Question

鉴于以下集合，将这些数据合并为按房间分组并使“名称”成为数组的最佳方法是什么;不包括重复项，并且如果有更多字段，则不希望使用键“name”来使其更具动态性：

var roomAssignments = [{
    room: 'Foo',
    name: 'Fooname',
    other: 'Other'
},{
    room: 'Bar',
    name: 'Barname',,
    other: 'OtherBar'
},{
    room: 'Foo',
    name: 'Baz',,
    other: 'Other'
},{
    room: 'Foo',
    name: 'Baz',,
    other: 'Other'
},{
    room: 'Foo',
    name: 'Bat',,
    other: 'Other'
}];

期望的输出：

[{
    room: 'Foo',
    name: [ 'Fooname', 'Baz', 'Bat' ],
    other: ['Other']
}, {
    room: 'Bar',
    name: [ 'Barname' ],
    other: ['OtherBar']
}]

我目前正在使用lodash并且更喜欢那个或普通的javascript。我想我已经看了太长时间，而且我有大约30个需要组合成数组的键，我正在寻找最有效的方法将所有键动态组合成数组。

Answer 1

您可以使用函数reduce和函数includes来放弃重复值。

＆＃13;

var roomAssignments = [{    room: 'Foo',    name: 'Fooname',    other: 'Other'},{    room: 'Bar',    name: 'Barname',    other: 'OtherBar'},{    room: 'Foo',    name: 'Baz',    other: 'Other'},{    room: 'Foo',    name: 'Baz',    other: 'Other'},{    room: 'Foo',    name: 'Bat',    other: 'Other'}],
    result = Object.values(roomAssignments.reduce(function(a, c) {
  if (a[c.room]) {
    Object.keys(c).forEach(function(k) {
      if (k === 'room') return;      
      if (a[c.room][k]) {
        if (!a[c.room][k].includes(c[k])) a[c.room][k].push(c[k]);
      } else a[c.room][k] = [c[k]];
    });
  } else { 
    a[c.room] = { room: c.room };
    Object.keys(c).forEach(function(k) {
      if (k === 'room') return;      
      a[c.room][k] = [c[k]];
    });
  }
  
  return a
}, {}));

console.log(result);

＆＃13;

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

＆＃13;

Answer 2

“高效”是一种主观的。以下代码相当有效，但更重要的是，它看起来更好。

var rooms = {};
roomAssignments.forEach(assignment => {
    var title = assignment.room;

    // grab existing room from our rooms map
    // if it doesn't exist, create it
    var room = rooms[title] || (rooms[title] = {room: title});

    for(var k in assignment){

      // skip the room key (it's our room title)
      if(k === 'room') continue;

      var value = assignment[k];

      // grab the existing field from the room object
      // if it doesn't exist, create it
      var field = room[k] || (room[k] = []);

      // see if the value already exists
      // if not, push it
      if(field.indexOf(value) === -1) field.push(value);
    }
})

结合这个系列最有效的方法是什么？

2 个答案: