Question

我希望通过比较两个文件中的电子邮件，在两个文件中找到匹配的电子邮件并发送日期。我有两个文件1）maillog.txt（postfix maillog）和2）testmail.txt（包含由换行符分隔的电子邮件）我使用re从maillog.txt文件中提取电子邮件和发送日期，如下所示，

Nov  3 10:08:43 server postfix/smtp[150754]: 78FA8209EDEF: to=<adamson@example.com>, relay=aspmx.l.google.com[74.125.24.26]:25, delay=3.2, delays=0.1/0/1.6/1.5, dsn=2.0.0, status=sent (250 2.0.0 OK 1509718076 m11si5060862pls.447 - gsmtp)
Nov  3 10:10:45 server postfix/smtp[150754]: 7C42A209EDEF: to=<addison@linux.com>, relay=mxa-000f9e01.gslb.pphosted.com[67.231.152.217]:25, delay=5.4, delays=0.1/0/3.8/1.5, dsn=2.0.0, status=sent (250 2.0.0 2dvkvt5tgc-1 Message accepted for delivery)
Nov  3 10:15:45 server postfix/smtp[150754]: 83533209EDE8: to=<johndoe@carchcoal.com>, relay=mxa-000f9e01.gslb.pphosted.com[67.231.144.222]:25, delay=4.8, delays=0.1/0/3.3/1.5, dsn=2.0.0, status=sent (250 2.0.0 2dvm8yww64-1 Message accepted for delivery)
Nov  3 10:16:42 server postfix/smtp[150754]: 83A5E209EDEF: to=<jackn@alphanr.com>, relay=aspmx.l.google.com[74.125.200.27]:25, delay=1.6, delays=0.1/0/0.82/0.69, dsn=2.0.0, status=sent (250 2.0.0 OK 1509718555 j186si6198120pgc.455 - gsmtp)
Nov  3 10:17:44 server postfix/smtp[150754]: 8636D209EDEF: to=<sbins@archcoal.com>, relay=mxa-000f9e01.gslb.pphosted.com[67.231.144.222]:25, delay=4.1, delays=0.11/0/2.6/1.4, dsn=2.0.0, status=sent (250 2.0.0 2dvm8ywwdh-1 Message accepted for delivery)
Nov  3 10:18:42 server postfix/smtp[150754]: 8A014209EDEF: to=<leo@adalphanr.com>, relay=aspmx.l.google.com[74.125.200.27]:25, delay=1.9, delays=0.1/0/0.72/1.1, dsn=2.0.0, status=sent (250 2.0.0 OK 1509718675 o2si6032950pgp.46 - gsmtp)

这是我的另一个文件testmail.txt：

adamson@example.com
jdswson@gmail.com

以下是我尝试过的方法，它也有效但我想知道是否有更有效的方法来处理大量的maillogs和电子邮件地址

import re
pattern=r'(?P<month>[A-Za-z]{3})\s{1,3}(?P<day>\d{1,2})\s{1,2}(?P<ts>\d+:\d+:\d+).*to=<(?P<email>([\w\.-]+)@([\w\.-]+))'
with open("testmail.txt") as fh1:
    for addr in fh1:
        if addr:
            with open("maillog.txt") as fh:
                for line in fh:
                    if line:
                        match=re.finditer(pattern,line)
                        for obj in match:
                            addr=addr.strip()
                            addr2=obj.group('email').strip()
                            if addr == addr2:
                                print(obj.groupdict('email'))

如果找到匹配，这将打印出来：

{'month': 'Nov', 'day': '3', 'ts': '10:08:43', 'email': 'adamson@example.com'}

Answer 1

这是我的解决方案

In [1]: import re

In [2]: pat = r'(?P<month>[A-Za-z]{3})\s{1,3}(?P<day>\d{1,2})\s{1,2}(?P<ts>\d+:\d+:\d+).*to=<(?P<email>([\w\.-]+)@([\w\.-]+))'

In [3]: emails = set()

In [4]: date_email = {}

In [6]: with open('maillog.txt', mode='r') as f:
   ...:     for line in f:
   ...:         month, day, ts, email = re.search(pat, line).group('month', 'day', 'ts', 'email')
   ...:         date_email[email] = (month, day, ts)
   ...:         

In [7]: date_email
Out[7]: 
{'adamson@example.com': ('Nov', '3', '10:08:43'),
 'addison@linux.com': ('Nov', '3', '10:10:45'),
 'jackn@alphanr.com': ('Nov', '3', '10:16:42'),
 'johndoe@carchcoal.com': ('Nov', '3', '10:15:45'),
 'leo@adalphanr.com': ('Nov', '3', '10:18:42'),
 'sbins@archcoal.com': ('Nov', '3', '10:17:44')}

