Question

我正在尝试将此嵌套数组数据转换为对象数组。这是数据。

  const tableData = [
    ["first_name", "last_name", "city", "state"],
    ["June", "Gregory", "Hammond", "IN"],
    ["James", "Wynn", "Gary", "IN"],
    ["Craig", "Norman", "Schererville", "IN"]
  ]

这是输出内容

[
  { first_name : "June", last_name : "Gregory", city: "Hammond", state : "IN" },
  { first_name : "James", last_name : "Wynn", city: "Gary", state : "IN" },
  { first_name : "Craig", last_name : "Norman", city: "Schererville", state : "IN" }
]

这是我的代码。

    function convertTable(table_data) {
var result = []
var key, value
for (var i=0; i<table_data.length; i++) {
  var employee = {}
  for (var j=0; j<table_data[i].length; j++) {
    key = table_data[i][j][0]
    value = table_data[i][j][1]
    employee[key] = value
   }
   result.push(employee)
  }
  return result
 }

  var input = [
    ["first_name", "last_name", "city", "state"],
    ["June", "Gregory", "Hammond", "IN"],
    ["James", "Wynn", "Gary", "IN"],
    ["Craig", "Norman", "Schererville", "IN"]
  ]

console.log(convertTable(input));

解决方案是打印出单词的前字母而不是整个单词。我很难理解为什么。我问是否有评论者可以逐行留下评论，以帮助我理解该程序。我在研究类似问题的帮助下编写了该书，但仍然很难理解。我一直在阅读有关w3schools，MDN上的Javascript的内容，并观看有关解释的Youtube视频。任何其他帮助，将不胜感激。

Answer 1

首先，您将两个值都用一个字符代替了，

class
<select name="dnn$ctr6707$TimeTableView$ClassesList" id="dnn_ctr6707_TimeTableView_ClassesList" class="HeaderClasses">
<option selected="selected" value="14">a</option>
<option value="15">b</option>
<option value="16">c</option>
<option value="17">d</option>
<option value="49">e</option>
<option value="60">f</option>
</select>
<hr/>
<input id="inp" type="text" value="0"/>
<button id="select">Select</button>

然后您需要进行两项更改，其中一项用于使外循环省略键，而内部则使第一行用于键。请在代码中查看注释。

另一个变化是在函数顶部声明所有变量。

key = table_data[i][j][0]
//                    ^^^
value = table_data[i][j][1]
//                      ^^^

function convertTable(table_data) {
    var result = [],
        key, value,
        employee,
        i, j;

    for (i = 1; i < table_data.length; i++) {        // start from index 1
        employee = {};
        for (j = 0; j < table_data[i].length; j++) {
            key = table_data[0][j];                  // take the value form index zero
                                                     // no following index for the letter
            value = table_data[i][j];                // no following index for the letter
            employee[key] = value;
        }
        result.push(employee);
    }
    return result;
}

var input = [["first_name", "last_name", "city", "state"], ["June", "Gregory", "Hammond", "IN"], ["James", "Wynn", "Gary", "IN"], ["Craig", "Norman", "Schererville", "IN"]]

console.log(convertTable(input));

不同的方法

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

function convertTable(table) {
    return table
        .slice(1)
        .map(a => Object.assign(...table[0].map((k, i) => ({ [k]: a[i] }))));
}

var input = [["first_name", "last_name", "city", "state"], ["June", "Gregory", "Hammond", "IN"], ["James", "Wynn", "Gary", "IN"], ["Craig", "Norman", "Schererville", "IN"]]

console.log(convertTable(input));

Answer 2

尝试一下

您只是放错了一些索引，只是一切都很好。只需与您的功能匹配，您就可以轻松找到错过的地方。

  const tableData = [
    ["first_name", "last_name", "city", "state"],
    ["June", "Gregory", "Hammond", "IN"],
    ["James", "Wynn", "Gary", "IN"],
    ["Craig", "Norman", "Schererville", "IN"]
  ]


    function convertTable(table_data) {
var result = []
var key, value
for (var i=1; i<table_data.length; i++) {
  var employee = {}
  for (var j=0; j<table_data[i].length; j++) {
    key = table_data[0][j]
    value = table_data[i][j]
    employee[key] = value
   }
   result.push(employee)
  }
  return result
 }

  var input = [
    ["first_name", "last_name", "city", "state"],
    ["June", "Gregory", "Hammond", "IN"],
    ["James", "Wynn", "Gary", "IN"],
    ["Craig", "Norman", "Schererville", "IN"]
  ]

console.log(convertTable(input));

