java - 迭代两个 hashMap 并比较键并比较形成键的元素以填充最终的 hashMap

Question

Class OpsRecord{
 private String id;
 private long sequenceNumber;

 //implemented hashcode and equals method considering both id and version.
 //So two OpsRecord are equal only if two OpsRecord has same id and sequenceNumber.
}

Class OpsCompleteRecord{
 // This class has 20 class variables including id and sequenceNumber.
}

我有一个现有的Map，它具有以下值。假设这些值来自随机数据源。

Map<OpsRecord, OpsCompleteRecord> sourceMap = new HashMap<>();



sourceMap.put(new OpsRecord("1","1"), new OpsCompleteRecord());
        sourceMap.put(new OpsRecord("1","3"), new OpsCompleteRecord());
        sourceMap.put(new OpsRecord("1","2"), new OpsCompleteRecord());
        sourceMap.put(new OpsRecord("2","2"), new OpsCompleteRecord());
        sourceMap.put(new OpsRecord("2","1"), new OpsCompleteRecord());

我想要一个目标 map ，该 map 应仅包含 OpsCompleteRecord ，其 id 为 1 且 sequenceNumber 为 3 ，以及另一个包含 id=2 和 sequenceNumber=2。[id 的最新 sequenceNumber]

目标 map

new OpsRecord("1","3"), new OpsCompleteRecord()
new OpsRecord("2","2"), new OpsCompleteRecord()

我正在尝试使用以下逻辑和代码来实现。需要您帮助填充 targetMap。我想我搞乱了逻辑。

public static void main(String[] args) {

        Map<OpsRecord, OpsCompleteRecord> sourceMap = new HashMap<>();

        sourceMap.put(new OpsRecord("1","1"), new OpsCompleteRecord());
        sourceMap.put(new OpsRecord("1","3"), new OpsCompleteRecord());
        sourceMap.put(new OpsRecord("1","2"), new OpsCompleteRecord());
        sourceMap.put(new OpsRecord("2","2"), new OpsCompleteRecord());
        sourceMap.put(new OpsRecord("2","1"), new OpsCompleteRecord());

        // Populate the target map 
        Map<OpsRecord, OpsCompleteRecord> targetMap = new HashMap<>();

        Iterator<OpsRecord> itr = sourceMap.keySet().iterator();

        while(itr.hasNext()){
            OpsRecord opsRecord = itr.next();
            // Iterate over targetMap.keySet(). 
            //     Take each key.getId()
            //     Compare targetMap key.getId() with opsRecord.getId() 
            //         and if equal then Compare targetMap key.getSequenceNumber() with opsRecord.getSequenceNumber()
             //                          if getSequenceNumber is small in targetMap, overwrite the object else ignore.
             //            else if all key.getId() is not equal to opsRecord.getId() then add that key and value to targetMap           
        }       

    }

我使用的是Java7。

Answer 1

在Java7中，也许你可以尝试使用guava Maps.filterKeys，例如:

    Map<OpsRecord, OpsCompleteRecord> sourceMap = new HashMap();

    sourceMap.put(new OpsRecord("1", 1), new OpsCompleteRecord());
    sourceMap.put(new OpsRecord("1", 3), new OpsCompleteRecord());
    sourceMap.put(new OpsRecord("1", 2), new OpsCompleteRecord());
    sourceMap.put(new OpsRecord("2", 2), new OpsCompleteRecord());
    sourceMap.put(new OpsRecord("2", 1), new OpsCompleteRecord());
    Map<OpsRecord, OpsCompleteRecord> targetMap = Maps.filterKeys(sourceMap, new Predicate<OpsRecord>() {
        @Override
        public boolean apply(OpsRecord input) {
            return input.id.equals(Long.toString(input.sequenceNumber));
        }
    });

在Java8中，您可以将stream与filter一起使用，collector可以轻松做到这一点:

    sourceMap.put(new OpsRecord("1", 1), new OpsCompleteRecord());
    sourceMap.put(new OpsRecord("1", 3), new OpsCompleteRecord());
    sourceMap.put(new OpsRecord("1", 2), new OpsCompleteRecord());
    sourceMap.put(new OpsRecord("2", 2), new OpsCompleteRecord());
    sourceMap.put(new OpsRecord("2", 1), new OpsCompleteRecord());
    Map<OpsRecord, OpsCompleteRecord> targetmap =
            sourceMap.entrySet().stream()
                    .filter(i -> i.getKey().id.equals(Long.toString(i.getKey().sequenceNumber)))
                    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));

还有一点需要注意:sequenceNumber是long类型，所以应该使用new OpsRecord("2", 1)。

希望对您有所帮助。