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答案

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

标签： javascript arrays array.prototype.map

【解决方案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);

【讨论】：

【解决方案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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【讨论】：

【解决方案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。

【解决方案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));

【讨论】：