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检查合并数组是否有来自两个数组的项,并更新合并数组

lodash object arrays javascript

arunmmanoharan · 技术社区 · 6 年前

我有两个数组。我将其合并并删除重复项。合并是通过考虑数组的所有键来完成的(数组1中对象的所有键值必须与数组2中的键值匹配)。

var result1 = [{
    name: 'Sandra',
    email: 'sandra@example.com'
  },
  {
    name: 'John',
    email: 'johnny@example.com'
  },
  {
    name: 'Peter',
    email: 'peter@example.com'
  },
  {
    name: 'Bobby',
    email: 'bobby@example.com'
  },
  {
    name: 'Arun',
    email: 'arun@gmail.com'
  },
];

var result2 = [{
    name: 'John',
    email: 'johnny@example.com'
  },
  {
    name: 'Bobby',
    email: 'bobby@example.com'
  },
  {
    name: 'Arun',
    email: 'arun@example.com'
  }
];

var result= _.uniqWith(_.concat(result1, result2), _.isEqual)

所以我的最终结果应该是这样。

var result = [{
    name: 'Sandra',
    email: 'sandra@example.com',
    presentInA: true,
    presentInB: false
  },
  {
    name: 'John',
    email: 'johnny@example.com',
    presentInA: true,
    presentInB: true
  },
  {
    name: 'Peter',
    email: 'peter@example.com',
    presentInA: true,
    presentInB: false
  },
  {
    name: 'Bobby',
    email: 'bobby@example.com',
    presentInA: true,
    presentInB: true
  },
  {
    name: 'Arun',
    email: 'arun@example.com',
    presentInA: false,
    presentInB: true
  },
  {
    name: 'Arun',
    email: 'arun@gmail.com',
    presentInA: true,
    presentInB: false
  }
];

请给我建议。

3 回复 | 直到 6 年前

German Burgardt 6 年前

你可以这样做

result.map(
    per => ({
        name: per.name,
        email: per.email,
        presentInA: _.find(
            result1, (o) => o.nombre === per.nombre && o.email === per.email
        ) ? true : false,
        presentInB: _.find(
            result2, (o) => o.nombre === per.nombre && o.email === per.email
        ) ? true : false,
    })
)

lealceldeiro VonC 6 年前

你可以的 iterate result 使用 _.some 为了检查里面的每个物体 result1 和 result2 (并设置相应的属性 presentInA 和 presentInB ). 像这样:

_.forEach(result, (obj) => {
   let presentInA = _.some(result1, obj);
   let presentInB = _.some(result2, obj);
   obj.presentInA = presentInA;
   obj.presentInB = presentInB;
});

var result1 = [{
    name: 'Sandra',
    email: 'sandra@example.com'
  },
  {
    name: 'John',
    email: 'johnny@example.com'
  },
  {
    name: 'Peter',
    email: 'peter@example.com'
  },
  {
    name: 'Bobby',
    email: 'bobby@example.com'
  },
];

var result2 = [{
    name: 'John',
    email: 'johnny@example.com'
  },
  {
    name: 'Bobby',
    email: 'bobby@example.com'
  },
  {
    name: 'Arun',
    email: 'arun@example.com'
  }
];

var result = _.uniqWith(_.concat(result1, result2), _.isEqual);

_.forEach(result, (obj) => {
   let presentInA = _.some(result1, obj);
   let presentInB = _.some(result2, obj);
   obj.presentInA = presentInA;
   obj.presentInB = presentInB;
});

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.js"></script>

Jonas Wilms 6 年前

 // Merges objects with the same name and email, sets the [arrName] property of the merged object to true
 // also takes an optional settings object for chaining
 function merge(arr, arrName, { hash = {}, result = [] } = {}) {
   for(const { name, email } of arr) {
      // Build up a hopefully unique key
     const key = name + "@" + email;
     // If its not in the hash yet, add it to the hash and the result
     if(!hash[key]) 
        result.push(hash[key] = { name, email });
     // then set the key
     hash[key][arrName] = true;
  }
  // allow chaining by exposing both hash and result
  return { hash, result, merge: (arr, arrName) => merge(arr, arrName, { hash, result }) };
 }

可用于:

 const { result } = merge(result1, "presentInA").merge(result2, "presentInB");

这是O(n),但假设电子邮件不包含两个@s。