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
类似的:
lines = 0
with open(path) as f:
for line in f:
lines += 1
其他回答
print open('file.txt', 'r').read().count("\n") + 1
简单的方法:
1)
>>> f = len(open("myfile.txt").readlines())
>>> f
430
>>> f = open("myfile.txt").read().count('\n')
>>> f
430
>>>
num_lines = len(list(open('myfile.txt')))
这个呢?
import sys
sys.stdin=open('fname','r')
data=sys.stdin.readlines()
print "counted",len(data),"lines"
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。
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。