javascript - 递归比较对象并将重复的键值放入数组

Question

我需要迭代三个对象以检查重复项以最终得到一个对象。如果有重复项，我想将它们放入一个数组中，以便用户可以选择最佳值，以便它们最终都以字符串或整数(或带有字符串/整数的对象)结束。它们可能看起来像这样:

const a = {
  first_name: 'Tom',
  height: {
    feet: 5,
    inches: 0
  }
}
const b = {
  first_name: 'Thomas',
  last_name: 'Walsh',
  email: 'tomwalsh@domain.com',
  height: {
    feet: 6,
    inches: 0
  }
}
const c = {
  email: 'tomwalsh@sample.edu'
}

结果是这样的:

const result = {
  first_name: ['Tom', 'Thomas'],
  last_name: 'Walsh',
  email: ['tomwalsh@domain.com', 'tomwalsh@sample.edu'],
  height: {
    feet: [5, 6],
    inches: 0
  }
}

我不知道 a、b 或 c 是否是 key 的权限，所以我必须假设一组未知的键/值对，但绝对很小(不超过 20 对，只有 2-3 对超过 3 层深，并且每个层最多有 4-5 个键/值对。

我可以创建一个新对象，递归地迭代这三个对象，然后转储值或创建一个值数组，但这似乎效率不高。有什么建议吗？

如果需要，我在项目中加入了 lodash。谢谢!

Answer 1

由于某些对象中可能缺少嵌套键，您可以通过 lodash 的 _.mergeWith() 合并它们，并将重复项收集到数组中:

const a = {"first_name":"Tom","height":{"feet":5,"inches":0}}
const b = {"first_name":"Thomas","last_name":"Walsh","email":"tomwalsh@domain.com","height":{"feet":6,"inches":0}}
const c = {"email":"tomwalsh@sample.edu"}

const shouldCollect = (s) => _.negate(_.overSome([
  _.isUndefined,
  _.isObject,
  _.partial(_.eq, s)
]))


const mergeDupes = (...args) => _.mergeWith({}, ...args, (o, s) => {
  if(_.isArray(o)) return _.uniq([...o, s]);
  if(shouldCollect(s)(o)) return [o, s];
})

const result = mergeDupes(a, b, c)

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>

如果你想删除不重复的属性，你可以通过递归 _.transform() 清理对象:

const a = {"first_name":"Tom","height":{"feet":5,"inches":0}}
const b = {"first_name":"Thomas","last_name":"Walsh","email":"tomwalsh@domain.com","height":{"feet":6,"inches":0}}
const c = {"email":"tomwalsh@sample.edu"}

const shouldCollect = (s) => _.negate(_.overSome([
  _.isUndefined,
  _.isObject,
  _.partial(_.eq, s)
]))

const omitNonDuplicates = obj =>
  _.transform(obj, (a, v, k) => {
    if (_.isArray(v)) a[k] = v;
    else if (_.isObject(v)) {
      const clean = omitNonDuplicates(v);
      if(!_.isEmpty(clean)) a[k] = clean;
      return;
    }
  });
  
const mergeDupes = (...args) => omitNonDuplicates(_.mergeWith({}, ...args, (o, s) => {
  if(_.isArray(o)) return _.uniq([...o, s]);
  if(shouldCollect(s)(o)) return [o, s];
}))

const result = mergeDupes(a, b, c)

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.min.js"></script>