比较下划线js中的两个对象数组

Posted 2023-02-19

【中文标题】比较下划线js中的两个对象数组

【英文标题】：Compare two arrays of objects in underscorejs

【发布时间】：2016-08-02 16:31:21

【问题描述】：

我有两个对象数组

var arr1 =
    [
    
        "lastInteracted": "2016-03-31T11:13:09.000Z",
        "email": "[e-mail]",
        "interactionCount": 2
    ,
    
        "lastInteracted": "2016-03-31T21:06:19.000Z",
        "email": "[e-mail]",
        "interactionCount": 1
    ,
    
        "lastInteracted": "2016-03-29T11:15:41.000Z",
        "email": "[e-mail]",
        "interactionCount": 1
    ,
    
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "[e-mail]",
        "interactionCount": 1
    
    ]

和

var arr2 =
[
    
        "lastInteracted": "2016-03-31T11:13:09.000Z",
        "email": "[e-mail]",
        "interactionCount": 2
    ,
    
        "lastInteracted": "2016-03-31T21:06:19.000Z",
        "email": "[e-mail]",
        "interactionCount": 4
    ,
    
        "lastInteracted": "2016-03-29T11:15:41.000Z",
        "email": "[e-mail]",
        "interactionCount": 1
    ,
    
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "[e-mail]",
        "interactionCount": 1
    ,
    
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "[e-mail]",
        "interactionCount": 10
    ,
    
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "[e-mail]",
        "interactionCount": 18
    
]

我想合并这两个数组，这样如果两个数组中都存在一个对象的电子邮件，则将 arr1 中的 interactionCount 与 arr2 进行比较，否则返回 arr1 或 arr2 的交互计数。

结果将是

var result = [
    
        "lastInteracted": "2016-03-31T11:13:09.000Z",
        "email": "[e-mail]",
        "interactionCount": 0
    ,
    
        "lastInteracted": "2016-03-31T21:06:19.000Z",
        "email": "[e-mail]",
        "interactionCount": -4
    ,
    
        "lastInteracted": "2016-03-29T11:15:41.000Z",
        "email": "[e-mail]",
        "interactionCount": 1
    ,
    
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "[e-mail]",
        "interactionCount": 1
    ,
    
        "lastInteracted": "2016-03-29T11:15:41.000Z",
        "email": "[e-mail]",
        "interactionCount": 1
    ,
    
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "[e-mail]",
        "interactionCount": 1
    ,
    
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "[e-mail]",
        "interactionCount": 10
    ,
    
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "[e-mail]",
        "interactionCount": 18
    
]

【问题讨论】：

你试过了吗？ ***.com/a/13514962/1702612

Merging two collections using Underscore.JS的可能重复

我已经完成了这些解决方案，但它们并没有满足我的需求。如果您可以查看结果数组，也许它会更清晰，因为我不仅想消除重复项，而且还希望在存在重复项的地方获得值的差异。

【参考方案1】：

使用下划线可以这样做：

var arr1 = [
  "lastInteracted": "2016-03-31T11:13:09.000Z",
  "email": "[e-mail]",
  "interactionCount": 2
, 
  "lastInteracted": "2016-03-31T21:06:19.000Z",
  "email": "[e-mail]",
  "interactionCount": 1
, 
  "lastInteracted": "2016-03-29T11:15:41.000Z",
  "email": "[e-mail]",
  "interactionCount": 1
, 
  "lastInteracted": "2016-03-24T10:02:29.000Z",
  "email": "[e-mail]",
  "interactionCount": 1
]
var arr2 = [
  "lastInteracted": "2016-03-31T11:13:09.000Z",
  "email": "[e-mail]",
  "interactionCount": 2
, 
  "lastInteracted": "2016-03-31T21:06:19.000Z",
  "email": "[e-mail]",
  "interactionCount": 4
, 
  "lastInteracted": "2016-03-29T11:15:41.000Z",
  "email": "[e-mail]",
  "interactionCount": 1
, 
  "lastInteracted": "2016-03-24T10:02:29.000Z",
  "email": "[e-mail]",
  "interactionCount": 1
, 
  "lastInteracted": "2016-03-24T10:02:29.000Z",
  "email": "[e-mail]",
  "interactionCount": 10
, 
  "lastInteracted": "2016-03-24T10:02:29.000Z",
  "email": "[e-mail]",
  "interactionCount": 18
]

var ary = _.chain(arr1.concat(arr2))//use chain
  .groupBy(function(d) 
    return d.email;
  )//grouping by email
  .map(function(d) 
    var last = _.last(d);//take the last in the group
    var k = 
      email: last.email,
      lastInteracted: last.lastInteracted,
      interactionCount: _.reduce(d, function(memo, d1) 
        return memo + d1.interactionCount;//sum up interactionCount
      , 0)
    ;
    return k;
  ).value()

document.write('<pre>' + JSON.stringify(ary, 0, 4) + '</pre>');

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore.js"></script>

工作小提琴here

【讨论】：

Jsfiddle 不起作用。它只是打印库中的一些函数。

感谢您的快速更新。如果尝试减去interactionCounts，为什么它只是在interactionCounts的总和后面加上'-'？

我认为问题是关于添加interactionCounts 你为什么要减去？

我的错我完全错过了写预期的结果。我实际上应该得到两个数组的interactionCount 的差异。我需要跟踪这两个数组的interactionCount 是否增加或减少。

非常感谢@cyril！在我基本上改变了要求之后，我真的不敢相信你真的帮助了我。再次感谢 :-) :-)