在 Lodash 中按条件合并两个 JSON 数组答案

【问题标题】：Merge two JSON array on condition in Lodash在 Lodash 中按条件合并两个 JSON 数组
【发布时间】：2018-06-05 05:02:29
【问题描述】：

我有两个 JSON 数组，例如：

var a = [{_id:1, name: "Bhavin"},{_id:2, name: "Raj"},{_id:3, name: "Rahul"}];    
var b = [{_id:1, post: "Developer"},{_id:2, post: "Quality Analyst"}];

现在，我想合并 Like：

var c = [{_id:1, name: "Bhavin", post: "Developer"},{_id:2, name: "Raj", post: "Quality Analyst"},{_id:3, name: "Rahul"}];

我知道我可以通过使用两个 for 循环在纯 JavaScript 中轻松完成...但这需要 n*n 时间。

我只想在n 时间内解决这个问题。

我怎样才能做到这一点？

【问题讨论】：

标签： javascript ecmascript-6 lodash

【解决方案1】：

你应该使用lodashmergeWith函数。

var a = [{_id:1, name: "Bhavin"},{_id:2, name: "Raj"},{_id:3, name: "Rahul"}];

var b = [{_id:1, post: "Developer"},{_id:2, post: "Quality Analyst"}];


// ouput [{_id:1, name: "Bhavin", post: "Developer"},{_id:2, name: "Raj", post: "Quality Analyst"},{_id:3, name: "Rahul"}];

function customizer(firstValue, secondValue) {
  return Object.assign({}, firstValue, secondValue);
}

console.log(_.mergeWith(a, b, customizer));

&lt;script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.js"&gt;&lt;/script&gt;

【讨论】：

【解决方案2】：

您可以通过将较小的数组映射到键值组合对象来轻松解决此问题，其中对象的键是 id，它们的值是关联的值本身。完成关联后，我们可以通过分配映射对象中的缺失值来迭代和转换更大的数组。

// ensure that b has more items than a
// to prevent data loss
if(a.length > b.length) {
  var t = b;
  b = a;
  a = t;
}

// create an object composed of of ids as key associated with each object.
// a loash alternative is: 
// var map = _.keys(a, '_id');
var map = a.reduce((r, v) => (r[v._id] = v, r), {});

// assign missing properties from each object from the associated mapped object.
var result = b.map(v => ({...map[v._id], ...v}));

var a = [{_id:1, name: "Bhavin"},{_id:2, name: "Raj"},{_id:3, name: "Rahul"}];
var b = [{_id:1, post: "Developer"},{_id:2, post: "Quality Analyst"}];

// ensure that b has more items than a
// to prevent data loss
if(a.length > b.length) {
  var t = b;
  b = a;
  a = t;
}

// create an object composed of of ids as key associated with each object.
// a loash alternative is: 
// var map = _.keys(a, '_id');
var map = a.reduce((r, v) => (r[v._id] = v, r), {});

// assign missing properties from each object from the associated mapped object.
var result = b.map(v => ({...map[v._id], ...v}));

console.log(result);

.as-console-wrapper{min-height:100%;top:0}

【讨论】：

【解决方案3】：

在O(n) 时间用纯 Javascript 实现这一点是完全可能的：

var a = [{_id:1, name: "Bhavin"},{_id:2, name: "Raj"},{_id:3, name: "Rahul"}];
var b = [{_id:1, post: "Developer"},{_id:2, post: "Quality Analyst"}];

const aById = a.reduce((map, { _id, name }) => map.set(_id, name), new Map());
const bById = b.reduce((map, { _id, post }) => map.set(_id, post), new Map());
const combined = [...aById.entries()].map(([_id, name]) => {
  const person = { _id, name };
  const post = bById.get(_id);
  if (post) person.post = post;
  return person;
});

console.log(combined);

var c = [{_id:1, name: "Bhavin", post: "Developer"},{_id:2, name: "Raj", post: "Quality Analyst"},{_id:3, name: "Rahul"}];

Maps 保证有O(1) 查找时间。对象哈希图（如{'1': 'Developer', '2': 'Quality Analyst'}）不是。

您的原始代码中有几个拼写错误，我假设这些是无意的 - 我在上面的 sn-p 中修复了它们。

【讨论】：

【解决方案4】：

您好，您可以使用以下代码实现它

_.map(a, function (el){
   return _.merge(el , _.find(b, {_id: el._id }))
})

// using chain, i would prefer first it looks clean :-p
_.chain(a).map(function (o){
   return _.merge(o, _.find(b, {_id: o._id }))
})
.value()

我会用链尝试它并更新答案

【讨论】：

这还不是 O(n*n) 吗？