我修改了缓冲区的情况如下:

def CountLines(filename):
    f = open(filename)
    try:
        lines = 1
        buf_size = 1024 * 1024
        read_f = f.read # loop optimization
        buf = read_f(buf_size)

        # Empty file
        if not buf:
            return 0

        while buf:
            lines += buf.count('\n')
            buf = read_f(buf_size)

        return lines
    finally:
        f.close()

现在空文件和最后一行(不带\n)也被计算在内。

如果文件能放进内存，那么

with open(fname) as f:
    count = len(f.read().split(b'\n')) - 1

您可以执行子进程并运行wc -l filename

import subprocess

def file_len(fname):
    p = subprocess.Popen(['wc', '-l', fname], stdout=subprocess.PIPE, 
                                              stderr=subprocess.PIPE)
    result, err = p.communicate()
    if p.returncode != 0:
        raise IOError(err)
    return int(result.strip().split()[0])

计数= max(开放(文件))[0]

创建一个可执行脚本文件count.py:

#!/usr/bin/python

import sys
count = 0
for line in sys.stdin:
    count+=1

然后将文件的内容导入python脚本:cat huge.txt | ./count.py。管道也适用于Powershell，因此您将最终计算行数。

对我来说，在Linux上它比简单的解决方案快30%:

count=1
with open('huge.txt') as f:
    count+=1

一句话解决方案:

import os
os.system("wc -l  filename")  

我的代码片段:

>>> os.system('wc -l *.txt')

0 bar.txt
1000 command.txt
3 test_file.txt
1003 total