基于键值对比较两个TreeMaps

Question

public static void compareTravelBalance(Map<String, Integer> travelCosts, Map<String, Integer> travellerBalances){

    Set set = travelCosts.entrySet();
    Iterator iterator = set.iterator();
    Set set1 = travellerBalances.entrySet();
    Iterator iterator1 = set1.iterator();

    while(iterator.hasNext()) {

        Map.Entry mentry = (Map.Entry)iterator.next();
        Map.Entry mapEntry= (Map.Entry) iterator1.next();
        int cost = (Integer) mentry.getValue();
        int balance = (Integer) mapEntry.getValue();

        if(travellerBalances.containsKey(mentry.getKey())){
            if(cost>balance){
                System.out.println("Insufficient funds for "+mentry.getKey()+": Cost: "+cost+" Balance: "+balance);
            }
            else if(cost<=balance){
                System.out.println("Approved! for: "+mentry.getKey()+": Cost: "+cost+" Balance: "+balance);
            }
        }
        else{
            System.out.println("Traveller ID "+mentry.getKey()+" does not exist");

        }
    }
}

上面是我的代码。我有两个CSV，我将它们的数据保存到两个不同的TreeMaps中- travelCosts和travellerBalances。键值对是作为键的travellerId和作为第一个TreeMap中的值的travelCost以及作为第二个TreeMap中的值的travelBalance。

我正在尝试比较来自不同映射的两个值，具体取决于travellerId。因此，如果cost>balance，它应该打印不足的资金。如果为cost<=balance，则应打印已批准。如果没有匹配的key，则说明旅行者不存在。

我附加了CSV，它将显示我在地图中的数据。

问题是:如果你看一下数据，它是比较travellerid 2001到2002,2002到2003，等等。相反，它应该说2001不存在。我希望我说的有道理。如果您需要更多的信息，请询问。

谢谢。Travel Costs-first col is travellerId and the third col is travelCost

Travel Balances - the first col is travellerId and third col is travelBalance

Answer 1

在设计程序时，使用并行数据结构通常很麻烦。在这种情况下，使用id、costs和balance创建Traveller类会更好。将这些存储在以traveller.id为关键字的映射中。在读取CSV时，必要时添加一个旅行者，否则更新现有的旅行者costs或balance。id不一定要在Traveller中，因为它在Map key中，但通常是个好主意。

Answer 2

由于travellerId可能只存在于其中一个映射中，因此要做的第一件事是收集键的超集：

Map<String, Integer> travelCosts = new TreeMap<>();
travelCosts.put("2001", 2000);
travelCosts.put("2002", 500);
Map<String, Integer> travellerBalances = new TreeMap<>();
travellerBalances.put("2002", 499);
travellerBalances.put("2003", 1155);

Set<String> travellerIds = new TreeSet<>();
travellerIds.addAll(travelCosts.keySet());
travellerIds.addAll(travellerBalances.keySet());

测试

System.out.println(travellerIds);

输出

[2001, 2002, 2003]

你现在可以做你的逻辑了：

for (String travellerId : travellerIds) {
    Integer cost = travelCosts.get(travellerId);
    Integer balance = travellerBalances.get(travellerId);
    if (cost == null) {
        System.out.println("Traveller ID " + travellerId + " is missing 'cost': " +
                           "Balance: " + balance);
    } else if (balance == null) {
        System.out.println("Traveller ID " + travellerId + " is missing 'balance': " +
                           "Cost: " + cost);
    } else if (cost > balance) {
        System.out.println("Insufficient funds for " + travellerId + ": " +
                           "Cost: " + cost + " Balance: " + balance);
    } else {
        System.out.println("Approved! for: " + travellerId + ": " +
                           "Cost: " + cost + " Balance: " + balance);
    }
}

输出

Traveller ID 2001 is missing 'balance': Cost: 2000
Insufficient funds for 2002: Cost: 500 Balance: 499
Traveller ID 2003 is missing 'cost': Balance: 1155

Answer 3

    Map.Entry mentry = (Map.Entry)iterator.next();
    Map.Entry mapEntry= (Map.Entry) iterator1.next();
    int cost = (Integer) mentry.getValue();
    int balance = (Integer) mapEntry.getValue();

entrySet().iterator()以键的升序返回您的映射。

第一个映射将返回{2001-x, 2002-y, 2003-z}

第二个映射将返回{2002-x1, 2003-y1, 2004-z1}

您的代码不是通过相同的键来比较映射，而是在相同类型的“索引”中比较映射(这会导致将2001-x与2002-x1进行比较)

你想要的是按键旅行，而不是两者都用：

public static void compareTravelBalance(Map<String, Integer> travelCosts, Map<String, Integer> travellerBalances){

     Iterator<String> keys = travelCosts.keySet().iterator();

     while (keys.hasNext()) {
         String key = keys.next();
         if (!travellerBalances.containsKey(key)) {
            System.out.println("Traveller ID "+mentry.getKey()+" does not exist");
         } else {
             int cost = travelCosts.get(key);
             int balance = travellerBalances.get(key);
             // ... compare cost and balance
         }
     }
}