如何使用季节和年份对字符串的 ArrayList 进行排序？

Question

我有一个未排序的季节和年份字符串数组列表，如下所示：

["Summer 2020", "Autumn 2018", "Spring 2019", "Autumn 2019", "Winter 2018", "Summer 2018", "Winter 2020", "Spring 2018"]

此列表的正确排序如下：

["Winter 2018", "Spring 2018", "Summer 2018", "Autumn 2018", "Spring 2019", "Autumn 2019", "Winter 2020", "Summer 2020"]

我怎样才能达到这个顺序？一年中的季节顺序如Winter 2021, Spring 2021, Summer 2021, Autumn 2021，不一定每年的所有季节都在列表中。

Answer 1

您需要为此类字符串编写自己的 comparator。我建议你使用 arrow 函数。另外，要对季节进行排序，您需要给它们一些优先级，我会使用 Map<String, Integer>。

试试这个代码：

public static void main(String[] args) {
    Map<String, Integer> seasonsPriority = new HashMap<>();
    seasonsPriority.put("Winter", 1);
    seasonsPriority.put("Spring", 2);
    seasonsPriority.put("Summer", 3);
    seasonsPriority.put("Autumn", 4);

    ArrayList<String> seasons = new ArrayList<>();
    seasons.add("Summer 2020");
    seasons.add("Autumn 2018");
    seasons.add("Spring 2019");
    seasons.add("Autumn 2019");
    seasons.add("Winter 2018");
    seasons.add("Summer 2018");
    seasons.add("Winter 2020");
    seasons.add("Spring 2018");


    seasons.sort((x, y) -> {
        int firstYear = Integer.parseInt(x.split(" ")[1]);
        int secondYear = Integer.parseInt(y.split(" ")[1]);

        if(firstYear < secondYear) {
            return -1;
        }

        if(firstYear > secondYear) {
            return 1;
        }

        String firstSeason = x.split(" ")[0];
        String secondSeason = y.split(" ")[0];

        return Integer.compare(seasonsPriority.get(firstSeason), seasonsPriority.get(secondSeason));
    });

    System.out.println(seasons);
}

输出：

[Winter 2018, Spring 2018, Summer 2018, Autumn 2018, Spring 2019, Autumn 2019, Winter 2020, Summer 2020]

Answer 2

您需要实施自定义 Comparator，先按年份排序，然后再按季节排序。 enum 可用于轻松指定季节的顺序。

private static class SeasonComparator implements Comparator<String> {
    
    private enum Season {
        WINTER, SPRING, SUMMER, AUTUMN
    }

    @Override
    public int compare(String s1, String s2) {
        String[] arr1 = s1.split(" ");
        String[] arr2 = s2.split(" ");
        
        int year1 = Integer.parseInt(arr1[1]);
        int year2 = Integer.parseInt(arr2[1]);
        
        if (year1 != year2) {
            return year1 - year2;
        } else {
            Season season1 = Season.valueOf(arr1[0].toUpperCase());
            Season season2 = Season.valueOf(arr2[0].toUpperCase());
            return season1.ordinal() - season2.ordinal();
        }
    }
}

public static void main(String[] args) {
    List<String> seasons = new ArrayList<>(Arrays.asList("Summer 2020", "Autumn 2018",
            "Spring 2019", "Autumn 2019", "Winter 2018", "Summer 2018", "Winter 2020", "Spring 2018"));

    System.out.println("Unordered list:");
    System.out.println(seasons);
    
    seasons.sort(new SeasonComparator());
    
    System.out.println("Sorted list:");
    System.out.println(seasons);
}

输出

Unordered list:
[Summer 2020, Autumn 2018, Spring 2019, Autumn 2019, Winter 2018, Summer 2018, Winter 2020, Spring 2018]

Sorted list:
[Winter 2018, Spring 2018, Summer 2018, Autumn 2018, Spring 2019, Autumn 2019, Winter 2020, Summer 2020]

Answer 3

考虑编写自定义比较器以使用自定义逻辑执行排序。

public static void main(String[] args) {
    List<String> list = Arrays.asList("Summer 2020", "Autumn 2018", "Spring 2019", "Autumn 2019", "Winter 2018", "Summer 2018", "Winter 2020", "Spring 2018");

    List<String> listOfSeasonsInAscendingOrder = Arrays.asList("Winter", "Spring", "Summer", "Autumn");

    Collections.sort(list, new Comparator<String>() {
        @Override
        public int compare(String o1, String o2) {
            String firstStringSplit[] = o1.split(" ");
            String season1 = firstStringSplit[0];
            int year1 = Integer.valueOf(firstStringSplit[1]);

            String secondStringSplit[] = o2.split(" ");
            String season2 = secondStringSplit[0];
            int year2 = Integer.valueOf(secondStringSplit[1]);

            if (year1 != year2) {
                return year1 < year2 ? -1 : 1;
            }

            int indexOfSeason1 = listOfSeasonsInAscendingOrder.indexOf(season1);
            int indexOfSeason2 = listOfSeasonsInAscendingOrder.indexOf(season2);

            return indexOfSeason1 == indexOfSeason2 ? 0 : indexOfSeason1 < indexOfSeason2 ? -1 : 1;
        }
    });

    System.out.println("Sorted List = " + list);
}

以上代码的输出是：-

Sorted List = [Winter 2018, Spring 2018, Summer 2018, Autumn 2018, Spring 2019, Autumn 2019, Winter 2020, Summer 2020]

Answer 4

试试这个。

static final Map<String, Integer> SEASON_ORDER = Map.of(
    "Winter", 0, "Spring", 1, "Summer", 2, "Autumn", 3);

public static void main(String[] args) {
    List<String> list = new ArrayList<>(List.of(
        "Summer 2020", "Autumn 2018", "Spring 2019", "Autumn 2019",
        "Winter 2018", "Summer 2018", "Winter 2020", "Spring 2018"));
    list.sort(Comparator.comparing(s -> {
        String[] a = s.split("\s+");
        return a[1] + SEASON_ORDER.get(a[0]);
    }));
    System.out.println(list);
}

输出：

[Winter 2018, Spring 2018, Summer 2018, Autumn 2018, Spring 2019, Autumn 2019, Winter 2020, Summer 2020]

Answer 5

前面的回答都是在comparator函数中进行字符串拆分解析。这工作得很好，但可能会导致大规模的性能问题 - 拆分和解析可能是相对昂贵的操作，并且在排序时比较器需要平均检查每个元素不止一次。让比较器为自然排序正确的每个元素计算一个代码，并使用映射记住这些代码，以避免对每个元素执行多次这些操作，可能会提高性能：

public class Scratch {
    public static void main(String[] args) {
        List<String> list = Arrays.asList("Summer 2020", "Autumn 2018", "Spring 2019", "Autumn 2019", "Winter 2018", "Summer 2018", "Winter 2020", "Spring 2018");
        list.sort(new SeasonYearComparator());
        System.out.println(list);
    }

    private enum Season {
        Winter, Spring, Summer, Autumn;
    }

    private static class SeasonYearComparator implements Comparator<String> {
        private final Map<String, String> codes = new HashMap<>();

        @Override
        public int compare(String s1, String s2) {
            String code1 = codes.computeIfAbsent(s1, this::computeCode);
            String code2 = codes.computeIfAbsent(s2, this::computeCode);
            return code1.compareTo(code2);
        }

        private String computeCode(String seasonYear) {
            String[] parts = seasonYear.split(" ");
            return parts[1] + Season.valueOf(parts[0]).ordinal();
        }
    }
}

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