Question

我有两个对象数组

var arr1 =
    [
    {
        "lastInteracted": "2016-03-31T11:13:09.000Z",
        "email": "concierge@inbound.com",
        "interactionCount": 2
    },
    {
        "lastInteracted": "2016-03-31T21:06:19.000Z",
        "email": "jbi@salesforce.com",
        "interactionCount": 1
    },
    {
        "lastInteracted": "2016-03-29T11:15:41.000Z",
        "email": "abc@insightsquared.com",
        "interactionCount": 1
    },
    {
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "diana@hubspot.com",
        "interactionCount": 1
    }
    ]

和

var arr2 =
[
    {
        "lastInteracted": "2016-03-31T11:13:09.000Z",
        "email": "concierge@inbound.com",
        "interactionCount": 2
    },
    {
        "lastInteracted": "2016-03-31T21:06:19.000Z",
        "email": "jbi@salesforce.com",
        "interactionCount": 4
    },
    {
        "lastInteracted": "2016-03-29T11:15:41.000Z",
        "email": "kstachowski@insightsquared.com",
        "interactionCount": 1
    },
    {
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "hammer@hubspot.com",
        "interactionCount": 1
    },
    {
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "life@hubspot.com",
        "interactionCount": 10
    },
    {
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "mike@hubspot.com",
        "interactionCount": 18
    }
]

我想合并这两个数组，这样如果两个对象的电子邮件都存在，那么使用arr2从arr1传递interactionCount，否则返回arr1或arr2的interactionCount。

结果将是

var result = [
    {
        "lastInteracted": "2016-03-31T11:13:09.000Z",
        "email": "concierge@inbound.com",
        "interactionCount": 0
    },
    {
        "lastInteracted": "2016-03-31T21:06:19.000Z",
        "email": "jbi@salesforce.com",
        "interactionCount": -4
    },
    {
        "lastInteracted": "2016-03-29T11:15:41.000Z",
        "email": "abc@insightsquared.com",
        "interactionCount": 1
    },
    {
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "diana@hubspot.com",
        "interactionCount": 1
    },
    {
        "lastInteracted": "2016-03-29T11:15:41.000Z",
        "email": "kstachowski@insightsquared.com",
        "interactionCount": 1
    },
    {
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "hammer@hubspot.com",
        "interactionCount": 1
    },
    {
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "life@hubspot.com",
        "interactionCount": 10
    },
    {
        "lastInteracted": "2016-03-24T10:02:29.000Z",
        "email": "mike@hubspot.com",
        "interactionCount": 18
    }
]

Answer 1

使用下划线你可以这样做：

＆＃13;

var arr1 = [{
  "lastInteracted": "2016-03-31T11:13:09.000Z",
  "email": "concierge@inbound.com",
  "interactionCount": 2
}, {
  "lastInteracted": "2016-03-31T21:06:19.000Z",
  "email": "jbi@salesforce.com",
  "interactionCount": 1
}, {
  "lastInteracted": "2016-03-29T11:15:41.000Z",
  "email": "abc@insightsquared.com",
  "interactionCount": 1
}, {
  "lastInteracted": "2016-03-24T10:02:29.000Z",
  "email": "diana@hubspot.com",
  "interactionCount": 1
}]
var arr2 = [{
  "lastInteracted": "2016-03-31T11:13:09.000Z",
  "email": "concierge@inbound.com",
  "interactionCount": 2
}, {
  "lastInteracted": "2016-03-31T21:06:19.000Z",
  "email": "jbi@salesforce.com",
  "interactionCount": 4
}, {
  "lastInteracted": "2016-03-29T11:15:41.000Z",
  "email": "kstachowski@insightsquared.com",
  "interactionCount": 1
}, {
  "lastInteracted": "2016-03-24T10:02:29.000Z",
  "email": "hammer@hubspot.com",
  "interactionCount": 1
}, {
  "lastInteracted": "2016-03-24T10:02:29.000Z",
  "email": "life@hubspot.com",
  "interactionCount": 10
}, {
  "lastInteracted": "2016-03-24T10:02:29.000Z",
  "email": "mike@hubspot.com",
  "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>');

＆＃13;

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

＆＃13;

工作小提琴here

比较underscorejs中的两个对象数组

1 个答案: