java 如何删除地图中重复的键值对
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How to remove duplicate key-value pairings in a map
提问by nitiger
I'm stuck on how to transfer key-value pairs from map1 into map2 only if each key has a unique value in map1.
只有当每个键在 map1 中都有一个唯一值时，我才知道如何将键值对从 map1 传输到 map2。
Let's say I have the following maps:
假设我有以下地图：
- map1: [1,2] [2,4] [4,4]
- map2: [1,2] [2,4]
- 地图1：[1,2] [2,4] [4,4]
- 地图2：[1,2] [2,4]
I suppose the algorithm would be:
我想算法是：
- Loop through entries in the first map.
- Add a key to map2.
- Add a value to a set which checks against the values of map2
- If the values are duplicate the value doesn't get added to the set and disregard adding its corresponding key to map2.
- 循环遍历第一个地图中的条目。
- 向 map2 添加一个键。
- 将一个值添加到一个检查 map2 值的集合中
- 如果值重复，则该值不会添加到集合中，并且忽略将其相应的键添加到 map2。
Code snippet:
代码片段：
public static <K,V> Map<K,V> unique (Map<K,V> m) {
Map<K,V> newMap = new ArrayMap<K,V>();
//Remember all values in the newMap.
Set<V> holding = new ArraySet<V>(newMap.values());
for (Map.Entry<K, V> graphEntry : m.entries()) {
//not sure.
}
return newMap;
}
Is my idea of how its supposed to be done on the right track? Quite lost here.
我的想法是如何在正确的轨道上完成？在这里迷路了。
回答by Luiggi Mendoza
From a Map<K, V>create a Map<V, K>that will add the item if and only if the key is not in the map. Using this Map<V, K>, recreate your Map<K, V>.
从Map<K, V>创建一个Map<V, K>当且仅当键不在地图中时才会添加项目。使用它Map<V, K>，重新创建您的Map<K, V>.
public static <K, V> Map<K, V> createMap(Map<K, V> m) {
Map<K, V> map = new HashMap<K, V>();
Map<V, K> tmpMap = new HashMap<V, K>();
for(Map.Entry<K, V> entry : m.entrySet()) {
if (!tmpMap.containsKey(entry.getValue())) {
tmpMap.put(entry.getValue(), entry.getKey());
}
}
for(Map.Entry<V, K> entry : tmpMap.entrySet()) {
map.put(entry.getValue(), entry.getKey());
}
return map;
}
If you need to keep the preserver order of the data, use LinkedHashMapinstead of HashMap.
如果您需要保持数据的保存顺序，请使用LinkedHashMap代替HashMap。
回答by Rohit Jain
Check out Guava BiMap.. This is what you need..
查看Guava BiMap.. 这就是您所需要的..
Although your problem is solved, you can take a look at the below code, to use GuavaAPI for what you want to do: -
虽然您的问题已解决，但您可以查看以下代码，以使用GuavaAPI 执行您想要执行的操作：-
public void removeDuplicateValue() {
Map<Integer, String> existingMap = new HashMap<Integer, String>();
existingMap.put(1, "a");
existingMap.put(2, "b");
// Create a new BiMap
BiMap<Integer, String> biMap = HashBiMap.create();
for (Integer val: existingMap.keySet()) {
// forcePut will add a key-value pair, and overwrite the duplicate value.
biMap.forcePut(val, existingMap.get(val));
}
// Create Inverse Map for newly created BiMap.
BiMap<String, Integer> inverseBiMap = biMap.inverse();
for(String val: inverseBiMap.keySet()) {
System.out.println(val + ":" + biMap.get(val));
}
}
回答by Santhanam
Try this one..
试试这个..
Map<String, String> myMap1 = new TreeMap<String, String>();
myMap1.put("1", "One");
myMap1.put("2", "Two");
myMap1.put("3", "One");
myMap1.put("4", "Three");
myMap1.put("5", "Two");
myMap1.put("6", "Three");
Set<String> mySet = new HashSet<String>();
for (Iterator itr = myMap1.entrySet().iterator(); itr.hasNext();)
{
Map.Entry<String, String> entrySet = (Map.Entry) itr.next();
String value = entrySet.getValue();
if (!mySet.add(value))
{
itr.remove();
}
}
Map<String, String> myMap2 = new TreeMap<String, String>(myMap1);
System.out.println("Result :"+myMap2);
Result :{1=One, 2=Two, 4=Three}
结果：{1=一，2=二，4=三}
