ValueError: dictionary update sequence element #0 has length 3; 2 is required when attempting to coerce generator function into dictionary
ValueError: dictionary update sequence element #0 has length 3; 2 is required when attempting to coerce generator function into dictionary
这是我正在使用的 CSV 文件：
"A","B","C","D","E","F","G","H","I","J"
"88",18,1,"<Req TID=""34"" ReqType=""MS""><IISO /><CID>2</CID><MemID>0000</MemID><MemPass /><RequestData><S>[REMOVED]</S><Na /><La /><Card>[REMOVED]</Card><Address /><HPhone /><Mail /></ReqData></Req>","<Response T=""3"" RequestType=""MS""><MS><Memb><PrivateMembers /><Ob>0-12-af</Ob><Locator /></Memb><S>[REMOVED]</S><CNum>[REMOVED]</CNum><FName /><LaName /><Address /><HPhone /><Email /><IISO /><MemID /><MemPass /><T /><CID /><T /></MS></Response>",0-JAN-10 12.00.02 AM,27-JUN-15 12.00.00 AM,"26",667,0
"22",22,1,"<Req TID=""45"" ReqType=""MS""><IISO /><CID>4</CID><MemID>0000</MemID><MemPass /><RequestData><S>[REMOVED]</S><Na /><La /><Card>[REMOVED]</Card><Address /><HPhone /><Mail /></ReqData></Req>","<Response T=""10"" RequestType=""MS""><MS><Memb><PrivateMembers /><Ob>0-12-af</Ob><Locator /></Memb><S>[REMOVED]</S><CNum>[REMOVED]</CNum><FName /><LaName /><Address /><HPhone /><Email /><IISO /><MemID /><MemPass /><T /><CID /><T /></MS></Response>",0-JAN-22 12.00.02 AM,27-JUN-22 12.00.00 AM,"26",667,0
"32",22,1,"<Req TID=""15"" ReqType=""MS""><IISO /><CID>45</CID><MemID>0000</MemID><MemPass /><RequestData><S>[REMOVED]</S><Na /><La /><Card>[REMOVED]</Card><Address /><HPhone /><Mail /></ReqData></Req>","<Response T=""10"" RequestType=""MS""><MS><Memb><PrivateMembers /><Ob>0-12-af</Ob><Locator /></Memb><S>[REMOVED]</S><CNum>[REMOVED]</CNum><FName /><LaName /><Address /><HPhone /><Email /><IISO /><MemID /><MemPass /><T /><CID /><T /></MS></Response>",0-JAN-20 12.00.02 AM,27-JUN-34 12.00.00 AM,"26",667,0
到目前为止，我已经编写了两个生成器函数来解析 列 E 中的 XML 数据，以便将 XML 标签及其文本转换为Python 字典。具体来说，flatten_dict() 函数 returns 一个可迭代的（键，值）对序列。可以将其转换为 list(flatten_dict(root)).
对列表
生成以下元组列表：
[('ResponseRequestType', 'MS'),
('ResponseT', '10'),
('PrivateMembers', None),
('Ob', '0-12-af'),
('Locator', None),
('S', '[REMOVED]'),
('CNum', '[REMOVED]'),
('FName', None),
('LaName', None),
('Address', None),
('HPhone', None),
('Email', None),
('IISO', None),
('MemID', None),
('MemPass', None),
('T', None),
('CID', None),
('T', None)]
实施块，从第 75 行 开始，returns 将 csv 文件的列作为键，将所有行实例作为值的列表对象
对列表 L（或可迭代的对）也可以用 zip(*L)
转置
当我尝试在生成器函数上构造字典时，我的错误发生在 行 79 上。我已经审查了几个帖子，即 this and this。
我意识到我需要传入一组元组，但为什么我收到此错误对我来说是一个悖论。我正在使用 Python 3.4 和 Jupyter notebooks 进行实验。
我欢迎建设性的（这里强调建设性的）反馈。
# In[37]:
import xml.etree.ElementTree as ET
def flatten_list(aList, prefix=''):
for i, element in enumerate(aList, 1):
eprefix = "{}{}".format(prefix, i)
if element:
# treat like dict
if len(element) == 1 or element[0].tag != element[1].tag:
yield flatten_dict(element, eprefix)
# treat like list
elif element[0].tag == element[1].tag:
yield flatten_list(element, eprefix)
elif element.text:
text = element.text.strip()
if text:
yield eprefix[:].rstrip('.'), element.text
def flatten_dict(parent_element, prefix=''):
prefix = prefix + parent_element.tag
if parent_element.items():
for k, v in parent_element.items():
yield prefix + k, v
for element in parent_element:
eprefix = element.tag
if element:
