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

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

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。

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

计数= max(开放(文件))[0]

我发现你可以。

f = open("data.txt")
linecout = len(f.readlines())

会给你答案吗

这个呢

def file_len(fname):
  counts = itertools.count()
  with open(fname) as f: 
    for _ in f: counts.next()
  return counts.next()