如何读取 javascript 数组并将其重新格式化为 JSON 对象？答案

【问题标题】：How to read a javascript array and reformat it as JSON objects?如何读取 javascript 数组并将其重新格式化为 JSON 对象？
【发布时间】：2018-07-05 19:31:34
【问题描述】：

我正在根据我给定的参数从 Google Analytics API 获取以下数据。这是一个 JavaScript 数组对象。

[ [ '201801', '(Other)', '129' ],
  [ '201801', 'Direct', '2236' ],
  [ '201801', 'Email', '2' ],
  [ '201801', 'Organic Search', '6263' ],
  [ '201801', 'Referral', '185' ],
  [ '201801', 'Social', '669' ],
  [ '201802', '(Other)', '371' ],
  [ '201802', 'Direct', '2037' ],
  [ '201802', 'Email', '3' ],
  [ '201802', 'Organic Search', '5790' ],
  [ '201802', 'Referral', '162' ],
  [ '201802', 'Social', '515' ],
  [ '201803', '(Other)', '213' ],
  [ '201803', 'Direct', '2465' ],
  [ '201803', 'Organic Search', '8596' ],
  [ '201803', 'Referral', '238' ],
  [ '201803', 'Social', '356' ],
  [ '201804', '(Other)', '65' ],
  [ '201804', 'Direct', '1872' ],
  [ '201804', 'Email', '1' ],
  [ '201804', 'Organic Search', '9275' ],
  [ '201804', 'Referral', '170' ],
  [ '201804', 'Social', '307' ],
  [ '201805', '(Other)', '35' ],
  [ '201805', 'Direct', '2429' ],
  [ '201805', 'Email', '2' ],
  [ '201805', 'Organic Search', '8995' ],
  [ '201805', 'Referral', '234' ],
  [ '201805', 'Social', '341' ],
  [ '201806', 'Direct', '51' ],
  [ '201806', 'Organic Search', '282' ],
  [ '201806', 'Referral', '1' ],
[ '201806', 'Social', '3' ] ]

如果不清楚，请查看此 GIST：https://gist.github.com/chanakaDe/3ad4e2a51c99386a2737b65a82f034b1

在这个数组中，你可以看到像这样的'201801'。这意味着 YEAR 是 2018 年，MONTH 是 1 日。它同样继续。

通常在某一特定月份，我们会获得 6 个值，例如 (Other) 、 Direct 、 Email 、 Organic Search 。推荐和社交。在这个数组中，同一日期重复 6 次。

我想做的是从中创建一个简单的 JSON 对象。这是我要创建的格式，以便在我的 AngularJS 前端显示所有这些数据。

[
    {
        "date" : "201801",
        "(Other)" : "129",
        "Direct" : "2236",
        "Email" : "2",
        "OrganicSearch" : "6263",
        "Referral" : "185",
        "Social" : "669"
    },
    {
        "date" : "201802",
        "(Other)" : "371",
        "Direct" : "2037",
        "Email" : "3",
        "OrganicSearch" : "5790",
        "Referral" : "162",
        "Social" : "515"
    }
]

我想要一个这样的 JSON 数组。我试图找到重复的值，然后尝试将它们合并到一个 JSON 对象中（如果附近的元素相同）。很多这样的尝试。但是还是不行。请各位，此刻真的很想得到你们的帮助。我不擅长数据分析和这类排序。请帮帮我。该项目使用 Node.js 8.10.0 完成。

【问题讨论】：

标签： javascript arrays json node.js google-analytics

【解决方案1】：

您可以将日期作为哈希表的键并收集所有键值对。稍后仅获取对象数组的值。

var array = [['201801', '(Other)', '129'], ['201801', 'Direct', '2236'], ['201801', 'Email', '2'], ['201801', 'Organic Search', '6263'], ['201801', 'Referral', '185'], ['201801', 'Social', '669'], ['201802', '(Other)', '371'], ['201802', 'Direct', '2037'], ['201802', 'Email', '3'], ['201802', 'Organic Search', '5790'], ['201802', 'Referral', '162'], ['201802', 'Social', '515'], ['201803', '(Other)', '213'], ['201803', 'Direct', '2465'], ['201803', 'Organic Search', '8596'], ['201803', 'Referral', '238'], ['201803', 'Social', '356'], ['201804', '(Other)', '65'], ['201804', 'Direct', '1872'], ['201804', 'Email', '1'], ['201804', 'Organic Search', '9275'], ['201804', 'Referral', '170'], ['201804', 'Social', '307'], ['201805', '(Other)', '35'], ['201805', 'Direct', '2429'], ['201805', 'Email', '2'], ['201805', 'Organic Search', '8995'], ['201805', 'Referral', '234'], ['201805', 'Social', '341'], ['201806', 'Direct', '51'], ['201806', 'Organic Search', '282'], ['201806', 'Referral', '1'], ['201806', 'Social', '3']],
    result = Object.values(
        array.reduce((r, [date, key, value]) => {
            r[date] = r[date] || { date };
            r[date][key] = value;
            return r;
        }, {})
    );

console.log(result);

