你可以使用操作系统。路径模块如下所示:

import os
import subprocess
Number_lines = int( (subprocess.Popen( 'wc -l {0}'.format( Filename ), shell=True, stdout=subprocess.PIPE).stdout).readlines()[0].split()[0] )

，其中Filename是文件的绝对路径。

计数= max(开放(文件))[0]

大文件的另一种选择是使用xreadlines():

count = 0
for line in open(thefilepath).xreadlines(  ): count += 1

对于Python 3，请参阅:在Python 3中什么替代xreadlines() ?

在perfplot分析之后，必须推荐缓冲读取解决方案

def buf_count_newlines_gen(fname):
    def _make_gen(reader):
        while True:
            b = reader(2 ** 16)
            if not b: break
            yield b

    with open(fname, "rb") as f:
        count = sum(buf.count(b"\n") for buf in _make_gen(f.raw.read))
    return count

它速度快，内存效率高。大多数其他解决方案大约要慢20倍。

代码重现情节:

import mmap
import subprocess
from functools import partial

import perfplot


def setup(n):
    fname = "t.txt"
    with open(fname, "w") as f:
        for i in range(n):
            f.write(str(i) + "\n")
    return fname


def for_enumerate(fname):
    i = 0
    with open(fname) as f:
        for i, _ in enumerate(f):
            pass
    return i + 1


def sum1(fname):
    return sum(1 for _ in open(fname))


def mmap_count(fname):
    with open(fname, "r+") as f:
        buf = mmap.mmap(f.fileno(), 0)

    lines = 0
    while buf.readline():
        lines += 1
    return lines


def for_open(fname):
    lines = 0
    for _ in open(fname):
        lines += 1
    return lines


def buf_count_newlines(fname):
    lines = 0
    buf_size = 2 ** 16
    with open(fname) as f:
        buf = f.read(buf_size)
        while buf:
            lines += buf.count("\n")
            buf = f.read(buf_size)
    return lines


def buf_count_newlines_gen(fname):
    def _make_gen(reader):
        b = reader(2 ** 16)
        while b:
            yield b
            b = reader(2 ** 16)

    with open(fname, "rb") as f:
        count = sum(buf.count(b"\n") for buf in _make_gen(f.raw.read))
    return count


def wc_l(fname):
    return int(subprocess.check_output(["wc", "-l", fname]).split()[0])


def sum_partial(fname):
    with open(fname) as f:
        count = sum(x.count("\n") for x in iter(partial(f.read, 2 ** 16), ""))
    return count


def read_count(fname):
    return open(fname).read().count("\n")


b = perfplot.bench(
    setup=setup,
    kernels=[
        for_enumerate,
        sum1,
        mmap_count,
        for_open,
        wc_l,
        buf_count_newlines,
        buf_count_newlines_gen,
        sum_partial,
        read_count,
    ],
    n_range=[2 ** k for k in range(27)],
    xlabel="num lines",
)
b.save("out.png")
b.show()

已经有很多答案了，但不幸的是，它们中的大多数只是一个几乎不可优化的问题上的微型经济……

在我参与的几个项目中，行数是软件的核心功能，以最快的速度处理大量文件是至关重要的。

行数的主要瓶颈是I/O访问，因为您需要读取每一行以检测行返回字符，因此没有其他方法。第二个潜在的瓶颈是内存管理:一次加载的内存越多，处理的速度就越快，但与第一个瓶颈相比，这个瓶颈可以忽略不计。

因此，除了禁用gc收集和其他微管理技巧等微小优化外，还有3种主要方法可以减少行计数函数的处理时间:

Hardware solution: the major and most obvious way is non-programmatic: buy a very fast SSD/flash hard drive. By far, this is how you can get the biggest speed boosts. Data preparation solution: if you generate or can modify how the files you process are generated, or if it's acceptable that you can pre-process them, first convert the line return to unix style (\n) as this will save 1 character compared to Windows or MacOS styles (not a big save but it's an easy gain), and secondly and most importantly, you can potentially write lines of fixed length. If you need variable length, you can always pad smaller lines. This way, you can calculate instantly the number of lines from the total filesize, which is much faster to access. Often, the best solution to a problem is to pre-process it so that it better fits your end purpose. Parallelization + hardware solution: if you can buy multiple hard disks (and if possible SSD flash disks), then you can even go beyond the speed of one disk by leveraging parallelization, by storing your files in a balanced way (easiest is to balance by total size) among disks, and then read in parallel from all those disks. Then, you can expect to get a multiplier boost in proportion with the number of disks you have. If buying multiple disks is not an option for you, then parallelization likely won't help (except if your disk has multiple reading headers like some professional-grade disks, but even then the disk's internal cache memory and PCB circuitry will likely be a bottleneck and prevent you from fully using all heads in parallel, plus you have to devise a specific code for this hard drive you'll use because you need to know the exact cluster mapping so that you store your files on clusters under different heads, and so that you can read them with different heads after). Indeed, it's commonly known that sequential reading is almost always faster than random reading, and parallelization on a single disk will have a performance more similar to random reading than sequential reading (you can test your hard drive speed in both aspects using CrystalDiskMark for example).

如果这些都不是选择，那么你只能依靠微观管理技巧来提高行数函数的速度，但不要指望有什么真正重要的东西。相反，您可以预期，与您将看到的速度改进回报相比，您花费在调整上的时间将是不均衡的。

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