Find deduplication

Since I do not know what title, what is the cause of an exchange of a small group of partners encountered such a demand:

The original words are as follows:

There are two arrays, m = [a, b], n = [{key: a, name: 'tom'}, {key: b, name: 'jerry'}, {key: c, name: 'marry' }, {key: b, name: 'jerry'}], define a variable, let newArry = n.filter ((e) => {}); will be how to achieve, so newArry equal to [{key: a, name : 'tom'}, {key: b, name: 'jerry'}], and the need to re newArry, seek guidance about ideas, thank

See through answers chiefs do not read the article pure water

The first line and a line and demand:

• Deduplication Array
• Returns the first item in the array corresponds to

See this question, my first idea was to use nested loops to traverse, but ES6 give us a lot more convenient syntax and features, such as Array.prototype.map(), , Array.prototype.filter(), Array.prototype.reduce(), Array.prototype.includes()these api can greatly improve our efficiency to solve the problem, use the method see Getting ES6 MDN or Ruan Yifeng teacher

Deduplication Array

let hash = {}; 
let newN = n.reduce(function(item, next) { 
hash[next.key] ? '' : hash[next.key] = true && item.push(next); 
return item 
}, [])
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Find the corresponding item

let newArray = newN.filter(el=>{
  if(m.includes(el.key)) {
    return el
  }
})
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Complete code:

let m=['a','b'];
let n=[{key:'a', name: 'tom'},{key:'b', name: 'jerry' }, {key:'c', name: 'marry' },{key:'b', name: 'jerry' }];
// 新数组满足两个条件:
//每个元素具有唯一key(先去重)
//查找key与m数组中的元素匹配

let hash = {}; 
let newN = n.reduce(function(item, next) { 
hash[next.key] ? '' : hash[next.key] = true && item.push(next); 
return item 
}, [])
let newArray = newN.filter(el=>{
  if(m.includes(el.key)) {
    return el
  }
})
console.log(newArray)
//[{key:'a', name: 'tom'},{key:'b', name: 'jerry' }]
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Water End

Reproduced in: https: //juejin.im/post/5d0bbc0cf265da1b904be871