从对象数组中返回名称数组

Question

我知道如何遍历一个对象并打印出我想要的一组值，但是我无法确定按我想要的顺序打印它。

**问题是：**

给定一组游戏结果记录，通过返回一组玩家姓名来确定所有玩家是谁。

该数组应按照遇到名称的方式排序。

示例输入：

[
  { winner: 'Alishah', loser: 'Bob',    loser_points: 3 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 1 },
  { winner: 'Elise',   loser: 'Bob',    loser_points: 2 },
  { winner: 'Elise',   loser: 'Maria',  loser_points: 4 },
  { winner: 'Alishah', loser: 'Maria',  loser_points: 2 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 3 },
  { winner: 'Xu Jin',  loser: 'Elise',  loser_points: 2 }
]

预期结果：

['Alishah', 'Bob', 'Maria', '徐瑾', 'Elise']

**我目前的代码：**

let data = [
  { winner: 'Alishah', loser: 'Bob',    loser_points: 3 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 1 },
  { winner: 'Elise',   loser: 'Bob',    loser_points: 2 },
  { winner: 'Elise',   loser: 'Maria',  loser_points: 4 },
  { winner: 'Alishah', loser: 'Maria',  loser_points: 2 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 3 },
  { winner: 'Xu Jin',  loser: 'Elise',  loser_points: 2 }
];

   
console.log(main(data));

Answer 1

您可以使用Array.reduce() 和Set 获取名称并删除重复项

Array.reduce() 遍历数组。所以你可以将所有winners 和losers 推送到初始的空数组中。

然后你可以创建一个新的Set。这样，您可以删除重复项（因为所有项目都是string）。之后，您可以使用spread syntax 将其转换回数组：[...new Set(array)]

const arr = [
  { winner: 'Alishah', loser: 'Bob',    loser_points: 3 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 1 },
  { winner: 'Elise',   loser: 'Bob',    loser_points: 2 },
  { winner: 'Elise',   loser: 'Maria',  loser_points: 4 },
  { winner: 'Alishah', loser: 'Maria',  loser_points: 2 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 3 },
  { winner: 'Xu Jin',  loser: 'Elise',  loser_points: 2 }
]

const names = [...new Set(arr.reduce((acc, cur) => [...acc, cur.winner, cur.loser], []))]

console.log(names)

Answer 2

您可以为此使用.flatMap() 和Set()：

let data = [
  { winner: 'Alishah', loser: 'Bob',    loser_points: 3 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 1 },
  { winner: 'Elise',   loser: 'Bob',    loser_points: 2 },
  { winner: 'Elise',   loser: 'Maria',  loser_points: 4 },
  { winner: 'Alishah', loser: 'Maria',  loser_points: 2 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 3 },
  { winner: 'Xu Jin',  loser: 'Elise',  loser_points: 2 }
];

const res = [...new Set(data.flatMap(x=>[x.winner, x.loser]))]
console.log( res )

说明：

• 使用.flatMap() 方法，我们将首先得到一个数组数组。这里内部数组将是 winner 和 loser 名称的数组。
• 然后我们将数组展平以获得所有玩家姓名的单个数组。
• 最后使用[...new Set(array)]，我们将在数组中获得不同的名称以实现所需的结果。

Answer 3

var players = collection.reduce((acc, player) => {
    if(!acc.inStore[player.winner]) {
        acc.players.push(player.winner)
        acc.inStore[player.winner] = true
    }
    if(!acc.inStore[player.loser]) {
        acc.players.push(player.loser)
        acc.inStore[player.loser] = true
    }
    return acc;
}, {players: [], inStore: {}}).players

// ["Alishah", "Bob", "Maria", "Xu Jin", "Elise"]

Answer 4

使用reduce 和includes

console.clear();

"use strict";

const score = [
  { winner: 'Alishah', loser: 'Bob',    loser_points: 3 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 1 },
  { winner: 'Elise',   loser: 'Bob',    loser_points: 2 },
  { winner: 'Elise',   loser: 'Maria',  loser_points: 4 },
  { winner: 'Alishah', loser: 'Maria',  loser_points: 2 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 3 },
  { winner: 'Xu Jin',  loser: 'Elise',  loser_points: 2 }
]

function reduce(coll, {winner, loser}) {
  if (!coll.includes(winner)) {
    coll.push(winner)
  }
  if (!coll.includes(loser)) {
    coll.push(loser)
  }
  return coll;
}

var players = score.reduce(reduce, [])
console.log(players)

Answer 5

查看您的代码和您的预期结果，我认为您缺少推入arr 以及looser 名称。要修复它，我最喜欢的方法是 js 数组的 reduce 方法：

const arr = outcomes.reduce((completeList,{winner, looser}) => {
    const extraNames = [winner, looser]
        .filter(x => !completeList.includes(x));
    return [
        ...completeList,
        ...extraNames
    ];
}, []);

Answer 6

使用地图和过滤器

let array = [
  { winner: 'Alishah', loser: 'Bob',    loser_points: 3 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 1 },
  { winner: 'Elise',   loser: 'Bob',    loser_points: 2 },
  { winner: 'Elise',   loser: 'Maria',  loser_points: 4 },
  { winner: 'Alishah', loser: 'Maria',  loser_points: 2 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 3 },
  { winner: 'Xu Jin',  loser: 'Elise',  loser_points: 2 }
];

let winners = array.map(i => i.winner).filter((x, i, a) => a.indexOf(x) == i)
console.log(winners);

Answer 7

const outcomes = [
  { winner: 'Alishah', loser: 'Bob',    loser_points: 3 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 1 },
  { winner: 'Elise',   loser: 'Bob',    loser_points: 2 },
  { winner: 'Elise',   loser: 'Maria',  loser_points: 4 },
  { winner: 'Alishah', loser: 'Maria',  loser_points: 2 },
  { winner: 'Maria',   loser: 'Xu Jin', loser_points: 3 },
  { winner: 'Xu Jin',  loser: 'Elise',  loser_points: 2 }
];

const result = outcomes.reduce((acc, cur) => {
  if(!acc.includes(cur.winner))
  acc.push(cur.winner);
  if(!acc.includes(cur.loser))
  acc.push(cur.loser);
  return acc;
}, []).join(",");

console.log(result);