比较两个地图的键和值，并将差异存储到结果地图中。

huangapple
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英文:

Compare two maps by key and value and storing difference into resultant map

问题

I have two maps:

Map<Long, Long> map1 = new HashMap<>();
Map<Long, Long> map2 = new HashMap<>();

这两个地图的大小不同。
地图将如下所示：

map1 map2
(1,1) (1,1)
(2,1) (2,1)
(3,1)

或：

map1 map2
(1,1) (1,1)
(2,1) (2,1)
(3,1)
(4,1)

现在，我想创建map3，它将只包含map1和map2之间的差异，即(K,V)-->(3,1) ( (K,V)-->(3,1);(4,1) )。然后，我将获取(K,V)的值并使用它们。
在这个类似的问题中有解决方案，但对我不起作用。传递给方法compareKeysAndValues(map1,map2)，收到：提供的Map<Long, Long>要求的Map<Long, Boolean>。

有什么想法？
1: https://stackoverflow.com/questions/21559811/how-to-compare-two-maps-by-both-key-and-value-and-storing-difference-map-in-resu

英文:

I know that some of you will mark it as duplicate, but didn't want to ask similar question in someone else thread.
I have two maps:

  1. Map&lt;Long, Long&gt; map1 = new HashMap&lt;&gt;();
  2. Map&lt;Long, Long) map2 = new HashMap&lt;&gt;();

This two maps are different in size.
Maps will look like this:

  1. map1        map2
  2. (1,1)       (1,1)
  3. (2,1)       (2,1)
  4.             (3,1)

or:

  1. map1        map2
  2. (1,1)       (1,1)
  3. (2,1)       (2,1)
  4.             (3,1) 
  5.             (4,1)

Now, I wan to create map3 that will contain only difference between map1 and map2, and that is (K,V)-->(3,1) ( (K,V)-->(3,1);(4,1) ). I will then get (K,V) values and use them.
In this thread similar question there is solution but not works for me. Passing to method compareKeysAndValues(map1,map2), receiving: Required Map<Long, Long> provided Map<Long, Boolean)

Any Ideas?

答案1

得分: 0

以下是您要翻译的代码部分：

  1. 显而易见的解决方案是在两个映射中遍历键，并将条目添加到新映射中，如果在另一个映射中不存在该键：
  3. import java.util.HashMap;
  4. import java.util.Map;
  6. public class test {
  7.     public static void main(String[] args){
  8.         Map<Long, Long> map1 = new HashMap<>();
  9.         Map<Long, Long> map2 = new HashMap<>();
  10.         map1.put(1L, 1L);
  11.         map1.put(2L, 1L);
  12.         map1.put(5L, 1L);
  13.         map2.put(1L, 1L);
  14.         map2.put(2L, 1L);
  15.         map2.put(3L, 1L);
  16.         map2.put(4L, 1L);
  18.         Map<Long, Long> map3 = new HashMap<>();
  20.         for(Map.Entry<Long, Long> entry : map1.entrySet()){
  21.             if(!map2.containsKey(entry.getKey())){
  22.                 map3.put(entry.getKey(), entry.getValue());
  23.             }
  24.         }
  26.         for(Map.Entry<Long, Long> entry : map2.entrySet()){
  27.             if(!map1.containsKey(entry.getKey())){
  28.                 map3.put(entry.getKey(), entry.getValue());
  29.             }
  30.         }
  32.         for(Map.Entry<Long, Long> entry : map3.entrySet()){
  33.             System.out.println(entry.getKey() + ", " + entry.getValue());
  34.         }
  35.     }
  36. }
  38. 输出:
  39. 3, 1
  40. 4, 1
  41. 5, 1

希望这能帮助您。

英文:

An obvious solution is to traverse over the keys in both maps and add the entries to the new map if the key doesn't exist in the other map:

  1. import java.util.HashMap;
  2. import java.util.Map;
  3. public class test {
  4. public static void main(String[] args){
  5. Map&lt;Long, Long&gt; map1 = new HashMap&lt;&gt;();
  6. Map&lt;Long, Long&gt; map2 = new HashMap&lt;&gt;();
  7. map1.put(1L, 1L);
  8. map1.put(2L, 1L);
  9. map1.put(5L, 1L);
  10. map2.put(1L, 1L);
  11. map2.put(2L, 1L);
  12. map2.put(3L, 1L);
  13. map2.put(4L, 1L);
  14. Map&lt;Long, Long&gt; map3 = new HashMap&lt;&gt;();
  15. for(Map.Entry&lt;Long, Long&gt; entry : map1.entrySet()){
  16. if(!map2.containsKey(entry.getKey())){
  17. map3.put(entry.getKey(), entry.getValue());
  18. }
  19. }
  20. for(Map.Entry&lt;Long, Long&gt; entry : map2.entrySet()){
  21. if(!map1.containsKey(entry.getKey())){
  22. map3.put(entry.getKey(), entry.getValue());
  23. }
  24. }
  25. for(Map.Entry&lt;Long, Long&gt; entry : map3.entrySet()){
  26. System.out.println(entry.getKey() + &quot;, &quot; + entry.getValue());
  27. }
  28. }
  29. }

Output:

  1. 3, 1
  2. 4, 1
  3. 5, 1

huangapple
  • 本文由 发表于 2020年7月29日 06:22:23
  • 转载请务必保留本文链接：https://go.coder-hub.com/63143640.html
  • hashmap
  • java
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