如何使用两个属性对对象数组进行排序，但根据条件按排序顺序有条件地分组？

Question

这个问题是这个问题的延续。

假设我有一个如下所示的数组：

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 }, 

];

如何编写代码对其进行排序，以便按日期和组对对象进行排序。'no-group'但是，如果：假设组 1 作为第二个对象出现，则以下对象应为按日期排序的组 1，直到列出所有组 1 对象，并且对于除具有组属性值的对象之外的所有其他对象都相同。

因此，对于上面的数组，我应该得到如下输出：

[

  {_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.

Answer 1

您可以通过date对相同组取消对象的排序、分组并获得平面数组。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; }

Answer 2

sort数组由datereduce带有累加器的数组Map。如果它是一个项目，请使用唯一的键no-group添加一个新条目。Map_id如果它是一个有效的项目group，则使用group作为键和具有相同的项目数组group。每个都no-group将作为键添加到 Map 中。其余项目将根据该项目的第一个条目添加group。获取values地图并将其击倒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 )

Answer 3

我们可以date先对它进行排序，然后对对象进行分组，group并将组保存在一个对象中，同时保存组的索引。然后我们最终可以将组插入到正确的索引处以获得最终的数组：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))