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[英]How do I create a new dict of dicts from a dict with nested dicts in Python
[英]How do I create a nested dict from a dict by splitting the keys on a delimiter in python?
我有一个像下面这样的字典:
a = {
'customer_name': 'bob',
'customer_phone': '555-1212',
'order_0_number': 'A33432-24',
'order_0_date': '12/12/12',
'order_1_number': 'asd24222',
'order_1_date': '12/14/12'
}
我需要它在下划线上拆分并放入这样的嵌套字典中:
b = {
'customer': {
'name': 'bob',
'phone': '555-1212'
},
'order': {
'0': {
'date': '12/12/12',
'number': '...' },
'1': { ... etc.
我拥有的实际数据比这更深层嵌套。 我已经很深入了,但是我一直在思考如何在 Python 中做到这一点:
def expand_data(field, value):
split_field = field.split('_', 1)
#base case, end of string
if len(split_field) == 1:
child_element[split_field[0] = value
return child_element
else:
child_element[split_field[0]] = expand_data(split_field[1],value)
return child_element
b = {}
for k,v in a.iteritems():
b += expand_data(k, v) # stuck here because I can't add nested dicts together
但我不完全确定这是否是正确的方法。 我还没有运行这段代码,现在只是想考虑一下。
此外,dict 键将来可能会改变,所以我只能依赖“_”下划线在那里拆分。 我也不知道它需要嵌套多深。
通用解决方案:
def nest_dict(flat_dict, sep='_'):
"""Return nested dict by splitting the keys on a delimiter.
>>> from pprint import pprint
>>> pprint(nest_dict({'title': 'foo', 'author_name': 'stretch',
... 'author_zipcode': '06901'}))
{'author': {'name': 'stretch', 'zipcode': '06901'}, 'title': 'foo'}
"""
tree = {}
for key, val in flat_dict.items():
t = tree
prev = None
for part in key.split(sep):
if prev is not None:
t = t.setdefault(prev, {})
prev = part
else:
t.setdefault(prev, val)
return tree
if __name__ == '__main__':
import doctest
doctest.testmod()
这是您可以使用的步骤,但由于您没有向我们展示您的任何代码,因此我不会提供代码。
for key in yourDico
使用for key in yourDico
customer
并order
以拆分值以获得所需的值。 _
重复拆分,这似乎是您的拆分字符。 b = {}
b["customer"] = {"name": a["customer_name"], "phone": a["customer_phone"]}
for key in a.keys():
if key.startswith("order_"):
o, i, f = key.split("_")
order = b.get("order", {})
order_i = order.get(i, {})
order_i[f] = a[key]
b["order"] = order
b["order"][i] = order_i
print(b)
输出:
{'customer': {'name': 'bob', 'phone': '555-1212'},
'order': {'0': {'date': '12/12/12', 'number': 'A33432-24'},
'1': {'date': '12/14/12', 'number': 'asd24222'}}}
另一个带有一些评论的通用解决方案:
def expand_flatdict_to_tree(dict, sep = '_', to_lower = True):
tree = {}
for src, val in dict.items():
ref = tree
if to_lower:
src = src.lower()
for i, part in enumerate(src.split(sep)):
if part not in ref:
ref[part] = {}
if i == len(src.split(sep)) -1:
ref[part] = val # we cannot do ref = val after loop, as assignment to the ref itself will be passed by assignment
break
ref = ref[part] # update nest reference
return tree
flat_dict = {
'LOGLEVEL': 'INFO',
'DB_ENABLED': True,
'DB_CONNECTOR_HOST': 'localhost',
'DB_CONNECTOR_USER': 'root',
}
print (expand_flatdict_to_tree(flat_dict))
# {
# 'loglevel': 'INFO',
# 'db': {
# 'enabled': True,
# 'connector': {
# 'host': 'localhost',
# 'user': 'root'
# }
# }
# }
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