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

【讨论】：

感谢回答的小伙伴

【解决方案2】：

这里有一个可行的解决方案，而且很容易阅读。

var example = [
  ['201801', '(Other)', '129'],
  ['201801', 'Direct', '2236'],
  ['201801', 'Email', '2'],
  ['201801', 'Organic Search', '6263'],
  ['201801', 'Referral', '185'],
  ['201801', 'Social', '669'],
  ['201802', '(Other)', '371'],
  ['201802', 'Direct', '2037'],
  ['201802', 'Email', '3'],
  ['201802', 'Organic Search', '5790'],
  ['201802', 'Referral', '162'],
  ['201802', 'Social', '515'],
  ['201803', '(Other)', '213'],
  ['201803', 'Direct', '2465'],
  ['201803', 'Organic Search', '8596'],
  ['201803', 'Referral', '238'],
  ['201803', 'Social', '356'],
  ['201804', '(Other)', '65'],
  ['201804', 'Direct', '1872'],
  ['201804', 'Email', '1'],
  ['201804', 'Organic Search', '9275'],
  ['201804', 'Referral', '170'],
  ['201804', 'Social', '307'],
  ['201805', '(Other)', '35'],
  ['201805', 'Direct', '2429'],
  ['201805', 'Email', '2'],
  ['201805', 'Organic Search', '8995'],
  ['201805', 'Referral', '234'],
  ['201805', 'Social', '341'],
  ['201806', 'Direct', '51'],
  ['201806', 'Organic Search', '282'],
  ['201806', 'Referral', '1'],
  ['201806', 'Social', '3']
]
//using a forEach to map to object and then a map to map it into an array.
var obj = {};
example.forEach(item => {
  if (obj[item[0]]) {
    obj[item[0]][item[1]] = item[2]
  } else {
    obj[item[0]] = {};
    obj[item[0]]["date"] = item[0]
  }
})

var result = Object.keys(obj).map(k => obj[k])
console.log(result)

【讨论】：

这是有效的，但缺少“（其他）”列。为什么是兄弟？

【解决方案3】：

这是获得所需输出的方法 -

let data = [
  ['201801', '(Other)', '129'],
  ['201801', 'Direct', '2236'],
  ['201801', 'Email', '2'],
  ['201801', 'Organic Search', '6263'],
  ['201801', 'Referral', '185'],
  ['201801', 'Social', '669'],
  ['201802', '(Other)', '371'],
  ['201802', 'Direct', '2037'],
  ['201802', 'Email', '3'],
  ['201802', 'Organic Search', '5790'],
  ['201802', 'Referral', '162'],
  ['201802', 'Social', '515'],
  ['201803', '(Other)', '213'],
  ['201803', 'Direct', '2465'],
  ['201803', 'Organic Search', '8596'],
  ['201803', 'Referral', '238'],
  ['201803', 'Social', '356'],
  ['201804', '(Other)', '65'],
  ['201804', 'Direct', '1872'],
  ['201804', 'Email', '1'],
  ['201804', 'Organic Search', '9275'],
  ['201804', 'Referral', '170'],
  ['201804', 'Social', '307'],
  ['201805', '(Other)', '35'],
  ['201805', 'Direct', '2429'],
  ['201805', 'Email', '2'],
  ['201805', 'Organic Search', '8995'],
  ['201805', 'Referral', '234'],
  ['201805', 'Social', '341'],
  ['201806', 'Direct', '51'],
  ['201806', 'Organic Search', '282'],
  ['201806', 'Referral', '1'],
  ['201806', 'Social', '3']
];

let output = {};
data.map((value) => { if(!output[value[0]]) { output[value[0]] = {}; } output[value[0]][value[1]] = value[2]; })

console.log(Object.keys(output).map(key => { output[key]["date"] = key; return output[key]; }));

【讨论】：

这是对map 的不当使用。 let output = {}; data.map(value => /* do something to output */) 更好地表达为let output = data.reduce((output, value) => /* do something to output and return it */, {})。
感谢 Vivek 的回答和努力
我仍然对你是如何在一条信号线上做到这一点感到困惑 :-)
使用这个：data.map((value) => { if(!output[value[0]]) { output[value[0]] = {}; } output[value[0 ]][value[1]] = value[2]; }) ---> 能否请您解释一下，如果您可以的话.... 抱歉问这个问题。我知道这太过分了。 :-)
@ChanakaDeSilva 因此，如果您查看它的输出，它将一次只使用数组中的一项 - ['201806', 'Social', '3'] 检查日期是否已在 output 中存在或不存在，如果存在则只需在对象中添加Social: 3，以201806 作为键。在第二条语句Object.keys... 中，它只需将键201806 添加到它自己的对象中，例如{ date:201806, Social: 3 }。希望这会有所帮助

