Python 如何处理“问题”;“非类型”;错误?
我正在写一个脚本,应该删除重复的条目。数据中的一些人在姓名中输入了两次,因为他们有两个电话号码,并且因为电话号码字段不是数组,要输入多个,他们输入了多个条目 我的脚本将条目更改为具有与列名对应的键的字典,然后遍历每一行。有一个主for循环遍历每一行,然后有一个嵌套for循环遍历每个元素的所有元素,比较它们以检测重复项。当我点击一个重复的,我的代码应该比较电话,电子邮件和网站,然后将它们附加到一个区域,如果它们是唯一的/不匹配的 这是我的密码:Python 如何处理“问题”;“非类型”;错误?,python,Python,我正在写一个脚本,应该删除重复的条目。数据中的一些人在姓名中输入了两次,因为他们有两个电话号码,并且因为电话号码字段不是数组,要输入多个,他们输入了多个条目 我的脚本将条目更改为具有与列名对应的键的字典,然后遍历每一行。有一个主for循环遍历每一行,然后有一个嵌套for循环遍历每个元素的所有元素,比较它们以检测重复项。当我点击一个重复的,我的代码应该比较电话,电子邮件和网站,然后将它们附加到一个区域,如果它们是唯一的/不匹配的 这是我的密码: import csv # This functio
import csv
# This function takes a tab-delim csv and merges the ones with the same name but different phone / email / websites.
def merge_duplicates(sheet):
myjson = [] # myjson = list of dictionaries where each dictionary
with(open("ieca_first_col_fake_text.txt", "rU")) as f:
sheet = csv.DictReader(f,delimiter="\t")
for row in sheet:
myjson.append(row)
write_file = csv.DictWriter(open('duplicates_deleted.csv','w'), ['name','phone','email','website'], restval='', delimiter = '\t')
for row in myjson:
# convert phone, email, and web to lists so that extra can be appended
row['phone'] = row['phone'].split()
row['email'] = row['email'].split()
row['website'] = row['website'].split()
print row
for i in len(myjson):
# if the names match, check to see if phone, em, web match. If any match, append to first row.
try:
if myjson[i]['name'] == myjson[i+1]['name']:
if myjson[i]['phone'] != myjson[i+1]['phone']:
myjson[i]['phone'].append(myjson[i+1]['phone'])
# if row['email'] != myjson[rowvalue+1]['email']:
# row['email'].append(myjson[rowvalue+1]['email'])
# if row['website'] != myjson[rowvalue+1]['website']:
# row['website'].append(myjson[rowvalue+1]['website'])
except IndexError:
print("We're at the end now")
write_file.writerow(row)
merge_duplicates('ieca_first_col_fake_text.txt')
因此,我的代码中的一切都很顺利,然后它遇到了第一个副本,我得到了以下错误:
{'website': [], 'phone': [], 'name': 'Diane Grant Albrecht M.S.', 'email': []}
{'website': ['www.got.com'], 'phone': ['111-222-3333'], 'name': 'Lannister G. Cersei M.A.T., CEP', 'email': ['cersei@got.com']}
{'website': [], 'phone': [], 'name': 'Argle D. Bargle Ed.M.', 'email': []}
{'website': ['www.daManWithThePlan.com'], 'phone': ['000-000-1111'], 'name': 'Sam D. Man Ed.M.', 'email': ['dman123@gmail.com']}
Traceback (most recent call last):
File "/Users/samuelfinegold/Documents/noodle/delete_duplicates.py", line 40, in <module>
merge_duplicates('ieca_first_col_fake_text.txt')
File "/Users/samuelfinegold/Documents/noodle/delete_duplicates.py", line 20, in merge_duplicates
row['email'] = row['email'].split()
AttributeError: 'NoneType' object has no attribute 'split'
logout
错误是,如果
行['phone']
为无
,则无法拆分它
你可以这样做
row['phone'] = row['phone'].split() if row['phone'] else []
row['email'] = row['email'].split() if row['email'] else []
row['website'] = row['website'].split() if row['website'] else []
[]
可以由您想要指定的任何默认值替换(例如:无
或“
)
更干净的方法是
row['phone'] = row['phone'].split() if row.get('phone') else []
row['email'] = row['email'].split() if row.get('email') else []
row['website'] = row['website'].split() if row.get('website') else []
错误是,如果
行['phone']
为无
,则无法拆分它
你可以这样做
row['phone'] = row['phone'].split() if row['phone'] else []
row['email'] = row['email'].split() if row['email'] else []
row['website'] = row['website'].split() if row['website'] else []
[]
可以由您想要指定的任何默认值替换(例如:无
或“
)
更干净的方法是
row['phone'] = row['phone'].split() if row.get('phone') else []
row['email'] = row['email'].split() if row.get('email') else []
row['website'] = row['website'].split() if row.get('website') else []
就我个人而言,我会使用
和
来做到这一点:
row['email'] = row.get('email',[]) and row['email'].split()
# you could also use hasattr(row['email'],'split')
if 'email' in row and isinstance(row['email'],str):
row['email'] = row['email'].split()
逻辑与以下内容相同:
if row.get('email'):
row['email'] = row['email'].split()
虽然严格来说,如果密钥丢失(或电子邮件已被编入列表),这会重新分配,因此您可能需要执行以下操作:
row['email'] = row.get('email',[]) and row['email'].split()
# you could also use hasattr(row['email'],'split')
if 'email' in row and isinstance(row['email'],str):
row['email'] = row['email'].split()
就我个人而言,我会使用
和
来做到这一点:
row['email'] = row.get('email',[]) and row['email'].split()
# you could also use hasattr(row['email'],'split')
if 'email' in row and isinstance(row['email'],str):
row['email'] = row['email'].split()
逻辑与以下内容相同:
if row.get('email'):
row['email'] = row['email'].split()
虽然严格来说,如果密钥丢失(或电子邮件已被编入列表),这会重新分配,因此您可能需要执行以下操作:
row['email'] = row.get('email',[]) and row['email'].split()
# you could also use hasattr(row['email'],'split')
if 'email' in row and isinstance(row['email'],str):
row['email'] = row['email'].split()
只是检查确认一下,但我不确定这是不是错误,因为输出显示有没有电话号码的条目。但错误并未发生。如果有帮助的话,我已经用数据更新了帖子。你的stacktrace清楚地显示了
行['email']=行['email'].split()
,意思是行['email']
是无
也是,“
和无
是不同的。所以,空电话号码不一定意味着它是None
对不起,卡蒂克,你说得对!我当时很困惑,但现在在检查打印出的是哪一行['phone']之后,我看到了区别。此外,我认为.get()可以工作,如果在.get()中出现错误,我可以像这样返回“”:row.get('website','')更简单,尽管如果有很多空字段,它可能会更昂贵:对于输入('phone','email','website'):row[key]=row.get(key),.split()
只是检查一下以确保,但我不确定这是不是错误,因为输出显示有没有电话号码的条目。但错误并未发生。如果有帮助的话,我已经用数据更新了帖子。你的stacktrace清楚地显示了行['email']=行['email'].split()
,意思是行['email']
是无
也是,“
和无
是不同的。所以,空电话号码不一定意味着它是None
对不起,卡蒂克,你说得对!我当时很困惑,但现在在检查打印出的是哪一行['phone']之后,我看到了区别。此外,我认为.get()可以工作,如果在.get()中出现错误,我可以像这样返回“”:row.get('website','')更简单,尽管如果有很多空字段,它可能会更昂贵:对于输入('phone','email','website'):row[key]=row.get(key),.split()