如何根據javascript中另一個數組的值映射和過濾值數組？

Question

這里我有兩個數組。

一個數組將保存 ID，另一個數組保存數據。 我想通過使用 Id 數組中的值進行過濾來映射數據中的值。

例如：如果 Ids 數組中的 6 與 data.id 匹配，則意味着我需要獲取 assets 數組中的值。 我已經完成了那個解決方案。 但是我的解決方案似乎很難理解。

const ids = [
    6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6,
    6, 5, 4, 4, 4
  ];
  
  const data = [
    {
      id: 10,
      assets: [
        {
          type: 'First Section',
          value: 60
        },
        {
          type: 'Second Section',
          value: 40
        }
      ]
    },
    {
      id: 11,
      assets: [
        {
          type: 'First Section',
          value: 65
        },
        {
          type: 'Second Section',
          value: 35
        }
      ]
    },
    {
      id: 12,
      assets: [
        {
          type: 'First Section',
          value: 70
        },
        {
          type: 'Second Section',
          value: 30
        }
      ]
    },
    {
      id: 13,
      assets: [
        {
          type: 'First Section',
          value: 75
        },
        {
          type: 'Second Section',
          value: 25
        }
      ]
    },
    {
      id: 14,
      assets: [
        {
          type: 'First Section',
          value: 80
        },
        {
          type: 'Second Section',
          value: 20
        }
      ]
    },
    {
      id: 15,
      assets: [
        {
          type: 'First Section',
          value: 85
        },
        {
          type: 'Second Section',
          value: 15
        }
      ]
    },
    {
      id: 16,
      assets: [
        {
          type: 'First Section',
          value: 90
        },
        {
          type: 'Second Section',
          value: 10
        }
      ]
    },
    {
      id: 2,
      assets: [
        {
          type: 'First Section',
          value: 20
        },
        {
          type: 'Second Section',
          value: 80
        }
      ]
    },
    {
      id: 3,
      assets: [
        {
          type: 'First Section',
          value: 25
        },
        {
          type: 'Second Section',
          value: 75
        }
      ]
    },
    {
      id: 4,
      assets: [
        {
          type: 'First Section',
          value: 30
        },
        {
          type: 'Second Section',
          value: 70
        }
      ]
    },
    {
      id: 5,
      assets: [
        {
          type: 'First Section',
          value: 35
        },
        {
          type: 'Second Section',
          value: 65
        }
      ]
    },
    {
      id: 6,
      assets: [
        {
          type: 'First Section',
          value: 40
        },
        {
          type: 'Second Section',
          value: 60
        }
      ]
    },
    {
      id: 7,
      assets: [
        {
          type: 'First Section',
          value: 45
        },
        {
          type: 'Second Section',
          value: 55
        }
      ]
    },
    {
      id: 8,
      assets: [
        {
          type: 'First Section',
          value: 50
        },
        {
          type: 'Second Section',
          value: 50
        }
      ]
    },
    {
      id: 9,
      assets: [
        {
          type: 'First Section',
          value: 55
        },
        {
          type: 'Second Section',
          value: 45
        }
      ]
    },
    {
      id: 1,
      assets: [
        {
          type: 'First Section',
          value: 15
        },
        {
          type: 'Second Section',
          value: 85
        }
      ]
    }
  ];

在這里，我只需要獲取 assets 數組中的第一個部分值。

我已經通過使用下面的代碼做到了這一點

 let res = ids.map((val) => data.filter(v => v.id === val)[0]).map(v => v.assets.filter(v => v.type === 'First Section')[0]).map(v => v.value)

console.log(res)

輸出

[
  40, 40, 40, 40, 40, 40, 40, 40, 40, 40,
  40, 40, 40, 40, 40, 40, 40, 40, 40, 40,
  40, 40, 40, 40, 40, 40, 40, 40, 40, 40,
  40, 40, 40, 40, 35, 30, 30, 30
]

但是有什么方法可以進一步優化功能。 請指導我

Answer 1

您正確地感覺到您的算法雖然有效，但效率極低。 您必須在整個數據數組中搜索 ids 中的每個項目，這使其成為 O(n^m) 時間復雜度，根本無法很好地擴展。

我認為最好和最有效的方法是為數據列表中的每個項目創建一個 id -> first section 值的映射。 從那里，獲取任何 ID 的部分值是一個非常簡單且快速的查找。 這是可以做到的。 它只需要 1 次通過數據數組，1 次通過 ids 數組 - O(n + m)，如下所示：

 const ids = [ 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 5, 4, 4, 4 ]; const data = [{"id":10,"assets":[{"type":"First Section","value":60},{"type":"Second Section","value":40}]},{"id":11,"assets":[{"type":"First Section","value":65},{"type":"Second Section","value":35}]},{"id":12,"assets":[{"type":"First Section","value":70},{"type":"Second Section","value":30}]},{"id":13,"assets":[{"type":"First Section","value":75},{"type":"Second Section","value":25}]},{"id":14,"assets":[{"type":"First Section","value":80},{"type":"Second Section","value":20}]},{"id":15,"assets":[{"type":"First Section","value":85},{"type":"Second Section","value":15}]},{"id":16,"assets":[{"type":"First Section","value":90},{"type":"Second Section","value":10}]},{"id":2,"assets":[{"type":"First Section","value":20},{"type":"Second Section","value":80}]},{"id":3,"assets":[{"type":"First Section","value":25},{"type":"Second Section","value":75}]},{"id":4,"assets":[{"type":"First Section","value":30},{"type":"Second Section","value":70}]},{"id":5,"assets":[{"type":"First Section","value":35},{"type":"Second Section","value":65}]},{"id":6,"assets":[{"type":"First Section","value":40},{"type":"Second Section","value":60}]},{"id":7,"assets":[{"type":"First Section","value":45},{"type":"Second Section","value":55}]},{"id":8,"assets":[{"type":"First Section","value":50},{"type":"Second Section","value":50}]}]; // for each item in data, map id to first section value for easy lookup let sectionMap = data.reduce((res, curr) => { res[curr.id] = curr.assets.find(asset => asset.type === 'First Section').value; return res; }, {}); console.log(sectionMap); // map the id to its respective first section value with the lookup map we created let result = ids.map(id => sectionMap[id]); console.log(result);