平面嵌套字典，其中新鍵將是所有嵌套鍵的元組

Question

我有這樣的命令：

previous_dict = {
    'dict_1': 'dict_1',
    'dict_2': {
        'dict_2_1': 'dict_2_1',
        'dict_2_2': 'dict_2_2'
    },
    'dict_3': 3,
    'dict_4': None,
    'dict_5': dict()
}

我編寫了一個函數，將所有鍵作為元組平整字典，其輸出為：

previously_expected_dict = {}
for key, value in previous_dict.items():
    if type(value) == dict:
        for k, v in value.items():
            previously_expected_dict[(key, k)] = v
    else:
        previously_expected_dict[(key,)] = value

輸出：

print(previously_expected_dict)
{
    ('dict_1',): 'dict_1',
    ('dict_2', 'dict_2_1'): 'dict_2_1',
    ('dict_2', 'dict_2_2'): 'dict_2_2',
    ('dict_3',): 3,
    ('dict_4',): None
}

dict_5由於沒有任何值而被丟棄

---

現在需求已更改，並且dict可以具有任意數量的嵌套

new_dict = {
    'dict_1': {
        'dict_1_1': {
            'dict_1_1_1': 'dict_1_1_1',
            'dict_1_1_2': 'dict_1_1_2'
        }
    },
    'dict_2': {
        'dict_2_1': 'dict_2_1',
        'dict_2_2': 'dict_2_2'
    },
    'dict_3': 'dict_3',
    'dict_4': dict()
}

到目前為止我嘗試過的代碼

def make_flat(my_dict):
    nd = dict()
    keys = []

    def loop_me(value):
        nonlocal keys
        if isinstance(value, dict):
            for k, v in value.items():
                keys.append(k)
                loop_me(v)
        else:
            nd[tuple(keys)] = value
            keys.pop(-1)

    loop_me(my_dict)
    return nd


print(make_flat(new_dict))

但是我在元組中收到了額外的密鑰

{
    ('dict_1', 'dict_1_1', 'dict_1_1_1'): 'dict_1_1_1',  # Perfect
    ('dict_1', 'dict_1_1', 'dict_1_1_2'): 'dict_1_1_2',  # Perfect
    ('dict_1', 'dict_1_1', 'dict_2', 'dict_2_1'): 'dict_2_1',  # Error, Expected is: ('dict_2', 'dict_2_1')
    ('dict_1', 'dict_1_1', 'dict_2', 'dict_2_2'): 'dict_2_2',  # Error, Expected is: ('dict_2', 'dict_2_2')
    ('dict_1', 'dict_1_1', 'dict_2', 'dict_3'): 'dict_3'  # Error, Expected is: ('dict_3',)
}

最終預期產出：

output = {
    ('dict_1', 'dict_1_1', 'dict_1_1_1'): 'dict_1_1_1',
    ('dict_1', 'dict_1_1', 'dict_1_1_2'): 'dict_1_1_2',
    ('dict_2', 'dict_2_1'): 'dict_2_1',
    ('dict_2', 'dict_2_2'): 'dict_2_2',
    ('dict_3',): 'dict_3'
}

我嘗試使用for循環和遞歸函數Failed編寫。

Answer 1

您可以使用遞歸：

def flatten(d, c = []):
  for a, b in d.items():
    if not isinstance(b, dict):
       yield (tuple(c+[a]), b)
    else:
       yield from flatten(b, c+[a])

print(dict(flatten(previous_dict)))

輸出：

{('dict_1',): 'dict_1', ('dict_2', 'dict_2_1'): 'dict_2_1', ('dict_2', 'dict_2_2'): 'dict_2_2', ('dict_3',): 3, ('dict_4',): None}

隨着new_dict ：

{('dict_1', 'dict_1_1', 'dict_1_1_1'): 'dict_1_1_1', ('dict_1', 'dict_1_1', 'dict_1_1_2'): 'dict_1_1_2', ('dict_2', 'dict_2_1'): 'dict_2_1', ('dict_2', 'dict_2_2'): 'dict_2_2', ('dict_3',): 'dict_3'}

Answer 2

您可以使用一個函數來遍歷dict項，並將鍵添加到遞歸調用返回的子鍵元組中：

def flatten(d):
    for k, v in d.items():
        if isinstance(v, dict):
            for s, i in flatten(v):
                yield (k, *s), i
        else:
            yield (k,), v

這樣dict(flatten(new_dict))返回：

{('dict_1', 'dict_1_1', 'dict_1_1_1'): 'dict_1_1_1', ('dict_1', 'dict_1_1', 'dict_1_1_2'): 'dict_1_1_2', ('dict_2', 'dict_2_1'): 'dict_2_1', ('dict_2', 'dict_2_2'): 'dict_2_2', ('dict_3',): 'dict_3'}

Answer 3

我使用了deep變量來確定和更正鍵，並且make_flat函數返回所需的輸出，但是，@ Ajax1234更清楚地做到了。

def make_flat(dict_):
    new_dict = dict()
    keys = []

    def loop_recursively(value, deep=0):
        nonlocal keys
        if isinstance(value, dict):
            deep += 1
            for k, v in value.items():
                keys.append(k)
                loop_recursively(v, deep)
            else:
                deep -= 1
        else:
            keys = keys[-deep:]
            new_dict[tuple(keys)] = value
            keys.pop(-1)

    loop_recursively(dict_)
    return new_dict