[英]Get a list of all keys in nested dictionary
我想获取包含列表和字典的嵌套字典中所有键的列表。
我目前有这段代码,但似乎缺少向列表中添加一些键,并且重复添加了一些键。
keys_list = []
def get_keys(d_or_l, keys_list):
if isinstance(d_or_l, dict):
for k, v in iter(sorted(d_or_l.iteritems())):
if isinstance(v, list):
get_keys(v, keys_list)
elif isinstance(v, dict):
get_keys(v, keys_list)
else:
keys_list.append(k)
elif isinstance(d_or_l, list):
for i in d_or_l:
if isinstance(i, list):
get_keys(i, keys_list)
elif isinstance(i, dict):
get_keys(i, keys_list)
else:
print "** Skipping item of type: {}".format(type(d_or_l))
return keys_list
这应该可以完成这项工作:
def get_keys(dl, keys_list):
if isinstance(dl, dict):
keys_list += dl.keys()
map(lambda x: get_keys(x, keys_list), dl.values())
elif isinstance(dl, list):
map(lambda x: get_keys(x, keys_list), dl)
为避免重复,您可以使用 set,例如:
keys_list = list( set( keys_list ) )
示例测试用例:
keys_list = []
d = {1: 2, 3: 4, 5: [{7: {9: 1}}]}
get_keys(d, keys_list)
print keys_list
>>>> [1, 3, 5, 7, 9]
就目前而言,您的代码会忽略导致list
或dict
值的键。 删除第一个for
循环中的else
块,无论值是什么,您都希望添加键。
keys_list = []
def get_keys(d_or_l, keys_list):
if isinstance(d_or_l, dict):
for k, v in iter(sorted(d_or_l.iteritems())):
if isinstance(v, list):
get_keys(v, keys_list)
elif isinstance(v, dict):
get_keys(v, keys_list)
keys_list.append(k) # Altered line
elif isinstance(d_or_l, list):
for i in d_or_l:
if isinstance(i, list):
get_keys(i, keys_list)
elif isinstance(i, dict):
get_keys(i, keys_list)
else:
print "** Skipping item of type: {}".format(type(d_or_l))
return keys_list
get_keys({1: 2, 3: 4, 5: [{7: {9: 1}}]}, keys_list)
返回[1, 3, 9, 7, 5]
为避免重复,您可以使用set
数据类型而不是list
。
更新@MackM 对 Python 3 的响应,因为dict.iteritems
已被弃用(我更喜欢在.format{}
样式上使用 f 字符串):
keys_list = []
def get_keys(d_or_l, keys_list):
if isinstance(d_or_l, dict):
for k, v in iter(sorted(d_or_l.items())): # Altered line to update deprecated method
if isinstance(v, list):
get_keys(v, keys_list)
elif isinstance(v, dict):
get_keys(v, keys_list)
keys_list.append(k)
elif isinstance(d_or_l, list):
for i in d_or_l:
if isinstance(i, list):
get_keys(i, keys_list)
elif isinstance(i, dict):
get_keys(i, keys_list)
else:
print(f'** Skipping item of type: {type(d_or_l)}') # Altered line to use f-strings
return keys_list
unique_keys = list(set(get_keys(my_json_dict, keys_list))) # Added line as example use case
这是一个简单的解决方案:
def get_nested_keys(d, keys):
for k, v in d.items():
if isinstance(v, dict):
get_nested_keys(v, keys)
else:
keys.append(k)
keys_list = []
get_nested_keys(test_listing, keys_list)
print(keys_list)
如果您还想知道键的层次结构,可以像这样修改函数:
def get_nested_keys(d, keys, prefix):
for k, v in d.items():
if isinstance(v, dict):
get_nested_keys(v, keys, f'{prefix}:{k}')
else:
keys.append(f'{prefix}:{k}')
我将通过 python 2 & 3 友好版本扩展@pm007 答案:
def get_keys(dl, keys=None):
keys = keys or []
if isinstance(dl, dict):
keys += dl.keys()
list(map(lambda x: get_keys(x, keys), dl.values()))
elif isinstance(dl, list):
list(map(lambda x: get_keys(x, keys), dl))
return list(set(keys))
d = {1: 2, 3: 4, 5: {7: {1: 1}}}
get_keys(d)
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