将结果数组缩减为具有总和和用户 ID 的对象数组

Question

我正在尝试获取一组结果并将它们缩减为一个数组。

数据为：

const data = [
    { profileid: '1', points: 25, winner: true },
    { profileid: '2', points: 15, winner: false },
    { profileid: '3', points: 18, winner: false },
    { profileid: '1', points: 15, winner: true },
    { profileid: '2', points: 18, winner: false },
    { profileid: '3', points: 25, winner: false },
]

我试图通过用户 ID 总结所有点（我已经做到了这一点）并且还总结了每个对象中的次数 winner === true

预期结果

[
    { profileid: '1', points: 40, wins: 1 },
    { profileid: '2', points: 33, wins: 0 },
    { profileid: '3', points: 43, wins: 1 },
]

目前我无法计算每个对象中“真实”结果的总和。此外，我返回一个对象而不是数组。

片段

const data = [
    { profileid: '1', points: 25, winner: true },
    { profileid: '2', points: 15, winner: false },
    { profileid: '3', points: 18, winner: false },
    { profileid: '1', points: 15, winner: false },
    { profileid: '2', points: 18, winner: false },
    { profileid: '3', points: 25, winner: true },
]

const result = data.reduce((acc, cur) => ({
            ...acc,
            [cur.profileid]: {
                profileid: cur.profileid,
                points: cur.points + (acc[cur.profileid] ? acc[cur.profileid].points : 0),
                wins: cur.winner
                    ? acc[cur.profile]
                        ? acc[cur.profile].wins + 1
                        : 0
                    : acc[cur.profile]
                    ? acc[cur.profile].wins
                    : 0,
            },
        }),
        []
    )
    
console.log(result)

Answer 1

对 points 应用相同的逻辑可以得出如下结果：

wins: cur.winner ? ((acc[cur.profileid]?.wins || 0) + 1) : (acc[cur.profileid]?.wins || 0),

我们是否首先检查当前 winner 是否为 true

如果是，请在 wins 不存在的地方第一次添加 1 (|| 0)
如果没有，就使用acc[cur.profileid]?.wins（如果第一次是false也是一个回退）

const data = [
    { profileid: '1', points: 25, winner: true },
    { profileid: '2', points: 15, winner: false },
    { profileid: '3', points: 18, winner: false },
    { profileid: '1', points: 15, winner: false },
    { profileid: '2', points: 18, winner: false },
    { profileid: '3', points: 25, winner: true },
]

const result = data.reduce((acc, cur) => ({
        ...acc,
        [cur.profileid]: {
            wins: cur.winner ? ((acc[cur.profileid]?.wins || 0) + 1) : (acc[cur.profileid]?.wins || 0),
            profileid: cur.profileid,
            points: cur.points + (acc[cur.profileid] ? acc[cur.profileid].points : 0),
        },
    }),
    []
)
    
console.log(result)

也就是说，我个人更喜欢一个简单的 for 和一个 if 来获得更具可读性的代码：

请考虑这个例子：

创建空result对象
遍历 data 数组中的每个对象
如果需要，在 data 中创建空对象
提高 points 和 wins 值

const data = [
    { profileid: '1', points: 25, winner: true },
    { profileid: '2', points: 15, winner: false },
    { profileid: '3', points: 18, winner: false },
    { profileid: '1', points: 15, winner: false },
    { profileid: '2', points: 18, winner: false },
    { profileid: '3', points: 25, winner: true },
];
const result = {};

for (let i in data) {
  
    // Create current profileId object
    if (!result[data[i].profileid]) {
        result[data[i].profileid] = {
            wins: 0,
            points: 0,
            profileid: +data[i].profileid
        };
    }

    // Bump points/wins values
    result[data[i].profileid].points += data[i].points;
    result[data[i].profileid].wins += +data[i].winner;
}

console.log(result);

Answer 2

您可以像对 points 求和一样对 wins 求和：

const data = [
    { profileid: '1', points: 25, winner: true },
    { profileid: '2', points: 15, winner: false },
    { profileid: '3', points: 18, winner: false },
    { profileid: '1', points: 15, winner: true },
    { profileid: '2', points: 18, winner: false },
    { profileid: '3', points: 25, winner: true },
]

const result = data.reduce((acc, cur) => ({
            ...acc,
            [cur.profileid]: {
                profileid: cur.profileid,
                points: cur.points + (acc[cur.profileid] ? acc[cur.profileid].points : 0),
                wins: cur.winner + (acc[cur.profileid] ? acc[cur.profileid].wins : 0)
            },
        }),
        []
    )
    
console.log(result)

您也可以像这样简化 wins：
wins: !!(acc[cur.profileid]?.wins) + cur.winner

如果 acc[cur.profileid]?.wins returns undefined,
您可以通过 [=17= 将其转换为 Boolean ] 或 Boolean() 以避免得到 NaN

Boolean
Optional chaining (?.)

将结果数组缩减为具有总和和用户 ID 的对象数组

Reduce array of results in to array of objects with sums and user id

javascript

arrays

reduce