使用过滤连接两个数组答案

【问题标题】：Concat two arrays with filtering使用过滤连接两个数组
【发布时间】：2018-11-23 07:50:29
【问题描述】：

我有两个数组。我需要将它们结合起来并创建一个具有dayOfWeek 2、3、4、5、6 的新数组。这意味着dayOfWeek 的优先级在array1 中。表示需要保留dayOfWeek 3、4、5 与array1。

array1 = [
  {dayOfWeek: 2, home1: "01:30"},
  {dayOfWeek: 3, home1: "02:30"},
  {dayOfWeek: 4, home1: "03:30"},
  {dayOfWeek: 5, home1: "04:30"},
]

array2 = [
  {dayOfWeek: 3, home1: "05:30"},
  {dayOfWeek: 4, home1: "06:30"},
  {dayOfWeek: 5, home1: "07:30"},
  {dayOfWeek: 6, home1: "08:30"},
]

输出应该是

finalArray = [
  {dayOfWeek: 2, home1: "01:30"},
  {dayOfWeek: 3, home1: "02:30"},
  {dayOfWeek: 4, home1: "03:30"},
  {dayOfWeek: 5, home1: "04:30"},
  {dayOfWeek: 6, home1: "08:30"},
]

我试过这个，但它会从两个数组中推送dayOfWeek。如何过滤它们？

const finalArray = []
array1.map((a) => {
    array2.map((a2) => {
        if (a.dayOfWeek === a2.dayOfWeek) {
          finalArray.push(a)
        }
        if (a.dayOfWeek === a2.dayOfWeek) {
          finalArray.push(a2)
        }
    })
})

提前致谢！！！

【问题讨论】：

标签： javascript arrays loops for-loop lodash

【解决方案1】：

您也可以简单地过滤第二个数组中缺少的项目，然后将它们连接到第一个数组而不使用 lodash：

const a1 = [ {dayOfWeek: 2, home1: "01:30"}, {dayOfWeek: 3, home1: "02:30"}, {dayOfWeek: 4, home1: "03:30"}, {dayOfWeek: 5, home1: "04:30"}, ]
const a2 = [ {dayOfWeek: 3, home1: "05:30"}, {dayOfWeek: 4, home1: "06:30"}, {dayOfWeek: 5, home1: "07:30"}, {dayOfWeek: 6, home1: "08:30"}, ]

const r = a1.concat(a2.filter(x => !a1.some(y => y.dayOfWeek == x.dayOfWeek)))

console.log(r)

这是通过Array.concat、Array.filter 和Array.some 完成的

【讨论】：

【解决方案2】：

使用 lodash 的 _.unionBy()。主要数组应该是传递给函数的第一个数组。

const array1 = [{"dayOfWeek":2,"home1":"01:30"},{"dayOfWeek":3,"home1":"02:30"},{"dayOfWeek":4,"home1":"03:30"},{"dayOfWeek":5,"home1":"04:30"}]
const array2 = [{"dayOfWeek":3,"home1":"05:30"},{"dayOfWeek":4,"home1":"06:30"},{"dayOfWeek":5,"home1":"07:30"},{"dayOfWeek":6,"home1":"08:30"}]

const result = _.unionBy(array1, array2, 'dayOfWeek')

console.log(result)

&lt;script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"&gt;&lt;/script&gt;

如果需要组合多个属性作为联合值，可以使用：

_.unionBy(array1, array2, o => `${o.id}-${o.dayOfWeek}`)

【讨论】：

我可以为 _.unionBy 使用两个字段吗？我需要通过dayOfWeek 和id 联合？
查看答案。

【解决方案3】：

您也可以使用Set 和Array.filter

let array1 = [
  {dayOfWeek: 2, home1: "01:30"},
  {dayOfWeek: 3, home1: "02:30"},
  {dayOfWeek: 4, home1: "03:30"},
  {dayOfWeek: 5, home1: "04:30"},
]

let array2 = [
  {dayOfWeek: 3, home1: "05:30"},
  {dayOfWeek: 4, home1: "06:30"},
  {dayOfWeek: 5, home1: "07:30"},
  {dayOfWeek: 6, home1: "08:30"},
]

let s = new Set()
console.log([...array1, ...array2].filter(d => {
    let avail = s.has(d.dayOfWeek)
    !avail && s.add(d.dayOfWeek)
    return !avail
  }
) )

【讨论】：

【解决方案4】：

您可以使用数组连接方法，然后进行过滤。

var c = array1.concat(array2);

【讨论】：

对不起，我不想连接。顺便说一句，谢谢你的回答。

【解决方案5】：

教授。您可以使用 Map 对象来获取唯一元素

const array1 = [
  {dayOfWeek: 2, home1: "01:30"},
  {dayOfWeek: 3, home1: "02:30"},
  {dayOfWeek: 4, home1: "03:30"},
  {dayOfWeek: 5, home1: "04:30"},
];

const array2 = [
  {dayOfWeek: 3, home1: "05:30"},
  {dayOfWeek: 4, home1: "06:30"},
  {dayOfWeek: 5, home1: "07:30"},
  {dayOfWeek: 6, home1: "08:30"},
];

function getFinalArray(array, uniqueProperty) {
  return array
    .filter(value => value)
    .reduce(
      (arrayMap, item) => {
        return arrayMap.set(item[uniqueProperty], item)
      }, new Map()
    );
}

const result = Array.from(
  getFinalArray([...array2, ...array1], 'dayOfWeek').values()
);

【讨论】：

如果您不希望订单被破坏，您可以这样做：.reduce( (arrayMap, item) => !arrayMap.get(item[uniqueProperty]) ? arrayMap.set(item[uniqueProperty], item) : arrayMap, new Map(), 将数组传递为 [...array1,...array2]`
感谢您的加入。最好以arrayMap.has(item[uniqueProperty]) 的方式检查而不是

【解决方案6】：

您可以结合使用Array#slice()、Array#forEach() 和Array#some() 方法：

使用.slice(0)获取array1在finalArray中的所有元素。
然后使用.forEach()遍历array2，得到finalArray中不存在的所有元素。
使用.some()检查迭代元素（dayOfWeek）是否存在。

这应该是你的代码：

var finalArray = array1.slice(0);
array2.forEach(function(a){
      if(!finalArray.some(e => e.dayOfWeek == a.dayOfWeek))
          finalArray.push(a);
});

演示：

let array1 = [
  {dayOfWeek: 2, home1: "01:30"},
  {dayOfWeek: 3, home1: "02:30"},
  {dayOfWeek: 4, home1: "03:30"},
  {dayOfWeek: 5, home1: "04:30"},
];

let array2 = [
  {dayOfWeek: 3, home1: "05:30"},
  {dayOfWeek: 4, home1: "06:30"},
  {dayOfWeek: 5, home1: "07:30"},
  {dayOfWeek: 6, home1: "08:30"},
];

var finalArray = array1.slice(0);
array2.forEach(function(a){
      if(!finalArray.some(e => e.dayOfWeek == a.dayOfWeek))
          finalArray.push(a);
});

console.log(finalArray);

【讨论】：