如何将特定问题排序到特定部分？

Question

我从API获取数据，并能够将从API接收到的大量数据排序到两个数组中。一个名为sections的数组和一个名为questions的数组。因此sections数组看起来是这样的：

const section = section-1: some_title，section-2:some_title，section-3:some_title，问题数组如下：

常量问题=Q1-1:某问题，Q1-2:某问题，Q1-3:某问题，Q2-1:某问题，Q2-2:某问题，Q2-3:某问题，Q3-1:某问题，Q3-2:某问题，Q3-3:某问题，Q3-3:某问题

我想要做的是将两个数组合并在一起，并按如下方式对它们进行排序：const MergedArray = [{section-1: some_title, questions: [ Q1-1:some_question,Q1-2:some_question,Q1-3:some_question]},{section-2:some_title, questions: [Q2-1:some_question,Q2-2:some_question,Q2-3:some_question]},{section-3:some_title, questions: [Q3-1:some_question,Q3-2:some_question,Q3-3:some_question]}]如果有任何帮助，我们将不胜感激

Answer 1

从你添加的评论中，我希望section，questions和MergedArray的结构如下

const section =  [
    'section-1:some_title',
    'section-2:some_title',
    'section-3:some_title',
];
const questions = [
    'Q1-1:some_question',
    'Q1-2:some_question',
    'Q1-3:some_question',
    'Q2-1:some_question',
    'Q2-2:some_question',
    'Q2-3:some_question',
    'Q3-1:some_question',
    'Q3-2:some_question',
    'Q3-3:some_question',
]
const MergedArray = [
    {
        'section-1': 'some_title',
        questions: [ 'Q1-1:some_question','Q1-2:some_question','Q1-3:some_question']
    },
    {
        'section-2':'some_title',
        questions: ['Q2-1:some_question','Q2-2:some_question','Q2-3:some_question']
    },
    {
        'section-3':'some_title',
        questions: ['Q3-1:some_question','Q3-2:some_question','Q3-3:some_question']
    }
];

请在下面找到生成它们的解决方案。我已经为代码功能添加了注释。

​

const section = [
    'section-1:some_title',
    'section-2:some_title',
    'section-3:some_title',
];
const questions = [
    'Q1-1:some_question',
    'Q1-2:some_question',
    'Q1-3:some_question',
    'Q2-1:some_question',
    'Q2-2:some_question',
    'Q2-3:some_question',
    'Q3-1:some_question',
    'Q3-2:some_question',
    'Q3-3:some_question',
]

const MergedArray = section.reduce((accumulator, currentValue) => {
    const node = {};

    const [key, value] = currentValue.split(':');
    // This will generate
    // key = 'section-1' and value='some_title' for first itration
    // key = 'section-2' and value='some_title' for second itration
    // key = 'section-3' and value='some_title' for third itration

    node[key] = value;
    // this will update node object as
    // node = {'section-1': 'some_title'} for first itration
    // node = {'section-2': 'some_title'} for second itration
    // node = {'section-3': 'some_title'} for third itration

    const sectionIndex = key.split('-')[1];
    // sectionIndex will be
    // 1 for first itration
    // 2 for second itration
    // 3 for third itration

    // Now you have to pick questions from questions array.
    // Questions For section-1 are 'Q1-1:some_question', 'Q1-2:some_question', 'Q1-3:some_question',
    // Questions For section-2 are 'Q2-1:some_question', 'Q2-2:some_question', 'Q2-3:some_question',
    // Questions For section-3 are 'Q3-1:some_question', 'Q3-2:some_question', 'Q3-3:some_question',
    // So you have to flter out the questions from the array.
    // `Q${sectionIndex}-` is string manipulation. This will generate 'Q1-', 'Q2-', 'Q3-'.
    // You have to filter out questions staring with these strings.
    // i.e, their respective index should be zero in the matching words from that array
    // The below expression will filter out questions starting with
    // 'Q1-' in the first itration
    // 'Q2-' in the second itration
    // 'Q3-' in the third itration
    const questionList = questions.filter((ques) => ques.indexOf(`Q${sectionIndex}-`) === 0);
    // this will make
    // questionList = [ 'Q1-1:some_question', 'Q1-2:some_question', 'Q1-3:some_question'] in the first itration
    // questionList = [ 'Q2-1:some_question', 'Q2-2:some_question', 'Q2-3:some_question'] in the second itration
    // questionList = [ 'Q3-1:some_question', 'Q3-2:some_question', 'Q3-3:some_question'] in the third itration

    node.questions = questionList;
    // This will update node as
    // node = {'section-1': 'some_title' questions: [ 'Q1-1:some_question', 'Q1-2:some_question', 'Q1-3:some_question']} in the first itration
    // node = {'section-2': 'some_title' questions: [ 'Q2-1:some_question', 'Q2-2:some_question', 'Q2-3:some_question']} in the second itration
    // node = {'section-3': 'some_title' questions: [ 'Q3-1:some_question', 'Q3-2:some_question', 'Q3-3:some_question']} in the third itration

    accumulator.push(node);
    return accumulator;
}, []);
console.log(MergedArray);

