按另一个数组过滤一个数组[重复]

Posted 2023-02-14

【中文标题】按另一个数组过滤一个数组[重复]

【英文标题】：Filter one array by another array [duplicate]

【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

【参考方案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

【参考方案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)

【参考方案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)