从下面的嵌套字典中,如何获取
'value4ac'
的每个父字典键?通过启动 'value4ac'
值,我想得到 'key4'
,'key4a'
,'Key4ac'
。
example_dict = { 'key1' : 'value1',
'key2' : 'value2',
'key3' : { 'key3a': 'value3a' },
'key4' : { 'key4a': {
'key4aa': 'value4aa',
'key4ab': 'value4ab',
'key4ac': 'value4ac'
},
'key4b': 'value4b'
}
}
递归来救援!
example_dict = { 'key1' : 'value1',
'key2' : 'value2',
'key3' : { 'key3a': 'value3a' },
'key4' : { 'key4a': { 'key4aa': 'value4aa',
'key4ab': 'value4ab',
'key4ac': 'value4ac'},
'key4b': 'value4b'}
}
def find_key(d, value):
for k,v in d.items():
if isinstance(v, dict):
p = find_key(v, value)
if p:
return [k] + p
elif v == value:
return [k]
print find_key(example_dict,'value4ac')
它是如何工作的
它检查项目并检查 2 个案例
这是 @mgilson 解决方案的更广泛变体,适用于 JSON:
example_dict_with_list = { 'key1' : 'value1',
'key2' : 'value2',
'key3' : { 'key3a': 'value3a' },
'key4' : { 'key4a': [{ 'key4aa': 'value4aa',
'key4ab': 'value4ab',
'key4ac': 'value4ac'}],
'key4b': 'value4b'}
}
def breadcrumb(json_dict_or_list, value) -> list:
if value == json_dict_or_list:
return [json_dict_or_list]
elif isinstance(json_dict_or_list, str) and isinstance(value, str):
if value.casefold() in json_dict_or_list.casefold():
return [json_dict_or_list]
elif isinstance(json_dict_or_list, dict):
for k, v in json_dict_or_list.items():
match = breadcrumb(k, value)
if match:
return match + [v]
match = breadcrumb(v, value)
if match:
return [k] + match
elif isinstance(json_dict_or_list, list):
lst = json_dict_or_list
for i in range(len(lst)):
match = breadcrumb(lst[i], value)
if match:
return [str(i)] + match
测试一下:
# search by exact value:
print(breadcrumb(example_dict_with_list, 'value4aa'))
# search by a substring:
print(breadcrumb(example_dict_with_list, 'value4a'))
# search by key:
print(breadcrumb(example_dict_with_list, 'key4aa'))
结果:
['key4', 'key4a', '0', 'key4aa', 'value4aa']
['key4', 'key4a', '0', 'key4aa', 'value4aa']
['key4', 'key4a', '0', 'key4aa', 'value4aa']
如果您需要将其很好地打印出来,就像一串面包屑一样,请这样做
print(
' > '.join(
breadcrumb(example_dict, 'value4aa')
)
)
将会回归
'key4 > key4a > 0 > key4aa > value4aa'
您可以使用
NestedDict
。
>>> from ndicts.ndicts import NestedDict
>>> example_dict = { 'key1' : 'value1',
... 'key2' : 'value2',
... 'key3' : { 'key3a': 'value3a' },
... 'key4' : { 'key4a': {
... 'key4aa': 'value4aa',
... 'key4ab': 'value4ab',
... 'key4ac': 'value4ac'
... },
... 'key4b': 'value4b'
... }
... }
>>> nd = NestedDict(example_dict)
>>> for key, value in nd.items():
... if value == "value4ac":
... print(key)
...
('key4', 'key4a', 'key4ac')
安装ndicts
pip install ndicts
对@KiriSakow 解决方案进行了轻微修改(受到@mgilson 的启发)。
它不会生成列表索引,并且不仅为最后一个孩子返回正确的父母,而且还为子父母返回正确的父母。
def breadcrumb(nested_dict, value):
if nested_dict == value:
return [nested_dict]
elif isinstance(nested_dict, dict):
for k, v in nested_dict.items():
if k == value:
return [k]
p = breadcrumb(v, value)
if p:
return [k] + p
elif isinstance(nested_dict, list):
lst = nested_dict
for i in range(len(lst)):
p = breadcrumb(lst[i], value)
if p:
return p
example_dict_with_list = { 'key1' : 'value1',
'key2' : 'value2',
'key3' : { 'key3a': 'value3a' },
'key4' : { 'key4a': [{ 'key4aa': 'value4aa',
'key4ab': 'value4ab',
'key4ac': 'value4ac'}],
'key4b': 'value4b'}
}
print(breadcrumb(example_dict_with_list, "value4aa"))
print(breadcrumb(example_dict_with_list, "key4aa"))
['key4', 'key4a', 'key4aa', 'value4aa']
['key4', 'key4a', 'key4aa']