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}];

我想要的是检索具有 elec > 50 且与 list1 的一个项目 id 相同的 uid 的 list2 对象，仅当 list1 的项目具有状态 ==“就绪”时。另外，我想将 list1 对象中的参数“comment”添加到此项。

在这个例子中，我的结果值为：{uid: '2',lect: 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/或更复杂的方式在 Javascript 中执行此操作。

Answer 1

您可以从 list1 收集想要的 cmets 并通过检查 elec 的值以及是否存在另一个列表中的项目来减少 list2。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

试试这个：

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);

Answer 4

这应该可以完成工作：

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

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

const filteredList = list2.filter(item => {
  const readyItemsList1 = list1.filter(item => item.status === 'ready').map(item => item.id);
  return item.elec > 50 && readyItemsList1.indexOf(item.uid) > -1
}).map(item => {
  const comment = list1.find(it => it.id === item.uid).comment;
  item.comment = comment;
  return item;
});

console.log(filteredList);

Answer 5

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 6

filter 和 map 会帮助你。

filter 可以选择符合某些条件的项目，map 让您更改要选择的数据。

对于list1，选择status等于'ready'的项目，然后只取id。

var ready_id_array = list1.filter(item=>item.status == 'ready').map(item=>item.id);

对于list2，检查其uid包含在ready_id_array中的项目，并且elec大于50。

var result = list2 .filter(item => ready_id_array.indexOf(item.uid) > -1 && item.elec > 50);

要附加评论，创建一个字典，然后将评论放回结果

var comment_dictionary = list1.reduce((a,x) => ({...a, [x.id]: x.comment}), {});
result.forEach(item => item.comment = comment_dictionary[item.uid]);

你会得到结果。

[{uid: "2", elec: 60, comment: "yyyy"}]