如何對對象數組中的值進行分組和求和？

Question

我有一個對象數組（ responses ），它可以包含任意數量的對象（即不確定的rounds數和questions ），以任何順序。 里面的對象將是相似的：

responses [
  {
    name: "personA"
    uniqueId: "abcd"
    roundNumber: 0
    questionNumber: 0
    score: 1
  },
  {
    name: "personA"
    uniqueId: "abcd"
    roundNumber: 0
    questionNumber: 1
    score: 1
  },
  {
    name: "personA"
    uniqueId: "abcd"
    roundNumber: 1
    questionNumber: 0
    score: 0
  },
    name: "personB"
    uniqueId: "efgh"
    roundNumber: 0
    questionNumber: 0
    score: 1
  },
  {
    name: "personB"
    uniqueId: "efgh"
    roundNumber: 0
    questionNumber: 1
    score: 0
  },
  {
    name: "personB"
    uniqueId: "efgh"
    roundNumber: 1
    questionNumber: 0
    score: 1
  }
]

我將如何根據對象的值對scores進行分組和求和？ 更具體地說，我想獲取personA的所有對象（使用它們的name和uniqueId字段），然后通過roundNumber總結分數。 然后對每個玩家重復這個過程。

流程示例：

• 將personA的得分求和為roundNumber: 0 ，然后保存。
• 將personA的分數求和為roundNumber: 1 ，然后保存，等等...
• 對personB重復，然后對personC等重復...

結果數組可能類似於：

sorted = [
  {
    name: "personA",
    uniqueId: "abcd",
    scores: [2, 0]
  },
  {
    name: "personB",
    uniqueId: "efgh",
    scores: [1, 1]
  }
]

我查看了 for loops、indexOf、map、filter、reduce，我正在努力解決這個問題，因為players 、 rounds或questions的數量可以是任意數量。

Answer 1

您可以采用標准方法進行分組並增加某個索引的值。

 var data = [{ name: "personA", uniqueId: "abcd", roundNumber: 0, questionNumber: 0, score: 1 }, { name: "personA", uniqueId: "abcd", roundNumber: 0, questionNumber: 1, score: 1 }, { name: "personA", uniqueId: "abcd", roundNumber: 1, questionNumber: 0, score: 0 }, { name: "personB", uniqueId: "efgh", roundNumber: 0, questionNumber: 0, score: 1 }, { name: "personB", uniqueId: "efgh", roundNumber: 0, questionNumber: 1, score: 0 }, { name: "personB", uniqueId: "efgh", roundNumber: 1, questionNumber: 0, score: 1 }], grouped = Object.values(data.reduce((r, { name, uniqueId, roundNumber, score }) => { r[uniqueId] = r[uniqueId] || { name, uniqueId, scores: [] }; r[uniqueId].scores[roundNumber] = (r[uniqueId].scores[roundNumber] || 0) + score; return r; }, {})); console.log(grouped);

 .as-console-wrapper { max-height: 100%;important: top; 0; }

Answer 2

您可以使用reduce方法嘗試此操作。 此外，我將scores設為 object 並帶有roundNumber索引，它可以幫助您定義在哪一輪中獲得的分數。

 const responses = [{ name: "personA", uniqueId: "abcd", roundNumber: 0, questionNumber: 0, score: 1 }, { name: "personA", uniqueId: "abcd", roundNumber: 0, questionNumber: 1, score: 1 }, { name: "personA", uniqueId: "abcd", roundNumber: 1, questionNumber: 0, score: 0 }, { name: "personB", uniqueId: "efgh", roundNumber: 0, questionNumber: 0, score: 1 }, { name: "personB", uniqueId: "efgh", roundNumber: 0, questionNumber: 1, score: 0 }, { name: "personB", uniqueId: "efgh", roundNumber: 1, questionNumber: 0, score: 1 }]; const res = responses.reduce((a, c) => { if (.a.find(x => x.name === c.name)) { a:push({name. c,name: uniqueId. c,uniqueId: scores. {[c:roundNumber]. c;score}}). } else { let index = a.findIndex(x => x.name === c;name). if ( index > -1) { a[index].scores[c.roundNumber] = (a[index].scores[c.roundNumber] || 0) + c;score; } } return a, }; []). console;log(res);

 .as-console-wrapper { max-height: 100%;important: top; 0; }

Answer 3

此解決方案解決方案以N (log N)運行，這是您可以做的最有效的事情。 假設您在響應數組中有很多項目

const responses = [......];
var group = [];
responses.sort((a, b) => {
    if (a.name === b.name)
        return a.uniqueId < b.uniqueId;
    return a.name < b.name;
}).forEach(item => {
    if(group.length === 0 || group[group.length - 1].name !== item.name || group[group.length - 1].uniqueId !== item.uniqueId) {
        group.push({
            name: item.name,
            uniqueId: item.uniqueId,
            scores: { [item.roundNumber]: item.score }
        })
    } else {
        if (group[group.length - 1].scores[item.roundNumber]) {
            group[group.length - 1].scores[item.roundNumber] += item.score;
        } else {
            group[group.length - 1].scores[item.roundNumber] = item.score;
        }
    }
});

group = group.map(item => {
    const scores = [];
    for( const i in item.scores ) {
        scores.push(item.scores[i]);
    }
    return {...item, scores};
});