​

Answer 2

一个通用的解决方案：

​

const section = [{"section-1": "some_title"}, {"section-2":"some_title"}, {"section-3":"some_title"}];

const questions = [{"Q1-1":"some_question"}, {"Q1-2":"some_question"}, {"Q1-3":"some_question"}, {"Q2-1":"some_question"}, {"Q2-2":"some_question"}, {"Q2-3":"some_question"}, {"Q3-1":"some_question"}, {"Q3-2":"some_question"}, {"Q3-3":"some_question"}];

let obj = {};

for(let ques of questions){
  let key =  Object.keys(ques)[0].split("-")[0].toLowerCase();
  obj[key] = (obj[key] || []).concat(ques);
}

let result = [];

for(let sec of section){
    let key =  "q" + Object.keys(sec)[0].split("-")[1];
    result.push({...sec, questions: obj[key]})
}

console.log(result);

​

Answer 3

假设对象是@Nitheesh设置的，这里有一种通过对象操作以及Array.reduce()和Array.filter()方法获得答案的方法。

const reduced = Object.values(
      Object.keys(section).reduce((acc, cur) => {
        const num = cur.split('-')[1]
        acc[num] = acc[num] || {};
        acc[num][cur] = section[cur];
        acc[num].questions = Object.fromEntries(
          Object.entries(questions)
          .filter(([key, value]) => key.substr(1, 1) == num));
        return acc;
      }, {})
    );

我将详细介绍每一步，以帮助可视化正在发生的事情

1. 创建

的截面对象键的数组

const sectionKeys = Object.keys(section);
// [ 'section-1', 'section-2', 'section-3'] 

1. 使用Array.reduce()创建一个对象，其中的键等于每个节号(注释中包含了其中的步骤)

const reducedObject = sectionKeys.reduce((acc, cur) => {

//Get the section number from the current value
  const num = cur.split('-')[1]

//Check if the key exists in the accumulator obj. If not return an empty object
  acc[num] = acc[num] || {}; //acc = { 1: {}, 2: {}, 3: {} }

//Get from the 'section' obj the value where the Key is the current value
  acc[num][cur] = section[cur]; //acc = { "1":{"section-1":"some_title"},
  //      "2":{"section-2":"some_title"}, "3":{"section-3":"some_title"} }

//Convert the 'questions' object variable into an array of [key, val] arrays
  const arrayKeyVals = Object.entries(questions); //arrayKeyVals = [ 
  // ["Q1-1","some_question"],["Q1-2","some_question"],["Q1-3","some_question"],
  // ["Q2-1","some_question"],["Q2-2","some_question"],["Q2-3","some_question"],
  // ["Q3-1","some_question"],["Q3-2","some_question"],["Q3-3","some_question"]]   

//Filter the array of arrays where the 2nd char of the key == the 'num' variable
  const filteredArray = arrayKeyVals
    .filter(([key, value]) => key.substr(1, 1) == num); //When num==1:  
  // [["Q1-1","some_question"],["Q1-2","some_question"],["Q1-3","some_question"]]

//Convert filteredArray back into an object
  const objQuestions= Object.fromEntries(filteredArray); //When num==1: 
  // {"Q1-1":"some_question","Q1-2":"some_question","Q1-3":"some_question"}

//Assign the filtered questions object
  acc[num].questions = objQuestions;
  return acc; //return the accumulator
}, {})

步骤2输出：

{
    "1": {
        "section-1": "some_title",
        "questions": {
            "Q1-1": "some_question",
            "Q1-2": "some_question",
            "Q1-3": "some_question"
        }
    },
    "2": {
        "section-2": "some_title",
        "questions": {
            "Q2-1": "some_question",
            "Q2-2": "some_question",
            "Q2-3": "some_question"
        }
    },
    "3": {
        "section-3": "some_title",
        "questions": {
            "Q3-1": "some_question",
            "Q3-2": "some_question",
            "Q3-3": "some_question"
        }
    }
}

1. 现在我们不再需要密钥了，所以我们只得到

的值

const final = Object.values(reducedObject);

最终输出：

[
    {
        "section-1": "some_title",
        "questions": {
            "Q1-1": "some_question",
            "Q1-2": "some_question",
            "Q1-3": "some_question"
        }
    },
    {
        "section-2": "some_title",
        "questions": {
            "Q2-1": "some_question",
            "Q2-2": "some_question",
            "Q2-3": "some_question"
        }
    },
    {
        "section-3": "some_title",
        "questions": {
            "Q3-1": "some_question",
            "Q3-2": "some_question",
            "Q3-3": "some_question"
        }
    }
]

我希望这不是TMI。我花了一段时间才掌握了Array.reduce()方法，所以我想打破这些步骤，希望能帮助其他可能需要它的人。

查看下面的工作代码片段：

​

const section = {
  'section-1': 'some_title',
  'section-2': 'some_title',
  'section-3': 'some_title'
};

const questions = {
  'Q1-1': 'some_question',
  'Q1-2': 'some_question',
  'Q1-3': 'some_question',
  'Q2-1': 'some_question',
  'Q2-2': 'some_question',
  'Q2-3': 'some_question',
  'Q3-1': 'some_question',
  'Q3-2': 'some_question',
  'Q3-3': 'some_question'
};

const reduced = Object.values(
  Object.keys(section).reduce((acc, cur) => {
    const num = cur.split('-')[1]
    acc[num] = acc[num] || {};
    acc[num][cur] = section[cur];
    acc[num].questions = Object.fromEntries(
      Object.entries(questions)
      .filter(([key, value]) => key.substr(1, 1) == num));
    return acc;
  }, {})
);

const pre = document.createElement('pre');
pre.innerText = JSON.stringify(reduced, null, 2);
document.querySelector('body').appendChild(pre);

​