如何使用lodash合并两个对象数组？

Question

我有两个数组,一个共同的字段成员.如何轻松合并主题？

例如:

var arr1 = [{
  "member" : ObjectId("57989cbe54cf5d2ce83ff9d6"),
  "bank" : ObjectId("575b052ca6f66a5732749ecc"),
  "country" : ObjectId("575b0523a6f66a5732749ecb")
},
{
  "member" : ObjectId("57989cbe54cf5d2ce83ff9d8"),
  "bank" : ObjectId("575b052ca6f66a5732749ecc"),
  "country" : ObjectId("575b0523a6f66a5732749ecb")
}];

var arr2 = [{
    "member" : ObjectId("57989cbe54cf5d2ce83ff9d6"),
    "name" : 'xxxxxx',
    "age" : 25
},
{
    "member" : ObjectId("57989cbe54cf5d2ce83ff9d8"),
    "name" : 'yyyyyyyyyy',
    "age" : 26
}];

预期:

var merge = [{
  "member" : ObjectId("57989cbe54cf5d2ce83ff9d6"),
  "bank" : ObjectId("575b052ca6f66a5732749ecc"),
  "country" : ObjectId("575b0523a6f66a5732749ecb"),
  "name" : 'xxxxxx',
  "age" : 25
},
{
  "member" : ObjectId("57989cbe54cf5d2ce83ff9d8"),
  "bank" : ObjectId("575b052ca6f66a5732749ecc"),
  "country" : ObjectId("575b0523a6f66a5732749ecb"),
  "name" : 'yyyyyyyyyy',
  "age" : 26
}];

我试过了

var merge = _.unionBy(arr1, arr2, 'member');

但没有按预期合并.显示了array1值.谁能帮我？

Answer 1

如果两个数组的顺序正确; 其中每个项目对应其关联的成员标识符,您只需使用.

var merge = _.merge(arr1, arr2);

哪个是简短版本:

var merge = _.chain(arr1).zip(arr2).map(function(item) {
    return _.merge.apply(null, item);
}).value();

或者,如果数组中的数据不是任何特定顺序,您可以按成员值查找关联项.

var merge = _.map(arr1, function(item) {
    return _.merge(item, _.find(arr2, { 'member' : item.member }));
});

您可以轻松将其转换为mixin.请参阅以下示例:

_.mixin({
  'mergeByKey' : function(arr1, arr2, key) {
    var criteria = {};
    criteria[key] = null;
    return _.map(arr1, function(item) {
      criteria[key] = item[key];
      return _.merge(item, _.find(arr2, criteria));
    });
  }
});

var arr1 = [{
  "member": 'ObjectId("57989cbe54cf5d2ce83ff9d6")',
  "bank": 'ObjectId("575b052ca6f66a5732749ecc")',
  "country": 'ObjectId("575b0523a6f66a5732749ecb")'
}, {
  "member": 'ObjectId("57989cbe54cf5d2ce83ff9d8")',
  "bank": 'ObjectId("575b052ca6f66a5732749ecc")',
  "country": 'ObjectId("575b0523a6f66a5732749ecb")'
}];

var arr2 = [{
  "member": 'ObjectId("57989cbe54cf5d2ce83ff9d8")',
  "name": 'yyyyyyyyyy',
  "age": 26
}, {
  "member": 'ObjectId("57989cbe54cf5d2ce83ff9d6")',
  "name": 'xxxxxx',
  "age": 25
}];

var arr3 = _.mergeByKey(arr1, arr2, 'member');

document.body.innerHTML = JSON.stringify(arr3, null, 4);

body { font-family: monospace; white-space: pre; }

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.14.0/lodash.min.js"></script>

Answer 2

使用_.keyBy(),合并字典并将结果转换为数组,为两个数组创建字典_.values().这样,数组的顺序无关紧要.此外,它还可以处理不同长度的数组.

const ObjectId = (id) => id; // mock of ObjectId
const arr1 = [{"member" : ObjectId("57989cbe54cf5d2ce83ff9d8"),"bank" : ObjectId("575b052ca6f66a5732749ecc"),"country" : ObjectId("575b0523a6f66a5732749ecb")},{"member" : ObjectId("57989cbe54cf5d2ce83ff9d6"),"bank" : ObjectId("575b052ca6f66a5732749ecc"),"country" : ObjectId("575b0523a6f66a5732749ecb")}];
const arr2 = [{"member" : ObjectId("57989cbe54cf5d2ce83ff9d6"),"name" : 'xxxxxx',"age" : 25},{"member" : ObjectId("57989cbe54cf5d2ce83ff9d8"),"name" : 'yyyyyyyyyy',"age" : 26}];

const merged = _(arr1) // start sequence
  .keyBy('member') // create a dictionary of the 1st array
  .merge(_.keyBy(arr2, 'member')) // create a dictionary of the 2nd array, and merge it to the 1st
  .values() // turn the combined dictionary to array
  .value(); // get the value (array) out of the sequence

console.log(merged);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.14.0/lodash.min.js"></script>

使用ES6 地图

Concat数组,并将组合数组减少为Map.使用Object#assign将具有相同对象的对象组合member到新对象,并存储在地图中.将地图转换为具有Map#值和spread的数组:

const ObjectId = (id) => id; // mock of ObjectId
const arr1 = [{"member" : ObjectId("57989cbe54cf5d2ce83ff9d8"),"bank" : ObjectId("575b052ca6f66a5732749ecc"),"country" : ObjectId("575b0523a6f66a5732749ecb")},{"member" : ObjectId("57989cbe54cf5d2ce83ff9d6"),"bank" : ObjectId("575b052ca6f66a5732749ecc"),"country" : ObjectId("575b0523a6f66a5732749ecb")}];
const arr2 = [{"member" : ObjectId("57989cbe54cf5d2ce83ff9d6"),"name" : 'xxxxxx',"age" : 25},{"member" : ObjectId("57989cbe54cf5d2ce83ff9d8"),"name" : 'yyyyyyyyyy',"age" : 26}];

const merged = [...arr1.concat(arr2).reduce((m, o) => 
  m.set(o.member, Object.assign(m.get(o.member) || {}, o))
, new Map()).values()];

console.log(merged);