javascript 使用 ids 列表从嵌套数组中提取对象

Question

我有一个 id 列表作为引用，并且我有一个对象，其中包含多个具有对象数组的对象。

我想以最快的方式在列表中创建一个具有相应 id 的对象数组。

const data = {
  "items": {
    "item1": [{
        "id": "id1",
        "info": "info1"
      },
      {
        "id": "id2",
        "info": "info22"
      }
    ],
    "item20": [{
      "id": "id3",
      "info": "info5"
    }],
    "item5": [{
        "id": "id4",
        "info": "info6"
      },
      {
        "id": "id5",
        "info": "info7"
      }
    ]
  }
};

const keep = ['id4', 'id2'];
const results = [];
keep.forEach(function(val) {

  const match = Object.keys(data.items).map(item => {
    return data.items[item].find(obj => obj.id === val)
  });
  results.push(match)
})
console.log('final: ', results)

当前没有返回我想要的。预期结果将是:

[
    {
      "id": "id2",
      "info": "info22"
    },
    {
        "id": "id4",
        "info": "info6"
    }
]

更新:

如果数据本身就是一个对象数组，并且我们希望对每个对象执行相同的操作，情况又如何？

    const data = [{
      "otherStuff": "otherB",
      "items": {
        "item1": [{
            "id": "id1",
            "info": "info1"
          },
          {
            "id": "id2",
            "info": "info22"
          }
        ],
        "item20": [{
          "id": "id3",
          "info": "info5"
        }],
        "item5": [{
            "id": "id4",
            "info": "info6"
          },
          {
            "id": "id5",
            "info": "info7"
          }
        ]
      }
    }, {
      "otherStuff": "otherA",
      "items": {
        "item1": [{
            "id": "id1",
            "info": "info10000"
          },
          {
            "id": "id2",
            "info": "info220000"
          }
        ],
        "item20": [{
          "id": "id3",
          "info": "info5000"
        }],
        "item5": [{
            "id": "id4",
            "info": "info60000"
          },
          {
            "id": "id5",
            "info": "info7000"
          }
        ]
      }
    }];


const keep = ['id4', 'id2'];
const results = [];

    keep.forEach(function(val) {
      data.forEach(function(entry){
          Object.keys(entry.items).forEach(item => {
            var match = entry.items[item].find(obj => obj.id === val);
            if (match) {
              results.push(match)
            }
          });
      });
    })
    
    console.log(results)

输出应该是:

[
    {
        "otherStuff": "otherB",
        "items": [
                {
                      "id": "id2",
                      "info": "info22"
                },
                {
                    "id": "id4",
                    "info": "info6"
            }
        ]
    },
    {
        "otherStuff": "otherA",
        "items": [
                {
                      "id": "id2",
                      "info": "info220000"
                },
                {
                    "id": "id4",
                    "info": "info60000"
            }
        ]
    }
]

结果并不一样。

Answer 1

如果使用 Set 来保存 id，则可以节省一次 O(n) 遍历:

 const keep = new Set(['id4', 'id2']);

 const result = [];

 for(const items of Object.values(data.items))
   for(const item of items)
     if(keep.has(item.id))
       result.push(item);