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
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。
其他回答
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))
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。
这个呢
def file_len(fname):
counts = itertools.count()
with open(fname) as f:
for _ in f: counts.next()
return counts.next()
我会使用Python的文件对象方法readlines,如下所示:
with open(input_file) as foo:
lines = len(foo.readlines())
这将打开文件,在文件中创建一个行列表,计算列表的长度,将其保存到一个变量中,然后再次关闭文件。
为了完成上述方法,我尝试了fileinput模块的一个变体:
import fileinput as fi
def filecount(fname):
for line in fi.input(fname):
pass
return fi.lineno()
并将一个60mil行文件传递给上述所有方法:
mapcount : 6.1331050396
simplecount : 4.588793993
opcount : 4.42918205261
filecount : 43.2780818939
bufcount : 0.170812129974
这让我有点惊讶,fileinput是如此糟糕,比所有其他方法都要糟糕得多…