遍历文本文件中的每一行以提取唯一的lis

Question

我一直试图从一个文本文件中提取一个唯一的数据名列表，但我似乎做不到，因为我对regex一无所知。在

如果我们有一个例子：

[Friday 17/10/2014 @ 07:30:55] The user user01 | account01 | namename1 has been granted access.
[Friday 17/10/2014 @ 07:30:57] The user user two | account_two | name2 has been granted access.
[Friday 17/10/2014 @ 07:30:59] The user user_three | account_ | name3 here3 has been granted access.
[Friday 17/10/2014 @ 07:31:41] The user user01 | account01 | namename1 has been granted access.

我希望它基本上找到两个管道之间的帐户信息|，并去掉管道和空白，这样它就可以在文本文件中输出一个包含以下内容的列表，在它遍历并删除任何重复项之后，它将严格地说是一个纯列表

^{pr2}$

它必须做的一项检查是确保它只在行中包含短语has been granted access.时获取帐户信息，因为数据可能看起来像：

[Friday 17/10/2014 @ 07:30:55] The user user01 | account01 | namename1 has been granted access.
[Friday 17/10/2014 @ 07:30:57] The user user two | account_two | name2 has been granted access.
[Friday 17/10/2014 @ 07:30:59] Details Granted | user two | access number 01239
[Friday 17/10/2014 @ 07:30:59] The user user_three | account_ | name3 here3 has been granted access.
[Friday 17/10/2014 @ 07:31:41] The user user01 | account01 | namename1 has been granted access.

我不希望它从该示例的第3行获取帐户信息user two。在

有谁能帮我做一些代码的例子吗？我们将不胜感激。在

Answer 1

def get_granted_accounts(filename):
    with open(filename) as f:
      return set(
               s.split('|')[1].strip() 
               for s in f.readlines() 
               if "has been granted access" in s) 

这段代码需要注意：

• 管道不能出现在第一个或第二个字段中（引用、转义）
• “已被授予访问权限”应仅出现在预期字段中（例如，不作为帐户名）

Answer 2

>>> granted_accounts = [line.split('|')[1].strip() for line in open('file.txt') if 'has been granted access' in line]
>>> print(granted_accounts)
['account01', 'account_two', 'account_', 'account01']

如果您想在命令行上执行它，只需将这两行与shebang放在一个.py文件中，如下所示(搜索.py)公司名称：

^{pr2}$

然后像这样跑：

$ python search.py

或者：

$ chmod +x search.py
$ ./search.py

如果您有很多帐户，您可能希望每个帐户只打印一次，并在一行单独打印：

>>> granted_accounts = [line.split('|')[1].strip() for line in open('file.txt') if 'has been granted access' in line]
>>> print('\n'.join(sorted(set(granted_accounts))))
account01
account_
account_two

Answer 3

我完全忽视了斯普利特。。。但以下是一个基于使用split的完全有效的版本：

按|拆分并选择拆分的第二部分，然后去掉所有空白，然后通过检查帐户是否不在列表中生成一个accountlist，这样可以删除重复项

最后但并非最不重要的是，它会将所有帐户输出到输出.txt在

accountlist = []
with open('mydatafile.txt', 'r') as infile: 
    for line in infile:
        if "has been granted access." in line:
            if line.strip().split('|')[1].strip(" ") not in accountlist:
                accountlist.append(line.strip().split('|')[1].strip(" "))
    print accountlist

    with open('output.txt', 'w') as outfile:
        for account in accountlist:
            outfile.write("%s\n" % account)