# treat like dict - we assume that if the first two tags
# in a series are different, then they are all different.
if len(element) == 1 or element[0].tag != element[1].tag:
yield flatten_dict(element, prefix=prefix)
# treat like list - we assume that if the first two tags
# in a series are the same, then the rest are the same.
else:
# here, we put the list in dictionary; the key is the
# tag name the list elements all share in common, and
# the value is the list itself
yield flatten_list(element, prefix=eprefix)
# if the tag has attributes, add those to the dict
if element.items():
for k, v in element.items():
yield eprefix+k
# this assumes that if you've got an attribute in a tag,
# you won't be having any text. This may or may not be a
# good idea -- time will tell. It works for the way we are
# currently doing XML configuration files...
elif element.items():
for k, v in element.items():
yield eprefix+k
# finally, if there are no child tags and no attributes, extract
# the text
else:
yield eprefix, element.text
# In[75]:
from glob import iglob
import csv
from collections import OrderedDict
from xml.etree.ElementTree import ParseError
from parsexml2 import flatten_dict, flatten_list
import xml.etree.cElementTree as ElementTree
import csv
headers = set()
results = []
with open('s.csv', 'rU') as infile:
reader = csv.DictReader(infile)
data = {}
for item in reader:
for header, value in item.items():
try:
data[header].append(value)
except KeyError:
data[header] = [value]
client_responses = data['E'] #returns a list of values
for client_response in client_responses:
print('\n' + client_response)
xml_string = (''.join(client_response)) #may be not necessary
print(xml_string)
xml_string = xml_string.replace('&', '')
xml_string = xml_string.replace('�','')
print(xml_string)
try:
roots = ElementTree.fromstring(xml_string) #serialization step here
except ET.ParseError:
print("catastrophic failure")
continue
# In[79]:
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-79-7053ec7639d9> in <module>()
----> 1 dict(flatten_dict(root))
ValueError: dictionary update sequence element #0 has length 3; 2 is required
有时您会产生 eprefix + k（例如一个元素），有时您会产生 eprefix, element.text（例如两个元素）。
可能生成器首先生成的是一个长度为 3 的字符串。
您的 flatten_dict 代码中存在一些问题。
正如@PatrickMaupin 所指出的那样，您有时会产生这样的值 - yield eprefix+k - 如果生成器确实产生了这个，它 dict() 将无法工作，这个可能是导致问题的原因。我相信你想要的是 - yield eprefix+k,v .
您有时会屈服 - flatten_dict(element, prefix=prefix) -（或 flatten_list() 对应），这也不起作用。让我们举一个简单的例子-
>>> def a():
... yield a()
...
>>> for i in a():
... print(i)
...
<generator object a at 0x00593B48>
如您所见，这生成了生成器对象，它没有迭代该生成器对象并生成实际结果。为此，您需要迭代并手动生成结果。示例 -
if len(element) == 1 or element[0].tag != element[1].tag:
for k,v in flatten_dict(element, prefix=prefix):
yield k,v
或者从 Python 3.3 开始，您可以使用 yield from 构造，从另一个可迭代对象中生成值。示例 -
if len(element) == 1 or element[0].tag != element[1].tag:
yield from flatten_dict(element, prefix=prefix)
同样适用于flatten_list()。
这是我正在使用的 CSV 文件：
"A","B","C","D","E","F","G","H","I","J"
"88",18,1,"<Req TID=""34"" ReqType=""MS""><IISO /><CID>2</CID><MemID>0000</MemID><MemPass /><RequestData><S>[REMOVED]</S><Na /><La /><Card>[REMOVED]</Card><Address /><HPhone /><Mail /></ReqData></Req>","<Response T=""3"" RequestType=""MS""><MS><Memb><PrivateMembers /><Ob>0-12-af</Ob><Locator /></Memb><S>[REMOVED]</S><CNum>[REMOVED]</CNum><FName /><LaName /><Address /><HPhone /><Email /><IISO /><MemID /><MemPass /><T /><CID /><T /></MS></Response>",0-JAN-10 12.00.02 AM,27-JUN-15 12.00.00 AM,"26",667,0
"22",22,1,"<Req TID=""45"" ReqType=""MS""><IISO /><CID>4</CID><MemID>0000</MemID><MemPass /><RequestData><S>[REMOVED]</S><Na /><La /><Card>[REMOVED]</Card><Address /><HPhone /><Mail /></ReqData></Req>","<Response T=""10"" RequestType=""MS""><MS><Memb><PrivateMembers /><Ob>0-12-af</Ob><Locator /></Memb><S>[REMOVED]</S><CNum>[REMOVED]</CNum><FName /><LaName /><Address /><HPhone /><Email /><IISO /><MemID /><MemPass /><T /><CID /><T /></MS></Response>",0-JAN-22 12.00.02 AM,27-JUN-22 12.00.00 AM,"26",667,0
"32",22,1,"<Req TID=""15"" ReqType=""MS""><IISO /><CID>45</CID><MemID>0000</MemID><MemPass /><RequestData><S>[REMOVED]</S><Na /><La /><Card>[REMOVED]</Card><Address /><HPhone /><Mail /></ReqData></Req>","<Response T=""10"" RequestType=""MS""><MS><Memb><PrivateMembers /><Ob>0-12-af</Ob><Locator /></Memb><S>[REMOVED]</S><CNum>[REMOVED]</CNum><FName /><LaName /><Address /><HPhone /><Email /><IISO /><MemID /><MemPass /><T /><CID /><T /></MS></Response>",0-JAN-20 12.00.02 AM,27-JUN-34 12.00.00 AM,"26",667,0
到目前为止，我已经编写了两个生成器函数来解析 列 E 中的 XML 数据，以便将 XML 标签及其文本转换为Python 字典。具体来说，flatten_dict() 函数 returns 一个可迭代的（键，值）对序列。可以将其转换为 list(flatten_dict(root)).
生成以下元组列表：
[('ResponseRequestType', 'MS'),
('ResponseT', '10'),
('PrivateMembers', None),
('Ob', '0-12-af'),
('Locator', None),
('S', '[REMOVED]'),
('CNum', '[REMOVED]'),
('FName', None),
('LaName', None),
('Address', None),
('HPhone', None),
('Email', None),
('IISO', None),
('MemID', None),
('MemPass', None),
('T', None),
('CID', None),
('T', None)]
实施块，从第 75 行 开始，returns 将 csv 文件的列作为键，将所有行实例作为值的列表对象
对列表 L（或可迭代的对）也可以用 zip(*L)
当我尝试在生成器函数上构造字典时，我的错误发生在 行 79 上。我已经审查了几个帖子，即 this and this。
我意识到我需要传入一组元组，但为什么我收到此错误对我来说是一个悖论。我正在使用 Python 3.4 和 Jupyter notebooks 进行实验。
我欢迎建设性的（这里强调建设性的）反馈。
# In[37]:
import xml.etree.ElementTree as ET
def flatten_list(aList, prefix=''):
for i, element in enumerate(aList, 1):
eprefix = "{}{}".format(prefix, i)
if element:
# treat like dict
if len(element) == 1 or element[0].tag != element[1].tag:
yield flatten_dict(element, eprefix)
# treat like list
elif element[0].tag == element[1].tag:
yield flatten_list(element, eprefix)
elif element.text:
text = element.text.strip()
if text:
yield eprefix[:].rstrip('.'), element.text
def flatten_dict(parent_element, prefix=''):
prefix = prefix + parent_element.tag
if parent_element.items():
for k, v in parent_element.items():
yield prefix + k, v
for element in parent_element:
eprefix = element.tag
if element:
