Question

返回只有重复数组的数组的有效方法是什么？

array = [
{ name: 'name1', description:'desc1', place:'place1' },
{ name: 'name1', description:'desc1', place:'place2' },
{ name: 'name3', description:'desc3', place:'place3' },
{ name: 'name4', description:'desc4', place:'place4' },
{ name: 'name4', description:'desc4', place:'place4' },
{ name: 'name5', description:'desc5', place:'place5' }
]

在这种情况下，我们只需要检查重复项的名称和描述属性。结果应该返回：

duplicates = [
{ name: 'name1', description:'desc1', place:'place1' },
{ name: 'name4', description:'desc4', place:'place4' },
]

这样做的简单方法是什么？

Answer 1

Filter数组，并使用Set检查name和description是否已经出现，如果是，则添加到输出中。如果没有添加到Set：

const array = [{"name":"name1","description":"desc1","place":"place1"},{"name":"name1","description":"desc1","place":"place2"},{"name":"name3","description":"desc3","place":"place3"},{"name":"name4","description":"desc4","place":"place4"},{"name":"name4","description":"desc4","place":"place4"},{"name":"name5","description":"desc5","place":"place5"}]

const result = array.filter(function({ name, description }) {
  const key = `${name}-${description}`
  
  return this.has(key) || !this.add(key)
}, new Set())

console.log(result)

要仅显示每个name和description组的一个副本，请改用Map，并计算出现次数。仅在数字为2时显示。

const array = [{"name":"name1","description":"desc1","place":"place1"},{"name":"name1","description":"desc1","place":"place2"},{"name":"name3","description":"desc3","place":"place3"},{"name":"name4","description":"desc4","place":"place4"},{"name":"name4","description":"desc4","place":"place4"},{"name":"name4","description":"desc4","place":"place4"},{"name":"name5","description":"desc5","place":"place5"}]

const result = array.filter(function({ name, description }) {
  const key = `${name}-${description}`
  const status = (this.get(key) || 0) + 1
  this.set(key, status)
  
  return status === 2
}, new Map())

console.log(result)

基于Map解决方案的IE11解决方案：

const array = [{"name":"name1","description":"desc1","place":"place1"},{"name":"name1","description":"desc1","place":"place2"},{"name":"name3","description":"desc3","place":"place3"},{"name":"name4","description":"desc4","place":"place4"},{"name":"name4","description":"desc4","place":"place4"},{"name":"name4","description":"desc4","place":"place4"},{"name":"name5","description":"desc5","place":"place5"}]

const result = array.filter(function(o) {
  const key = o.name + '-' + o.description
  const status = (this.get(key) || 0) + 1
  this.set(key, status)
  
  return status === 2
}, new Map())

console.log(result)

Answer 2

这适用于2的重复。

＆＃13;

let array = [
{ name: 'name1', description:'desc1', place:'place1' },
{ name: 'name1', description:'desc1', place:'place2' },
{ name: 'name3', description:'desc3', place:'place3' },
{ name: 'name4', description:'desc4', place:'place4' },
{ name: 'name4', description:'desc4', place:'place4' },
{ name: 'name5', description:'desc5', place:'place5' }
]

   
let duplicates = array.reverse()
            .filter((el,index)=> {
                let isDuplicate = index !== array.findIndex(({name,description})=> el.name === name && el.description === description);
               return isDuplicate});
              

              
console.log(duplicates)

＆＃13;

返回匹配重复属性的对象数组

2 个答案: