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
def file_len(fname):
counts = itertools.count()
with open(fname) as f:
for _ in f: counts.next()
return counts.next()
其他回答
print open('file.txt', 'r').read().count("\n") + 1
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。
def count_text_file_lines(path):
with open(path, 'rt') as file:
line_count = sum(1 for _line in file)
return line_count
凯尔的回答
num_lines = sum(1 for line in open('my_file.txt'))
最好的替代方案是什么
num_lines = len(open('my_file.txt').read().splitlines())
这里是两者的性能比较
In [20]: timeit sum(1 for line in open('Charts.ipynb'))
100000 loops, best of 3: 9.79 µs per loop
In [21]: timeit len(open('Charts.ipynb').read().splitlines())
100000 loops, best of 3: 12 µs per loop
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])