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.

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')))

一行，可能很快:

num_lines = sum(1 for line in open('myfile.txt'))

我不得不在类似的问题上发表这篇文章，直到我的声誉分数上升了一点(感谢那些撞了我的人!)。

所有这些解决方案都忽略了一种使其运行得更快的方法，即使用无缓冲(原始)接口，使用字节数组，并进行自己的缓冲。(这只适用于Python 3。在Python 2中，原始接口在默认情况下可以使用，也可以不使用，但在Python 3中，您将默认使用Unicode。)

使用一个修改版本的计时工具，我相信下面的代码比任何提供的解决方案都更快(并且稍微更python化):

def rawcount(filename):
    f = open(filename, 'rb')
    lines = 0
    buf_size = 1024 * 1024
    read_f = f.raw.read

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

    return lines

使用单独的生成器函数，运行速度会快一点:

def _make_gen(reader):
    b = reader(1024 * 1024)
    while b:
        yield b
        b = reader(1024*1024)

def rawgencount(filename):
    f = open(filename, 'rb')
    f_gen = _make_gen(f.raw.read)
    return sum( buf.count(b'\n') for buf in f_gen )

这完全可以用itertools内嵌的生成器表达式来完成，但它看起来非常奇怪:

from itertools import (takewhile,repeat)

def rawincount(filename):
    f = open(filename, 'rb')
    bufgen = takewhile(lambda x: x, (f.raw.read(1024*1024) for _ in repeat(None)))
    return sum( buf.count(b'\n') for buf in bufgen )

以下是我的时间安排:

function      average, s  min, s   ratio
rawincount        0.0043  0.0041   1.00
rawgencount       0.0044  0.0042   1.01
rawcount          0.0048  0.0045   1.09
bufcount          0.008   0.0068   1.64
wccount           0.01    0.0097   2.35
itercount         0.014   0.014    3.41
opcount           0.02    0.02     4.83
kylecount         0.021   0.021    5.05
simplecount       0.022   0.022    5.25
mapcount          0.037   0.031    7.46

我使用的最简单和最短的方法是:

f = open("my_file.txt", "r")
len(f.readlines())

打开一个文件的结果是一个迭代器，它可以转换为一个序列，它有一个长度:

with open(filename) as f:
   return len(list(f))

这比显式循环更简洁，并避免了枚举。