python - 根据字符串类型（样式）转换列表元素

Question

我有一个如下列表，

['Message-ID: <5525962.1075855679785.JavaMail.evans@thyme>\r\n',
'Date: Wed, 13 Dec 2000 07:04:00 -0800 (PST)\r\n',
'From: [e-mail]\r\n',
'To: [e-mail], [e-mail], [e-mail], \r\n',
'\[e-mail], [e-mail], [e-mail],\r\n',
'\[e-mail], [e-mail], \r\n',
'\[e-mail], [e-mail], \r\n',
'\[e-mail], [e-mail], \r\n',
'\[e-mail]\r\n',
'Subject: \r\n',
'Mime-Version: 1.0\r\n']

在该To属性中仅包含 3 个电子邮件 ID，然后是该元素，一些元素以\t. 实际上那些列表元素是属性\t的延续。To我的目标是我想合并To属性中所有缺失的元素。

到目前为止，我使用下面的代码来解决我的问题。

l=['Message-ID: <5525962.1075855679785.JavaMail.evans@thyme>\r\n',
'Date: Wed, 13 Dec 2000 07:04:00 -0800 (PST)\r\n',
'From: [e-mail]\r\n',
'To: [e-mail], [e-mail], [e-mail], \r\n',
'\[e-mail], [e-mail], [e-mail],\r\n',
'\[e-mail], [e-mail], \r\n',
'\[e-mail], [e-mail], \r\n',
'\[e-mail], [e-mail], \r\n',
'\[e-mail]\r\n',
'Subject: \r\n',
'Mime-Version: 1.0\r\n']
act= [ele.rstrip('\r\n') for ele in l if ele.startswith('To: ')]
rem=[ele.lstrip('\t').rstrip('\r\n') for ele in l if ele.startswith('\t')]
act.extend(rem)
act=[''.join(act)]

l=[ele for ele in l if not ele.startswith('To: ') and not ele.startswith('\t')]
l.extend(act)
print l

输出：

['Message-ID: <5525962.1075855679785.JavaMail.evans@thyme>\r\n',
'Date: Wed, 13 Dec 2000 07:04:00 -0800 (PST)\r\n',
'From: [e-mail]\r\n',
'Subject: \r\n',
'Mime-Version: 1.0\r\n',
'To: [e-mail], [e-mail], [e-mail], [e-mail], [e-mail], [e-mail],[e-mail], [e-mail], [e-mail], [e-mail], [e-mail], [e-mail], [e-mail]']

我认为我的代码变得更加复杂。

是否有任何简单的方法或任何其他更好的方法来解决这个问题？或者我在哪里可以提高我的代码效率？

任何努力都将是非常可观的。

提前致谢。

标签： pythonlistlist-comprehension

Answer 1

您正在解析电子邮件，这非常棘手，因为有很多极端情况。您应该查看 python电子邮件模块以避免许多陷阱。

import email

headers = ['Message-ID: <5525962.1075855679785.JavaMail.evans@thyme>\r\n',
  'Date: Wed, 13 Dec 2000 07:04:00 -0800 (PST)\r\n',
  'From: [e-mail]\r\n',
  'To: [e-mail], [e-mail], [e-mail], \r\n',
  '\[e-mail], [e-mail], [e-mail],\r\n',
  '\[e-mail], [e-mail], \r\n',
  '\[e-mail], [e-mail], \r\n',
  '\[e-mail], [e-mail], \r\n',
  '\[e-mail]\r\n',
  'Subject: \r\n',
  'Mime-Version: 1.0\r\n']

mail = email.message_from_string("".join(headers)+"\r\n"+"foo body") # rebuild mail message and parse

for to in email.utils.getaddresses(mail.get_all("to")):
    print(to[1])

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