如何以最有效的内存和时间方式获取大文件的行数?

def file_len(filename):
    with open(filename) as f:
        for i, _ in enumerate(f):
            pass
    return i + 1

当前回答

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()

其他回答

这个呢

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

这是我用纯python发现的最快的东西。 你可以通过设置buffer来使用任意大小的内存,不过在我的电脑上2**16似乎是一个最佳位置。

from functools import partial

buffer=2**16
with open(myfile) as f:
        print sum(x.count('\n') for x in iter(partial(f.read,buffer), ''))

我在这里找到了答案为什么在c++中从stdin读取行要比Python慢得多?稍微调整了一下。这是一个非常好的阅读来理解如何快速计数行,尽管wc -l仍然比其他任何方法快75%。

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])
def count_text_file_lines(path):
    with open(path, 'rt') as file:
        line_count = sum(1 for _line in file)
    return line_count

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