  { first_name : "June", last_name : "Gregory", city: "Hammond", state : "IN" },
  { first_name : "James", last_name : "Wynn", city: "Gary", state : "IN" },
  { first_name : "Craig", last_name : "Norman", city: "Schererville", state : "IN" }
] */

Answer 3

您正在访问根数组上的3-level深度，因此您可以在2-level数组中索引字符串。

使用此方法：key = table_data[i][j][0]您正在访问索引j-placed上的数组内i字符串的第一个字符。

并使用以下命令：value = table_data[i][j][1]正在访问索引j-placed上的数组内i字符串的第二个字符。

此外，您已经知道第一个数组是键（或标头），因此您不想在其上循环。检查下一个示例：

function convertTable(table_data)
{
    var result = [];
    var key, value;
    
    for (var i = 1; i < table_data.length; i++)
    {
        var employee = {};

        for (var j = 0; j < table_data[i].length; j++)
        {
            key = table_data[0][j];
            value = table_data[i][j];
            employee[key] = value;
        }

        result.push(employee);
     }
     
     return result;
 }

  var input = [
    ["first_name", "last_name", "city", "state"],
    ["June", "Gregory", "Hammond", "IN"],
    ["James", "Wynn", "Gary", "IN"],
    ["Craig", "Norman", "Schererville", "IN"]
  ]

console.log(convertTable(input));

Answer 4

您的代码包含三个错误，可以通过以下方式修复：

从i = 1开始迭代您的外循环以跳过标头（索引为0）。
使用tableData[0][j]而不是tableData[i][j][0]进入标头数组的索引，这是一个维度过多（访问字符），并且错误地包含了行索引i。
使用tableData[i][j]而不是tableData[i][j][1]索引数据表，它再次访问字符串中的字符。

除此以外，我建议注意间距，分号和驼峰式惯例。这将使您的代码更易于理解和调试。

将其放在一起将产生：

function convertTable(tableData) {
  var result = [];
  var key;
  var value;
  
  for (var i = 1; i < tableData.length; i++) { // start at 1 to skip headers
    var employee = {};
    
    for (var j = 0; j < tableData[i].length; j++) {
      key = tableData[0][j];   // remove character index and use [0][j]
      value = tableData[i][j]; // remove character index
      employee[key] = value;
    }
    
    result.push(employee);
  }
  
  return result;
}

var input = [
  ["first_name", "last_name", "city", "state"],
  ["June", "Gregory", "Hammond", "IN"],
  ["James", "Wynn", "Gary", "IN"],
  ["Craig", "Norman", "Schererville", "IN"]
];

console.log(convertTable(input));

除此之外，对于数组函数map和reduce来说，执行CSV到对象的转换是一项艰巨的任务。 shift()将标题从数组的前面弹出，map遍历行并将reduce应用于每个行。 reduce使用i和j行/列索引来构建对象，以与函数相同的方式引用标头数组。

const tableData = [
  ["first_name", "last_name", "city", "state"],
  ["June", "Gregory", "Hammond", "IN"],
  ["James", "Wynn", "Gary", "IN"],
  ["Craig", "Norman", "Schererville", "IN"]
];

const headers = tableData.shift();
const result = tableData.map((e, i) => 
  e.reduce((a, f, j) => {
    a[headers[j]] = f;
    return a;
  }, {})
);

console.log(result);

Answer 5

以下是使用Array.reduce的简洁版本：

const data = [["first_name", "last_name", "city", "state"], ["June", "Gregory", "Hammond", "IN"], ["James", "Wynn", "Gary", "IN"], ["Craig", "Norman", "Schererville", "IN"]]

const result = data.reduce((r,c,i,arr) => {
  if(i) r.push(c.reduce((a,b,j) => (a[arr[0][j]] = b, a), {}))
  return r
}, [])
console.log(result)

想法是通过跳过第一行/标题行的数据来处理数据，然后继续将内部reduce的组合对象推入累加器，内部merge commit只是用标题中的值装饰和清空对象（作为键））和当前数组（作为值）

将嵌套数组转换为对象数组问题

5 个答案: