如何基于分隔符将 List<String> 转换为 Map<String,List<String>>

Question

我有一个字符串列表，例如：

List<String> locations = Arrays.asList("US:5423","US:6321","CA:1326","AU:5631");

我想在Map<String, List<String>> 中转换为：

AU = [5631]
CA = [1326]
US = [5423, 6321]

我已经尝试过这段代码，它可以工作，但在这种情况下，我必须创建一个新类GeoLocation.java。

List<String> locations=Arrays.asList("US:5423", "US:6321", "CA:1326", "AU:5631");
Map<String, List<String>> locationMap = locations
        .stream()
        .map(s -> new GeoLocation(s.split(":")[0], s.split(":")[1]))
        .collect(
                Collectors.groupingBy(GeoLocation::getCountry,
                Collectors.mapping(GeoLocation::getLocation, Collectors.toList()))
        );

locationMap.forEach((key, value) -> System.out.println(key + " = " + value));

GeoLocation.java

private class GeoLocation {
    private String country;
    private String location;

    public GeoLocation(String country, String location) {
        this.country = country;
        this.location = location;
    }

    public String getCountry() {
        return country;
    }

    public void setCountry(String country) {
        this.country = country;
    }

    public String getLocation() {
        return location;
    }

    public void setLocation(String location) {
        this.location = location;
    }
}

但我想知道，有什么方法可以在不引入新类的情况下将List<String> 转换为Map<String, List<String>>。

Answer 1

你可以这样做：

Map<String, List<String>> locationMap = locations.stream()
        .map(s -> s.split(":"))
        .collect(Collectors.groupingBy(a -> a[0],
                Collectors.mapping(a -> a[1], Collectors.toList())));

一个更好的方法是，

private static final Pattern DELIMITER = Pattern.compile(":");

Map<String, List<String>> locationMap = locations.stream()
    .map(s -> DELIMITER.splitAsStream(s).toArray(String[]::new))
        .collect(Collectors.groupingBy(a -> a[0], 
            Collectors.mapping(a -> a[1], Collectors.toList())));

更新

根据以下评论，这可以进一步简化为，

Map<String, List<String>> locationMap = locations.stream().map(DELIMITER::split)
    .collect(Collectors.groupingBy(a -> a[0], 
        Collectors.mapping(a -> a[1], Collectors.toList())));

Answer 2

试试这个

Map<String, List<String>> locationMap = locations.stream()
            .map(s ->  new AbstractMap.SimpleEntry<String,String>(s.split(":")[0], s.split(":")[1]))
            .collect(Collectors.groupingBy(Map.Entry::getKey,
                     Collectors.mapping(Map.Entry::getValue, Collectors.toList())));

Answer 3

您可以将代码按部分分组，将第一组作为键，第二组作为值，而不是先映射它

Map<String, List<String>> locationMap = locations
            .stream()
            .map(s -> s.split(":"))
            .collect( Collectors.groupingBy( s -> s[0], Collectors.mapping( s-> s[1], Collectors.toList() ) ) );

Answer 4

POJO 怎么样。与流相比，它看起来并不复杂。

public static Map<String, Set<String>> groupByCountry(List<String> locations) {
    Map<String, Set<String>> map = new HashMap<>();

    locations.forEach(location -> {
        String[] parts = location.split(":");
        map.compute(parts[0], (country, codes) -> {
            codes = codes == null ? new HashSet<>() : codes;
            codes.add(parts[1]);
            return codes;
        });
    });

    return map;
}

Answer 5

好像你的位置图需要按key排序，你可以试试下面的

List<String> locations = Arrays.asList("US:5423", "US:6321", "CA:1326", "AU:5631");

    Map<String, List<String>> locationMap = locations.stream().map(str -> str.split(":"))
            .collect(() -> new TreeMap<String, List<String>>(), (map, parts) -> {
                if (map.get(parts[0]) == null) {
                    List<String> list = new ArrayList<>();
                    list.add(parts[1]);
                    map.put(parts[0], list);
                } else {
                    map.get(parts[0]).add(parts[1]);
                }
            }, (map1, map2) -> {
                map1.putAll(map2);
            });

    System.out.println(locationMap); // this outputs {AU=[5631], CA=[1326], US=[5423, 6321]}