我如何在 java 中使用 groupingby

Question

我有一个来自 Web 服务的电影列表，我需要使用数据中的流派属性对其进行分组。我的问题与此非常相似，但在 java 中。

来自网络服务的电影列表如下所示

[
    {
        "title": "AAAAAAAA",
        "genres": [
          "Comedy"
        ]
    },
    {
        "title": "BBBBBBBBBBBB",
        "genres": [
          "Action", "Adventure"
        ]
    },
    {
        "title": "CCCCCCCCCCCCC",
        "genres": [
          "Action"
        ]
    },
    {
        "title": "DDDDDDDDDDDD",
        "genres": [
          "Comedy", "Adventure"
        ]
    },
    {
        "title": "EEEEEEEEEEEEEEEEEEE",
        "genres": [
          "Horror"
        ]
    }
]

这是我正在努力实现的目标，但我似乎做不对

[
    {
        "Action": [
            {
                "title": "BBBBBBBBBBBB",
                "genres": [
                  "Action", "Adventure"
                ]
            },
            {
                "title": "CCCCCCCCCCCCC",
                "genres": [
                  "Action"
                ]
            }
        ],
        "Adventure": [
            {
                "title": "BBBBBBBBBBBB",
                "genres": [
                  "Action", "Adventure"
                ]
            },
            {
                "title": "DDDDDDDDDDDD",
                "genres": [
                  "Comedy", "Adventure"
                ]
            }
        ],
        "Comedy": [
            {
                "title": "AAAAAAAA",
                "genres": [
                  "Comedy"
                ]
            },
            {
                "title": "DDDDDDDDDDDD",
                "genres": [
                  "Comedy", "Adventure"
                ]
            }
        ],
        "Horror": [
            {
                "title": "EEEEEEEEEEEEEEEEEEE",
                "genres": [
                  "Horror"
                ]
            }
        ]
    }
]

我试过从中的答案复制 java脚本解决方案但我卡住了

    ItemArray.stream().reduce((identity, accumulator) -> {    
        accumulator.getGenres().forEach((k) -> {});
        return identity;
    });

Answer 1

我认为不使用流会更容易：

Map<String, List<Item>> itemsByGenre = new HashMap<>();
for (Item item : items) {
  for (String genre : item.genres()) {
    itemsByGenre.computeIfAbsent(genre, g -> new ArrayList<>()).add(item);
  }
}

这里的重点是，你需要按流派来展开，因为每个项目都可以在多个流派中。所以，你可以这样做：

Map<String, List<Item>> itemsByGentre = items.stream()
    // Make a stream of map entries where the key is the genre, and the value is the item.
    .flatMap(i -> i.genres().stream().map(g -> new AbstractMap.SimpleEntry<>(g, i))
    .collect(
        // Now, group by key (genre)...
        Collectors.groupingBy(
            Map.Entry::getKey,
            // and map the entry to just the value (the item),
            // and collect the items into a list.
            Collectors.mapping(Map.Entry::getValue, Collectors.toList())));

我如何在 java 中使用 groupingby

How do i make use of groupingby in java

java

arrays

spring-boot

java-stream