将嵌套字典的键转换为大写

Question

我有一本字典：

data = {'Common': {'height': 165, 'weight': 70, 'measure': ['cm', 'kg']},
         'Man': 'handsome',
         'Woman': {'feature': 'pretty', 'weight': 50},
         'Dog': {'feature': 'barks', 'height': 10, 'weight': 20}}

只想将字典键转换为大写

尝试了以下代码：

d = {}
d1 = {}
for k, v in data.items():
    if isinstance(v, dict):
        for i, j in v.items():
            d1[i.upper()] = j
        d[k.upper()] = d1
    else:
        d[k.upper()] = v

print(d)

…产生的输出带有不必要的键，高度和重量合理化如下：

{'COMMON': {'HEIGHT': 10, 'WEIGHT': 20, 'MEASURE': ['cm', 'kg'], 'FEATURE': 'barks'}, 
    'MAN': 'handsome', 
  'WOMAN': {'HEIGHT': 10, 'WEIGHT': 20, 'MEASURE': ['cm', 'kg'], 'FEATURE': 'barks'}, 
    'DOG': {'HEIGHT': 10, 'WEIGHT': 20, 'MEASURE': ['cm', 'kg'], 'FEATURE': 'barks'}}

我的预期产出是：

{'COMMON': {'HEIGHT': 165, 'WEIGHT': 70, 'MEASURE': ['cm', 'kg']},
 'MAN': 'handsome',
 'WOMAN': {'FEATURE': 'pretty', 'WEIGHT': 50},
 'DOG': {'FEATURE': 'barks', 'HEIGHT': 10, 'WEIGHT': 20}}

1. 我哪里做错了
2. 正确的词典理解是什么样的

Answer 1

您可以执行类似操作，使用所需的键复制到其他dict：

data = {'Common': {'height': 165, 'weight': 70, 'measure': ['cm', 'kg']},
        'Man': 'handsome',
        'Woman': {'feature': 'pretty', 'weight': 50},
        'Dog': {'feature': 'barks', 'height': 10, 'weight': 20}}
data2 = {}
for k in data.keys():
    data2[k.upper()] = data[k]

更新： 如果您不仅要更改级别1上的键，还应使用递归函数：

data = {'Common': {'height': 165, 'weight': 70, 'measure': ['cm', 'kg']},
        'Man': 'handsome',
        'Woman': {'feature': 'pretty', 'weight': 50},
        'Dog': {'feature': 'barks', 'height': 10, 'weight': 20}}

def keys_to_upper(dict1):
    dict2 = {}
    for k in dict1.keys():
        if isinstance(dict1[k], dict):
            dict2[k.upper()] = keys_to_upper(dict1[k])
        else:
            dict2[k.upper()] = dict1[k]
    return dict2

d2 = keys_to_upper(data)

Answer 2

代码的问题是，如果值是字典，则重新分配d1。您可以使用^{}解决此问题：

代码：

from copy import deepcopy

d = {}
d1 = {}
for k, v in data.items():
    if isinstance(v, dict):
        for i, j in v.items():
            d1[i.upper()] = j
        d[k.upper()] = deepcopy(d1)
    else:
        d[k.upper()] = v

输出：

>>> d
{'COMMON': {'HEIGHT': 165, 'WEIGHT': 70, 'MEASURE': ['cm', 'kg']},
 'MAN': 'handsome',
 'WOMAN': {'FEATURE': 'pretty', 'WEIGHT': 50},
 'DOG': {'FEATURE': 'barks', 'HEIGHT': 10, 'WEIGHT': 20}}

或者，作为词典理解：

>>> {k.upper(): {i.upper(): j for i, j in v.items()} if isinstance(v, dict) else v for k, v in data.items()}
{'COMMON': {'HEIGHT': 165, 'WEIGHT': 70, 'MEASURE': ['cm', 'kg']},
 'MAN': 'handsome',
 'WOMAN': {'FEATURE': 'pretty', 'WEIGHT': 50},
 'DOG': {'FEATURE': 'barks', 'HEIGHT': 10, 'WEIGHT': 20}}

Answer 3

列表理解更快，因为它针对Python interpreter进行了优化，以便在looping期间发现可预测的模式。除了list comprehensions的语法优势外，它们通常与map的等效用法一样快或更快

data = {'Common': {'height': 165, 'weight': 70, 'measure': ['cm', 'kg']},
        'Man': 'handsome',
        'Woman': {'feature': 'pretty', 'weight': 50},
        'Dog': {'feature': 'barks', 'height': 10, 'weight': 20}}

    for k, v in data.items():
      if isinstance(v, dict):
        data[k.upper()] = data.pop(k)

输出>

data = {'COMMON': {'height': 165, 'weight': 70, 'measure': ['cm', 'kg']},
        'Man': 'handsome',
        'WOMAN': {'feature': 'pretty', 'weight': 50},
        'DOG': {'feature': 'barks', 'height': 10, 'weight': 20}}