如何从对象列表中获取唯一值？答案

【问题标题】：how to get unique values from a list of objects?如何从对象列表中获取唯一值？
【发布时间】：2022-06-22 23:21:14
【问题描述】：

数据：

[
  {
    "name": "Ankh of Anubis",
    "rank": {
      "_type": "medal",
      "current": "ankh-of-anubis"
    }
  },
  {
    "name": "Bonus Roulette",
    "rank": {
      "_type": "medal",
      "current": "bonus-roulette"
    }
  },
  {
    "name": "jetx",
    "rank": {
      "_type": "medal",
      "current": "jetx"
    }
  },
  {
    "name": "Gates of Olympus",
    "rank": {
      "_type": "trophy",
      "current": "gates-of-olympus"
    }
  },
]

如何只过滤唯一值，

uniqueValues = ["medal","trophy"]

我试过了，

const uniqueTitles = new Set(games.category.title);
const uniqueTitles = [...new Set(games.category.title)] //打字错误。

 useEffect(() => {
    const uniqueTitles = games.filter((game:any) => {
      return new Set(game.category.title);
    })
    setTitles(uniqueTitles);
  },[])

【问题讨论】：

标签： javascript reactjs arrays

【解决方案1】：

您正在使用Set 作为过滤器函数的返回值。真的是这样打算的吗？给定数据：

const data = [
  {
    "name": "Ankh of Anubis",
    "rank": {
      "_type": "medal",
      "current": "ankh-of-anubis"
    }
  },
  {
    "name": "Bonus Roulette",
    "rank": {
      "_type": "medal",
      "current": "bonus-roulette"
    }
  },
  {
    "name": "jetx",
    "rank": {
      "_type": "medal",
      "current": "jetx"
    }
  },
  {
    "name": "Gates of Olympus",
    "rank": {
      "_type": "trophy",
      "current": "gates-of-olympus"
    }
  },
]

你可以这样做：

const uniqueValues = new Set();
data.forEach(record => uniqueValues.add(record.rank._type));
console.log(uniqueValues);

这是link。

【讨论】：

【解决方案2】：

与您的第一次尝试类似的解决方案：

const data = [{"name":"Ankh of Anubis","rank":{"_type":"medal","current":"ankh-of-anubis"}},{"name":"Bonus Roulette","rank":{"_type":"medal","current":"bonus-roulette"}},{"name":"jetx","rank":{"_type":"medal","current":"jetx"}},{"name":"Gates of Olympus","rank":{"_type":"trophy","current":"gates-of-olympus"}},];

const uniqueValues = new Set(data.map(elem => elem.rank._type));
console.log(uniqueValues);

【讨论】：