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

如果你想在Linux下的Python中廉价地获取行数,我推荐这个方法:

import os
print os.popen("wc -l file_path").readline().split()[0]

File_path可以是抽象文件路径,也可以是相对路径。希望这能有所帮助。

其他回答

这个呢

def file_len(fname):
  counts = itertools.count()
  with open(fname) as f: 
    for _ in f: counts.next()
  return counts.next()
def file_len(full_path):
  """ Count number of lines in a file."""
  f = open(full_path)
  nr_of_lines = sum(1 for line in f)
  f.close()
  return nr_of_lines
def line_count(path):
    count = 0
    with open(path) as lines:
        for count, l in enumerate(lines, start=1):
            pass
    return count

为什么不读取前100行和后100行,然后估计平均行长,然后用这些数字除以总文件大小呢?如果你不需要一个确切的值,这可以工作。

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