嵌套的字典：解压路径叶

Question

我有一个Python嵌套字典如下所示：

{'dist_river': 
    {'high': 
    {'wind_speed': 
     {'1': 
     {'population': 
      {'high': 
      {'school': 
       {'high':'T', 'medium':'T', 'low':'F'} 
      }, 
      'medium': 
       {'land_cover': 
       {'Mix_garden': 
        {'income_source': 
        {'Plantation':'T', 'Agriculture':'F'} 
        } 
       } 
       } 
      } 
      } 
     } 
     }, 
    'low': 'F' 
    } 
} 

我怎样才能从分字典嵌套词典？例如，来自DIC分字典：

results = [ 
    {'dist_river': 
    {'high': 
     {'wind_speed': 
     {'1': 
     {'population': 
     {'high': 
      {'school': 
      {'high': 'T', 'medium': 'T', 'low': 'F'} 
      }}}}}}}, 

    {'dist_river': 
    {'high': 
     {'wind_speed': 
     {'1': 
     {'population': 
     {'medium': 
      {'land_cover': 
      {'Mix_garden': 
      {'income_source': 
      {'Plantation': 'T', 'Agriculture': 'F'} 
      }}}}}}}}}, 

    {'dist_river': 
    {'low': 'F'} 
    } 
] 

lengths(results) == 3 

谢谢您的帮助

社区编辑：看来，每个结果字典只能有一个嵌套的每级条目。换句话说，每个结果都包含字典树中每个叶子的整个路径。 - 蒂姆Pietzcker 13小时前

Answer 1

它的完成方式与从字典中“获取”任何其他值完全相同。

print dic1['dist_river']['high'] 

等等。

编辑：

在情况下，我误解了这个问题，它的的确确是越来越一次所有类型的字典列表，这里的情况下，有一个在他们每个人只需一个键为例：

def get_nested_dicts(d): 
    dicts = [] 
    probe = d 
    while type(probe) == dict: 
     dicts.append(probe) 
     probe = probe.values()[0] 
    return dicts 

Answer 2

也许这：

def enum_paths(p): 
    if not hasattr(p, 'items'): 
     yield p 
    else: 
     for k, v in p.items(): 
      for x in enum_paths(v): 
       yield {k: x} 


for x in enum_paths(dic): 
    print x 

Answer 3

import collections 

def isDict(d): 
    return isinstance(d, collections.Mapping) 

def isAtomOrFlat(d): 
    return not isDict(d) or not any(isDict(v) for v in d.values()) 

def leafPaths(nestedDicts, noDeeper=isAtomOrFlat): 
    """ 
     For each leaf in NESTEDDICTS, this yields a 
     dictionary consisting of only the entries between the root 
     and the leaf. 
    """ 
    for key,value in nestedDicts.items(): 
     if noDeeper(value): 
      yield {key: value} 
     else: 
      for subpath in leafPaths(value): 
       yield {key: subpath} 

演示：

>>> pprint.pprint(list(leafPaths(dic))) 
[{'dist_river': {'high': {'wind_speed': {'1': {'population': {'high': {'school': {'high': 'T', 
                        'low': 'F', 
                        'medium': 'T'}}}}}}}}, 
{'dist_river': {'high': {'wind_speed': {'1': {'population': {'medium': {'land_cover': {'Mix_garden': {'income_source': {'Agriculture': 'F', 
                                 'Plantation': 'T'}}}}}}}}}}, 
{'dist_river': {'low': 'F'}}] 

---

旁注1：但是，除非这个格式是必要由于某种原因，我个人觉得它会更好，以产生节点元组的方式，例如是这样的：

...noDeeper=lambda x:not isDict(x)... 
...yield tuple(value) 
...yield (key,)+subpath 

[('dist_river', 'high', 'wind_speed', '1', 'population', 'high', 'school', 'high', 'T'), 
('dist_river', 'high', 'wind_speed', '1', 'population', 'high', 'school', 'medium', 'T'), 
('dist_river', 'high', 'wind_speed', '1', 'population', 'high', 'school', 'low', 'F'), 
('dist_river', 'high', 'wind_speed', '1', 'population', 'medium', 'land_cover', 'Mix_garden', 'income_source', 'Plantation', 'T'), 
('dist_river', 'high', 'wind_speed', '1', 'population', 'medium', 'land_cover', 'Mix_garden', 'income_source', 'Agriculture', 'F'), 
('dist_river', 'low', 'F')] 

---

阿里纳斯2（从“简单”的答案，这恰好是thg435的答案很容易提取。）：请注意，幼稚的做法也不是什么OP正在寻找。天真的实施将有noDeeper=lambda x:not isDict(x)，与结果：

>>> pprint.pprint(list(leafPaths(dic))) 
[{'dist_river': {'high': {'wind_speed': {'1': {'population': {'high': {'school': {'high': 'T'}}}}}}}}, 
{'dist_river': {'high': {'wind_speed': {'1': {'population': {'high': {'school': {'medium': 'T'}}}}}}}}, 
{'dist_river': {'high': {'wind_speed': {'1': {'population': {'high': {'school': {'low': 'F'}}}}}}}}, 
{'dist_river': {'high': {'wind_speed': {'1': {'population': {'medium': {'land_cover': {'Mix_garden': {'income_source': {'Plantation': 'T'}}}}}}}}}}, 
{'dist_river': {'high': {'wind_speed': {'1': {'population': {'medium': {'land_cover': {'Mix_garden': {'income_source': {'Agriculture': 'F'}}}}}}}}}}, 
{'dist_river': {'low': 'F'}}] 

---

编辑：这是一个低效率的算法。每片叶片L重新产生depth(L)次。更有效的方法是将生成器链接到自定义数据结构，或手动模拟堆栈。