javascript - 重构对象数组

Question

我一直在努力重组这个特定的对象；

[
  {
    "userId": 1,
    "name": "Breaker of Sky",
    "rank": 1
  },
  {
    "userId": 1,
    "name": "Slayer of Mountain",
    "rank": 2
  },
  {
    "userId": 1,
    "name": "Balthromaw",
    "rank": 3
  },
  {
    "userId": 1,
    "name": "Death",
    "rank": -1
  },
  {
    "userId": 3,
    "name": "Breaker of Sky",
    "rank": 2
  },
  {
    "userId": 3,
    "name": "Slayer of Mountain",
    "rank": 3
  },
  {
    "userId": 3,
    "name": "Balthromaw",
    "rank": 1
  },
  {
    "userId": 3,
    "name": "Death",
    "rank": -1
  },
  {
    "userId": 4,
    "name": "Breaker of Sky",
    "rank": 3
  },
  {
    "userId": 4,
    "name": "Slayer of Mountain",
    "rank": 2
  },
  {
    "userId": 4,
    "name": "Balthromaw",
    "rank": 1
  },
  {
    "userId": 4,
    "name": "Death",
    "rank": -1
  },
  {
    "userId": 8,
    "name": "Breaker of Sky",
    "rank": 2
  },
  {
    "userId": 8,
    "name": "Slayer of Mountain",
    "rank": 3
  },
  {
    "userId": 8,
    "name": "Balthromaw",
    "rank": 4
  },
  {
    "userId": 8,
    "name": "Death",
    "rank": 1
  }
]

让它看起来像下面这个对象。

[
     { ranking: [['Death'], ['Breaker of Sky'], ['Slayer of Mountain'], ['Balthromaw']], count: 2 },
     { ranking: [['Death'], ['Balthromaw'], ['Breaker of Sky'], ['Slayer of Mountain']], count: 1 },
     { ranking: [['Death'], ['Balthromaw'], ['Slayer of Mountain'], ['Breaker of Sky']], count: 1 },
]

为了简单地解释这个过程，我们首先需要查找 userId 并将具有相同 userId 的排名的名称分组。然后它应该通过查看他们的排名来排列名字(-1 总是排在第一位)。之后，我们应该寻找序列，计数并记下。例如，此序列 [['Death'], ['Breaker of Sky'], ['Slayer of Mountain'], ['Balthromaw']] 已被发现 2 次并且因此计数为 2。

非常感谢您对这个问题的帮助。

我的代码:

这是我到目前为止所做的。它仍然缺少通过排序排名排列名称的部分，并且排名属性的值也不是数组。此外，我也无法计算序列部分。你能帮我解决一下吗？

const merged = data.reduce((r, { userId, ...rest }) => {
  const key = `${userId}`;
  r[key] = r[key] || { ranking: [] };
  r[key]['ranking'].push(rest);
  return r;
}, {});

const rankArray = Object.values(merged);
console.log(rankArray);

我得到的输出:

[
  {
    "ranking": [
      {
        "name": "Breaker of Sky",
        "rank": 1
      },
      {
        "name": "Slayer of Mountain",
        "rank": 2
      },
      {
        "name": "Balthromaw",
        "rank": 3
      },
      {
        "name": "Death",
        "rank": -1
      }
    ]
  },
  {
    "ranking": [
      {
        "name": "Breaker of Sky",
        "rank": 2
      },
      {
        "name": "Slayer of Mountain",
        "rank": 3
      },
      {
        "name": "Balthromaw",
        "rank": 1
      },
      {
        "name": "Death",
        "rank": -1
      }
    ]
  },
  {
    "ranking": [
      {
        "name": "Breaker of Sky",
        "rank": 3
      },
      {
        "name": "Slayer of Mountain",
        "rank": 2
      },
      {
        "name": "Balthromaw",
        "rank": 1
      },
      {
        "name": "Death",
        "rank": -1
      }
    ]
  },
  {
    "ranking": [
      {
        "name": "Breaker of Sky",
        "rank": 2
      },
      {
        "name": "Slayer of Mountain",
        "rank": 3
      },
      {
        "name": "Balthromaw",
        "rank": 4
      },
      {
        "name": "Death",
        "rank": 1
      }
    ]
  }
]

Answer 1

有几个步骤要做:

1. 按 userId 分组，您已经完成了。
2. 按等级对每个用户的值进行排序。
3. 为每个排名项数组构建一个计数图。我假设排名数字无关紧要，重要的是字符串名称。
4. trim map 中 count > 1 的任何额外元素。 (我认为这是您想要的逻辑；如果不是，您可以删除 delete byUser[userId] 行)
5. 迭代并将最终计数附加到结果数组中的每个对象。

这是解决所有这些问题的一种方法:

const data = [ { "userId": 1, "name": "Breaker of Sky", "rank": 1 }, { "userId": 1, "name": "Slayer of Mountain", "rank": 2 }, { "userId": 1, "name": "Balthromaw", "rank": 3 }, { "userId": 1, "name": "Death", "rank": -1 }, { "userId": 3, "name": "Breaker of Sky", "rank": 2 }, { "userId": 3, "name": "Slayer of Mountain", "rank": 3 }, { "userId": 3, "name": "Balthromaw", "rank": 1 }, { "userId": 3, "name": "Death", "rank": -1 }, { "userId": 4, "name": "Breaker of Sky", "rank": 3 }, { "userId": 4, "name": "Slayer of Mountain", "rank": 2 }, { "userId": 4, "name": "Balthromaw", "rank": 1 }, { "userId": 4, "name": "Death", "rank": -1 }, { "userId": 8, "name": "Breaker of Sky", "rank": 2 }, { "userId": 8, "name": "Slayer of Mountain", "rank": 3 }, { "userId": 8, "name": "Balthromaw", "rank": 4 }, { "userId": 8, "name": "Death", "rank": 1 } ];

const byUser = data.reduce((a, e) => {
  a[e.userId] = a[e.userId] || [];
  a[e.userId].push(e);
  return a;
}, {});
const counts = {};

for (const [userId, ranking] of Object.entries(byUser)) {
  ranking.sort((a, b) => a.rank - b.rank);
  byUser[userId] = ranking.map(e => e.name);
  const k = byUser[userId];
  counts[k] = ++counts[k] || 1;
  
  if (counts[k] > 1) {
    delete byUser[userId];
  }
}

const result = Object.values(byUser).map(v => ({
  ranking: v,
  count: counts[v],
}));
console.log(result);