即使列表 x 比 y 长,如何使 str x 列表与列表 y 中的列表完全匹配?我想要额外的 x 值与 None 配对

How can I make a list of str x match completely those in list y even when list x is longer than y? I want extra x values paired with None

我正在尝试设置一种方法来匹配电子邮件列表和名称列表作为元组。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(“”)?

基本上,我有 excel 行格式,我将其设置到 pandas 数据帧中:

cust_ID buyer_names buyer_emails
1234 name 1; name 2; name 3 email1; email2; email3; email 4
..... ..... ......

我试过这个:

# 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)]

最终,在遍历此元组并将它们放入查询格式后,如下所示:

  # 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'。我想到了集合模块中的容器,但我并没有真正看到为此使用 defaultdict 的明确方法。

如何使元组包含 email4 并仅将“”值作为名称与之配对?

提前致谢。

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名称的电子邮件,我将它们与空字符串配对,如下所示:

("", email4)