按另一个数组过滤一个数组[重复]答案

【问题标题】：Filter one array by another array [duplicate]按另一个数组过滤一个数组[重复]
【发布时间】：2022-01-16 08:40:59
【问题描述】：

const items = [[{name:"p2"},{name:"p3"}, {name:"p7"},{name:"p9"},{name:"p1"}],[{name:"p6"}, {name:"p3"},{name:"p7"}, {name:"p9"},{name:"p2"}],[{name:"p3"},{name:"p6"}, {name:"p7"},{name:"p9"},{name:"p4"}],[{name:"p2"}, {name:"p3"},{name:"p1"}, {name:"p9"},{name:"p6"}]]

const findObj = [{name:"p1"},{name:"p2"},{name:"p6"}]

从 findobj 对象中找到包含所有三个元素的集合的子数组

【问题讨论】：

或Filter array of objects based on another array in javascript

标签： javascript arrays sorting filter include

【解决方案1】：

您可以使用filter、Set 和reduceas 轻松实现结果：

const items = [
  [
    { name: "p2" },
    { name: "p3" },
    { name: "p7" },
    { name: "p9" },
    { name: "p1" },
  ],
  [
    { name: "p6" },
    { name: "p3" },
    { name: "p7" },
    { name: "p9" },
    { name: "p2" },
  ],
  [
    { name: "p3" },
    { name: "p6" },
    { name: "p7" },
    { name: "p9" },
    { name: "p4" },
  ],
  [
    { name: "p2" },
    { name: "p3" },
    { name: "p1" },
    { name: "p9" },
    { name: "p6" },
  ],
];
const findObj = [{ name: "p1" }, { name: "p2" }, { name: "p6" }];
const set = new Set(findObj.map((o) => o.name));

const result = items.filter((arr) => {
  const remain = arr.reduce((acc, curr) => {
    if (set.has(curr.name)) acc.add(curr.name);
    return acc;
  }, new Set());
  return remain.size === set.size;
});
console.log(result);

【讨论】：

还有一个明显的重复

【解决方案2】：

const items = [[{name:"p2"},{name:"p3"}, {name:"p7"},{name:"p9"},{name:"p1"}],[{name:"p6"}, {name:"p3"},{name:"p7"}, {name:"p9"},{name:"p2"}],[{name:"p3"},{name:"p6"}, {name:"p7"},{name:"p9"},{name:"p4"}],[{name:"p2"}, {name:"p3"},{name:"p1"}, {name:"p9"},{name:"p6"}]]

const findObj = [{name:"p1"},{name:"p2"},{name:"p6"}]


const result = items.filter(item=>{
  const childItem = item.map(childItem=>childItem.name);
  let allExist=true;
  findObj.forEach(obj=>{
  if(!childItem.includes(obj.name)){
  allExist=false;
  }
  })
  return allExist;
})
console.log(result)

【讨论】：

【解决方案3】：

使用filter 并通过使用map .map(prop=>prop.name) 或更简单的解构 .map(({name})=>name) 和过滤items 从每个item 获取所有名称名称包含在findObj

const items = [[{name:"p2"},{name:"p3"}, {name:"p7"},{name:"p9"},{name:"p1"}],[{name:"p6"}, {name:"p3"},{name:"p7"}, {name:"p9"},{name:"p2"}],[{name:"p3"},{name:"p6"}, {name:"p7"},{name:"p9"},{name:"p4"}],[{name:"p2"}, {name:"p3"},{name:"p1"}, {name:"p9"},{name:"p6"}]]

const findObj = [{name:"p1"},{name:"p2"},{name:"p6"}]
const result = items.filter(item=>{
    const arrProp  = item.map(prop=>prop.name)
    const filtProp = findObj.map(({name})=>name)
    return filtProp.every(x=>arrProp.includes(x));
})
console.log(result)

【讨论】：