Java 8-如何使用lambda、groupingBy和summingInt以及使用按位运算符

Question

public class CheckIn {
    public int id;
    public String day;
    public int workingHours;
    public boolean inProgress;

    public CheckIn(String day, in hours, boolean inProgress) {
        this.day = day;
        this.workingHours = hours;
        this.inProgress = inProgress;
    }
}

// Suppose this is the inputs 
List<CheckIn> checkinsList = new ArrayList<>();
checkinsList.add(new CheckIn("26-11-2015",6,true));
checkinsList.add(new CheckIn("27-11-2015",6,false));
checkinsList.add(new CheckIn("26-11-2015",6,false));
checkinsList.add(new CheckIn("27-11-2015",4,false));
checkinsList.add(new CheckIn("26-11-2015",1,false));
checkinsList.add(new CheckIn("28-11-2015",6,false));
checkinsList.add(new CheckIn("28-11-2015",6,false));
checkinsList.add(new CheckIn("28-11-2015",6,true));


List<CheckIn> summary = new ArrayList<>();

checkinsList.stream().collect(Collectors.groupingBy(Function.identity(),
    () -> new TreeMap<>(
        Comparator.<CheckIn, String>comparing(entry -> entry.day)),
        Collectors.summingInt(entry -> entry.duration))).forEach((e, sumTargetDuration) -> {
        CheckIn entry = new CheckIn();
        entry.day = e.day;
        entry.duration = sumTargetDuration;
        // Here my something like what I need?
        entry.inProgress = e.inProgress;
        summary.add(entry);
});

• 第一项“26-11-2015”，13,true<--true，因为“26-11-2015”一天有1项为true。
• 第二项“27-11-2015”，10，假
• 第三项“28-11-2015”，18，真

我希望总结带有inprogresstrue，如果当天有任何条目haveinprogress==true，它适用于lambda吗？

Answer 1

您需要为此任务生成自定义收集器。问题是没有内置收集器将多个收集器组合在一起并应用一个整理操作。在本例中，我们需要组合3个收集器:summingint、reducation，它们将或每个inprogress，以及mapping，它们将映射到当天。

下面的类将在收集过程中保存汇总，并相应地计算结果。我们可以添加关于当天的额外检查，因为在整个过程中它应该是相同的，但为了简单起见，我省略了这些检查。

public class CheckInSummary {

    private int workingHours;
    private boolean inProgress;
    private String day;

    public void accept(CheckIn checkIn) {
        workingHours += checkIn.workingHours;
        inProgress = inProgress || checkIn.inProgress;
        day = checkIn.day;
    }

    public CheckInSummary combine(CheckInSummary summary) {
        workingHours += summary.workingHours;
        inProgress = inProgress || summary.inProgress;
        return this;
    }

    public CheckIn finish() {
        return new CheckIn(day, workingHours, inProgress);
    }

}

然后，可以使用以下方法从该类创建收集器：

private static Collector<CheckIn, ?, CheckIn> summary() {
    return Collector.of(
                CheckInSummary::new,
                CheckInSummary::accept,
                CheckInSummary::combine,
                CheckInSummary::finish
           );
}

List<CheckIn> result = 
      new ArrayList<>(checkinsList.stream()
                                  .collect(groupingBy(c -> c.day, summary()))
                                  .values());

[[27-11-2015, 10, false], [26-11-2015, 13, true], [28-11-2015, 18, true]]