从数组中删除重复对象但合并嵌套对象答案

【问题标题】：Remove duplicate objects from an array but merge nested objects从数组中删除重复对象但合并嵌套对象
【发布时间】：2019-11-15 08:13:36
【问题描述】：

目前有一个包含游戏版本的对象数组。然而，游戏发布可能发生在多个平台上，并且这些在数组中显示为单独的对象。我希望通过比较游戏 ID 来删除重复的游戏，但合并平台对象

我已尝试使用 reduce 函数，该函数成功地通过游戏 ID 删除重复对象，但我无法将其调整为合并平台

const filteredArr = data.reduce((acc, current) => {
  const x = acc.find(item => item.game.id === current.game.id);

  if (!x) {
    return acc.concat([current]);
  } else {
    return acc;
  }
}, []);

当前数组：

const data = [{
  "id": 157283,
  "date": 1553212800,
  "game": {
    "id": 76882,
    "name": "Sekiro: Shadows Die Twice",
    "popularity": 41.39190295640344
  },
  "human": "2019-Mar-22",
  "m": 3,
  "platform": {"id": 48, "name": "PlayStation 4"},
  "region": 8,
  "y": 2019
}, {
  "id": 12,
  "date": 1553212800,
  "game": {
    "id": 76832,
    "name": "Spiderman",
    "popularity": 41.39190295640344
  },
  "human": "2019-Mar-22",
  "m": 3,
  "platform": {"id": 6, "name": "PC (Microsoft Windows)"},
  "region": 8,
  "y": 2019
}, {
  "id": 157283,
  "date": 1553212800,
  "game": {
    "id": 76882,
    "name": "Sekiro: Shadows Die Twice",
    "popularity": 41.39190295640344
  },
  "human": "2019-Mar-22",
  "m": 3,
  "platform": {"id": 48, "name": "Xbox"},
  "region": 8,
  "y": 2019
}]

合并后的预期格式：

[{
  "id": 157283,
  "date": 1553212800,
  "game": {
    "id": 76882,
    "name": "Sekiro: Shadows Die Twice",
    "popularity": 41.39190295640344
  },
  "human": "2019-Mar-22",
  "m": 3,
  "platforms": ["PlayStation", "Xbox"],
  "region": 8,
  "y": 2019
}, {
  "id": 12,
  "date": 1553212800,
  "game": {
    "id": 76832,
    "name": "Spiderman",
    "popularity": 41.39190295640344
  },
  "human": "2019-Mar-22",
  "m": 3,
  "platforms": ["Playstation"],
  "region": 8,
  "y": 2019
}]

【问题讨论】：

不同平台发布日期不同怎么办？在这种情况下，《只狼：Shadows Die Twice》同时在多个平台上发布了他们的游戏。但稍后为某些平台发布游戏的情况并不少见。以《暗黑破坏神3》为例，2012 年在 Windows 和 OS X 上发布，2013 年在 PS3 上发布，2014 年在 PS4 上发布，2018 年在 Switch 上发布。
好点，我认为一个简单的操作员会解决这个用例。 const x = acc.find(item => item.game.id === current.game.id && item.releaseDate == current.releaseDate);

标签： javascript arrays object ecmascript-6

【解决方案1】：

你真的很接近，你只需要稍微改变一下逻辑。您可以尝试以下方法；一个例子 - https://repl.it/@EQuimper/ScaryBumpyCircle

const filteredArr = data.reduce((acc, current) => {
  const x = acc.find(item => item.game.id === current.game.id);

  if (!x) {
    current.platform = [current.platform]
    acc.push(current);
  } else {
    x.platform.push(current.platform);
  }

  return acc;
}, []);

返回值为

[
  {
    "id": 157283,
    "date": 1553212800,
    "game": {
      "id": 76882,
      "name": "Sekiro: Shadows Die Twice",
      "popularity": 41.39190295640344
    },
    "human": "2019-Mar-22",
    "m": 3,
    "platform": [
      {
        "id": 48,
        "name": "PlayStation 4"
      },
      {
        "id": 48,
        "name": "Xbox"
      }
    ],
    "region": 8,
    "y": 2019
  },
  {
    "id": 12,
    "date": 1553212800,
    "game": {
      "id": 76832,
      "name": "Spiderman",
      "popularity": 41.39190295640344
    },
    "human": "2019-Mar-22",
    "m": 3,
    "platform": [
      {
        "id": 6,
        "name": "PC (Microsoft Windows)"
      }
    ],
    "region": 8,
    "y": 2019
  }
]

如果您只想拥有一个平台字符串数组，请使用

const filteredArr = data.reduce((acc, current) => {
  const x = acc.find(item => item.game.id === current.game.id);

  if (!x) {
    current.platform = [current.platform.name]
    acc.push(current);
  } else {
    x.platform.push(current.platform.name);
  }

  return acc;
}, []);

