Question

如何合并此对象数组

[
    {
        substoreUuid: '1450215d-e1a4-454d-8042-014d0f70f01e',
        storeUuid: '6627a93e-1e16-4e08-9057-44ce701ba169'
    }
]

到这个对象数组，其中键名是相同的

[
    {
        acceptanceTaskDetailsUuid: '080d7831-9cab-4bf4-af92-bae7b75bd50d',
        storeUuid: '6627a93e-1e16-4e08-9057-44ce701ba169',
        quantity: 2,
        uuid: '799898cc-f74a-42d9-9fa2-84da238d6b39'
    },
    {
        acceptanceTaskDetailsUuid: '080d7831-9cab-4bf4-af92-bae7b75bd50d',
        storeUuid: 'a0e64311-2410-48be-b97d-17c68a0ad2a1',
        quantity: 1,
        uuid: '211c626c-e8f7-40d7-b42a-690e2fa0082b'
    }
]

如您所见，我需要合并来自两个不同数组的对象。

预期产出：

[
    {
        substoreUuid: '1450215d-e1a4-454d-8042-014d0f70f01e', 
        quantity: 2, 
        uuid: '799898cc-f74a-42d9-9fa2-84da238d6b39',
        storeUuid: '6627a93e-1e16-4e08-9057-44ce701ba169',
    }
]

这是我到目前为止所拥有的

const substoreArr = [
    {
        substoreUuid: '1450215d-e1a4-454d-8042-014d0f70f01e',
        storeUuid: '6627a93e-1e16-4e08-9057-44ce701ba169'
    }
];

const distributionStore = [
    {
        acceptanceTaskDetailsUuid: '080d7831-9cab-4bf4-af92-bae7b75bd50d',
        storeUuid: '6627a93e-1e16-4e08-9057-44ce701ba169',
        quantity: 2,
        uuid: '799898cc-f74a-42d9-9fa2-84da238d6b39'
    },
    {
        acceptanceTaskDetailsUuid: '080d7831-9cab-4bf4-af92-bae7b75bd50d',
        storeUuid: 'a0e64311-2410-48be-b97d-17c68a0ad2a1',
        quantity: 1,
        uuid: '211c626c-e8f7-40d7-b42a-690e2fa0082b'
     }
];

const newArr = distributionStore.map(diststore => {
   for (const sub of substoreArr) {
     if (sub.store_uuid === diststore.storeUuid) {
        return diststore;
     }
   }
}); //output [undefined, undefined]

输出将删除不相关的storeUuid并删除一些密钥。如何以更快的方式将其合并到vanilla js中。

Answer 1

这是我到目前为止所学到的。我使用g++ -dM -E foo.cpp > defines.h映射到相同的密钥

reduce

欢迎任何改进

Answer 2

我已经为您的问题找到了可行的解决方案。您可能应该重构此代码并将操作拆分为单独的函数。

 const result = substoreArr.reduce((acc, val) => {
    const filtered = distributionStore
        .filter(e => e.storeUuid === val.storeUuid)
        .map(e => ({...e, ...val}));

    if (filtered.length > 0) {
        return [...acc, ...filtered];
    } else {
        return acc;
    }
}, []);

Answer 3

首先按storeUuid过滤结果，然后映射结果，只获取必需的属性。所有代码都是vanilla Javascript ES6。

const substoreArr = [
    {
        substoreUuid: '1450215d-e1a4-454d-8042-014d0f70f01e',
        storeUuid: '6627a93e-1e16-4e08-9057-44ce701ba169'
    }
];

const distributionStore = [
    {
        acceptanceTaskDetailsUuid: '080d7831-9cab-4bf4-af92-bae7b75bd50d',
        storeUuid: '6627a93e-1e16-4e08-9057-44ce701ba169',
        quantity: 2,
        uuid: '799898cc-f74a-42d9-9fa2-84da238d6b39'
    },
    {
        acceptanceTaskDetailsUuid: '080d7831-9cab-4bf4-af92-bae7b75bd50d',
        storeUuid: 'a0e64311-2410-48be-b97d-17c68a0ad2a1',
        quantity: 1,
        uuid: '211c626c-e8f7-40d7-b42a-690e2fa0082b'
     }
];

const newArr = distributionStore
  .filter(x => substoreArr.find(i => i.storeUuid === x.storeUuid))
  .map(({ storeUuid, quantity, uuid }) => ({ storeUuid, quantity, uuid, ...substoreArr.find(i => i.storeUuid === storeUuid) }));

console.log(newArr);

Answer 4

通过使用storeUuid作为键来构建子目录查找，可以实现有效的解决方案，以避免嵌套循环。

具有嵌套循环意味着您必须比较第一个数组中的每个项目，以及第二个数组中的每个项目，这将花费与n1 * n2成比例的时间。但是如果你首先构建了substore的hashmap，你最终会做两个独立的循环，所以所花费的时间与n1 + n1成正比，对于更大的数组来说，这将更快。

＆＃13;

const substoreArr = [
    {
        substoreUuid: '1450215d-e1a4-454d-8042-014d0f70f01e',
        storeUuid: '6627a93e-1e16-4e08-9057-44ce701ba169'
    }
];

const distributionStore = [
    {
        acceptanceTaskDetailsUuid: '080d7831-9cab-4bf4-af92-bae7b75bd50d',
        storeUuid: '6627a93e-1e16-4e08-9057-44ce701ba169',
        quantity: 2,
        uuid: '799898cc-f74a-42d9-9fa2-84da238d6b39'
    },
    {
        acceptanceTaskDetailsUuid: '080d7831-9cab-4bf4-af92-bae7b75bd50d',
        storeUuid: 'a0e64311-2410-48be-b97d-17c68a0ad2a1',
        quantity: 1,
        uuid: '211c626c-e8f7-40d7-b42a-690e2fa0082b'
     }
];

// build an object (or hash) for substores, where the key is the storeUuid for fast lookup
const substoresByStoreUuid = {}

substoreArr.forEach(substore => 
  substoresByStoreUuid[substore.storeUuid]=substore
)


const result = distributionStore
  // remove items that don't have a matching substore
  .filter(store => substoresByStoreUuid[store.storeUuid])
  // return the merged properties
  .map(({quantity, uuid, storeUuid}) => ({ 
    quantity, 
    uuid, 
    storeUuid, 
    substoreUuid: substoresByStoreUuid[storeUuid].substoreUuid
  }))

＆＃13;

使用相同的键合并来自不同数组的对象

4 个答案: