如何通过ID删除重复项并计算数组中的总记录

时间：2018-07-03 19:33:29

标签： javascript ecmascript-6 reduce

练习如下：其中键是银行名称和值的对象是仅在该银行中拥有帐户的客户数量。

发问这个数组：

const accounts = [ 
   { clientId: 6, bankId: 1, balance: 15000 }, 
   { clientId: 1, bankId: 3, balance: 18000 },
   { clientId: 5, bankId: 3, balance: 135000 },
   { clientId: 2, bankId: 2, balance: 5600 },
   { clientId: 3, bankId: 1, balance: 23000 },
   { clientId: 5, bankId: 2, balance: 15000 },
   { clientId: 3, bankId: 3, balance: 45900 },
   { clientId: 2, bankId: 3, balance: 19000 },
   { clientId: 4, bankId: 3, balance: 51000 },
   { clientId: 5, bankId: 1, balance: 89000 },
   { clientId: 1, bankId: 2, balance: 1600 },
   { clientId: 5, bankId: 3, balance: 37500 },
   { clientId: 6, bankId: 1, balance: 19200 },
   { clientId: 2, bankId: 3, balance: 10000 },
   { clientId: 3, bankId: 2, balance: 5400 },
   { clientId: 3, bankId: 1, balance: 9000 },
   { clientId: 4, bankId: 3, balance: 13500 },
   { clientId: 2, bankId: 1, balance: 38200 },
   { clientId: 5, bankId: 2, balance: 17000 },
   { clientId: 1, bankId: 3, balance: 1000 },
   { clientId: 5, bankId: 2, balance: 600 },
   { clientId: 6, bankId: 1, balance: 16200 },
   { clientId: 2, bankId: 2, balance: 10000 }
];

我现在有这个：

function banksFidelity() {
  var map = accounts.reduce(function(map, account) {
    var bankByid = account.bankId;
    var clientByid = account.clientId;

    map[bankByid] = map[clientByid] || 0 + clientByid

    return map

  }, {})

  return map

}

当前结果是： 

{1: 6, 2: 5, 3: 5}

期望的结果是： 

{1: 4, 2: 4, 3: 4}

3 个答案:

答案 0 :(得分：1)

您可以使用.foreach方法遍历帐户对象数组，并通过检查客户是否已经拥有该银行的帐户来填充另一个对象。您剩下一个对象（clients），它将每个客户与他使用的银行阵列相关联：

{
  "1": [3, 2],
  "2": [2, 3, 1],
  "3": [1, 3, 2],
  "4": [3],
  "5": [3, 2, 1],
  "6": [1]
}

使用另一个.foreach循环填充另一个对象（exclusiveClients），该对象将把银行与仅使用该银行以获得所需结果的客户数量相关联：

{
  "1": 1,
  "3": 1
}

代码如下：

const accounts=[{clientId:6,bankId:1,balance:15000},{clientId:1,bankId:3,balance:18000},{clientId:5,bankId:3,balance:135000},{clientId:2,bankId:2,balance:5600},{clientId:3,bankId:1,balance:23000},{clientId:5,bankId:2,balance:15000},{clientId:3,bankId:3,balance:45900},{clientId:2,bankId:3,balance:19000},{clientId:4,bankId:3,balance:51000},{clientId:5,bankId:1,balance:89000},{clientId:1,bankId:2,balance:1600},{clientId:5,bankId:3,balance:37500},{clientId:6,bankId:1,balance:19200},{clientId:2,bankId:3,balance:10000},{clientId:3,bankId:2,balance:5400},{clientId:3,bankId:1,balance:9000},{clientId:4,bankId:3,balance:13500},{clientId:2,bankId:1,balance:38200},{clientId:5,bankId:2,balance:17000},{clientId:1,bankId:3,balance:1000},{clientId:5,bankId:2,balance:600},{clientId:6,bankId:1,balance:16200},{clientId:2,bankId:2,balance:10000}]

let clients = {};

accounts.forEach(({bankId, clientId}) => {

  if(clients[clientId] && !clients[clientId].includes(bankId)) {
  
    clients[clientId].push(bankId);
  
  } else if(!clients[clientId]) {
  
    clients[clientId] = [bankId];
  
  }
  
});


let exclusiveClients = {};

Object.entries(clients).forEach(([client, banks]) => {

  if(banks.length == 1) {
  
    const bank = banks[0];
    exclusiveClients[bank] = (exclusiveClients[bank] + 1) || 1;
    
