Question

我需要从过滤器列表中获取该类别中有多少个规则的特定规则。

我试图从过滤器列表中提取这种类型的规则。规则模式如下

“ /example.com $ script，domain = example.com”

第二个例外规则是

“ @@ / example.com $ script，domain = example.com”

具有域锚的第三条规则是

“ ||| example.com

带有锚点和域标记的第四条规则是

“ || jizz.best ^ $ popup，domain = vivo.sx

第五个是

“ @@ || pagead2.googlesyndication.com/pagead/js/adsbygoogle.js$script,domain=quebeccoupongratuit.com

第6个具有域限制的地址如下

“ example.com ### examplebanner

第7个不受域限制的是

“ ### examplebanner

第8个元素隐藏很典型

example.com＃@ ## examplebanner

这些是我必须分别获取的规则的不同类别

a=open('1-19-16anti-adblock-killer-filters.txt','r')
text=a.read()
 line_starts_with_2pipes_no_domain = 0
  line_starts_with_2pipes_with_domain = 0
 line_starts_with_2ats_with_domain = 0
 line_with_domain = 0

 for line in text.split("\n"): 
   if line.startswith("||"):
    if ",domain" in line: 
        line_starts_with_2pipes_with_domain += 1
    else:
        line_starts_with_2pipes_no_domain += 1
  elif line.startswith("@@") and ",domain" in line:
    line_starts_with_2ats_with_domain += 1
   elif ",domain" in line: 
    line_with_domain += 1
   elif line.strip(): 
      print(f"No idea what to do with :{line}")

print("2pipes_no_group", line_starts_with_2pipes_no_domain ) 
print("2pipes_with_group", line_starts_with_2pipes_with_domain ) 
 print("2@_with_group", line_starts_with_2ats_with_domain ) 
 print("line_with_domain", line_with_domain)

我现在正在尝试获取第5，第6，第7和第8条规则。感谢您的答复。

Answer 1

您的正则表达式不适合域前的,：

"\/[a-zA-Z0-9-]+\.[a-zA-Z0-9-.]+domain="
#                  ^^^^^^^^^^^^ no , allowed

您还可以简化很多操作：

with open("easylist.txt") as f: 
    print('There are total Rule With Domain tag are =', f.read().count(",domain="))

应该给您答案',domain='发生的频率。如果文件很大，还可以按行计数：

domain_rule_count = 0
with open("easylist.txt") as f:
    for line in f:
        domain_rule_count += 1 if ",domain=" in line else 0

在评论中的问题后进行编辑：您只需测试所需的内容即可：

text = """ some text
/example.com $script,domain=example.com
@@/example.com $script,domain=example.com 
||example.com
||jizz.best^$popup,domain=vivo.sx
"""

line_starts_with_2pipes_no_domain = 0
line_starts_with_2pipes_with_domain = 0
line_starts_with_2ats_with_domain = 0
line_with_domain = 0

for line in text.split("\n"): 
    if line.startswith("||"):
        if ",domain" in line: 
            line_starts_with_2pipes_with_domain += 1
        else:
            line_starts_with_2pipes_no_domain += 1
    elif line.startswith("@@") and ",domain" in line:
        line_starts_with_2ats_with_domain += 1
    elif ",domain" in line: 
        line_with_domain += 1
    elif line.strip(): 
        print(f"No idea what to do with '{line}'")

print("2pipes_no_group", line_starts_with_2pipes_no_domain ) 
print("2pipes_with_group", line_starts_with_2pipes_with_domain ) 
print("2@_with_group", line_starts_with_2ats_with_domain ) 
print("line_with_domain", line_with_domain)

输出：

No idea what to do with ' some text'
2pipes_no_group 1
2pipes_with_group 1
2@_with_group 1
line_with_domain 1

正则表达式以获取特定模式

1 个答案: