ArrayList中类对象属性的java求和值

Question

我有一个ArrayList，它有许多对象。我想对相同属性名的值求和

示例Array List中的数据是ProfitAndLossDataDO的对象

      "Michel" , 5000
      "Alex" , 5000
      "Edvin" , 4000
      "Sosha" , 3000
      "Michel" , 2000
      "Alex" , 3000

结果将是（当人相同时的值之和）——

      "Michel" ,7000
      "Alex" ,  8000
      "Edvin" , 4000
      "Sosha" , 3000

逻辑类ProfitAndLossDataDO

public class ProfitAndLossDataDO {

String ledgerName;
double ledgerAmount;

public ProfitAndLossDataDO() {
}

public ProfitAndLossDataDO(String ledgerName, double ledgerAmount) {
    this.ledgerName = ledgerName;
    this.ledgerAmount = ledgerAmount;
}



public String getLedgerName() {
    return ledgerName;
}

public void setLedgerName(String ledgerName) {
    this.ledgerName = ledgerName;
}

public double getLedgerAmount() {
    return ledgerAmount;
}

public void setLedgerAmount(double ledgerAmount) {
    this.ledgerAmount = ledgerAmount;
}


@Override
public String toString()
{
            return ("ledger name : "+this.getLedgerName()+" Amount : "+this.getLedgerAmount());
}
}

谢谢

Answer 1

您可以按照aleroot的建议使用哈希映射。使用name作为键，sum作为值。每当需要插入值时，只需检查具有相同键的条目是否存在并更新它

public class ProfitAndLossDataDO {

HashMap<String, Double> data = new HashMap<String, Double>();

public ProfitAndLossDataDO() {

}

public void updateLedger(String ledgerName, double ledgerAmount) {
    double temp;
    if(data.containsKey(ledgerName)) {
        temp = data.get(ledgerName)+ledgerAmount;
        data.put(ledgerName,temp);
    }
    else {
        data.put(ledgerName,ledgerAmount);
    }

}
}

Answer 2

为什么不把它们放在一个基于地图的集合中，并带有键/值对

我认为在这种情况下，基于地图的集合更适合您拥有的数据和您想要使用它的上下文

如果需要保留列表中的单个项目，可以使用链接列表作为映射的值

例如：

HashMap<String, LinkedList<Double>> cache = new HashMap<String, LinkedList<Double>>();

并对值LinkedList的每个元素求和，以找到总数。。。您甚至可以使用getTotal（）或Sum（）等辅助函数将双精度链接列表包装到容器对象中

Answer 3

我将迭代列表并在地图中收集结果：

public static List<ProfitAndLossDataDO> sumPerLedgerName
    (List<ProfitAndLossDataDO> list) {

    Map<String, ProfitAndLossDataDO> map = new HashMap<>();

    for (ProfitAndLossDataDO p : list) {
        String name = p.getLedgerName();
        ProfitAndLossDataDO sum = map.get(name);
        if (sum == null) {
            sum = new ProfitAndLossDataDO(name, 0.0);
            map.put(name, sum);
        }
        sum.setLedgerAmount(sum.getLedgerAmount() + p.getLedgerAmount());
    }
    return new ArrayList<>(map.values());
}

编辑：

Java 8对^{}接口的增强使我们能够以稍微优雅的方式实现此方法，中间是恼人的if块：

public static List<ProfitAndLossDataDO> sumPerLedgerName
    (List<ProfitAndLossDataDO> list) {

    Map<String, ProfitAndLossDataDO> map = new HashMap<>();

    for (ProfitAndLossDataDO p : list) {
        String name = p.getLedgerName();
        ProfitAndLossDataDO sum = map.computeIfAbsent(name, n -> new ProfitAndLossDataDO(n, 0.0));
        sum.setLedgerAmount(sum.getLedgerAmount() + p.getLedgerAmount());
    }
    return new ArrayList<>(map.values());
}