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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?
我有一個像下面這樣的字典:
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'
}
我需要它在下划線上拆分並放入這樣的嵌套字典中:
b = {
'customer': {
'name': 'bob',
'phone': '555-1212'
},
'order': {
'0': {
'date': '12/12/12',
'number': '...' },
'1': { ... etc.
我擁有的實際數據比這更深層嵌套。 我已經很深入了,但是我一直在思考如何在 Python 中做到這一點:
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
但我不完全確定這是否是正確的方法。 我還沒有運行這段代碼,現在只是想考慮一下。
此外,dict 鍵將來可能會改變,所以我只能依賴“_”下划線在那里拆分。 我也不知道它需要嵌套多深。
通用解決方案:
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()
這是您可以使用的步驟,但由於您沒有向我們展示您的任何代碼,因此我不會提供代碼。
for key in yourDico
使用for key in yourDico
customer
並order
以拆分值以獲得所需的值。 _
重復拆分,這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)
輸出:
{'customer': {'name': 'bob', 'phone': '555-1212'},
'order': {'0': {'date': '12/12/12', 'number': 'A33432-24'},
'1': {'date': '12/14/12', 'number': 'asd24222'}}}
另一個帶有一些評論的通用解決方案:
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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