使用 Lodash 进行深度合并

Question

我有两个对象数组，它们包含带有标签和实际地址的地址：

var originalAddresses = [
  {
    label: 'home',
    address: { city: 'London', zipCode: '12345' }
  },
  {
    label: 'work',
    address: { city: 'New York', zipCode: '54321' }
  }
];

var updatedAddresses = [
  {
    label: 'home',
    address: { city: 'London (Central)', country: 'UK' }
  },
  {
    label: 'spain',
    address: { city: 'Madrid', zipCode: '55555' }
  }
];

现在我想按 label 合并这些数组，并比较地址的各个属性，只合并新地址中实际存在的属性。因此，结果应该像这样：

var result = [
  {
    label: 'home',
    address: { city: 'London (Central)', zipCode: '12345', country: 'UK' }
  },
  {
    label: 'work',
    address: { city: 'New York', zipCode: '54321' }
  },
  {
    label: 'spain',
    address: { city: 'Madrid', zipCode: '55555' }
  }
]

如何使用 lodash 实现此操作？ 我尝试了unionBy()和merge()的组合。使用 unionBy() 函数，我可以按标签比较和合并数组，但这总是替换整个对象。虽然可以合并地址，但是这不是以标签为基础进行的。

- benjiman

Answer 1

您可以使用_.keyBy(arr, 'label')将这两个数组转换为对象，然后使用_.merge()进行深度合并：

var originalAddresses = [{
  label: 'home',
  address: {
    city: 'London',
    zipCode: '12345'
  }
}, {
  label: 'work',
  address: {
    city: 'New York',
    zipCode: '54321'
  }
}];

var updatedAddresses = [{
  label: 'home',
  address: {
    city: 'London (Central)',
    country: 'UK'
  }
}, {
  label: 'spain',
  address: {
    city: 'Madrid',
    zipCode: '55555'
  }
}];

var result = _.values(_.merge(
  _.keyBy(originalAddresses, 'label'),
  _.keyBy(updatedAddresses, 'label')
));

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.15.0/lodash.min.js"></script>

- Ori Drori

Answer 2

我尝试使用VanillaJS来处理这个问题。

const originalAddresses = [{
    label: 'home',
    address: {
        city: 'London',
        zipCode: '12345'
    }
}, {
    label: 'work',
    address: {
        city: 'New York',
        zipCode: '54321'
    }
}];

const updatedAddresses = [{
    label: 'home',
    address: {
        city: 'London (Central)',
        country: 'UK'
    }
}, {
    label: 'spain',
    address: {
        city: 'Madrid',
        zipCode: '55555'
    }
}];


const groupBy = (array, property) => {
    return array.reduce((acc, cur) => {
            let key = cur[property]
            if (!acc[key]) {
                acc[key] = []
            }
            acc[key].push(cur)

            return acc
        }
        , {})
}

const groupByLabel = groupBy([...originalAddresses, ...updatedAddresses], 'label')

const result  = Object.keys(groupByLabel).map((key) => {
        return {
            label: groupByLabel[key][0].label,
            address: groupByLabel[key].reduce((acc, cur) => {
                    return Object.assign(acc, cur.address)
                }
                , {})
        }
    }
)
console.log(result)

- panchihpeng

---

groupBy是通用的，可以接受任何输入，但结果不是。它与此数据组合相关联。建议将其制作成接受最终方法的任何输入。 - Imants Volkovs