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
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
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))
其他回答
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()
def count_text_file_lines(path):
with open(path, 'rt') as file:
line_count = sum(1 for _line in file)
return line_count
您可以执行子进程并运行wc -l filename
import subprocess
def file_len(fname):
p = subprocess.Popen(['wc', '-l', fname], stdout=subprocess.PIPE,
stderr=subprocess.PIPE)
result, err = p.communicate()
if p.returncode != 0:
raise IOError(err)
return int(result.strip().split()[0])
def file_len(full_path):
""" Count number of lines in a file."""
f = open(full_path)
nr_of_lines = sum(1 for line in f)
f.close()
return nr_of_lines
这是我用的,看起来很干净:
import subprocess
def count_file_lines(file_path):
"""
Counts the number of lines in a file using wc utility.
:param file_path: path to file
:return: int, no of lines
"""
num = subprocess.check_output(['wc', '-l', file_path])
num = num.split(' ')
return int(num[0])
更新:这比使用纯python略快,但以内存使用为代价。子进程在执行您的命令时将派生一个与父进程具有相同内存占用的新进程。