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问题描述
my_dict1 = {'a':1,'chk':{'b':2,'c':3},'e':{'chk' f':5,'g':6}}}
我想循环通过dict递归地,如果密钥为'chk',则将其拆分。
预期输出:
{'a':1,'b':2,'e':{'f':5}}
{'a':1,'c':3,'e' :5}}
{'a':1,'b':2,'e':{'g':6}}
{'a':1,'c' 'e':{'g':6}}
不知道如何实现这一点。请帮忙。
我试过的是下面的。
temp_list = []
for k,v in my_dict1.iteritems():
temp = {}
如果k是chk:
为key,val在v.iteritems()中:
temp [key] = val
my_dict1 [k] = {}
for ky,vl在temp.iteritems()中:
my_new_dict = copy.deepcopy(my_dict1)
k,v in my_new_dict.iteritems():
如果k是chk:
my_new_dict [k] = {ky:vl}
temp_list.append(my_new_dict)
print temp_list
输出:
'pre>
[{'a':1,'chk':('c',3),'e':{'chk':{'f' g':6}}},
{'a':1,'chk':('b',2),'e':{'chk':{'f':5,'g' :6}}}]
如何使其递归?
解决方案
from itertools import product
my_dict = {'a':1,'chk':{'b':2,'c':3},'e':{'chk':{'f' ,'g':6}}}
def process(d):
to_product = []#[[('a',1)],[(' b',2),('c',3)],...]
for k,v in d.items():
if k =='chk':
to_product.append([(k2,v2)
for d2 in process(v)
for k2,v2 in d2.items()])
elif isinstance(v,dict):
to_product.append(进程(v)中的d2的[(k,d2)])
else:
to_product.append([(k,v)])
lst = [dict(l)for l in product(* to_product)]
unique = []
[unique.append(item)for item in lst if item not in unique]
return unique
我正在处理(my_dict):
print(i)
#{'e':{'f':5},'b':2 ,'a':1}
#{'e':{'g':6},'b':2,'a':1}
#{'e':{'f ':5},' a':1,'c':3}
#{'e':{'g':6},'a':1,'c':3}
my_dict1 = {'a':1, 'chk':{'b':2, 'c':3}, 'e':{'chk':{'f':5, 'g':6}} }
I would like to loop through the dict recursively and if the key is 'chk', split it.
Expected output:
{'a':1, 'b':2, 'e':{'f':5}}
{'a':1, 'c':3, 'e':{'f':5}}
{'a':1, 'b':2, 'e':{'g':6}}
{'a':1, 'c':3, 'e':{'g':6}}
Not sure of how to achieve this. Please help.
What I have tried is below.
temp_list =[]
for k,v in my_dict1.iteritems():
temp ={}
if k is "chk":
for key,val in v.iteritems():
temp[key] = val
my_dict1[k]={}
for ky,vl in temp.iteritems():
my_new_dict = copy.deepcopy(my_dict1)
for k,v in my_new_dict.iteritems():
if k is "chk":
my_new_dict[k] = {ky:vl}
temp_list.append(my_new_dict)
print temp_list
output:
[{'a': 1, 'chk': ('c', 3), 'e': {'chk': {'f': 5, 'g': 6}}},
{'a': 1, 'chk': ('b', 2), 'e': {'chk': {'f': 5, 'g': 6}}}]
How to make it recursive?
解决方案
from itertools import product
my_dict = {'a':1, 'chk':{'b':2, 'c':3}, 'e':{'chk':{'f':5, 'g':6}} }
def process(d):
to_product = [] # [[('a', 1)], [('b', 2), ('c', 3)], ...]
for k, v in d.items():
if k == 'chk':
to_product.append([(k2, v2)
for d2 in process(v)
for k2, v2 in d2.items()])
elif isinstance(v, dict):
to_product.append([(k, d2) for d2 in process(v)])
else:
to_product.append([(k, v)])
lst = [dict(l) for l in product(*to_product)]
unique = []
[unique.append(item) for item in lst if item not in unique]
return unique
for i in process(my_dict):
print(i)
# {'e': {'f': 5}, 'b': 2, 'a': 1}
# {'e': {'g': 6}, 'b': 2, 'a': 1}
# {'e': {'f': 5}, 'a': 1, 'c': 3}
# {'e': {'g': 6}, 'a': 1, 'c': 3}
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