java8 流分组聚合

Question

给定一个 java class Something

class Something {
     String parent;
     String parentName;
     String child;
     Date at;
     int noThings;

     Something(String parent, String parentName, String child, Date at, int noThings) {
          this.parent = parent;
          this.parentName = parentName;
          this.child = child;
          this.at = at;
          this.noThings = noThings;
      }

      String getParent() { return parent; }
      String getChild() { return child; }
      int getNoThings() { return noThings; }
}

我有一些东西的清单 objects,

List<Something> hrlySomethings = Arrays.asList(
    new Something("parent1", "pname1", "child1", new Date("01-May-2015 10:00:00"), 4),
    new Something("parent1", "pname1", "child1", new Date("01-May-2015 12:00:00"), 2),
    new Something("parent1", "pname1", "child1", new Date("01-May-2015 17:00:00"), 8),
    new Something("parent1", "pname1", "child2", new Date("01-May-2015 07:00:00"), 12),
    new Something("parent1", "pname1", "child2", new Date("01-May-2015 17:00:00"), 14),
    new Something("parent2", "pname2", "child3", new Date("01-May-2015 11:00:00"), 3),
    new Something("parent2", "pname2", "child3", new Date("01-May-2015 16:00:00"), 2));

我想将objects按parent和children分组，然后找到"noThings"字段最近24小时的total/sum .

List<Something> dailySomethings = Arrays.asList(
    new Something("parent1", "pname1", "child1", new Date("01-May-2015 00:00:00"), 14),
    new Something("parent1", "pname1", "child2", new Date("01-May-2015 00:00:00"), 26),
    new Something("parent2", "pname2", "child3", new Date("01-May-2015 00:00:00"), 5))

我正在尝试使用流来执行此操作

我能想出如何使用分组得到地图的地图，总计

    Map<String,Map<String,IntSummaryStatistics>> daily =
        hrlySomethings.stream().collect(
    Collectors.groupingBy(Something ::getParent, 
    Collectors.groupingBy(ClientCollectionsReceived::getChild,
    Collectors.summarizingInt(ClientCollectionsReceived::getNoThings))));

我可以弄清楚如何根据 parent 和 child、

获得不同的列表

    Date startHour = "01-May-2015 00:00:00";
    int totalNoThings = 0; // don't know how to put sum in here
    List<Something> newList 
      = hrlySomethings.stream()
            .map((Something other) -> {
                    return new Something(other.getParent(),
                    other.getChild(), startHour, totalNoThings);
                })
            .distinct()
            .collect(Collectors.toList());

但我不知道如何将两者结合起来以获得不同的列表和总数。这可能吗？

Answer 1

首先，我假设您使用的是 java.util.Date（不过我建议您改用新的 java.time API）。其次，我假设 Something class 也正确地实现了 equals 和 hashCode。还需要更多吸气剂：

String getParentName() { return parentName; }
Date getAt() { return at; }

在这些假设下，您的任务可以这样解决：

List<Something> dailySomethings = hrlySomethings.stream().collect(
    Collectors.groupingBy(
        smth -> new Something(smth.getParent(), 
                              smth.getParentName(), 
                              smth.getChild(), 
                              new Date(smth.getAt().getYear(),
                                       smth.getAt().getMonth(), 
                                       smth.getAt().getDate()), 
                              0),
        Collectors.summingInt(Something::getNoThings)
    )).entrySet().stream()
                 .map(entry -> new Something(entry.getKey().getParent(),
                                             entry.getKey().getParentName(), 
                                             entry.getKey().getChild(), 
                                             entry.getKey().getAt(), 
                                             entry.getValue()))
                 .collect(Collectors.toList());

我们只使用了一次groupingBy，但创建了一个合适的分组键，即Something与parent、parentName和child设置为原来，at 更改为开始日期，noThings 设置为零。这样你就可以分组你想要的东西。如果你只需要总和，那么 summarizingInt 就不需要了， summingInt 就足够了。之后，我们将生成的地图转换为创建新 Something 对象的列表，其中 noThings 由地图值填充，其余由键填充。

java8 流分组聚合

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