【解决方案4】：

如果ary 是您的输入数组，应该这样做：

const outputObj = ary.reduce((obj, subAry) => {
  const [date, key, val] = subAry 
  obj[date] = obj[date] || {}
  obj[date]['date'] = date
  obj[date][key] = val
  return obj
}, {})

console.log(Object.values(outputObj))

【讨论】：

【解决方案5】：

使用reduce将输入数据按月份分组，键为日期，并将分组后的对象映射到所需的输出：

const data = [ [ '201801', '(Other)', '129' ],
  [ '201801', 'Direct', '2236' ],
  [ '201801', 'Email', '2' ],
  [ '201801', 'Organic Search', '6263' ],
  [ '201801', 'Referral', '185' ],
  [ '201801', 'Social', '669' ],
  [ '201802', '(Other)', '371' ],
  [ '201802', 'Direct', '2037' ],
  [ '201802', 'Email', '3' ],
  [ '201802', 'Organic Search', '5790' ],
  [ '201802', 'Referral', '162' ],
  [ '201802', 'Social', '515' ],
  [ '201803', '(Other)', '213' ],
  [ '201803', 'Direct', '2465' ],
  [ '201803', 'Organic Search', '8596' ],
  [ '201803', 'Referral', '238' ],
  [ '201803', 'Social', '356' ],
  [ '201804', '(Other)', '65' ],
  [ '201804', 'Direct', '1872' ],
  [ '201804', 'Email', '1' ],
  [ '201804', 'Organic Search', '9275' ],
  [ '201804', 'Referral', '170' ],
  [ '201804', 'Social', '307' ],
  [ '201805', '(Other)', '35' ],
  [ '201805', 'Direct', '2429' ],
  [ '201805', 'Email', '2' ],
  [ '201805', 'Organic Search', '8995' ],
  [ '201805', 'Referral', '234' ],
  [ '201805', 'Social', '341' ],
  [ '201806', 'Direct', '51' ],
  [ '201806', 'Organic Search', '282' ],
  [ '201806', 'Referral', '1' ],
[ '201806', 'Social', '3' ] ];

const dataObj = data.reduce((all, [month, key, val]) => {

  if (!all.hasOwnProperty(month)) all[month] = {};
  all[month][key] = val;
  return all;

}, {});

const result = Object.keys(dataObj).map(k => Object.assign({}, dataObj[k], {date: k}))

console.log(result);

【讨论】：

【解决方案6】：

const input = [ 
  [ '201801', '(Other)', '129' ],
  [ '201801', 'Direct', '2236' ],
  [ '201801', 'Email', '2' ],
  [ '201801', 'Organic Search', '6263' ],
  [ '201801', 'Referral', '185' ],
  [ '201801', 'Social', '669' ],
  [ '201802', '(Other)', '371' ],
  [ '201802', 'Direct', '2037' ],
  [ '201802', 'Email', '3' ],
  [ '201802', 'Organic Search', '5790' ],
  [ '201802', 'Referral', '162' ],
  [ '201802', 'Social', '515' ],
  [ '201803', '(Other)', '213' ],
  [ '201803', 'Direct', '2465' ],
  [ '201803', 'Organic Search', '8596' ],
  [ '201803', 'Referral', '238' ],
  [ '201803', 'Social', '356' ],
  [ '201804', '(Other)', '65' ],
  [ '201804', 'Direct', '1872' ],
  [ '201804', 'Email', '1' ],
  [ '201804', 'Organic Search', '9275' ],
  [ '201804', 'Referral', '170' ],
  [ '201804', 'Social', '307' ],
  [ '201805', '(Other)', '35' ],
  [ '201805', 'Direct', '2429' ],
  [ '201805', 'Email', '2' ],
  [ '201805', 'Organic Search', '8995' ],
  [ '201805', 'Referral', '234' ],
  [ '201805', 'Social', '341' ],
  [ '201806', 'Direct', '51' ],
  [ '201806', 'Organic Search', '282' ],
  [ '201806', 'Referral', '1' ],
  [ '201806', 'Social', '3' ]
];

let output = [];

for (let i = 0; i < input.length; i++) {
    let index = output.findIndex(obj => obj.date === input[i][0]);
    if (index === -1) {
        index = output.length;
        output[index] = {date: input[i][0]};
    }
    output[index][input[i][1]] = input[i][2];
}

【讨论】：