<p>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:OP发布的函数(<code>opcount</code>);对文件中的行进行简单的迭代(<code>simplecount</code>);带有内存映射文件(mmap)的readline(<code>mapcount</code>);以及Mykola Kharechko提供的缓冲区读取解决方案(<code>bufcount</code>)。</p>
<p>我运行每个函数五次,计算120万行文本文件的平均运行时间。</p>
<p>Windows XP、Python 2.5、2GB RAM、2 GHz AMD处理器</p>
<p>以下是我的结果:</p>
<pre><code>mapcount : 0.465599966049
simplecount : 0.756399965286
bufcount : 0.546800041199
opcount : 0.718600034714
</code></pre>
<p><strong>编辑</strong>:Python2.6的数字:</p>
<pre><code>mapcount : 0.471799945831
simplecount : 0.634400033951
bufcount : 0.468800067902
opcount : 0.602999973297
</code></pre>
<p>因此,缓冲区读取策略似乎是Windows/Python 2.6中最快的</p>
<p>代码如下:</p>
<pre><code>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))
</code></pre>