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

我相信内存映射文件将是最快的解决方案。我尝试了四个函数:由OP发布的函数(opcount);对文件中的行进行简单迭代(simplecount);带有内存映射字段(mmap)的Readline (mapcount);以及Mykola Kharechko (buffcount)提供的缓冲区读取解决方案。

我将每个函数运行五次，并计算出120万在线文本文件的平均运行时间。

Windows XP, Python 2.5, 2GB RAM, 2ghz AMD处理器

以下是我的结果:

mapcount : 0.465599966049
simplecount : 0.756399965286
bufcount : 0.546800041199
opcount : 0.718600034714

编辑:Python 2.6的数字:

mapcount : 0.471799945831
simplecount : 0.634400033951
bufcount : 0.468800067902
opcount : 0.602999973297

因此，对于Windows/Python 2.6，缓冲区读取策略似乎是最快的

代码如下:

from __future__ import with_statement
import time
import mmap
import random
from collections import defaultdict

def mapcount(filename):
    f = open(filename, "r+")
    buf = mmap.mmap(f.fileno(), 0)
    lines = 0
    readline = buf.readline
    while readline():
        lines += 1
    return lines

def simplecount(filename):
    lines = 0
    for line in open(filename):
        lines += 1
    return lines

def bufcount(filename):
    f = open(filename)                  
    lines = 0
    buf_size = 1024 * 1024
    read_f = f.read # loop optimization

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

    return lines

def opcount(fname):
    with open(fname) as f:
        for i, l in enumerate(f):
            pass
    return i + 1


counts = defaultdict(list)

for i in range(5):
    for func in [mapcount, simplecount, bufcount, opcount]:
        start_time = time.time()
        assert func("big_file.txt") == 1209138
        counts[func].append(time.time() - start_time)

for key, vals in counts.items():
    print key.__name__, ":", sum(vals) / float(len(vals))

这个呢

def file_len(fname):
  counts = itertools.count()
  with open(fname) as f: 
    for _ in f: counts.next()
  return counts.next()

我得到了一个小(4-8%)的改进，这个版本重用了一个常量缓冲区，所以它应该避免任何内存或GC开销:

lines = 0
buffer = bytearray(2048)
with open(filename) as f:
  while f.readinto(buffer) > 0:
      lines += buffer.count('\n')

您可以调整缓冲区大小，可能会看到一些改进。

下面这句话怎么样:

file_length = len(open('myfile.txt','r').read().split('\n'))

用这种方法在一个3900行的文件上计时只需要0.003秒

def c():
  import time
  s = time.time()
  file_length = len(open('myfile.txt','r').read().split('\n'))
  print time.time() - s

为什么下面的方法行不通呢?

import sys

# input comes from STDIN
file = sys.stdin
data = file.readlines()

# get total number of lines in file
lines = len(data)

print lines

在这种情况下，len函数使用输入行作为确定长度的方法。