In [11]: with open('testmail.txt', mode='r') as f:
    ...:     for line in f:
    ...:         emails.add(line.strip())
    ...:         

In [12]: emails
Out[12]: {'adamson@example.com', 'jdswson@gmail.com'}

In [15]: for email in emails:
    ...:     if email in date_email:
    ...:         print(email, date_email[email])
    ...:         
('adamson@example.com', ('Nov', '3', '10:08:43'))

您可以按照自己的方式格式化输出。

与＆＃34;以及＃34;一起公开声明关键字可以像这样组合

with open(file1, mode='r') as f1, open(file2, mode='r') as f2:
    # do something with f1
    # do something with f2

Answer 2

您可以尝试使用正则表达式并捕获该组：

让我们分三步解决您的解决方案：

首先从email.txt中捕获所有电子邮件地址：

emails=[]
    with open('emails.txt','r') as f:
        for line in f:
            emails.append(re.search(email_pattern,line).group())

从data.txt中捕获所需数据的第二步：

with open('data.txt','r') as f:
        month_day=[[find.group(4) if find.group(4) != None else [find.group(1), find.group(2), find.group(3)] for find in re.finditer(pattern,line)]for line in f]

第三步：现在我们拥有所有数据，只需检查该电子邮件是否在我们的数据列表然后将该组信息添加到dict：

for item in month_day:
        final_dict = {}
        if item[1] in emails:
            final_dict['month'] = item[0][0]
            final_dict['day'] = item[0][1]
            final_dict['ts'] = item[0][2]
            final_dict['email'] = item[1]
        if final_dict:
            print(final_dict)

完整代码：

 import re
pattern='^(\w{0,3})\s.(\d)\s(\d.+?\s)|<(\w+[@]\w+[.]\w+)>'
email_pattern='\w+[@]\w+[.]\w+'

emails=[]
with open('emails.txt','r') as f:
    for line in f:
        emails.append(re.search(email_pattern,line).group())
with open('data.txt','r') as f:
    month_day=[[find.group(4) if find.group(4) != None else [find.group(1), find.group(2), find.group(3)] for find in re.finditer(pattern,line)]for line in f]


for item in month_day:
    final_dict = {}
    if item[1] in emails:
        final_dict['month'] = item[0][0]
        final_dict['day'] = item[0][1]
        final_dict['ts'] = item[0][2]
        final_dict['email'] = item[1]
    if final_dict:
        print(final_dict)

输出：

{'ts': '10:08:43 ', 'month': 'Nov', 'email': 'adamson@example.com', 'day': '3'}

正则表达式信息：

^ asserts position at start of a line
\w{0,3} matches any word character (equal to [a-zA-Z0-9_])
\s matches any whitespace character (equal to [\r\n\t\f\v ])
\d matches a digit (equal to [0-9])

Answer 3

快速且未经测试但在概念上足够简单：编写一个大的whoppin＆＃39;带有所有地址的正则表达式。

import re

with open("testmail.txt") as fh1:
    emails = []
    for addr in fh1:
        emails.append(re.escape(addr.strip()))
    pattern=re.compile(
        r'(?P<month>[A-Za-z]{3})\s{1,3}(?P<day>\d{1,2})\s{1,2}(?P<ts>\d+:\d+:\d+).*to=<(?P<email>%s)' %
            '|'.join(emails))

with open("maillog.txt") as fh:
    for line in fh:
        for match in pattern.finditer(line):
            print(match.groupdict())

Answer 4

我的建议是将来自testmail.txt的所有电子邮件存储在一个集合中，编译正则表达式，然后迭代maillog.txt的行，并在邮件中搜索集合。这样，只有较短的文件必须驻留在内存中，正则表达式模式只编译一次，并且研究是在为这种访问优化的集合中完成的：

import re
pattern=r'(?P<month>[A-Za-z]{3})\s{1,3}(?P<day>\d{1,2})\s{1,2}(?P<ts>\d+:\d+:\d+).*to=<(?P<email>([\w\.-]+)@([\w\.-]+))'

# load the testmail file into a set
mails = set()
with open('testmail.txt') as fd:
    for line in fd:
        mails.add(line.strip())

#compile the regex once
rx = re.compile(pattern)

#process the maillog file:
with open('maillog.txt') as fd:
    for line in fd:
        m = rx.match(line)
        if m is not None and m.groupdict()['email'] in mails:
            print(m.groupdict())

示例数据的输出符合预期：

{'month': 'Nov', 'day': '3', 'ts': '10:08:43', 'email': 'adamson@example.com'}

如何使用python比较来自两个不同文件的行的字符串？

4 个答案: