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如何基于2个属性删除阵列中的重复对象?

ecmascript-6 filter arrays javascript

Stephan-v · 技术社区 · 7 年前

room_rate_type_id 属性:

const rooms = [{
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 189
  },
  {
    room_rate_type_id: 190,
    price: 200
  }
];

const newRooms = rooms.filter((room, index, array) => {
  const roomRateTypeIds = rooms.map(room => room.room_rate_type_id);
  // Returns the first index found.
  return roomRateTypeIds.indexOf(room.room_rate_type_id) === index;
});

console.log(newRooms);

但是,我还需要确保只有当对象的 房间价格类型id

6 回复 | 直到 7 年前

marvel308 7 年前

你能做到的

const rooms = [
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 189
  },
  {
    room_rate_type_id: 190,
    price: 200
  }
];

let result = rooms.filter((e, i) => {
    return rooms.findIndex((x) => {
    return x.room_rate_type_id == e.room_rate_type_id && x.price == e.price;}) == i;

});

console.log(result);

这将过滤除任何对象第一次出现之外的所有重复项

Ori Drori 7 年前

reduce 数组到 Map Map#values 返回阵列:

const rooms = [{
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 189
  },
  {
    room_rate_type_id: 190,
    price: 200
  }
];

const newRooms = [...rooms.reduce((m, r) => {
  const key = `${r.room_rate_type_id}-${r.price}`; // create the key by combining both props
  return m.has(key) ? m : m.set(key, r); // if key exists skip, if not add to map
}, new Map()).values()]; // get the map values and convert back to array

console.log(newRooms);

T.J. Crowder 7 年前

const newRooms = rooms.filter((room, index) => {
  // Only include this room if there isn't another room earlier
  // in the array that has the same values
  return !rooms.some((r, i) =>
    i < index &&
    r.room_rate_type_id == room.room_rate_type_id &&
    r.price == room.price
  );
});

const rooms = [{
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 189
  },
  {
    room_rate_type_id: 190,
    price: 200
  }
];

const newRooms = rooms.filter((room, index) => {
  // Only include this room if there isn't another room earlier
  // in the array that has the same values
  return !rooms.some((r, i) =>
    i < index &&
    r.room_rate_type_id == room.room_rate_type_id &&
    r.price == room.price
  );
});

console.log(newRooms);

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

如果数组真的很大,那么效率会很低,最好记住以前看到的组合,而不是不断地重新搜索数组:

const seenRooms = Object.create(null);
const newRooms = rooms.filter((room, index) => {
  const key = room.room_rate_type_id + "**" + room.price;
  if (seenRooms[key]) {
    return false;
  }
  seenRooms[key] = true;
  return true;
});

const rooms = [{
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 189
  },
  {
    room_rate_type_id: 190,
    price: 200
  }
];

const seenRooms = Object.create(null);
const newRooms = rooms.filter((room, index) => {
  const key = room.room_rate_type_id + "**" + room.price;
  if (seenRooms[key]) {
    return false;
  }
  seenRooms[key] = true;
  return true;
});

console.log(newRooms);

.作为控制台包装{
最大高度:100%!重要的
}

RomanPerekhrest 7 年前

简单方法:使用 room_rate_type_id price

const rooms = [
    {room_rate_type_id: 202,price: 200},{room_rate_type_id: 202,price: 200},{room_rate_type_id: 202,price: 189},{room_rate_type_id: 190,price: 200}
];

const roomRateKeys = [];
const newRooms = rooms.filter((r, i, a) => {
    var k = r.room_rate_type_id + "" + r.price;
    if (roomRateKeys.indexOf(k) === -1) {
        roomRateKeys.push(k);
	return r;
    }
});

console.log(newRooms);

Olian04 7 年前

这可以做到:

const rooms = [{
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 189
  },
  {
    room_rate_type_id: 190,
    price: 200
  }
];

const newRooms = rooms.reduce((rooms, room) => {
  let l = rooms.filter(r => {
    return r.room_rate_type_id === room.room_rate_type_id && r.price === room.price;
  });
  if (l.length === 0) {
    return [...rooms, room]
  }
  return rooms;
}, [rooms[0]]);

console.log(newRooms);

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

Marko Savic 7 年前

const rooms = [
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 200
  },
  {
    room_rate_type_id: 202,
    price: 189
  },
  {
    room_rate_type_id: 190,
    price: 200
  }
];

let newRooms = rooms.filter((x, i, arr) => arr.findIndex(y => y.room_rate_type_id === x.room_rate_type_id && y.price === x.price) === i);


console.log(newRooms);