Java 如何从与其他列表关联的映射中获取密钥

我需要通过将List与map>的键相关联来创建一个映射，如下所示

 Input Map< String,List< String>>  - {"Fruit" -> ["apple","orange"], "Animal" -> ["cat","dog"]}

Input List< String> - {"apple","dog","xyzzy"}

output map as {"Fruit" -> "apple","Animal" -> "dog"} by discarding unmatching entries like "xyzzy" in this case" 


```
Map <String, String> outputMap = new HashMap<>();
Map <String, String> tempMap = new HashMap<>();

for (Map.Entry<String, List<String>> entry : inputMap.entrySet()) {
      entry.getValue().forEach(value -> tempMap.put(value, entry.getKey()));
   }    

 inputList.forEach(value ->
            {   
             if(tempMap.get(value)!=null)
             outputMap.put(tempMap.get(value),value); });
            }

有没有一种更简洁、更精确的方法来实现这一点？

因为您的输出类型是Map，所以无法获得输出

{"Fruit" -> "apple","Animal" -> "dog", "Animal" -> "cat"}

字符串动物是键，不能复制

但是，如果要过滤输入映射并检查包含列表，可以执行以下操作：

final Map<String, List<String>> inputMap = new HashMap<>();
inputMap.put("Fruit", asList("apple", "orange"));
inputMap.put("Animal", asList("cat", "dog"));

final List<String> list = asList("apple", "dog", "cat");

final Map<String, List<String>> outputMap = inputMap.entrySet().stream()
        .filter(e -> e.getValue().stream().anyMatch(list::contains))
        .collect(toMap(e -> e.getKey(), e -> e.getValue().stream().filter(list::contains).collect(toList())));

outputMap.forEach((k, v) -> System.out.printf("%s: %s%n", k, v));

final Map<String, List<String>> outputMap = inputMap.entrySet().stream()
        .map(e -> new AbstractMap.SimpleEntry<>(e.getKey(), e.getValue().stream().filter(list::contains).collect(toList())))
        .filter(e -> !e.getValue().isEmpty())
        .collect(toMap(e -> e.getKey(), e -> e.getValue()));

双重过滤遍历将列出两次，这取决于输入贴图的方式。无论如何，不需要重新分组；如果输入匹配成高百分比，您可以执行以下操作：

final Map<String, List<String>> inputMap = new HashMap<>();
inputMap.put("Fruit", asList("apple", "orange"));
inputMap.put("Animal", asList("cat", "dog"));

final List<String> list = asList("apple", "dog", "cat");

final Map<String, List<String>> outputMap = inputMap.entrySet().stream()
        .filter(e -> e.getValue().stream().anyMatch(list::contains))
        .collect(toMap(e -> e.getKey(), e -> e.getValue().stream().filter(list::contains).collect(toList())));

outputMap.forEach((k, v) -> System.out.printf("%s: %s%n", k, v));

final Map<String, List<String>> outputMap = inputMap.entrySet().stream()
        .map(e -> new AbstractMap.SimpleEntry<>(e.getKey(), e.getValue().stream().filter(list::contains).collect(toList())))
        .filter(e -> !e.getValue().isEmpty())
        .collect(toMap(e -> e.getKey(), e -> e.getValue()));

---

一个不直接相关的问题是，为了提高性能，最好将过滤器列表定义为一个集合。

地图中不能有重复的键。所以这个问题是无效的。这里怎么能有两个动物的钥匙？

final Map<String, List<String>> outputMap = inputMap.entrySet().stream()
        .map(e -> new AbstractMap.SimpleEntry<>(e.getKey(), e.getValue().stream().filter(list::contains).collect(toList())))
        .filter(e -> !e.getValue().isEmpty())
        .collect(toMap(e -> e.getKey(), e -> e.getValue()));