如何获得两个地图Java之间的区别？

Question

我有两个地图，如下所示：

Map<String, Record> sourceRecords;
Map<String, Record> targetRecords;

我想获得与每个maps.ie不同的键

1. 它显示了在sourceRecords中可用但在targetRecords中不可用的映射键。
2. 它显示了targetRecords中可用的映射键，而sourceRecords中没有。

我做到了如下：

Set<String> sourceKeysList = new HashSet<String>(sourceRecords.keySet());
Set<String> targetKeysList = new HashSet<String>(targetRecords.keySet());

SetView<String> intersection = Sets.intersection(sourceKeysList, targetKeysList);
Iterator it = intersection.iterator();
while (it.hasNext()) {
    Object object = (Object) it.next();
    System.out.println(object.toString());
}

SetView<String> difference = Sets.symmetricDifference(sourceKeysList, targetKeysList);
ImmutableSet<String> immutableSet = difference.immutableCopy();

编辑

if(sourceKeysList.removeAll(targetKeysList)){
            //distinct sourceKeys
            Iterator<String> it1 = sourceKeysList.iterator();
            while (it1.hasNext()) {
                String id = (String) it1.next();
                String resultMessage = "This ID exists in source file but not in target file";
                System.out.println(resultMessage);
                values = createMessageRow(id, resultMessage);
                result.add(values);
            }
        }
        if(targetKeysList.removeAll(sourceKeysList)){
            //distinct targetKeys
            Iterator<String> it1 = targetKeysList.iterator();
            while (it1.hasNext()) {
                String id = (String) it1.next();
                String resultMessage = "This ID exists in target file but not in source file";
                System.out.println(resultMessage);
                values = createMessageRow(id, resultMessage);
                result.add(values);
            }
        }

我能够找到通用键，但找不到不同的键。请帮忙。

Answer 1

集允许您也 删除 元素。

如果生成“ helper”集对您来说不是问题（因为条目太多，那么：

Set<String> sources = get a copy of all source entries
Set<String> targets = get a copy of all source entries

然后：

sources.removeAll(targets) ... leaves only entries in sources that are only in sources, not in target

而

sources.retainAll(targets) ... leaves only entries that are in both sets

您可以从这里开始工作…