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。
最佳答案
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。
关于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,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39382447/