通過串列的字典遞回函式

鑒于這種資料結構

[{'name': 'Business Operations', 
  'team_edit': ['BusinessOperations'], 
  'team_view': ['Snaptest'], 
  'test_sub_subfolder': [
       {'name': 'test_sub', 
        'team_edit': ['BusinessOperations', 'Freddy'], 
        'team_view': ['hugo']}], 
  'test_sub2_subfolder': [
       {'name': 'test_sub2', 
        'test_sub_subfolder': [
            {'name': 'test_sub_sub', 
             'team_edit': ['BusinessOperations', 'Freddy'], 
             'team_view': ['hugo']}]}]}]

我正在嘗試撰寫一個遞回函式，它可以讓我訪問所有元素，基本上是將它們列印出來。嵌套結構可以是任意長度，但始終遵循這種格式。

我的第一次嘗試是

def recurse_over_object(dict_obj):
    for key, value in dict_obj.items():
        if 'subfolder' in key:
            for pair in recurse_over_object(value[0]):
                print(key, *pair)
                # yield (key, *pair)
        else:
            print(key, value)
            # yield (key, value)

這里的問題是，當我沿著樹向下走時，我會呼叫該值subfolder并沿著那條路線走下去，但在此示例中，我有兩個位于同一級別的子檔案夾（test_sub和test_sub2），我需要獲取它們的值。

有誰知道任何奇特的方法來做到這一點？如果需要，我絕對可以重組資料輸入以使其更容易，例如，我的一個想法是將子檔案夾放在同一級別上，list但后來我在一個遞回函式中努力迭代這些元素。

任何想法的幫助將不勝感激。

uj5u.com熱心網友回復：

你可以迭代地呼叫recurse_over_objectunder if 'subfolder' in key:。就像：

def recurse_over_object(dict_obj):
    for key, value in dict_obj.items():
        if 'subfolder' in key:
            for item in value:
                recurse_over_object(item)
        else:
            print(key, value)

然后

for d in data:
    recurse_over_object(d)

輸出：

name Business Operations
team_edit ['BusinessOperations']
team_view ['Snaptest']
name test_sub
team_edit ['BusinessOperations', 'Freddy']
team_view ['hugo']
name test_sub2
name test_sub_sub
team_edit ['BusinessOperations', 'Freddy']
team_view ['hugo']

uj5u.com熱心網友回復：

為了遵循您對使用yield()（以及print）的引導，這里是調整后的代碼，我認為它可以滿足您的問題。請注意，最后一條x = list( ... )陳述句旨在呼叫您的函式創建的生成器。

dataStructure = [
    {'name': 'Business Operations', 
     'team_edit': ['BusinessOperations'],
     'team_view': ['Snaptest'], 
     'test_sub_subfolder': [
         {'name': 'test_sub', 
          'team_edit': ['BusinessOperations', 'Freddy'], 
          'team_view': ['hugo']}], 
     'test_sub2_subfolder': [
         {'name': 'test_sub2',
          'test_sub_subfolder': [
              {'name': 'test_sub_sub', 
               'team_edit': ['BusinessOperations', 'Freddy'], 
               'team_view': ['hugo']}]}]}
]
def recurse_over_object(dict_obj):
    for key, value in dict_obj.items():
        if 'subfolder' in key:
            '''
            for pair in recurse_over_object(value[0]):
                print(key, *pair)
                yield (key, *pair)
            '''
            print(key, *list(recurse_over_object(value[0])))
            yield(key, *list(recurse_over_object(value[0])))
        else:
            print(key, value)
            yield (key, value)   

x = list(recurse_over_object(dataStructure[0]))

輸出：

name Business Operations
team_edit ['BusinessOperations']
team_view ['Snaptest']
name test_sub
team_edit ['BusinessOperations', 'Freddy']
team_view ['hugo']
test_sub_subfolder ('name', 'test_sub') ('team_edit', ['BusinessOperations', 'Freddy']) ('team_view', ['hugo'])
name test_sub
team_edit ['BusinessOperations', 'Freddy']
team_view ['hugo']
name test_sub2
name test_sub_sub
team_edit ['BusinessOperations', 'Freddy']
team_view ['hugo']
test_sub_subfolder ('name', 'test_sub_sub') ('team_edit', ['BusinessOperations', 'Freddy']) ('team_view', ['hugo'])
name test_sub_sub
team_edit ['BusinessOperations', 'Freddy']
team_view ['hugo']
test_sub2_subfolder ('name', 'test_sub2') ('test_sub_subfolder', ('name', 'test_sub_sub'), ('team_edit', ['BusinessOperations', 'Freddy']), ('team_view', ['hugo']))
name test_sub2
name test_sub_sub
team_edit ['BusinessOperations', 'Freddy']
team_view ['hugo']
test_sub_subfolder ('name', 'test_sub_sub') ('team_edit', ['BusinessOperations', 'Freddy']) ('team_view', ['hugo'])
name test_sub_sub
team_edit ['BusinessOperations', 'Freddy']
team_view ['hugo']

uj5u.com熱心網友回復：

您有一個包含串列、字典和字串的嵌套結構。遞回函式可以對其進行測驗以決定如何處理元素：

def get_branches(x, acc=()):
    if isinstance(x, dict):
        for k,v in x.items():
            yield from get_branches(v, acc (k,))
    elif isinstance(x, list):
        for k in x:
            yield from get_branches(k, acc)
    else:
        yield acc   (x,)

x = [{'name': 'Business Operations', 
  'team_edit': ['BusinessOperations'], 
  'team_view': ['Snaptest'], 
  'test_sub_subfolder': [
       {'name': 'test_sub', 
        'team_edit': ['BusinessOperations', 'Freddy'], 
        'team_view': ['hugo']}], 
  'test_sub2_subfolder': [
       {'name': 'test_sub2', 
        'test_sub_subfolder': [
            {'name': 'test_sub_sub', 
             'team_edit': ['BusinessOperations', 'Freddy'], 
             'team_view': ['hugo']}]}]}]

print(list( get_branches(x) ))

輸出：

[('name', 'Business Operations'),
 ('team_edit', 'BusinessOperations'),
 ('team_view', 'Snaptest'),
 ('test_sub_subfolder', 'name', 'test_sub'),
 ('test_sub_subfolder', 'team_edit', 'BusinessOperations'),
 ('test_sub_subfolder', 'team_edit', 'Freddy'),
 ('test_sub_subfolder', 'team_view', 'hugo'),
 ('test_sub2_subfolder', 'name', 'test_sub2'),
 ('test_sub2_subfolder', 'test_sub_subfolder', 'name', 'test_sub_sub'),
 ('test_sub2_subfolder', 'test_sub_subfolder', 'team_edit', 'BusinessOperations'),
 ('test_sub2_subfolder', 'test_sub_subfolder', 'team_edit', 'Freddy'),
 ('test_sub2_subfolder', 'test_sub_subfolder', 'team_view', 'hugo')
]

標籤：Python python-3.x 字典 递归 递归查询

上一篇：在減少嵌套字典的同時保持字典結構

下一篇：創建嵌套的subdictpython