流分组后如何聚合分组实体By

Question

我有一个简单的class人：

class Person {
    String firstName;
    String lastName;
    //getter, setter, constructor, toString
}

以及人员的输入列表，如：

List<Person> myList = List.of(
        new Person("Helena", "Graves"),
        new Person("Jasmine", "Knight"),
        new Person("Phoebe", "Reyes"),
        new Person("Aysha", "Graham"),
        new Person("Madeleine", "Jenkins"),
        new Person("Christina", "Johnson"),
        new Person("Melissa", "Carpenter"),
        new Person("Marie", "Daniel"),
        new Person("Robin", "French"),
        new Person("Tamara", "Wyatt"),
        new Person("Freya", "Montgomery"),
        new Person("Lacey", "Todd"),
        new Person("Heather", "Parker"),
        new Person("Lauren", "Wright"),
        new Person("Annie", "Bradley")
);

现在，我需要根据此人姓氏的第一个字符对上述列表进行分组，并再次对这些组进行分组，使所有从A-H开始的姓氏归为一组，下一组以I-N开始，最后以O-Z开始。

我已经可以按姓氏的第一个字符对列表进行分组：

myList.stream()
        .collect(Collectors.groupingBy(p -> String.valueOf(p.getLastName().charAt(0))))
        .entrySet()
        .forEach(System.out::println);

这给了我：

P=[Person{Heather, Parker}]
B=[Person{Annie, Bradley}]
R=[Person{Phoebe, Reyes}]
C=[Person{Melissa, Carpenter}]
T=[Person{Lacey, Todd}]
D=[Person{Marie, Daniel}]
F=[Person{Robin, French}]
W=[Person{Tamara, Wyatt}, Person{Lauren, Wright}]
G=[Person{Helena, Graves}, Person{Aysha, Graham}]
J=[Person{Madeleine, Jenkins}, Person{Christina, Johnson}]
K=[Person{Jasmine, Knight}]
M=[Person{Freya, Montgomery}]

我很难从这里开始，因为我需要进一步汇总以上内容，以获得包含三个条目/键的地图。期望输出：

Map<String, List<Person>> result = ...

A-H = [Person{Helena, Graves}, Person{Aysha, Graham}, Person{Melissa, Carpenter}, Person{Marie, Daniel}, Person{Robin, French}, Person{Annie, Bradley}]
I-N = [Person{Jasmine, Knight}, Person{Madeleine, Jenkins}, Person{Christina, Johnson}, Person{Freya, Montgomery}]
O-Z = [Person{Phoebe, Reyes}, Person{Tamara, Wyatt}, Person{Lacey, Todd}, Person{Heather, Parker}, Person{Lauren, Wright}]




Answer 1

匿名用户

基本上，您只需要使用收集器进行分组。groupBy（Function）和将每个人分配到正确组的函数：

/**
 * This method is null-friendly
 */
String group(Person person) {
    return Optional.ofNullable(person)
        .map(Person::getFirstName)
        .filter(name -> name.length() > 0)
        .map(name -> name.charAt(0))
        .map(ch -> {
            if (ch >= 'A' && ch <= 'H') {
                return "A-H";
            } else if (ch > 'H' && ch <= 'N') {
                return "I-N";
            } else if (ch > 'N' && ch <= 'Z') {
                return "O-Z";
            }
            return "*";   // In case the name starts with a character out of A-Z range
        })
        .orElse("none");  // In case there is empty/null firstName
}

Map<String, List<Person>> map = myList
        .stream()
        .collect(Collectors.groupingBy(this::group));

Answer 2

您应该稍微更改分类器函数以组合一系列字符。

此外，可能需要对entrySet（）进行排序（或在收集到地图时使用SortedMap/TreeMap）：

myList.stream()
      .collect(Collectors.groupingBy(
          p -> p.getLastName().charAt(0) < 'I' ? "A-H" : 
               p.getLastName().charAt(0) < 'O' ? "I-N" : "O-Z"
      ))
      .entrySet()
      .stream()
      .sorted(Map.Entry.comparingByKey())
      .forEach(System.out::println);

输出：

A-H=[Person {Helena Graves}, Person {Aysha Graham}, Person {Melissa Carpenter}, Person {Marie Daniel}, Person {Robin French}, Person {Annie Bradley}]
I-N=[Person {Jasmine Knight}, Person {Madeleine Jenkins}, Person {Christina Johnson}, Person {Freya Montgomery}]
O-Z=[Person {Phoebe Reyes}, Person {Tamara Wyatt}, Person {Lacey Todd}, Person {Heather Parker}, Person {Lauren Wright}]