我正在尝试创建一个通用函数来替换嵌套字典键中的点。我有一个非泛型函数,深度为 3 级,但必须有一种方法来实现这个泛型。任何帮助表示赞赏!到目前为止我的代码:
output = {'key1': {'key2': 'value2', 'key3': {'key4 with a .': 'value4', 'key5 with a .': 'value5'}}}
def print_dict(d):
new = {}
for key,value in d.items():
new[key.replace(".", "-")] = {}
if isinstance(value, dict):
for key2, value2 in value.items():
new[key][key2] = {}
if isinstance(value2, dict):
for key3, value3 in value2.items():
new[key][key2][key3.replace(".", "-")] = value3
else:
new[key][key2.replace(".", "-")] = value2
else:
new[key] = value
return new
print print_dict(output)
更新:为了回答我自己的问题,我使用 json object_hooks 做了一个解决方案:
import json
def remove_dots(obj):
for key in obj.keys():
new_key = key.replace(".","-")
if new_key != key:
obj[new_key] = obj[key]
del obj[key]
return obj
output = {'key1': {'key2': 'value2', 'key3': {'key4 with a .': 'value4', 'key5 with a .': 'value5'}}}
new_json = json.loads(json.dumps(output), object_hook=remove_dots)
print new_json