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
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
其他回答
与此答案类似的一行bash解决方案,使用了现代子进程。check_output功能:
def line_count(filename):
return int(subprocess.check_output(['wc', '-l', filename]).split()[0])
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
def line_count(path):
count = 0
with open(path) as lines:
for count, l in enumerate(lines, start=1):
pass
return count
计数= max(开放(文件))[0]
下面这句话怎么样:
file_length = len(open('myfile.txt','r').read().split('\n'))
用这种方法在一个3900行的文件上计时只需要0.003秒
def c():
import time
s = time.time()
file_length = len(open('myfile.txt','r').read().split('\n'))
print time.time() - s