如何高效地获取文件行数

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

mapcount : 0.471799945831
simplecount : 0.634400033951
bufcount : 0.468800067902
opcount : 0.602999973297

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('
')
 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))