重新组织一个包含对象的javascript数组

Question

我有一个问题和不同答案的数组（我动态构建，因此输入和不同答案的数量可能会有所变化）：

var qaArray = [
    { question: "Question", answer: "Answer A"},
    { question: "Question", answer: "Answer B"},
    { question: "Question", answer: "Answer B"},
    { question: "Question", answer: "Answer C"},
    { question: "Question", answer: "Answer A"},
    { question: "Question", answer: "Answer B"},
    { question: "Question", answer: "Answer C"},
    { question: "Question", answer: "Answer A"},
    { question: "Question", answer: "Answer B"},
]

我需要一个函数来构建另一个数组对象，其中包含“answer”和“count”。在这种情况下，它将是:

[
    { answer: "Answer A", count: 3 },
    { answer: "Answer B", count: 4 },
    { answer: "Answer C", count: 2 },
]

有没有一种简单的方法来做到这一点，而不需要使用直接的循环？

- Gintas K

---

我建议您在这里使用 lodash 并使用 groupBy 轻松执行它。 - Koushik Chatterjee

Answer 1

如果您已掌握ES6，可以这样实现：

var qaArray = [
    { question: "Question", answer: "Answer A"},
    { question: "Question", answer: "Answer B"},
    { question: "Question", answer: "Answer B"},
    { question: "Question", answer: "Answer C"},
    { question: "Question", answer: "Answer A"},
    { question: "Question", answer: "Answer B"},
    { question: "Question", answer: "Answer C"},
    { question: "Question", answer: "Answer A"},
    { question: "Question", answer: "Answer B"},
];

var result = [...new Set(qaArray.map(q => q.answer))].map(a => ({ answer: a, count: qaArray.filter(e => e.answer === a).length }));

console.log(result);

- Faly

Answer 2

你可以使用reduce来实现这个功能。其概念是：如果元素不存在，则将其添加到累加器中；如果存在，则增加count计数。

请参见下面的解决方案。

var qaArray = [
    { question: "Question", answer: "Answer A"},
    { question: "Question", answer: "Answer B"},
    { question: "Question", answer: "Answer B"},
    { question: "Question", answer: "Answer C"},
    { question: "Question", answer: "Answer A"},
    { question: "Question", answer: "Answer B"},
    { question: "Question", answer: "Answer C"},
    { question: "Question", answer: "Answer A"},
    { question: "Question", answer: "Answer B"},
]


result = qaArray.reduce((acc, item) => {
  var exist = acc.find(x => x.answer === item.answer);
  if (exist) {
    exist.count += 1;
  } else {
    acc.push({
      answer: item.answer,
      count: 1
    })
  }
  return acc;
}, []);


console.log(result);

- Vipin Kumar

Answer 3

如果您想要不使用reduce或forEach的解决方案，这里提供了一种使用lodash的冗长方式。 _.countBy可以给您一个接近的结果，但如果您想将其转换为所需的数组，请使用mapValues。

_.chain(qaArray)
.countBy('answer')
.mapValues(function(value, key) {
  return { answer: key, count: value }
})
.values()
.value()

- emil

Answer 4

你可以使用 Map 将答案作为键，然后对数据进行 reduce 操作以获取计数，如下所示：

var result = Array.from(qaArray.reduce((acc, { answer }) => {
  const current = acc.get(answer) || { answer, count: 0 };
  current.count++;
  return acc.set(answer, current);
}, new Map()).values());

console.log(result);

- dashton