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
我修改了缓冲区的情况如下:
def CountLines(filename):
f = open(filename)
try:
lines = 1
buf_size = 1024 * 1024
read_f = f.read # loop optimization
buf = read_f(buf_size)
# Empty file
if not buf:
return 0
while buf:
lines += buf.count('\n')
buf = read_f(buf_size)
return lines
finally:
f.close()
现在空文件和最后一行(不带\n)也被计算在内。
其他回答
如果文件能放进内存,那么
with open(fname) as f:
count = len(f.read().split(b'\n')) - 1
您可以执行子进程并运行wc -l filename
import subprocess
def file_len(fname):
p = subprocess.Popen(['wc', '-l', fname], stdout=subprocess.PIPE,
stderr=subprocess.PIPE)
result, err = p.communicate()
if p.returncode != 0:
raise IOError(err)
return int(result.strip().split()[0])
计数= max(开放(文件))[0]
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
一句话解决方案:
import os
os.system("wc -l filename")
我的代码片段:
>>> os.system('wc -l *.txt')
0 bar.txt
1000 command.txt
3 test_file.txt
1003 total