在javascript中同时映射和减少一个数组

Question

我有一个约 1800 个对象的数组，代表联赛中的比赛。我需要一个新数组，其中包含每个团队的对象，并将包括 4 个新字段（wins、losses、ties 和 points）。这是我正在使用的数组的示例：

[
  {
    "homeGoals": 2,
    "gameId": "12221",
    "homeTeam": {
      "id": "aasfdsf1",
      "teamName": "Team 1"
    },
    "awayTeam": {
      "id": "aasfdsf2",
      "teamName": "Team 2"
    },
    "id": "ggaew1",
    "awayGoals": 4
  },
  {
    "homeGoals": 5,
    "gameId": "12222",
    "homeTeam": {
      "id": "aasfdsf1",
      "teamName": "Team 1"
    },
    "awayTeam": {
      "id": "aasfdsf3",
      "teamName": "Team 3"
    },
    "id": "ggaew2",
    "awayGoals": 1
  },
  {
    "homeGoals": 4,
    "gameId": "12223",
    "homeTeam": {
      "id": "aasfdsf2",
      "teamName": "Team 2"
    },
    "awayTeam": {
      "id": "aasfdsf3",
      "teamName": "Team 3"
    },
    "id": "ggaew3",
    "awayGoals": 4
  },
  {
    "homeGoals": null,
    "gameId": "12223",
    "homeTeam": {
      "id": "aasfdsf2",
      "teamName": "Team 2"
    },
    "awayTeam": {
      "id": "aasfdsf3",
      "teamName": "Team 3"
    },
    "id": "ggaew4",
    "awayGoals": null
  }
]

这是我需要的结果的示例：

 [
  {
    "id": "aasfdsf1",
    "name": "Team 1",
    "wins": 1,
    "losses": 1,
    "ties": 0,
    "points": 2 
  },
  {
    "id": "aasfdsf2",
    "name": "Team 2",
    "wins": 1,
    "losses": 0,
    "ties": 1,
    "points": 3 
  },
  {
    "id": "aasfdsf3",
    "name": "Team 3",
    "wins": 0,
    "losses": 1,
    "ties": 1,
    "points": 1 
  }
]

有些游戏还没有玩过，所以homeGoals 和awayGoals 字段将为空。

到目前为止，我有一个独特的团队列表，只有那些游戏已经完成：

const completedGames = games.filter(x => x.homeGoals !== null)
const homeTeams = [...new Set(completedGames.map(x => x['homeTeam']))];
const awayTeams = [...new Set(completedGames.map(x => x['awayTeam']))];
const teams = [...new Set([...homeTeams, ...awayTeams])]

我知道我需要执行某种 reduce 功能，但我无法确定它。如果我有适当的 map reduce 功能，我很确定我之前所做的步骤将是无关紧要的。任何帮助将不胜感激！

Answer 1

这可以用flatMap 以更简单的方式表达。不是JS内置的，但是很容易实现：

let flatMap = (a, fn) => [].concat(...a.map(fn));

现在，在地图步骤中，您可以在每个游戏中发出两个“结果”对象（如果游戏不完整，则根本没有结果）：

results = flatMap(data, g => {

    if (g.homeGoals === null || g.awayGoals === null)
        return [];

    if (g.homeGoals > g.awayGoals)
        return [
            {id: g.homeTeam.id, r: 'win'},
            {id: g.awayTeam.id, r: 'loss'},
        ];

    if (g.homeGoals < g.awayGoals)
        return [
            {id: g.homeTeam.id, r: 'loss'},
            {id: g.awayTeam.id, r: 'win'},
        ];

    if (g.homeGoals === g.awayGoals)
        return [
            {id: g.homeTeam.id, r: 'tie'},
            {id: g.awayTeam.id, r: 'tie'},
        ];
});

这会创建一个类似的数组

{ id: 'aasfdsf1', r: 'loss' },
{ id: 'aasfdsf2', r: 'win' },
{ id: 'aasfdsf1', r: 'win' }, etc

容易减少的：

summary = results.reduce((m, {id, r}) => {
    let e = m[id] || {};
    e[r] = (e[r] || 0) + 1;
    return Object.assign(m, {[id]: e})
}, {});

您还可以通过将胜利、失败和平局分别编码为 1、-1、0 来减少冗长，在这种情况下，映射器变为：

results = flatMap(
    data.filter(g => g.homeGoals !== null),
    g => {
        let d = g.homeGoals - g.awayGoals;
        return [
            {id: g.homeTeam.id, r: Math.sign(+d)},
            {id: g.awayTeam.id, r: Math.sign(-d)},
        ]
    });

Answer 2

我认为您正在寻找这样的东西：

const hashMapTeams = games.filter(x => x.homeGoals !== null)
.reduce((res, match)=>{
   /* do the calculations here */
   /* put the values on the res object, using res as a HashMap*/
    res["/*the home team id*/"].id = /*id value*/
    res["/*the home team id*/"].name = /*name value*/
    res["/*the home team id*/"].wins= /* the right value */;
    res["/*the home team id*/"].losses= /* the right value */;
    res["/*the home team id*/"].ties= /* the right value */;
    res["/*the home team id*/"].points= /* the right value */;

    res["/*the away team id*/"].id = /*id value*/
    res["/*the away team id*/"].name = /*name value*/
    res["/*the away team id*/"].wins= /* the right value */;
    res["/*the away team id*/"].losses= /* the right value */;
    res["/*the away team id*/"].ties= /* the right value */;
    res["/*the away team id*/"].points= /* the right value */;
 },{});

/* This will convert again the object to an array */
const arrayTeams = Object.keys(hashMapTeams).map(function (key) { return hashMapTeams[key]; });

Answer 3

这将得到您正在寻找的确切结果：

{
  "id": "aasfdsf1",
  "name": "Team 1",
  "wins": 1,
  "losses": 1,
  "ties": 0,
  "points": 2 
},

我使用三元组和方括号向您展示了不止一种方法，您可以使用任何一种。

let result = [];

your1800ArrayObj.map(data => {      
    let wins = data.wins ? data.wins : 0;
    let losses = data.losses ? data.losses : 0;
    let ties = data['ties'] || 0;
    let points = data['points'] || 0;

    if (data.homeGoals === null && data.awayGoals === null) {
        console.log('game not played')
        } else {
            if (data.homeGoals > data.awayGoals) {
                wins += 1
                points += 1
            } else if (data.homeGoals < data.awayGoals) {
                losses += 1
            } else {
                ties += 1
            }
        }

    result.push({
        id: data.id,
        name: data.homeTeam.teamName ,
        wins: wins,
        losses: losses,
        ties: ties,
        points: points
    })
})

return result
}