我有一句话是这样的:
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()
}
我已经写了一个函数,用所有键作为元组来扁平化字典,输出如下:
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
输出:
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
被丢弃,因为它没有任何值
现在需求已经改变了,dict可以有任意数量的嵌套
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()
}
到目前为止我已经尝试过的代码
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))
但是我收到了额外的元组中的键
{
('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',)
}
最终预期输出:
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'
}
我尝试使用for循环和递归函数编写代码,但失败了。
发布于 2019-04-20 02:24:50
您可以使用递归:
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)))
输出:
{('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}
使用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'}
发布于 2019-04-20 02:37:46
您可以使用一个函数来迭代dict项,并将键预先添加到递归调用返回的子键的元组中:
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
因此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'}
发布于 2019-04-20 02:33:37
我已经使用了deep
变量来确定和纠正键,make_flat
函数返回您想要的输出,但是@Ajax1234做得更清楚。
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
https://stackoverflow.com/questions/55766114
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