如何获得两个地图Java之间的差异?

2023-10-14Java开发问题
2

本文介绍了如何获得两个地图Java之间的差异?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

问题描述

我有两张地图如下:

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

我想获得与每个地图不同的键.即

I want to get the keys differ from each of the maps.i.e.

  1. 它显示 sourceRecords 中可用但 targetRecords 中不可用的映射键.
  2. 它显示 targetRecords 中可用但 sourceRecords 中不可用的映射键.

我是这样做的:

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);
            }
        }

我能够找到公共键,但不能找到不同的键.请帮忙.

I am able to find the common keys but not distinct keys. Please help.

推荐答案

集合也允许您删除元素.

如果生成帮助"集对您来说不是问题(因为条目太多;那么:

If generating "helper" sets is not a problem for you (because of too many entries; what about:

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

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

您可以从这里开始工作......

You can work your way from here ...

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