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如何使用兩個屬性對對象數(shù)組進行排序，但根據(jù)條件按排序順序有條件地分組？

如何使用兩個屬性對對象數(shù)組進行排序，但根據(jù)條件按排序順序有條件地分組？

精慕HU 2023-11-11 21:47:54
這個問題是這個問題的延續(xù)。假設我有一個如下所示的數(shù)組：const questions = [  {_id: 1, q: 'why?', group: 'no-group', date: '8', selected: false },   {_id: 2, q: 'what?', group: 'group 1', date: '6', selected: false },   {_id: 3, q: 'when?', group: 'no-group', date: '7', selected: false },   {_id: 4, q: 'where?', group: 'group 1', date: '5', selected: false },   {_id: 5, q: 'which?', group: 'group 2', date: '3', selected: false },  {_id: 6, q: 'who?', group: 'no-group', date: '0', selected: false },  {_id: 7, q: 'why not?', group: 'group 2', date: '9', selected: false },   {_id: 8, q: 'who, me?', group: 'group 1', date: '4', selected: false },   {_id: 9, q: 'where is waldo?', group: 'group 1', date: '1', selected: false },   {_id: 10, q: 'which way is up?', group: 'no-group', date: '2', selected: false },  {_id: 11, q: 'when is lunch?', group: 'group-2', date: '10', selected: false }, ];如何編寫代碼對其進行排序，以便按日期和組對對象進行排序。'no-group'但是，如果：假設組 1 作為第二個對象出現(xiàn)，則以下對象應為按日期排序的組 1，直到列出所有組 1 對象，并且對于除具有組屬性值的對象之外的所有其他對象都相同。因此，對于上面的數(shù)組，我應該得到如下輸出：[  {_id: 6, q: 'who?', group: 'no-group', date: '0', selected: false },  {_id: 9, q: 'where is waldo?', group: 'group 1', date: '1', selected: false },   {_id: 8, q: 'who, me?', group: 'group 1', date: '4', selected: false },   {_id: 4, q: 'where?', group: 'group 1', date: '5', selected: false },   {_id: 2, q: 'what?', group: 'group 1', date: '6', selected: false },  {_id: 10, q: 'which way is up?', group: 'no-group', date: '2', selected: false },  {_id: 5, q: 'which?', group: 'group 2', date: '3', selected: false },  {_id: 7, q: 'why not?', group: 'group 2', date: '9', selected: false },   {_id: 11, q: 'when is lunch?', group: 'group 2', date: '10', selected: false },  {_id: 3, q: 'when?', group: 'no-group', date: '7', selected: false },   {_id: 1, q: 'why?', group: 'no-group', date: '8', selected: false }, ];// the spacing is just for easier readability.
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互換的青春

TA貢獻1797條經(jīng)驗 獲得超6個贊

您可以通過date對相同組取消對象的排序、分組并獲得平面數(shù)組。


const 

    questions = [{ _id: 1, q: 'why?', group: 'no-group', date: '8', selected: false }, { _id: 2, q: 'what?', group: 'group 1', date: '6', selected: false }, { _id: 3, q: 'when?', group: 'no-group', date: '7', selected: false }, { _id: 4, q: 'where?', group: 'group 1', date: '5', selected: false }, { _id: 5, q: 'which?', group: 'group 2', date: '3', selected: false }, { _id: 6, q: 'who?', group: 'no-group', date: '0', selected: false }, { _id: 7, q: 'why not?', group: 'group 2', date: '9', selected: false }, { _id: 8, q: 'who, me?', group: 'group 1', date: '4', selected: false }, { _id: 9, q: 'where is waldo?', group: 'group 1', date: '1', selected: false }, { _id: 10, q: 'which way is up?', group: 'no-group', date: '2', selected: false }, { _id: 11, q: 'when is lunch?', group: 'group 2', date: '10', selected: false }],

    result = questions

        .sort((a, b) => a.date - b.date)

        .map((groups => o => {

            if (o.group === 'no-group') return o;

            if (groups[o.group]) {

                groups[o.group].push(o);

                return [];

