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
这个呢?
import sys
sys.stdin=open('fname','r')
data=sys.stdin.readlines()
print "counted",len(data),"lines"
其他回答
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。
这个呢
def file_len(fname):
counts = itertools.count()
with open(fname) as f:
for _ in f: counts.next()
return counts.next()
创建一个可执行脚本文件count.py:
#!/usr/bin/python
import sys
count = 0
for line in sys.stdin:
count+=1
然后将文件的内容导入python脚本:cat huge.txt | ./count.py。管道也适用于Powershell,因此您将最终计算行数。
对我来说,在Linux上它比简单的解决方案快30%:
count=1
with open('huge.txt') as f:
count+=1
对我来说,这个变体是最快的:
#!/usr/bin/env python
def main():
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)
print lines
if __name__ == '__main__':
main()
原因:缓冲比逐行和逐字符串读取快。计数也非常快
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))