如何编写将字典拆分为字典数组的递归 Python 函数？答案

【问题标题】：How can I write a recursive python function that splits a dictionary into an array of dictionaries?如何编写将字典拆分为字典数组的递归 Python 函数？
【发布时间】：2016-01-21 01:52:56
【问题描述】：

我想写一个递归函数：

arguments: d, dictionary

result: list of dictionaries

def expand_dictionary(d):
    return []

该函数递归地遍历字典并使用 _ 展平嵌套对象，此外它还将嵌套列表展开到数组中，并包含父标签。

考虑从文档创建关系模型。

这是一个输入和输出示例：

original_object = {
  "id" : 1,
  "name" : {
    "first" : "Alice",
    "last" : "Sample"
  },
  "cities" : [
    {
      "id" : 55,
      "name" : "New York"
    },
    {
      "id" : 60,
      "name" : "Chicago"
    }
  ],
  "teachers" : [
    {
      "id" : 2
      "name" : "Bob",
      "classes" : [
        {
          "id" : 13,
          "name" : "math"
        },
        {
          "id" : 16,
          "name" : "spanish"
        }
      ]
    }
  ]
}

expected_output = [
  {
    "id" : 1,
    "name_first" : "Alice",
    "name_last" : "Sample"
  },
  {
    "_parent_object" : "cities",
    "id" : 55,
    "name" : "New York"  
  },
  {
    "_parent_object" : "cities",
    "id" : 60,
    "name" : "Chicago"  
  },
  {
    "parent_object" :"teachers",
    "id" : 2,
    "name" : "Bob"
  },
  {
    "parent_object" :"teachers_classes",
    "id" : 13,
    "name" : "math"
  },
  {
    "parent_object" :"teachers_classes",
    "id" : 16,
    "name" : "spanish"
  }
]

目前用于展平的代码是：

def flatten_dictionary(d):
  def expand(key, value):
      if isinstance(value, dict):
        return [ (key + '_' + k, v) for k, v in flatten_dictionary(value).items() ]
      else:
        #If value is null or empty array don't include it
        if value is None or value == [] or value == '':
          return []
        return [ (key, value) ]

  items = [ item for k, v in d.items() for item in expand(k, v) ]

  return dict(items)

【问题讨论】：

您的“original_obejct”是字典吗？那么第一行的 '"id" : 1' 后面应该有 'comma' 吗？
正确，为语法错误道歉。
看起来你在破坏信息：如果你有不止一位老师，你能从期望的输出中分辨出每个老师教什么课程吗？

标签： python arrays dictionary recursion

【解决方案1】：

这样就可以了

def expand_dictionary(d,name=None,l=None):
    obj = {}
    if l == None:
        l = [obj]
    else:
        l.append(obj)

    prefix = (name+'_'if name else '')
    if prefix: obj['_parent_object'] = name

    for i, v in d.iteritems():
        if isinstance(v, list):
            map(lambda x:expand_dictionary(x,prefix+i,l),v)
        elif isinstance(v, dict):
            obj.update(flatten_dictionary({i: v}))
        else:
            obj[i] = v
    return l

【讨论】：

没有得到正确的输出。我在下面发布了我一直在研究的解决方案。
除非您指的是排序，否则它的输出与所需的匹配。谢谢。

【解决方案2】：

经过一番努力，这就是我想出的。大概可以显着优化。根据@paulo-scardine 的评论，我添加了父主键以保留关系模型。很想听听优化的想法。

def expand_dictionary(original_object, object_name, objects=None):
  if objects is None:
    objects = []
  def flatten_dictionary(dictionary):

    def expand(key, value):
      if isinstance(value, dict):
        return [ (key + '_' + k, v) for k, v in flatten_dictionary(value).items() ]
      else:
        #If value is null or empty array don't include it
        if value is None or value == [] or value == '':
          return []
        return [ (key, value) ]

    items = [ item for k, v in dictionary.items() for item in expand(k, v) ]
    return dict(items)

  original_object_root = flatten_dictionary(original_object).copy()
  original_object_root['_meta_object_name'] = object_name

  for key,value in original_object_root.copy().items():
    if isinstance(value, dict):
      flatten_dictionary(value, objects)
    if isinstance(value, list):
      original_object_root.pop(key)
      for nested_object in value:
        nested_object['_meta_parent_foreign_key'] = original_object_root['id']
        nested_object['_meta_object_name'] = object_name + "_" + key
        expand_dictionary(nested_object, object_name + "_" + key, objects)

  objects.append(original_object_root)

  return objects

【讨论】：