消除重复行并写入文件

Question

file1.txt包含以下行：

[SUM]   0.00-34.53  sec  2.11 GBytes   524 Mbits/sec                  sender    
[SUM]   0.00-34.53  sec  2.11 GBytes   524 Mbits/sec                  sender    
[SUM]   0.00-34.62  sec  2.36 GBytes   586 Mbits/sec                  sender    
[SUM]   0.00-34.62  sec  2.36 GBytes   586 Mbits/sec                  sender 
[SUM]   0.00-34.75  sec  2.39 GBytes   591 Mbits/sec                  receiver  
[SUM]   0.00-34.75  sec  2.39 GBytes   591 Mbits/sec                  receiver  
[0]   0.00-34.53  sec  0.00 Bytes  0.00 bits/sec                  receiver 
[0]   0.00-34.75  sec  0.00 Bytes  0.00 bits/sec                  sender 

将以[SUM]开头、以sender和receiver结尾的行打印到另一个文本文件-file2.txt中

代码如下：

with open(r"C:\Users\file1.txt", 'r') as f:

 contents = f.read()
 s=contents

def my_function1():

 regex = "^\s*\[SUM\]\s*[0-9\-\.]+\s+sec(?!\s+0\.00 Bytes).*sender.*"   
 items=re.findall(regex,s,re.MULTILINE)
 for y in items:
   file=open('file2.txt', "a")
   file.write(str(y))
   file.write("\n")
   file.close()


def my_function2():

 regex = "^\s*\[SUM\]\s*[0-9\-\.]+\s+sec(?!\s+0\.00 Bytes).*receiver.*"   
 items=re.findall(regex,s,re.MULTILINE)
 for y in items:
   file=open('file2.txt', "a")
   file.write(str(y))
   file.write("\n")
   file.close()
   #print(y)

my_function1()

my_function2()

将输出写入file2.txt，如下所示：

[SUM]   0.00-34.53  sec  2.11 GBytes   524 Mbits/sec                  sender    
[SUM]   0.00-34.53  sec  2.11 GBytes   524 Mbits/sec                  sender    
[SUM]   0.00-34.62  sec  2.36 GBytes   586 Mbits/sec                  sender    
[SUM]   0.00-34.62  sec  2.36 GBytes   586 Mbits/sec                  sender
[SUM]   0.00-34.62  sec  2.36 GBytes   586 Mbits/sec                  sender 
[SUM]   0.00-34.75  sec  2.39 GBytes   591 Mbits/sec                  receiver  
[SUM]   0.00-34.75  sec  2.39 GBytes   591 Mbits/sec                  receiver

应为：仅打印一次

[SUM]   0.00-34.53  sec  2.11 GBytes   524 Mbits/sec                  sender
[SUM]   0.00-34.62  sec  2.36 GBytes   586 Mbits/sec                  sender
[SUM]   0.00-34.75  sec  2.39 GBytes   591 Mbits/sec                  receiver 

Answer 1

此处不需要re模块，也不必加载内存中的所有内容：

with open(r"C:\Users\file1.txt", 'r') as f, open('file2.txt', "w") as file:
    seen = set()     # use a set to only keep distinct lines
    for line in f:   # iterate the input file
        lr = line.rstrip()
        if line.startswith('one') and lr.endswith('apple'):
            if lr not in seen:
                seen.add(lr)
                _ = file.write(line)

---

如果搜索实际上更复杂并且需要re模块，我仍然坚持一次处理一行，并在循环之外编译正则表达式：

with open(r"C:\Users\file1.txt", 'r') as f, open('file2.txt', "w") as file:
    seen = set()     # use a set to only keep distinct lines
    rx = re.compile(pattern)
    for line in f:   # iterate the input file
        lr = line.rstrip()
        if rx.match(lr):
            if lr not in seen:
                seen.add(lr)
                _ = file.write(line)

---

如果需要搜索2种模式，并确保第一种模式的匹配在第二种模式的匹配之前写入，则可以使用：

patterns = ["^\s*\[SUM\]\s*[0-9\-\.]+\s+sec(?!\s+0\.00 Bytes).*sender.*",
            "^\s*\[SUM\]\s*[0-9\-\.]+\s+sec(?!\s+0\.00 Bytes).*receiver.*"]
rxs = [re.compile(pattern) for pattern in patterns]

with open(r"C:\Users\file1.txt", 'r') as f:
    data = [[], []]
    seen = set()     # use a set to only keep distinct lines
    for line in f:   # iterate the input file
        lr = line.rstrip()
        for i, rx in enumerate(rxs):
            if rx.match(lr):
                if lr not in seen:
                    seen.add(lr)
                    data[i].append(line)
with open('file2.txt', "w") as file:
    for lst in data:
        for line in lst:
            _ = file.write(line)
    print(file.getvalue())

它给出了预期的结果：

[SUM]   0.00-34.53  sec  2.11 GBytes   524 Mbits/sec                  sender    
[SUM]   0.00-34.62  sec  2.36 GBytes   586 Mbits/sec                  sender    
[SUM]   0.00-34.75  sec  2.39 GBytes   591 Mbits/sec                  receiver