过滤器和克隆对象属性

Question

我有2个对象数组：itemsList和itemsFetched。每个数组内的所有对象都具有相同的结构（键/值的数量）。其中一个按键具有相同的 '意义'，但不同的名称（ITEM_ID上itemsList，ID上itemsFetched）。他们的价值观是一样的。

我需要过滤itemsList阵列，只留下那有ITEM_ID值等于ID值上itemsFetched的对象。然后从itemsDetched数组（其与item_id = id匹配）复制（添加）键/值count到过滤数组。

我有一个工作代码，但我确定它不是解决此问题的最佳方法。我已经问过类似的问题了（关于'filter'部分）解决了我的问题，但由于我必须在过滤之后添加'count'部分，所以我最终重构了整个问题。

itemsList（样品）

[ 
    { 
     "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（样品）

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

resultArray（我想要的到底）

[ 
    { 
    "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" 
    } 
] 

我现在的代码（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：我可以使用ES6/7语法/功能。

Answer 1

您可以使用哈希表来减少时间complexitly，你的算法是O(m*n)，后续是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

是否这样？

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);

Answer 3

一种改善的方法是使用for..of声明，而不是forEach为内循环。一旦id匹配，这有助于打破循环。没有直接的方法可以打破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

结合使用for ... of环路（一个ES6功能）与Array#map。

第一次找到匹配项时returnreturn合并对象变得更容易，这是一个逻辑优化，因为这两个列表都不应包含给定的id以上的一个条目。

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 
     } 
    } 
    } 
}) 

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)