在具有复杂条件的两个对象数组之间检索项目

Question

我有两个对象列表：let list1 = [{id: '1', status: 'use', comment: 'xxxx'}, {id: '2', status: 'ready', comment: 'yyyy'}, {id: '3', status: 'ready', comment: 'zzzz'}];let list2 = [{uid: '1', elec: 60}, {uid: '2', elec: 60}, {uid: '10', elec: 60}, {uid: '3', elec: 40}];我想要的是检索 list2 的一个对象，该对象具有 elec > 50 且 uid 与 list1 的一项 id 相同，前提是 list1 的该项具有状态==“就绪”。另外，我想向此项目添加来自 list1 对象的参数“comment”。在此示例中，我的结果值为：{uid: '2', elect: 60, comment: 'yyyy'}。我这样做了：  let list1Filtered = list1.filter(itemList1 => itemList1.status == 'ready');  let list2Filtered = list2.filter(itemList2 => itemList2.elec > 50);  var result;  for ( let  itemList1Filtered of list1Filtered ) {    for ( let  itemList2Filtered of list2Filtered ) {      if (!result && itemList1Filtered.id == itemList2Filtered.uid) {        result = itemList2Filtered;        result.comment = itemList1Filtered.comment;      }    }  }    return result;我想知道是否有更优雅和/或更复杂的方法可以在 Javascript 中执行此操作。

Answer 1

您可以从 list1 中收集想要的评论，并通过检查其他列表中是否存在某个项目list2的值来减少。elec然后返回一个新对象。

这种方法只需要两个循环。

const

    list1 = [{ id: '1', status: 'use', comment: 'xxxx' }, { id: '2', status: 'ready', comment: 'yyyy' }, { id: '3', status: 'ready', comment: 'zzzz' }],

    list2 = [{ uid: '1', elec: 60 }, { uid: '2', elec: 60 }, { uid: '10', elec: 60 }, { uid: '3', elec: 40 }],

    l1 = list1.reduce((r, { id, status, comment }) => {

        if (status === 'ready') r[id] = { comment };

        return r;

    }, {}),

    result = list2.reduce((r, o) => {

        if (o.elec > 50 && o.uid in l1) r.push({ ...o, ...l1[o.uid]})

        return r;

    }, []);

console.log(result);

Answer 2

let result = {

    ...list2.filter(

    a => a.elec > 50 && a.uid === list1.filter(b => b.status === "ready")[0].id)[0],

    comments: list1.filter(b => b.status === "ready")[0].comment

}

Answer 3

you can restructure your data. List1 convert it in object. And now we can find solution in O(n).

let list1 = [{

  id: '1',

  status: 'use',

  comment: 'xxxx'

}, {

  id: '2',

  status: 'ready',

  comment: 'yyyy'

}, {

  id: '3',

  status: 'ready',

  comment: 'zzzz'

}];

let list2 = [{

  uid: '1',

  elec: 60

}, {

  uid: '2',

  elec: 60

}, {

  uid: '10',

  elec: 60

}, {

  uid: '3',

  elec: 40

}];

const itemList = {}

list1.forEach(item => {

  if (item.status === 'ready') {

    itemList[item.id] = item.comment

  }

});

const result = list2.filter(item => itemList[item.uid] && item.elec > 50).map(item => {

  item['comment'] = itemList[item.uid]

  return item

})

console.log(result)

Answer 4

尝试这个：

let list1 = [

  { id: '1', status: 'use', comment: 'xxxx' },

  { id: '2', status: 'ready', comment: 'yyyy' },

  { id: '3', status: 'ready', comment: 'zzzz' },

];

let list2 = [

  { uid: '1', elec: 60 },

  { uid: '2', elec: 60 },

  { uid: '10', elec: 60 },

  { uid: '3', elec: 40 },

];

let result = null;

let itemFound;

const filteredList = list2.filter((list2Item) => {

  if (!result) {

    itemFound =

      list2Item.elec > 50 &&

      list1.find(

        (list1Item) =>

          list2Item.uid === list1Item.id && list1Item.status === 'ready'

      );

    if (itemFound) {

      result = {

        uid: list2Item.uid,

        elect: list2Item.elec,

        comment: itemFound.comment,

      };

    }

  }

});

console.log(result);