged) { 
     if(map.containsKey(shopOrderItem.getItemcode())){ 
      int current = map.get(shopOrderItem.getItemcode()); 
      current+=shopOrderItem.getCount(); 
      map.remove(shopOrderItem.getItemcode()); 
      map.put(shopOrderItem.getItemcode(), current); 
     }else{ 
      map.put(shopOrderItem.getItemcode(), shopOrderItem.getCount()); 
     } 
    } 
    List<ShopOrderItem> result = new ArrayList<ShopOrderItem>(); 
    Set<Entry<String,Integer>> set = map.entrySet(); 
    for (Entry<String, Integer> entry : set) { 
     if(entry.getKey().equals(a.getItemcode())){ 
      ShopOrderItem t = new ShopOrderItem(a.getName(), a.getItemcode(), entry.getValue(), a.getPrice(), a.getTaxRate()); 
      result.add(t); 
     }else if(entry.getKey().equals(b.getItemcode())){ 
      ShopOrderItem t = new ShopOrderItem(b.getName(), b.getItemcode(), entry.getValue(), b.getPrice(), b.getTaxRate()); 
      result.add(t); 
     } 
     else if(entry.getKey().equals(c.getItemcode())){ 
      ShopOrderItem t = new ShopOrderItem(c.getName(), c.getItemcode(), entry.getValue(), c.getPrice(), c.getTaxRate()); 
      result.add(t); 
     } 
     else if(entry.getKey().equals(g.getItemcode())){ 
      ShopOrderItem t = new ShopOrderItem(g.getName(), g.getItemcode(), entry.getValue(), g.getPrice(), g.getTaxRate()); 
      result.add(t); 
     } 

    } 

Answer 3

：

Collection<ShopOrderItem> result = Stream.of(orderA, orderB, orderC) 
              .flatMap(Collection::stream) 
              .collect(Collectors.groupingBy(
                 ShopOrderItem::getItemcode, 
                 Collectors.reducing(new ShopOrderItem(), Use::merge) 
                )) 
              .values(); 

private static ShopOrderItem merge(ShopOrderItem s1, ShopOrderItem s2) 
{ 
    return new ShopOrderItem(s2.getName(), s2.getItemcode(), s1.getCount() + s2.getCount(), s2.getPrice(), s2.getTaxRate()); 
}