java - Java 8 是否可以一步创建一个列表或任何其他分组、排序和计数的数据结构？

Question

我有一个具有以下结构的文件:

a, City1, 2013-09-05 14:08:15b, City2, 2015-10-08 16:10:15c, City2, 2010-09-05 14:08:15d, City1, 2011-09-05 14:08:15

It's comma separated values and lines separated with end of line character.

I need to create a data structure in Java 8 where I have the rows grouped by city and within each such group sort by date in ascending order and counting the amount of rows within each group.

I tried:

1. Create a List<Row> from the file
2. Create a Map<String, List<Row>> which is grouped by city and sort by date withing each group
3. Create a Map<String, Long> for grouping by city and amount of rows

This is code I've tried:

public PhotoResponse processFile()  {
    //read each line of the file and create a new object PhotoIn for each one
    List<PhotoIn> lista = null;
    try {
        lista = Files.lines(Paths.get(file))
        .map(line -> line.split(","))
        .map(photo -> new PhotoIn(photo[0].substring(0, photo[0].lastIndexOf(".")), photo[0].substring(photo[0].lastIndexOf(".") + 1 ), photo[1].trim(), parseDate(photo[2]), index++))
        .collect(Collectors.toList());
    } catch (IOException e) {
        e.printStackTrace();
    }
    return generateOutput(lista); 
}


private PhotoResponse generateOutput(List<PhotoIn> photos) {

    //Grouping photos by city
    Map<String, List<PhotoIn>> photosByCity = photos.stream().collect(Collectors.groupingBy(PhotoIn::getCity));

    //Sorting photos by date into each city
    photosByCity.values().forEach(list -> list.sort(Comparator.comparing(PhotoIn::getDate)));

    //Grouping photos by city and amount
    Map<String, Long> numeroPorCiudades = photos.stream().collect(Collectors.groupingBy(PhotoIn::getCity, Collectors.counting()));

    List<PhotoOut> photoOutList = new ArrayList<PhotoOut>();

    photosByCity.forEach((key, list) -> {
        String digits = Integer.toString(Long.toString(numeroPorCiudades.get(key)).length());
        counter = 1;
        list.forEach(photoIn -> {
            photoOutList.add(new PhotoOut(photoIn.getName() + "." + photoIn.getExtension(), key + String.format("%0" + digits + "d", counter++) + "." + photoIn.getExtension(), photoIn.getIndex()));
        });
    });
    return sortOutput(photoOutList);
}

我正在解决这个问题，但我正在寻找一种更好、更有效的方法来使用 Java 8 来解决这个问题。是否可以一步完成这 3 个步骤？。我需要的是将所有信息分组到一个数据结构中。

Answer 1

使用该模型:

class Model {
  private String title;
  private String city;
  private LocalDateTime date;
}

可以这样完成:

List<Model> list = getFromFile();

Comparator<Model> comparator = Comparator.comparing(Model::getDate);

//grouping the list by City with lists sorted by date
Map<String, List<Model>> map = list.stream()
    .collect(
        Collectors.groupingBy( //grouping the list by City with lists 
            m -> m.getCity(),
            Collectors.collectingAndThen(Collectors.toList(), l->{
              l.sort(comparator);
              return l;
            })
        )
    );

//getting another map with counts
Map<String, Object> countMap = map.entrySet().stream()
    .collect(Collectors.toMap(Entry::getKey, entry -> entry.getValue().size()));