下划线：合并 2 个嵌套数组并替换特定字段答案

【问题标题】：Uderscore: Merge 2 nested Arrays and replace specific field下划线：合并 2 个嵌套数组并替换特定字段
【发布时间】：2017-11-09 19:59:00
【问题描述】：

合并两个数组（“target”、“updater”）并用作第二个更新器，以使用 uderscore 更新第一个数组中的单个字段或某些字段。

arr1 = [{"_id": 1, "name": "Name One", "surname": "Surname One", 
"Addresses": [{"_id": 1, "country":"Country One", "Town": "Town One" }, 
{"_id": 2, "country":"Country Two", "Town": "Town Two" }]
},
{"_id": 2, "name": "Name Two", "surname": "Surname Two", 
"Addresses": [{"_id": 4, "country":"Country Three"}, 
{"_id": 4, "country":"Country Four"}]
}]

arr2 = [{"_id": 2, "name": "Name Two", "surname": "Surname Two", 
"Addresses": [{"_id": 4, "country":"Country Three", "Town": "Town Three" }, 
{"_id": 4, "country":"Country Four", "Town": "Town Four" }]
}]

预期结果：

result = [{"_id": 1, "name": "Name One", "surname": "Surname One", 
"Addresses": [{"_id": 1, "country":"Country One", "Town": "Town One" }, 
{"_id": 2, "country":"Country Two", "Town": "Town Two" }]
},
{"_id": 2, "name": "Name Two", "surname": "Surname Two", 
"Addresses": [{"_id": 4, "country":"Country Three", "Town": "Town Three" }, 
{"_id": 4, "country":"Country Four", "Town": "Town Four" }]
}]

我已经检查了_.map、_.extend、findWhere 并尝试了事件_.values(_.extend(_.indexBy(array1, 'field'), _.indexBy(array2, 'field')))，但没有一个给出预期的结果。

【问题讨论】：

标签： javascript arrays merge underscore.js

【解决方案1】：

尝试使用扁平化

var result0 = _.flatten(_.map(data, '_id'));
var result1 = _.flatten(_.map(data, 'name'));

...

【讨论】：

【解决方案2】：

这不是最优的。但是您可以获得正确的格式。而且真的不使用下划线¯\_(ツ)_/¯

// log func
function log(name, msg) {
  if (_.isUndefined(msg)) {
    msg = name;
    name = null;
  }
  if (_.isObject(msg)) {
     msg = JSON.stringify(msg, null, 2);
  }
  msg = (name ? name + ': ' : '') + msg;
  $('#content').append($('<p></p>').html(msg));
}
var arr1 = [{"_id": 1, "name": "Name One", "surname": "Surname One", 
"Addresses": [{"_id": 1, "country":"Country One", "Town": "Town One" }, 
{"_id": 2, "country":"Country Two", "Town": "Town Two" }]
},
{"_id": 2, "name": "Name Two", "surname": "Surname Two", 
"Addresses": [{"_id": 3, "country":"Country Three"}, 
{"_id": 4, "country":"Country Four"}]
}]

var arr2 = [{"_id": 2, "name": "Name Two", "surname": "Surname Two", 
"Addresses": [{"_id": 3, "country":"Country Three", "Town": "Town Three" }, 
{"_id": 4, "country":"Country Four", "Town": "Town Four" }]
}]
var a = _.reduce(_.flatten(_.zip(arr1, arr2)), function (collection, obj) {
    var uniqueObject = _.find(collection, function (item) {
        return (item && item._id) === (obj && obj._id);
    });
    
    if (uniqueObject) {
      uniqueObject.Addresses = _.map(_.values(_.groupBy(obj.Addresses.concat(uniqueObject.Addresses), "_id")), function (value) {
          return _.reduce(value || [], function (a, b) {
              return _.extend(a, b);
          });
      });
    } else {
    	collection = collection.concat(obj)
    }
    
    return collection;
}, []);
log(_.compact(a))

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore.js"></script>
<div id="content">

</div>

这将返回

[
  {
    "_id": 1,
    "name": "Name One",
    "surname": "Surname One",
    "Addresses": [
      {
        "_id": 1,
        "country": "Country One",
        "Town": "Town One"
      },
      {
        "_id": 2,
        "country": "Country Two",
        "Town": "Town Two"
      }
    ]
  },
  {
    "_id": 2,
    "name": "Name Two",
    "surname": "Surname Two",
    "Addresses": [
      {
        "_id": 3,
        "country": "Country Three",
        "Town": "Town Three"
      },
      {
        "_id": 4,
        "country": "Country Four",
        "Town": "Town Four"
      }
    ]
  }
]

【讨论】：