过滤和克隆对象属性

Question

我有 2 个对象数组：itemsList 和 itemsFetched。每个数组中的所有对象都具有相同的结构（nr 个键/值）。其中一个键具有相同的“含义”，但名称不同（itemsList 上的 item_id，itemsFetched 上的 id ）。它们的值是相同的。

我需要过滤 itemsList 数组，只留下 item_id 值等于 上 id 值的对象itemsFetched。然后将 itemsFetched 数组（与 item_id=id 匹配）上每个对象的键/值 count 复制（添加）到过滤后的数组。

我有一个工作代码，但我确信这不是解决这个问题的最佳方法。我之前已经问过类似的问题（关于“过滤器”部分）解决了我的问题，但由于我必须在过滤后添加“计数”部分，我最终重构了整个事情。

itemsList（示例）

[
  {
      "id": 0,
      "name": "Egg",
      "img": "http://www.serebii.net/pokemongo/items/egg.png"
  },
  {
      "id": 1,
      "name": "Pokeball",
      "img": "http://www.serebii.net/pokemongo/items/20pokeballs.png"
  },
  {
      "id": 2,
      "name": "Greatball",
      "img": "http://www.serebii.net/pokemongo/items/greatball.png"
  },
  { 
      "id": 401,
      "name": "Incense",
      "img": "http://www.serebii.net/pokemongo/items/incense.png" 
  },
  { 
      "id": 901,
      "name": "Incubator (Unlimited)",
      "img": "http://www.serebii.net/pokemongo/items/eggincubator.png"
  }
]

itemsFetched（示例）

[ 
  {
    "item_id": 1, 
    "count": 50,
    "unseen": true
  },
  {
    "item_id": 401,
    "count": 2, 
    "unseen": true
  },
  { 
    "item_id": 901, 
    "count": 1, 
    "unseen": true 
  }
]

resultArray（到底想要什么）

[
  { 
    "id": 1,
    "name": "Pokeball",
    "count": 50,
    "img": "http://www.serebii.net/pokemongo/items/20pokeballs.png",
  },
  { 
    "id": 401,
    "name": "Incense",
    "count": 2,
    "img": "http://www.serebii.net/pokemongo/items/incense.png"
  },
  { 
    "id": 901,
    "name": "Incubator (Unlimited)",
    "count": 1,
    "img": "http://www.serebii.net/pokemongo/items/eggincubator.png"
  }
]

我当前的代码（工作）

let arr = [];
itemsFetched.forEach((item) => {
  itemsList.forEach((item2) => {
    if (item.item_id === item2.id) {
      arr.push({
        "id": item.item_id,
        "name": item2.name,
        "count": item.count,
        "img": item2.img
      });
    }
  });
});

PS：我能够使用 ES6/7 语法/功能。

Answer 1

你可以使用hash map来复杂地减少时间，你的算法是O(m*n)，下面是O(m+n+r)

const itemsMap = itemsList.reduce((map, item) => {
    map[item.id] = item
    return map
}, {})
const results = itemsFetched
    .filter((item) => itemsMap.hasOwnProperty(item.item_id))
    .map((item) => ({
        id: item.item_id,
        name: itemsMap[item.item_id].name,
        count: item.count,
        img: itemsMap[item.item_id].img,
    }))

Answer 2

将for ... of 循环（ES6 功能）与Array#map 结合使用。

这使您在第一次找到匹配项时更容易return 合并对象，这是一种逻辑优化，因为两个列表都不应包含多个具有给定id 的条目。

const result = itemsFetched.map(data => {
  for (let item of itemsList) {
    if (data.item_id === item.id) {
      return {
        id: item.id,
        name: item.name,
        count: data.count,
        img: item.img
      }
    }
  }
})

片段：

const itemsList = [{
  "id": 0,
  "name": "Egg",
  "img": "http://www.serebii.net/pokemongo/items/egg.png"
}, {
  "id": 1,
  "name": "Pokeball",
  "img": "http://www.serebii.net/pokemongo/items/20pokeballs.png"
}, {
  "id": 2,
  "name": "Greatball",
  "img": "http://www.serebii.net/pokemongo/items/greatball.png"
}, {
  "id": 401,
  "name": "Incense",
  "img": "http://www.serebii.net/pokemongo/items/incense.png"
}, {
  "id": 901,
  "name": "Incubator (Unlimited)",
  "img": "http://www.serebii.net/pokemongo/items/eggincubator.png"
}]

const itemsFetched = [{
  "item_id": 1,
  "count": 50,
  "unseen": true
}, {
  "item_id": 401,
  "count": 2,
  "unseen": true
}, {
  "item_id": 901,
  "count": 1,
  "unseen": true
}]


const result = itemsFetched.map(data => {
  for (let item of itemsList) {
    if (data.item_id === item.id) {
      return {
        id: item.id,
        name: item.name,
        count: data.count,
        img: item.img
      }
    }
  }
})

console.log(result)

Answer 3

一种改进方法是使用for..of 语句而不是forEach 用于内部循环。一旦 id 匹配，这有助于打破循环。没有直接的方法可以脱离 forEach 方法。

let arr = [];
itemsFetched.forEach((item) => {
  for (let item2 of itemsList) {
    if (itemsFetched.item_id === itemsList.id) {
      arr.push({
        "id": itemsFetched.item_id,
        "name": itemsList.name,
        "count": itemsFetched.count,
        "img": itemsList.img
      });
      break;
    }
  }
});

Answer 4

像这样？

var itemsList = [
  {
      "id": 0,
      "name": "Egg",
      "img": "http://www.serebii.net/pokemongo/items/egg.png"
  },
  {
      "id": 1,
      "name": "Pokeball",
      "img": "http://www.serebii.net/pokemongo/items/20pokeballs.png"
  },
  {
      "id": 2,
      "name": "Greatball",
      "img": "http://www.serebii.net/pokemongo/items/greatball.png"
  },
  { 
      "id": 401,
      "name": "Incense",
      "img": "http://www.serebii.net/pokemongo/items/incense.png" 
  },
  { 
      "id": 901,
      "name": "Incubator (Unlimited)",
      "img": "http://www.serebii.net/pokemongo/items/eggincubator.png"
  }
];

var itemsFetched = [ 
  {
    "item_id": 1, 
    "count": 50,
    "unseen": true
  },
  {
    "item_id": 401,
    "count": 2, 
    "unseen": true
  },
  { 
    "item_id": 901, 
    "count": 1, 
    "unseen": true 
  }
]



let arr = [];
itemsFetched.forEach((item) => {
  itemsList.forEach((item2) => {
    if (item.item_id == item2.id) {
      arr.push({
        "id": item.item_id,
        "name": item2.name,
        "count": item.count,
        "img": item2.img
      });
    }
  });
});

console.log(arr);