0

假设我有这些对象数组

selectedNames = [
   {label: 'Nicolas', value:'Nicolas'},
   {label: 'Michael', value:'Michael'},
   {label: 'Sean', value:'Sean'},
   {label: 'George', value:'George'}
]

selectedWhatever = [
   {label: 'Guitar', value: 'Guitar'},
   {label: 'Bass', value: 'Bass'}
]

然后是另一个像这样的对象数组：

list = [
   {name: 'Nicolas', surname: 'Smith', birthplace: 'NewYork', whatever: 'Guitar'},
   {name: 'Sean', surname: 'Narci', birthplace: 'LA', whatever: 'Bass'},
   {name: 'George', surname: 'Rossi', birthplace: 'NewYork', whatever: 'Bass'},
   {name: 'Fiona', surname: 'Gallagher', birthplace: 'Madrid', whatever: 'Drum'},
   {name: 'Michael', surname: 'Red', birthplace: 'London', whatever: 'Triangle'},
]

我想根据我在其他两个数组中的数据过滤列表并按出生地分组，所以我想要这个：

result = {
   LA: [
      {name: 'Sean', surname: 'Narci', birthplace: 'LA', whatever: 'Bass'},
   ],
   NewYork: [
      {name: 'Nicolas', surname: 'Smith', birthplace: 'NewYork', whatever: 'Guitar'},
      {name: 'George', surname: 'Rossi', birthplace: 'NewYork', whatever: 'Bass'},
   ]
}

我所做的是以下，它工作正常。有没有更聪明或更优雅的方法来做同样的事情？

const obj = {};

list.map((res) => {
   if ((data.selectedNames.map((r) => r.label).includes(res.name))
      && (data.selectedWhatever.map((r) => r.label).includes(res.whatever))
   ){
      result[res.birthplace] = result[res.birthplace] ?? []:
      result[res.birthplace].push(res);
   }
})
4

1 回答 1

0

与其为每个检查的条目创建和搜索一个新的名称数组和任何东西，不如只创建一次并将它们与他检查的键一起存储为集合：

 const filters = {
    name: new Set( selectedNames.map(it => it.label) )
 };

 const result = list.filter(it => 
    Object.entries(filters).some(([key, set]) =>
        set.has( it[key] )
     )
  );

分组部分实际上也应该是一个可重用的功能。

于 2021-09-16T22:15:07.780 回答