现在返回值为

[
  {
    "id": 157283,
    "date": 1553212800,
    "game": {
      "id": 76882,
      "name": "Sekiro: Shadows Die Twice",
      "popularity": 41.39190295640344
    },
    "human": "2019-Mar-22",
    "m": 3,
    "platform": [
      "PlayStation 4",
      "Xbox"
    ],
    "region": 8,
    "y": 2019
  },
  {
    "id": 12,
    "date": 1553212800,
    "game": {
      "id": 76832,
      "name": "Spiderman",
      "popularity": 41.39190295640344
    },
    "human": "2019-Mar-22",
    "m": 3,
    "platform": [
      "PC (Microsoft Windows)"
    ],
    "region": 8,
    "y": 2019
  }
]

【讨论】：

【解决方案2】：

您可以将对象的platform 分开，并查看是否有具有相同id 的对象并将平台添加到数组中，而不是创建新数据集。

const
    data = [{ id: 157283, date: 1553212800, game: { id: 76882, name: "Sekiro: Shadows Die Twice", popularity: 41.39190295640344 }, human: "2019-Mar-22", m: 3, platform: { id: 48, name: "PlayStation 4" }, region: 8, y: 2019 }, { id: 12, date: 1553212800, game: { id: 76832, name: "Spiderman", popularity: 41.39190295640344 }, human: "2019-Mar-22", m: 3, platform: { id: 6, name: "PC (Microsoft Windows)" }, region: 8, y: 2019 }, { id: 157283, date: 1553212800, game: { id: 76882, name: "Sekiro: Shadows Die Twice", popularity: 41.39190295640344 }, human: "2019-Mar-22", m: 3, platform: { id: 48, name: "Xbox" }, region: 8, y: 2019 }],
    result = data.reduce((r, { platform, ...o }) => {
        var temp = r.find(({ id }) => id === o.id);
        if (!temp) r.push(temp = { ...o, platforms: [] });
        temp.platforms.push(platform);
        return r;
    }, []);

console.log(result);

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【讨论】：

【解决方案3】：

请看一下：

const data = [    {        "id": 157283,
        "date": 1553212800,
        "game": {
            "id": 76882,
            "name": "Sekiro: Shadows Die Twice",
            "popularity": 41.39190295640344
        },
        "human": "2019-Mar-22",
        "m": 3,
        "platform": {
            "id": 48,
            "name": "PlayStation 4"
        },
        "region": 8,
        "y": 2019
    },
    {
        "id": 12,
        "date": 1553212800,
        "game": {
            "id": 76832,
            "name": "Spiderman",
            "popularity": 41.39190295640344
        },
        "human": "2019-Mar-22",
        "m": 3,
        "platform": {
            "id": 6,
            "name": "PC (Microsoft Windows)"
        },
        "region": 8,
        "y": 2019
    },{        "id": 157283,
    "date": 1553212800,
    "game": {
        "id": 76882,
        "name": "Sekiro: Shadows Die Twice",
        "popularity": 41.39190295640344
    },
    "human": "2019-Mar-22",
    "m": 3,
    "platform": {
        "id": 48,
        "name": "Xbox"
    },
    "region": 8,
    "y": 2019
},
]

const filteredArr = data.reduce((acc, current) => {
  const x = acc.find(item => item.game.id === current.game.id);
    if (!x) {
      current.platform = [current.platform.name]
      return acc.concat([current]);
    } else {
      x.platform.push(current.platform.name);
      return acc;
    }
  }, []);
console.log(filteredArr);

【讨论】：

【解决方案4】：

这是另一种解决方案，使用forEach 而不是reduce。这利用了查找哈希，如果处理大量数据比使用 find 更快。

const data = [{"id": 157283, "date": 1553212800, "game": {"id": 76882, "name": "Sekiro: Shadows Die Twice", "popularity": 41.39190295640344}, "human": "2019-Mar-22", "m": 3, "platform": {"id": 48, "name": "PlayStation 4"}, "region": 8, "y": 2019}, {"id": 12, "date": 1553212800, "game": {"id": 76832, "name": "Spiderman", "popularity": 41.39190295640344}, "human": "2019-Mar-22", "m": 3, "platform": {"id": 6, "name": "PC (Microsoft Windows)"}, "region": 8, "y": 2019}, {"id": 157283, "date": 1553212800, "game": {"id": 76882, "name": "Sekiro: Shadows Die Twice", "popularity": 41.39190295640344}, "human": "2019-Mar-22", "m": 3, "platform": {"id": 48, "name": "Xbox"}, "region": 8, "y": 2019}];

let result = {};

data.forEach(({platform, ...release}) => {
  release.platforms = [platform.name];

  const releaseLookup = result[release.game.id];
  if (!releaseLookup) {
    result[release.game.id] = release;
  } else {
    releaseLookup.platforms.push(...release.platforms);
  }
});

console.log(Object.values(result));

【讨论】：