比较具有不同属性数量的两个数组，并使用javascript在条件下将第三个数组映射到比较值

Posted 2023-02-14

【中文标题】比较具有不同属性数量的两个数组，并使用javascript在条件下将第三个数组映射到比较值

【英文标题】：compare two arrays having different number of properties and mapping a third array with compared values on condition using javascript

【发布时间】：2018-09-30 06:57:39

【问题描述】：

我有 2 个数组

var array1 = ["1", "2", "3", "4", "5", "6"]

和

var array2 = [
   offId: "4", offname: "four" ,
   offId: "9", offname: "nine" ,
   offId: "15", offname: "fifteen" ,
   offid: "3", offname: "three" ,
   offId: "1", offname: "one" ,
   offId: "0", offname: "zero" ,
   offId: "8", offname: "eight" ,
   offId: "10", offname: "ten" ,
]

我需要将两个数组与 offId 的值进行比较，结果数组应该是

var array3 = [
   offId: "1", offname: "one" ,
   offId: "2", offname: "" ,
   offId: "3", offname: "three" ,
   offId: "4", offname: "four" ,
   offId: "5", offname: "" ,
   offId: "6", offname: "" ,
]

我怎样才能做到这一点（数组的长度可能相同或不同）

【问题讨论】：

'offId' !== 'offid'

@NinaScholz 两者都相等 'offId' ='offId'..edited

@HKI345 检查我的回答我认为它会解决你的问题。

【参考方案1】：

您可以使用Array.reduce 从array1 和array2 创建array3。

步骤如下：

acc（累加数组）最初将包含空数组。 使用Array.find 检查curr（当前变量）是否在array2 中。 如果a.offId 匹配curr 值，Array.find 将返回对象。 如果找到obj，则将obj推入acc（累加数组） 否则将offId: curr, offname: '' 推送到数组中

var array1 = ["1", "2", "3", "4", "5", "6"]

var array2 = [
   offId: "4", offname: "four" ,
   offId: "9", offname: "nine" ,
   offId: "15", offname: "fifteen" ,
   offId: "3", offname: "three" ,
   offId: "1", offname: "one" ,
   offId: "0", offname: "zero" ,
   offId: "8", offname: "eight" ,
   offId: "10", offname: "ten" ,
]

var array3 = array1.reduce((acc, curr) => 
  var obj = array2.find(a => a.offId === curr);
  if (obj) 
    acc.push(obj);
   else 
    acc.push( offId: curr, offname: "");
  
  return acc;
, []);

console.log(array3);

【参考方案2】：

您可以使用Map 并使用存储的对象或新对象。

var array1 = ["1", "2", "3", "4", "5", "6"],
    array2 = [ offId: "4", offname: "four" ,  offId: "9", offname: "nine" ,  offId: "15", offname: "fifteen" ,  offId: "3", offname: "three" ,  offId: "1", offname: "one" ,  offId: "0", offname: "zero" ,  offId: "8", offname: "eight" ,  offId: "10", offname: "ten" ],
    map = array2.reduce((m, o) => m.set(o.offId, o), new Map),
    result = array1.map(offId => map.get(offId) ||  offId, offname: '' );

console.log(result);

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【参考方案3】：

var array1 = ["1", "2", "3", "4", "5", "6"];

var array2 = [
   offId: "4", offname: "four" ,
   offId: "9", offname: "nine" ,
   offId: "15", offname: "fifteen" ,
   offId: "3", offname: "three" ,
   offId: "1", offname: "one" ,
   offId: "0", offname: "zero" ,
   offId: "8", offname: "eight" ,
   offId: "10", offname: "ten" ,
];

var array3 = array1.map(val => array2.filter(obj => obj.offId == val)[0] ||  offId: val, offname: "" );

console.log(array3);

【讨论】：

该方法的性能最差，因为您使用过滤器并且这会迭代数组的所有元素。 find 迭代直到找到一个元素。或使用哈希表，如对象或地图，无需迭代即可快速访问。

【参考方案4】：

最简单的方法是简单地 filter() 第二个数组：

const array3 = array2.filter(v => array1.includes(v.offId));

这将遍历array2 中的每个元素，并且只将具有offId 的元素留在array1 中。

如果您需要对其进行排序，则只需在其末尾添加一个排序：

const array3 = array2.filter(v => array1.includes(v.offId))
  .sort((a, b) => b.offId > a.offId ? 1 : a.offId < b.offId ? -1 : 0);

如果您需要对它进行数字排序（即，您希望 2 在 10 之前），那么您需要转换它们的值，但是您可以减去它们：

const array3 = array2.filter(v => array1.includes(v.offId))
  .sort((a, b) => parseInt(b.offId) - parseInt(a.offId));