python 在文件中查找正则表达式匹配次数最多的部分

我每行读取一个文件行并检查代码部分结束的位置:出现特定的字符序列。该序列可能出现在代码部分中，因此我必须检查冗余:连续行中有多少次包含该序列。连续出现 10 次，我应该返回连续出现开始检测代码部分结尾的第一行。

regexp_dict_02 = {'Name': 'EMPTY_PAGES', 'Expr':  '(FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF)'} 
def FindEmptyPg(Inpfile,Section): 
 
 NbrLine = [] 
 Pos = [] 
 flag = 0 
 index = 0 
 Ln = 0 
 
 with open(Inpfile) as fp: 
  for i, line in enumerate(fp): 
   if i >= Section.startline and i < 30061 : 
    s=re.search(regexp_dict_02['Expr'],line) 
    if s:   
     NbrLine.append(i) 
 
 
  logging.info (NbrLine) 
  logging.info (len(NbrLine))  
  for index in range((len(NbrLine))-1): 
   if NbrLine[index+1] - NbrLine[index] == 1 :  
    logging.info (str (NbrLine[index+1]) + '  ' + str(NbrLine[index])) 
    Pos.append (index) 
    flag += 1    
    if flag == 5 :  
     Ln = NbrLine[Pos[0]] 
     break 
  logging.info (Pos) 
  return Ln 

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在上面的代码中，我只检查了两个连续的行，我得到了错误的行号。我避免使用状态机等复杂的东西，但我仍然卡住了。

请您参考如下方法:

这是一种解决方案。以下代码遍历每一行。每次找到匹配项时，它都会将行索引添加到 block 中。一旦找到没有任何匹配项的行，该 block 将被视为“关闭”并创建一个新的空 block ，但在此之前，它会保存 block 的长度和 results 中的第一个索引。这些是您唯一感兴趣的信息。最后，您对 results 进行排序并选择最后一项(元组列表的排序将默认按元组的第一项排序，在本例中为 len block 的)，一个元组，其中包含找到的最长 block 以及该 block 第一行的索引。

t = \ 
''' 
000010000000000000000000000000000000000011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
000010000000000000000000000000000000000011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
000010000000000000000000000000000000000011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
00001FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF00011111 
000010000000000000000000000000000000000011111 
000010000000000000000000000000000000000011111 
000010000000000000000000000000000000000011111 
''' 
 
pattern = 'FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF' 
block = [] 
results = [] 
for i, line in enumerate(t.split('\n')): 
    if pattern in line: 
        block.append(i) 
    else: 
        try: 
            results.append((len(block), block[0])) #save the len and the first index of each block 
            block = [] 
        except IndexError: 
            pass 
 
 
cons, index = sorted(results)[-1] #number of consecutive match, line index 
print(f'max consecutive matches found: {cons} , stating at line {index}') 

输出:

max consecutive matches found: 14 , stating at line 11 

解决评论:

I need the first sufficient successive occurrences: first 10 successive occurrences matched then I catch the line.

您可以改用以下代码。

pattern = 'FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF' 
block = [] 
for i, line in enumerate(t.split('\n')): 
    if pattern in line: 
        block.append(i) 
    else: 
        if len(block) >= 10: 
            print(f'found a block of at least 10 lines starting from line {block[0]}') 
            break 
        block = []