# treat like dict - we assume that if the first two tags
# in a series are different, then they are all different.
if len(element) == 1 or element[0].tag != element[1].tag:
yield flatten_dict(element, prefix=prefix)
# treat like list - we assume that if the first two tags
# in a series are the same, then the rest are the same.
else:
# here, we put the list in dictionary; the key is the
# tag name the list elements all share in common, and
# the value is the list itself
yield flatten_list(element, prefix=eprefix)
# if the tag has attributes, add those to the dict
if element.items():
for k, v in element.items():
yield eprefix+k
# this assumes that if you've got an attribute in a tag,
# you won't be having any text. This may or may not be a
# good idea -- time will tell. It works for the way we are
# currently doing XML configuration files...
elif element.items():
for k, v in element.items():
yield eprefix+k
# finally, if there are no child tags and no attributes, extract
# the text
else:
yield eprefix, element.text
# In[75]:
from glob import iglob
import csv
from collections import OrderedDict
from xml.etree.ElementTree import ParseError
from parsexml2 import flatten_dict, flatten_list
import xml.etree.cElementTree as ElementTree
import csv
headers = set()
results = []
with open('s.csv', 'rU') as infile:
reader = csv.DictReader(infile)
data = {}
for item in reader:
for header, value in item.items():
try:
data[header].append(value)
except KeyError:
data[header] = [value]
client_responses = data['E'] #returns a list of values
for client_response in client_responses:
print('\n' + client_response)
xml_string = (''.join(client_response)) #may be not necessary
print(xml_string)
xml_string = xml_string.replace('&', '')
xml_string = xml_string.replace('�','')
print(xml_string)
try:
roots = ElementTree.fromstring(xml_string) #serialization step here
except ET.ParseError:
print("catastrophic failure")
continue
# In[79]:
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-79-7053ec7639d9> in <module>()
----> 1 dict(flatten_dict(root))
ValueError: dictionary update sequence element #0 has length 3; 2 is required
有时您会产生 eprefix + k（例如一个元素），有时您会产生 eprefix, element.text（例如两个元素）。
可能生成器首先生成的是一个长度为 3 的字符串。
您的 flatten_dict 代码中存在一些问题。
正如@PatrickMaupin 所指出的那样，您有时会产生这样的值 -
yield eprefix+k- 如果生成器确实产生了这个，它dict()将无法工作，这个可能是导致问题的原因。我相信你想要的是 -yield eprefix+k,v.您有时会屈服 -
flatten_dict(element, prefix=prefix)-（或flatten_list()对应），这也不起作用。让我们举一个简单的例子->>> def a(): ... yield a() ... >>> for i in a(): ... print(i) ... <generator object a at 0x00593B48>如您所见，这生成了生成器对象，它没有迭代该生成器对象并生成实际结果。为此，您需要迭代并手动生成结果。示例 -
if len(element) == 1 or element[0].tag != element[1].tag: for k,v in flatten_dict(element, prefix=prefix): yield k,v或者从 Python 3.3 开始，您可以使用
yield from构造，从另一个可迭代对象中生成值。示例 -if len(element) == 1 or element[0].tag != element[1].tag: yield from flatten_dict(element, prefix=prefix)同样适用于
flatten_list()。