需要在python中猎取大文件（数十万行）的行数。
def file_len(fname):
with open(fname) as f:
for i, l in enumerate(f):
pass return i + 1
有用的要领（缓冲区读取战略）：
起首看下运转的效果：
mapcount : 0.471799945831 simplecount : 0.634400033951 bufcount : 0.468800067902 opcount : 0.602999973297
因而，关于Windows/Python2.6来讲，缓冲区读取战略似乎是最快的。
以下是代码：
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))
以上就是怎样高效地猎取文件行数的细致内容，更多请关注ki4网别的相干文章！
