比较文件中的文件模式

Question

如何将文件与另一个文件中的模式进行比较

• 带模式的文件（patterns.txt）

  pattern1
  pattern2
  pattern3
  

• 文件（file.txt）

  something1
  something3
  pattern1
  something4
  something5
  pattern2
  something6
  pattern3
  something7
  

• 脚本

  #!/usr/bin/python
  
  import sys
  
  patternFile = sys.argv[1]
  fileName = sys.argv[2]
  
  with open(patternFile, 'r') as file_open:
      for pattern in file_open:
              with open(fileName, 'r') as file_to_compare:
                      if pattern in file_to_compare:
                              print pattern
  

它不起作用;我错过了什么？

./ searchPattern.py patterns.txt file.txt

由于

Answer 1

你有没有尝试过这样的事情：

patternList = open(patternFile, 'r')
patternListLines = patternList.readlines() 
fileTXT = open(fileName , 'r')

def checkPattern(patternList ,x):
    if x in patternList:
        return True

matchPattern=[x for x in fileTXT.readlines() if checkPattern(patternList, x)]

Answer 2

• 我没有，但它现在正在运作。

• 我在file_to_compare中为添加了：

• 现在工作了！

Sctipt：

#!/usr/bin/python 
import sys 
patternFile = sys.argv[1] 
fileName = sys.argv[2] 
with open(patternFile, 'r') as file_open: 
    for pattern in file_open: 
        with open(fileName, 'r') as file_to_compare: 
            for line in file_to_compare: 
                if pattern in line: print pattern 

./searchPattern.py patterns.txt file.txt 
pattern1 
pattern2 
pattern3