Python 递归总是更新字典
如何从递归中获得中间结果 考虑如下所示的输入:Python 递归总是更新字典,python,python-3.x,list,dictionary,recursion,Python,Python 3.x,List,Dictionary,Recursion,如何从递归中获得中间结果 考虑如下所示的输入: def recursion(input_type): print('input_type ',input_type) if isinstance(input_type, dict): num = 0 for k,v in input_type.items(): if isinstance(v, dict): print('from recursi
def recursion(input_type):
print('input_type ',input_type)
if isinstance(input_type, dict):
num = 0
for k,v in input_type.items():
if isinstance(v, dict):
print('from recursion')
recursion(v)
elif isinstance(v, list):
for j in v:
if isinstance(j, dict):
print('from recursion level 2')
recursion(j)
else:
temp_dict = {k:v}
print('type: ',type(temp_dict), k, v)
print('num',num)
num = num+1
for i in list_:
recursion(i)
如果密钥是重复的,那么它应该更新,例如“f”到“f_new”或类似的内容
提前谢谢你 您可以迭代dict项,如果项是列表,则递归展平其中的dict:
input: [{'a':a, 'b':b, 'c':[{'d':d, 'e':e}]}]
Updated input: [ {'a':a, 'b':b, 'c': { 'd':d, 'e': [ {'f':f, 'g':g}, {'f':f1, 'g':g1} ] } } ]
desired output: [{'a':a, 'b':b, 'd':d, 'f':f, 'g':g, 'f_new':f1, 'g_new':g1}]
因此,鉴于:
def f(o):
return {a: b for k, v in o.items() for a, b in ((i for d in (v if isinstance(v, list)
else (v,)) for i in f(d).items()) if isinstance(v, (list, dict)) else ((k, v),))}
[{'a': 'a', 'b': 'b', 'd': 'd', 'e': 'e', 'g': 'g'}]
def f(o):
output = {}
for k, v in o.items():
for a, b in ((i for d in (v if isinstance(v, list) else (v,)) for i in f(d).items())
if isinstance(v, (list, dict)) else ((k, v),)):
while a in output:
a += '_new'
output[a] = b
return output
[lst中d的f(d)]
将返回:
lst = [{'a': 'a', 'b': 'b', 'c': [{'d': 'd', 'e': 'e'}, {'f': [{'g': 'g'}]}]}]
lst = [{'a': 'a', 'b': 'b', 'c': {'d': 'd', 'e': {'f': 'f', 'g': 'g'}}}]
lst = [{'a': 'a', 'b': 'b', 'c': {'d': 'd', 'e': [{'f': 'f', 'g': 'g'}, {'f': 'f1', 'g': 'g1'}]}}]
鉴于:
def f(o):
return {a: b for k, v in o.items() for a, b in ((i for d in (v if isinstance(v, list)
else (v,)) for i in f(d).items()) if isinstance(v, (list, dict)) else ((k, v),))}
[{'a': 'a', 'b': 'b', 'd': 'd', 'e': 'e', 'g': 'g'}]
def f(o):
output = {}
for k, v in o.items():
for a, b in ((i for d in (v if isinstance(v, list) else (v,)) for i in f(d).items())
if isinstance(v, (list, dict)) else ((k, v),)):
while a in output:
a += '_new'
output[a] = b
return output
[lst中d的f(d)]
将返回:
lst = [{'a': 'a', 'b': 'b', 'c': [{'d': 'd', 'e': 'e'}, {'f': [{'g': 'g'}]}]}]
lst = [{'a': 'a', 'b': 'b', 'c': {'d': 'd', 'e': {'f': 'f', 'g': 'g'}}}]
lst = [{'a': 'a', 'b': 'b', 'c': {'d': 'd', 'e': [{'f': 'f', 'g': 'g'}, {'f': 'f1', 'g': 'g1'}]}}]
要避免合并键中的冲突,请将\u new
附加到复制键,直到发现它不是预先存在的,在这种情况下,您不能使用:
[{'a': 'a', 'b': 'b', 'd': 'd', 'f': 'f', 'g': 'g'}]
因此,鉴于:
def f(o):
return {a: b for k, v in o.items() for a, b in ((i for d in (v if isinstance(v, list)
else (v,)) for i in f(d).items()) if isinstance(v, (list, dict)) else ((k, v),))}
[{'a': 'a', 'b': 'b', 'd': 'd', 'e': 'e', 'g': 'g'}]
def f(o):
output = {}
for k, v in o.items():
for a, b in ((i for d in (v if isinstance(v, list) else (v,)) for i in f(d).items())
if isinstance(v, (list, dict)) else ((k, v),)):
while a in output:
a += '_new'
output[a] = b
return output
[lst中d的f(d)]
将返回:
lst = [{'a': 'a', 'b': 'b', 'c': [{'d': 'd', 'e': 'e'}, {'f': [{'g': 'g'}]}]}]
lst = [{'a': 'a', 'b': 'b', 'c': {'d': 'd', 'e': {'f': 'f', 'g': 'g'}}}]
lst = [{'a': 'a', 'b': 'b', 'c': {'d': 'd', 'e': [{'f': 'f', 'g': 'g'}, {'f': 'f1', 'g': 'g1'}]}}]
演示:我正在迭代列表,但它只是一个for循环。所以函数的输入是字典,我知道它不是一个格式良好的结构。但是,如果要删除键
'c'
及其值,必须以任何方式提取吗?@Ch3steR,是的。我不想用“c”作为键anymore@blhsing,我已经编辑了代码。我对解决方案进行了计时,你的解决方案比我的要快得多。这是有道理的,因为我每次都在构建一个新字典。+1并删除了我的解决方案。我明白了。我已经相应地更新了我的答案。如果你认为这个答案是正确的,你能把它标记为被接受吗?@blhsing。你节省了我的时间@德凡舒霍卡尼我已经相应地更新了答案。@blhsing。这解决了所有问题。非常感谢你!!!