简单的方法:

1)

>>> f = len(open("myfile.txt").readlines())
>>> f

430

>>> f = open("myfile.txt").read().count('\n')
>>> f
430
>>>

num_lines = len(list(open('myfile.txt')))

没有比这更好的了。

毕竟，任何解决方案都必须读取整个文件，计算出有多少\n，并返回结果。

在不读取整个文件的情况下，你有更好的方法吗?不确定……最好的解决方案总是I/ o受限，你能做的最好的就是确保不使用不必要的内存，但看起来你已经覆盖了这个问题。

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))

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。

这是对其他一些答案的元评论。

The line-reading and buffered \n-counting techniques won't return the same answer for every file, because some text files have no newline at the end of the last line. You can work around this by checking the last byte of the last nonempty buffer and adding 1 if it's not b'\n'. In Python 3, opening the file in text mode and in binary mode can yield different results, because text mode by default recognizes CR, LF, and CRLF as line endings (converting them all to '\n'), while in binary mode only LF and CRLF will be counted if you count b'\n'. This applies whether you read by lines or into a fixed-size buffer. The classic Mac OS used CR as a line ending; I don't know how common those files are these days. The buffer-reading approach uses a bounded amount of RAM independent of file size, while the line-reading approach could read the entire file into RAM at once in the worst case (especially if the file uses CR line endings). In the worst case it may use substantially more RAM than the file size, because of overhead from dynamic resizing of the line buffer and (if you opened in text mode) Unicode decoding and storage. You can improve the memory usage, and probably the speed, of the buffered approach by pre-allocating a bytearray and using readinto instead of read. One of the existing answers (with few votes) does this, but it's buggy (it double-counts some bytes). The top buffer-reading answer uses a large buffer (1 MiB). Using a smaller buffer can actually be faster because of OS readahead. If you read 32K or 64K at a time, the OS will probably start reading the next 32K/64K into the cache before you ask for it, and each trip to the kernel will return almost immediately. If you read 1 MiB at a time, the OS is unlikely to speculatively read a whole megabyte. It may preread a smaller amount but you will still spend a significant amount of time sitting in the kernel waiting for the disk to return the rest of the data.

与此答案类似的一行bash解决方案，使用了现代子进程。check_output功能:

def line_count(filename):
    return int(subprocess.check_output(['wc', '-l', filename]).split()[0])