            }

            return groups[o.group] = [o];

        })({}))

        .flat();


console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }


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?
HUH函數(shù)

TA貢獻1836條經(jīng)驗 獲得超4個贊

  • sort數(shù)組由date

  • reduce帶有累加器的數(shù)組Map。

  • 如果它是一個項目，請使用唯一的鍵no-group添加一個新條目。Map_id

  • 如果它是一個有效的項目group，則使用group作為鍵和具有相同的項目數(shù)組group。

  • 每個都no-group將作為鍵添加到 Map 中。其余項目將根據(jù)該項目的第一個條目添加group。

  • 獲取values地圖并將其擊倒

const questions = [{ _id: 1, q: 'why?', group: 'no-group', date: '8', selected: false }, { _id: 2, q: 'what?', group: 'group 1', date: '6', selected: false }, { _id: 3, q: 'when?', group: 'no-group', date: '7', selected: false }, { _id: 4, q: 'where?', group: 'group 1', date: '5', selected: false }, { _id: 5, q: 'which?', group: 'group 2', date: '3', selected: false }, { _id: 6, q: 'who?', group: 'no-group', date: '0', selected: false }, { _id: 7, q: 'why not?', group: 'group 2', date: '9', selected: false }, { _id: 8, q: 'who, me?', group: 'group 1', date: '4', selected: false }, { _id: 9, q: 'where is waldo?', group: 'group 1', date: '1', selected: false }, { _id: 10, q: 'which way is up?', group: 'no-group', date: '2', selected: false }, { _id: 11, q: 'when is lunch?', group: 'group 2', date: '10', selected: false }]


const group = questions.sort((a, b) => a.date - b.date)

? ? ? ? ? ? ? ? ? ? ? ? .reduce((map, o) =>

? ? ? ? ? ? ? ? ? ? ? ? ? o.group === 'no-group'

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? map.set(o._id, o)

? ? ? ? ? ? ? ? ? ? ? ? ? ? : map.set(o.group, [...map.get(o.group) || [], o] )

? ? ? ? ? ? ? ? ? ? ? ? , new Map)


const output = Array.from(group.values()).flat()


console.log( output )


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?
肥皂起泡泡

TA貢獻1829條經(jīng)驗 獲得超6個贊

我們可以date先對它進行排序，然后對對象進行分組，group并將組保存在一個對象中，同時保存組的索引。


然后我們最終可以將組插入到正確的索引處以獲得最終的數(shù)組：


const 

    groups = [{ _id: 1, q: 'why?', group: 'no-group', date: '8', selected: false }, { _id: 2, q: 'what?', group: 'group 1', date: '6', selected: false }, { _id: 3, q: 'when?', group: 'no-group', date: '7', selected: false }, { _id: 4, q: 'where?', group: 'group 1', date: '5', selected: false }, { _id: 5, q: 'which?', group: 'group 2', date: '3', selected: false }, { _id: 6, q: 'who?', group: 'no-group', date: '0', selected: false }, { _id: 7, q: 'why not?', group: 'group 2', date: '9', selected: false }, { _id: 8, q: 'who, me?', group: 'group 1', date: '4', selected: false }, { _id: 9, q: 'where is waldo?', group: 'group 1', date: '1', selected: false }, { _id: 10, q: 'which way is up?', group: 'no-group', date: '2', selected: false }, { _id: 11, q: 'when is lunch?', group: 'group 2', date: '10', selected: false }];

    

const sort = (groups) => {

  const groupSorted = groups.sort((a, b) => a.date - b.date)

    .reduce((r, o, i) => {

      if (o.group !== 'no-group') {

        r[o.group] = (r[o.group] || []).concat(o);

        //insert the index of the group

        r.index[o.group] = r.index[o.group] || i;

      } else {

        r.res.push(o)

      }

      return r;

    }, {

      index: {},

      res: []

    });

  Object.entries(groupSorted.index).forEach(([group, idx]) => {

    groupSorted.res.splice(idx, 0, groupSorted[group])

  });

  return groupSorted.res.flat();

}



console.log(sort(groups))


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