    
  }
  
});

console.log(exclusiveClients)

答案 1 :(得分：1)

您可以reduce删除所有重复的数组，然后缩小该数组以形成最终对象。

const accounts = [ 
   { clientId: 6, bankId: 1, balance: 15000 }, 
   { clientId: 1, bankId: 3, balance: 18000 },
   { clientId: 5, bankId: 3, balance: 135000 },
   { clientId: 2, bankId: 2, balance: 5600 },
   { clientId: 3, bankId: 1, balance: 23000 },
   { clientId: 5, bankId: 2, balance: 15000 },
   { clientId: 3, bankId: 3, balance: 45900 },
   { clientId: 2, bankId: 3, balance: 19000 },
   { clientId: 4, bankId: 3, balance: 51000 },
   { clientId: 5, bankId: 1, balance: 89000 },
   { clientId: 1, bankId: 2, balance: 1600 },
   { clientId: 5, bankId: 3, balance: 37500 },
   { clientId: 6, bankId: 1, balance: 19200 },
   { clientId: 2, bankId: 3, balance: 10000 },
   { clientId: 3, bankId: 2, balance: 5400 },
   { clientId: 3, bankId: 1, balance: 9000 },
   { clientId: 4, bankId: 3, balance: 13500 },
   { clientId: 2, bankId: 1, balance: 38200 },
   { clientId: 5, bankId: 2, balance: 17000 },
   { clientId: 1, bankId: 3, balance: 1000 },
   { clientId: 5, bankId: 2, balance: 600 },
   { clientId: 6, bankId: 1, balance: 16200 },
   { clientId: 2, bankId: 2, balance: 10000 }
]

let result = accounts
  // Remove duplicates
  .reduce((arr, itm) => {
    let containsClient = !!arr.find(i => i.clientId == itm.clientId && i.bankId == itm.bankId)
    return !containsClient ? arr.concat(itm) : arr
  }, [])
  // Get the counts after duplicates are removed
  .reduce((obj, itm) => {
    if (obj[itm.bankId]) obj[itm.bankId]++
    else obj[itm.bankId] = 1
    return obj
  }, {})

console.log(result)

答案 2 :(得分：1)

变化：使用引用，因此当我们更新客户端一次时，我们可以从添加客户端/引用复制的任何地方检索信息

 const accounts = [ 
   { clientId: 6, bankId: 1, balance: 15000 }, 
   { clientId: 1, bankId: 3, balance: 18000 },
   { clientId: 5, bankId: 3, balance: 135000 },
   { clientId: 2, bankId: 2, balance: 5600 },
   { clientId: 3, bankId: 1, balance: 23000 },
   { clientId: 5, bankId: 2, balance: 15000 },
   { clientId: 3, bankId: 3, balance: 45900 },
   { clientId: 2, bankId: 3, balance: 19000 },
   { clientId: 4, bankId: 3, balance: 51000 },
   { clientId: 5, bankId: 1, balance: 89000 },
   { clientId: 1, bankId: 2, balance: 1600 },
   { clientId: 5, bankId: 3, balance: 37500 },
   { clientId: 6, bankId: 1, balance: 19200 },
   { clientId: 2, bankId: 3, balance: 10000 },
   { clientId: 3, bankId: 2, balance: 5400 },
   { clientId: 3, bankId: 1, balance: 9000 },
   { clientId: 4, bankId: 3, balance: 13500 },
   { clientId: 2, bankId: 1, balance: 38200 },
   { clientId: 5, bankId: 2, balance: 17000 },
   { clientId: 1, bankId: 3, balance: 1000 },
   { clientId: 5, bankId: 2, balance: 600 },
   { clientId: 6, bankId: 1, balance: 16200 },
   { clientId: 2, bankId: 2, balance: 10000 }
  ]

var clients = {}
var dupes = {}
var banks = {}
accounts.forEach(a=>{
    if(dupes[a.clientId+'_'+a.bankId]){
        return;
    }
    banks[a.bankId] = banks[a.bankId] || new Set;
    if(clients[a.clientId]){
        //this client has already a bank
        //different than the current one
        clients[a.clientId].multi = true;
    }else{
        clients[a.clientId]={}
    }
    banks[a.bankId].add(a.clientId)
    dupes[a.clientId+'_'+a.bankId] = 1;
}, {})
var res = Object.keys(banks).reduce((acc,bId)=>{
    acc[bId] = [...banks[bId]].filter(x=>!x.multi).length
    return acc;
},{})
console.log('res ', res)

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