将对象列表转换为属性映射

Question

我有一个Section具有如下属性的类

class Section {

    private String name;

    private String code;

    // respective getters and setters.

}

现在我有一个部分对象列表，我想将该列表转换为名称和代码的映射。我知道可以按以下常规方式完成。

List<Section> sections = getSections();

Map<String, String> nameCodeMap = new HashMap<>();

for (Section section : sections) {

    nameCodeMap.put(section.getCode(), section.getName());

}

我想知道 Java-8 流是否可以实现类似的功能。

Answer 1

不难。只需将toMap收集器与对吸气剂的适当方法引用一起使用：sections.stream().collect( &nbsp;&nbsp;&nbsp;&nbsp;Collectors.toMap(Section::getName,&nbsp;Section::getCode) );

Answer 2

如果您没有具有Section相同 getCode() 值的元素：Map<String,&nbsp;String>&nbsp;map&nbsp;=&nbsp;sections.stream() &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.collect(toMap(Section::getCode,&nbsp;Section::getName);如果您有Section具有相同 getCode() 值的元素，则前一个将引发，IllegalStateException因为它不接受该值。所以你必须合并它们。例如，要实现与您的实际代码相同的事情，即覆盖现有键的现有值，请使用此重载：toMap(Function<?&nbsp;super&nbsp;T,&nbsp;?&nbsp;extends&nbsp;K>&nbsp;keyMapper, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Function<?&nbsp;super&nbsp;T,&nbsp;?&nbsp;extends&nbsp;U>&nbsp;valueMapper, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;BinaryOperator<U>&nbsp;mergeFunction)并返回合并函数的第二个参数：Map<String,&nbsp;String>&nbsp;map&nbsp;=&nbsp;sections.stream() &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;.collect(toMap(Section::getCode,&nbsp;Section::getName,&nbsp;(a,&nbsp;b)&nbsp;->&nbsp;b);

Answer 3

请在下面找到女士的代码：List<Section> sections = Arrays.asList(new Section("Pratik", "ABC"),&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;new Section("Rohit", "XYZ"));Map<String, String> nameCodeMap = sections.stream().collect(&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; Collectors.toMap(section -> section.getName(),&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;section -> section.getCode()));nameCodeMap.forEach((k, v) -> System.out.println("Key " + k + " " + "Value " + v));