python列表匹配多种模式

Question

我有一个具有字符串元素的python 2-D列表my_list 。 我想找到与pattern_list匹配的嵌套列表，并要删除匹配的嵌套列表。

my_list = [['UserName', 'EmailID', 'NoofLogins', 'Logs'], 
           ['abrretrerfe2', 'abarrett@polarstar.com', 1.0, 'User : Abarrett has logged in successfully. \n'], 
           ['adminadminerfdDcm', 'user@comp-test.in', 2.0, 'User : Admin has Prince comp logged in successfully']
           ['adminadminerfdDcm', 'user@liveonlinecloud.info', 2.0, 'User : Admin Harry liveonline has logged in successfully']
          ]

pattern_list = ['comp-test.in', 'liveonlinecloud.info','alertdomain.in']

例如，从上面的列表中，我希望结果列表如下

result_list = [['UserName', 'EmailID', 'NoofLogins', 'Logs'], 
               ['abrretrerfe2', 'abarrett@polarstar.com', 1.0, 'User : Abarrett has logged in successfully. \n']
              ]

my_list中的第二和第三嵌套列表被删除，因为它们的值与模式之一匹配。 在第一种情况下，comp-test.in被匹配。 在第二种情况下，liveonlinecloud.info匹配。

谁能帮我这个？ -感谢您的帮助

另一种情况：

the_list = [['UserName', 'EmailID', 'NoofLogins', 'Logs']
           ['abariettq7g3sab9s2', 'abariett@polarstar.com', 1.0, 'User : Abariett has logged in successfully. $$'],
           ['adminadswdasdpj362xQ', 'admin@windacademy.us', 1.0, 'User : Admin has logged in successfully. $$ Scan Started Successfully for - Account Name : windacademy.us. $$']
          ]

valid_pattern = ['started', 'initiated', 'stopped', 'added', 'fetched', 'completed', 'deleted', 'updated', 'disabled']

从the_list ，我只想在比赛中的任何项目，该列表valid_pattern列表

output_list = [['adminadswdasdpj362xQ', 'admin@windacademy.us', 1.0, 'User : Admin has logged in successfully. $$ Scan Started Successfully for - Account Name : windacademy.us. $$']
              ]

见output_list ，只有1嵌套列表相匹配的图形元素开始 。 如何获得此输出？

Answer 1

my_list = [['UserName', 'EmailID', 'NoofLogins', 'Logs'], 
           ['abrretrerfe2', 'abarrett@polarstar.com', 1.0, 'User : Abarrett has logged in successfully. \n'], 
           ['adminadminerfdDcm', 'user@comp-test.in', 2.0, 'User : Admin has Prince comp logged in successfully']
           ['adminadminerfdDcm', 'user@liveonlinecloud.info', 2.0, 'User : Admin Harry liveonline has logged in successfully']
      ]
pattern_list = ['comp-test.in', 'liveonlinecloud.info','alertdomain.in']

result_list = []
for row in my_list:
    for pattern in pattern_list:
        if pattern not in row[1]:
            result_list.append(row)

Answer 2

在这种情况下，可以使用简单的列表理解：

result_list = [row for row in my_list if not any(p in row[1] for p in pattern_list)]

Answer 3

您可以在不使用正则表达式的情况下做到这一点，我认为模式可以位于嵌套列表中的任何索引上，而不仅仅是1：

result_list = my_list[:] 
for l in my_list:
    tmp = '_'.join(map(str,l))
    for pattern in pattern_list:
        if pattern in tmp:
            result_list.remove(l)
            break