从两个对象数组中过滤，然后合并它们

Question

我的要求是根据Phone 或Email 从两个数组中查找共同对象，即如果电话是null or ''，则使用电子邮件查找共同对象，反之亦然。从两个数组中找到共同点后，我想合并这两个对象，使它们都具有相同的键/值对。我设法找到了共同点，但未能合并它们。这是我到目前为止所做的。

数组1：

[
  {
    "Name": "Name1",
    "Phone": "",
    "Email": "name1@mail.com",
    "ID": 1
  },
  {
    "Name": "Name2",
    "Phone": "12345566",
    "Email": "name2@mail.com",
    "ID": 2
  }
]

数组2：

[
  {
    "Name": "Name1",
    "Phone": "987654321",
    "Email": "name1@mail.com"
  },
  {
    "Name": "Name2",
    "Phone": "12345566",
    "Email": "name2@mail.com"
  },
  {
    "Name": "Name3",
    "Phone": "23445677",
    "Email": ""
  }
]

这就是我设法从这两者中获得共同点的方法。

const common = array1
                  .filter(user=> 
                  array2.some(otherUser => 
                  (user.Email && (user.Email === otherUser.Email)) ||
                  (user.Phone && (user.Phone === otherUser.Phone))));

如何在不使用common 数组上的循环的情况下合并这两个，因为这将花费大量时间，因为我的数组中的数据可能包含数千条记录？ 使用ES6 实现这一目标的最佳方法是什么？

Answer 1

您的方法很明确。当您使用 some() 比较值时，只需覆盖过滤后数组的值即可。

示例：

const arr1 = [{ "Name": "Name1", "Phone": "", "Email": "name1@mail.com", "ID": 1 },{ "Name": "Name2", "Phone": "12345566", "Email": "name2@mail.com", "ID": 2 }];
const arr2 = [{"Name": "Name1","Phone": "987654321","Email": "name1@mail.com"},{"Name": "Name2","Phone": "12345566","Email": "name2@mail.com"},{"Name": "Name3","Phone": "23445677","Email": ""}];

const arr3 = arr1.filter(i => {
    return arr2.some(v => {
        if (i.Phone === v.Phone || i.Email === v.Email) {
            !i.Phone && (i.Phone = v.Phone), !i.Email && (i.Phone = v.Email)
            return !0
        }
    })
})

console.log(arr3)

Answer 2

使用for loop

const arra1 = [
  {
    Name: 'Name1',
    Phone: '',
    Email: 'name1@mail.com',
    ID: 1,
  },
  {
    Name: 'Name2',
    Phone: '12345566',
    Email: 'name2@mail.com',
    ID: 2,
  },
];


const arr2 = [
  {
    Name: 'Name1',
    Phone: '987654321',
    Email: 'name1@mail.com',
  },
  {
    Name: 'Name2',
    Phone: '12345566',
    Email: 'name2@mail.com',
  },
  {
    Name: 'Name3',
    Phone: '23445677',
    Email: '',
  },
];

const common = [];

for (let i = 0; i < arra1.length; i += 1) {
  const element = arra1[i];
  const { Phone, Email } = element;
  let searchKey = null;
  let searchValue = null;
  if (Phone) {
    searchKey = 'Phone';
    searchValue = Phone;
  } else if (Email) {
    searchKey = 'Email';
    searchValue = Email;
  }
  const index = arr2.findIndex((r) => {
    return searchKey ? r[searchKey] === searchValue : false;
  });
  if (index > -1) {
    common.push(element);
  }
}

console.log(common); 

使用filter

const arra1 = [
  {
    Name: 'Name1',
    Phone: '',
    Email: 'name1@mail.com',
    ID: 1,
  },
  {
    Name: 'Name2',
    Phone: '12345566',
    Email: 'name2@mail.com',
    ID: 2,
  },
];


const arr2 = [
  {
    Name: 'Name1',
    Phone: '987654321',
    Email: 'name1@mail.com',
  },
  {
    Name: 'Name2',
    Phone: '12345566',
    Email: 'name2@mail.com',
  },
  {
    Name: 'Name3',
    Phone: '23445677',
    Email: '',
  },
];

const common = arra1.filter((f) => {
  const { Phone, Email } = f;
  let searchKey = null;
  let searchValue = null;
  if (Phone) {
    searchKey = 'Phone';
    searchValue = Phone;
  } else if (Email) {
    searchKey = 'Email';
    searchValue = Email;
  }
  const index = arr2.findIndex((r) => {
    return searchKey ? r[searchKey] === searchValue : false;
  });
  return !!(index > -1);
});

console.log(common); 

Answer 3

使用地图并查找。但是，这只会将数组 1 中的缺失字段查找到数组 2 中。如果需要完全连接，则必须执行两次，然后使用 Object.assign()。

const common = array1.map(user => {
    if ((user.Email && user.Phone) || (!user.Email && !user.Phone)) {
        return user;
    }
    else if (!user.Phone) {
        const match = array2.find(otherUser => {
            return otherUser.Email === user.Email;
        });
        user.Phone = match.Phone;
    }
    else if (!user.Email) {
        const match = array2.find(otherUser => {
            return otherUser.Phone === user.Phone;
        });
        user.Email = match.Email;
    }
    return user;
})

如果两个字段都已填充或缺失，第一个“if”对于避免迭代第二个数组很有用。

Answer 4

这是一种两阶段方法，适用于要合并的任意数量的数组。

首先，所有条目都放在一个对象 (merged) 中，其中包含电子邮件和电话号码的组合。作为关键。然后我通过将具有部分键的条目与具有完整键标识的条目结合起来进一步巩固它。

var arr1= [
  { "Name": "Name1", "Phone": "", "Email": "name1@mail.com", "ID": 1 },
  { "Name": "Name2", "Phone": "12345566", "Email": "name2@mail.com", "ID": 2 },
  { "Name": "Name3", "Phone": "23445677", "Email": "name3@mail.com" } ],
     arr2=[
  { "Name": "Name1", "Phone": "987654321", "Email": "name1@mail.com" },
  { "Name": "Name2", "Phone": "12345566", "Email": "name2@mail.com" },
  { "Name": "Name3", "Phone": "23445677", "Email": "" , "ID": 3 }
];

console.log(merge(arr1,arr2))

function merge(arr) {
  var merged={};
  for (var i=1;i<arguments.length;i++) arr=arr.concat(arguments[i]);
  arr.forEach(function(o){ // first round of merging ...
    var k=(o.Email||'')+'|'+(o.Phone||'');
    if (!merged[k]) merged[k]=o;
    else for (p in o) merged[k][p]=o[p];
  });
  var keys=Object.keys(merged);
  keys.forEach((k,i)=>{
    var l=k.length,p,o=merged[k];
    // now look for partial matches (only phone no. given)
    if (k[0]=='|'){ keys.forEach((kk,j)=>{
      if(j!=i && kk.substr(kk.length-l)==k) {
        for (p in o) if(o[p]) merged[kk][p]=o[p];
        delete merged[k];
      }
      })
    }
    else if (k[l-1]=='|'){ keys.forEach((kk,j)=>{
    // partial matches (only email given)
      if(j!=i && kk.substr(0,l)==k) {
        for (p in o) if(o[p]) merged[kk][p]=o[p];
        delete merged[k];
      }
      })
    }
  })
  return Object.values(merged)
}