我正在尝试找到一种比以下更优雅的方法来创建一个映射,该映射使用 Java 8 按字段名称对字段值进行分组:
@Test
public void groupFieldValuesByFieldNames() {
Person lawrence = aPerson().withFirstName("Lawrence").withLastName("Warren").born();
Person gracie = aPerson().withFirstName("Gracie").withLastName("Ness").born();
Map<String, List<String>> valuesByFieldNames = new HashMap<>();
Stream.of(lawrence, gracie).forEach(person -> {
valuesByFieldNames.computeIfAbsent("lastName", s -> new ArrayList<>()).add(person.getLastName());
valuesByFieldNames.computeIfAbsent("firstName", s -> new ArrayList<>()).add(person.getFirstName());
});
assertThat(valuesByFieldNames, hasEntry("lastName", asList("Warren", "Ness")));
assertThat(valuesByFieldNames, hasEntry("firstName", asList("Lawrence", "Gracie")));
}
最佳答案
试试这个。
Map<String, List<String>> valuesByFieldNames = Stream.of(lawrence, gracie)
.flatMap(p -> Stream.of(new String[]{"firstName", p.getFirstName()},
new String[]{"lastName", p.getLastName()}))
.collect(Collectors.groupingBy(a -> a[0],
Collectors.mapping(a -> a[1], Collectors.toList())));
或者更一般地说
Map<String, List<String>> valuesByFieldNames = Stream.of(lawrence, gracie)
.flatMap(p -> Stream.of(new AbstractMap.SimpleEntry<>("firstName", p.getFirstName()),
new AbstractMap.SimpleEntry<>("lastName", p.getLastName())))
.collect(Collectors.groupingBy(e -> e.getKey(),
Collectors.mapping(e -> e.getValue(), Collectors.toList())));
关于Java 8 : grouping field values by field names,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/36948753/