Question

我有一个字符串列表，如：

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

我想像这样在Map<String, List<String>>中进行转换

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

我已经尝试了这段代码并且可以正常工作，但是在这种情况下，我必须创建一个新的类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;
    }
}

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

Answer 1

您可以这样做：

df <- structure(list(a = c("ABC,SZY", "XYZ", NA, NA, "ABC,BNM,JKL", 
"DEF", "XCV"), b = c("SZY", "XYZ,IOP", "QWE", NA, "BNM,JKL,STU", 
"DEF", "HJK")), row.names = c(NA, -7L), class = "data.frame")

一种更好的方法，

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

更新

根据以下评论，可以将其进一步简化为

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())));

Answer 2

尝试一下

let dateFormatter = DateFormatter()
dateFormatter.dateFormat = "MM/yyyy"
let enteredDate = dateFormatter.date(from: expiryDate.text!)

//Also convert the current date in the entered date format and then checks that the date is valid if enteredDate month or year is greater than current date. 
let currentDate = dateFormatter.date(from: dateFormatter.string(from: Date()))

if enteredDate.compare(now) != ComparisonResult.orderedAscending {
    print("Valid")   
} else {
    print("Not Valid")
}

Answer 3

您只需将代码按部分分组即可，将第一组作为键，第二组作为值，而不是先映射

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呢？与流相比，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

似乎您的位置图需要根据键进行排序，您可以尝试以下操作

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]}

Java 8：如何将String转换为Map <string，list <string >>？

5 个答案: