根据某些键合并驻留在 json 数组中的行答案

【问题标题】：Merge rows residing inside json array based on certain Keys根据某些键合并驻留在 json 数组中的行
【发布时间】：2019-07-11 16:02:20
【问题描述】：

寻找问题的解决方案：

需要合并以下数组，使得最终输出只有一行对应每个唯一的类别和子类别组合，其余字段（主题和个人）合并为一个数组：

输入：

[
   {
      "category":"cat2",
      "subcategory":"b",
      "personal":[
         "inclsn2"
      ],
      "topic":[
         "topic1",
         "topic3"
      ]
   },
   {
      "category":"cat1",
      "subcategory":"a",
      "personal":[
         "inclsn2"
      ],
      "topic":[
         "topic4"
      ]
   },
   {
      "category":"cat2",
      "subcategory":"b",
      "personal":[
         "inclsn2"
      ],
      "topic":[
         "topic5",
         "topic2"
      ]
   },
   {
      "category":"cat2",
      "subcategory":"b",
      "personal":[
         "inclsn2"
      ],
      "topic":[
         "topic1",
         "topic2"
      ]
   },
   {
      "category":"cat2",
      "subcategory":"b",
      "personal":[
         "inclsn2"
      ],
      "topic":[
         "topic1",
         "topic2"
      ]
   },
   {
      "category":"cat1",
      "subcategory":"a",
      "personal":[
         "inclsn2"
      ],
      "topic":[
         "topic1",
         "topic2"
      ]
   },
   {
      "category":"cat1",
      "subcategory":"a",
      "personal":[
         "inclsn2"
      ],
      "topic":[
         "topic1",
         "topic2"
      ]
   },
   {
      "category":"cat3",
      "subcategory":"c",
      "personal":[
         "inclsn3"
      ],
      "topic":[
         "topic1",
         "topic2"
      ]
   },
   {
      "category":"cat1",
      "subcategory":"a",
      "personal":[
         "inclsn2"
      ],
      "topic":[
         "topic1",
         "topic2"
      ]
   },
   {
      "category":"cat1",
      "subcategory":"a",
      "personal":null,
      "topic":[
         "topic6"
      ]
   }
]

预期输出：

[{"category": "cat2", "subcategory": "b", "personal": ["inclsn2"], "topic": ["topic2","topic5","topic4","topic1"]},
{"category": "cat1", "subcategory": "a", "personal": ["inclsn2"], "topic": ["topic6","topic2","topic4","topic1"]},
{"category": "cat3", "subcategory": "c", "personal": ["inclsn3"], "topic": ["topic1","topic2"]}]

对此的任何帮助将不胜感激。

谢谢

【问题讨论】：

标签： javascript node.js merge group-by

【解决方案1】：

您的问题的一个解决方案是过滤数组，条件是以前看不到类别/子类别组合。您可以使用一个对象，其键是类别/子类别的某种组合（例如，与某些分隔符连接）并查看该对象是否具有给定的键。

你可以这样做：

const input = [] // your array here

const d = '\t';
const merge = (a, c = {}) => a.filter((e, f, k) =>
              (k = e.category + d + e.subcategory, f = c[k], c[k] = 1, !f));

console.log(merge(input));

确保将d 设置为永远不会成为真实类别或子类别名称一部分的字符串 - 这里它是一个制表符，但您可以将其设置为其他任何值。

数组的输出：

[ { category: 'cat2',
    subcategory: 'b',
    personal: [ 'inclsn2' ],
    topic: [ 'topic1', 'topic3' ] },
  { category: 'cat1',
    subcategory: 'a',
    personal: [ 'inclsn2' ],
    topic: [ 'topic4' ] },
  { category: 'cat3',
    subcategory: 'c',
    personal: [ 'inclsn3' ],
    topic: [ 'topic1', 'topic2' ] } ]

如果您使用 Lodash，那么您可以进一步简化您的 merge() 函数：

const _ = require('lodash');

const d = '\t';
const merge = a => _.uniqBy(a, e => e.category + d + e.subcategory);

console.log(merge(input));

【讨论】：