问题描述
我正在尝试设置一种将电子邮件列表和名称列表作为元组进行匹配的方法。但是,我发现当它到达姓氏时,那些没有要配对的名字的电子邮件不包含在我的元组中,我怎样才能让这些额外的电子邮件简单地配对一个空字符串(“”)?
本质上,我有格式的 excel 行,我将其设置为 Pandas 数据帧:
我试过了:
# Set regular expression to catch emails
regex = r"[a-zA-Z0-9_.+-]*@[a-zA-Z0-9-]+.[a-zA-Z\.]*"
# Initialise empty list to add query ready emails
emails_query_format = []
# Iterate over retailer_id / emails template rows and append formatted emails to list
for i,row in df.iterrows():
# Put all emails in the row into a list
emails = re.findall(regex,df['additional_emails'][i])
emails = [email.strip() for email in emails]
# Put all additional buyers into a list
buyer_names = row['additional_buyers']
buyers = re.split(r";",buyer_names)
buyers = [buyer.strip() for buyer in buyers]
buyer_email_tuple = [*zip(emails,buyers)]
# For each pair I want to create a row with the formated
for email,buyer in buyer_email_tuple:
# Here I am just putting it into a specific format to copy paste to query template
query_format = "(" + str(row['retailer_id']) + "," + "'" + buyer + "'" + "," + "'" + \
email + "'" + ")" + ","
emails_query_format.append(query_format)
# New DataFrame to input query ready emails
query_df = pd.DataFrame(emails_query_format,columns=['query_ready'])
这样,元组就不会包含额外的“email4”。我想到了 collections 模块中的容器,但我并没有真正看到为此使用 defaultdict 的明确方法。
如何让元组包含 email4,并且只用一个 "" 值作为与之配对的名称?
提前致谢。
解决方法
解决了问题:
for idx in range(len(emails)):
if idx <= len(buyers) -1:
buyer_emails_tuple_list.append((buyers[idx],emails[idx]))
elif idx > len(buyers) -1:
buyer_emails_tuple_list.append(("",emails[idx]))
现在我可以确保对于那些没有相应买家姓名的电子邮件,我将它们与空字符串配对为:
("",email4)