如果“车库”中只有一个“汽车”,以下代码将产生一个错误:
import xmltodict
mydict = xmltodict.parse(xmlstringResults)
for carsInGarage in mydict['garage']['car']:
# do something...原因是,只有当“汽车”中有多个元素时,mydict['garage']['car']才是一个列表。所以我做了这样的事:
import xmltodict
mydict = xmltodict.parse(xmlstringResults)
if isinstance(mydict['garage']['car'], list):
for carsInGarage in mydict['garage']['car']:
# do something for each car...
else:
# do something for the car让代码运行。但对于更先进的行动来说,这不是解决办法。
有人知道某种函数可以使用吗,即使只有一个元素?
发布于 2016-05-19 09:42:02
当然,这不是一种优雅的方法,但这是我为运行代码所做的(如果有人在google上发现了同样的问题):
import xmltodict
def guaranteed_list(x):
if not x:
return []
elif isinstance(x, list):
return x
else:
return [x]
mydict = xmltodict.parse(xmlstringResults)
for carsInGarage in guaranteed_list(mydict['garage']['car']):
# do something...但我想,我会再次编写我的代码,并“直接使用XML”,正如其中一条评论所述。
发布于 2016-08-21 20:07:18
发布于 2020-11-06 03:38:58
我使用的组合
1)
json_dict = xmltodict.parse(s, force_list={'item'})和
2)
# Removes a level in python dict if it has only one specific key
#
# Examples:
# recursive_skip_dict_key_level({"c": {"a": "b"}}, "c") # -> {"a", "b"}
# recursive_skip_dict_key_level({"c": ["a", "b"]}, "c") # -> ["a", "b"]
#
def recursive_skip_dict_key_level(d, skipped_key):
if issubclass(type(d), dict):
if list(d.keys()) == [skipped_key]:
return recursive_skip_dict_key_level(d[skipped_key], skipped_key)
else:
for key in d.keys():
d[key] = recursive_skip_dict_key_level(d[key], skipped_key)
return d
elif issubclass(type(d), list):
new_list = []
for e in d:
new_list.append(recursive_skip_dict_key_level(e, skipped_key))
return new_list
else:
return d# Removes None values from a dict
#
# Examples:
# recursive_remove_none({"a": None}) # -> {}
# recursive_remove_none([None]) # -> []
#
def recursive_remove_none(d):
if issubclass(type(d), dict):
new_dict = {}
for key in d.keys():
if not (d[key] is None):
new_dict[key] = recursive_remove_none(d[key])
return new_dict
elif issubclass(type(d), list):
new_list = []
for e in d:
if not (e is None):
new_list.append(recursive_remove_none(e))
return new_list
else:
return d json_dict = recursive_skip_dict_key_level(json_dict, "item")
json_dict = recursive_remove_none(json_dict)将任何"item“XML-元素解释为列表。
https://stackoverflow.com/questions/37207353
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