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
在perfplot分析之后,必须推荐缓冲读取解决方案
def buf_count_newlines_gen(fname):
def _make_gen(reader):
while True:
b = reader(2 ** 16)
if not b: break
yield b
with open(fname, "rb") as f:
count = sum(buf.count(b"\n") for buf in _make_gen(f.raw.read))
return count
它速度快,内存效率高。大多数其他解决方案大约要慢20倍。
代码重现情节:
import mmap
import subprocess
from functools import partial
import perfplot
def setup(n):
fname = "t.txt"
with open(fname, "w") as f:
for i in range(n):
f.write(str(i) + "\n")
return fname
def for_enumerate(fname):
i = 0
with open(fname) as f:
for i, _ in enumerate(f):
pass
return i + 1
def sum1(fname):
return sum(1 for _ in open(fname))
def mmap_count(fname):
with open(fname, "r+") as f:
buf = mmap.mmap(f.fileno(), 0)
lines = 0
while buf.readline():
lines += 1
return lines
def for_open(fname):
lines = 0
for _ in open(fname):
lines += 1
return lines
def buf_count_newlines(fname):
lines = 0
buf_size = 2 ** 16
with open(fname) as f:
buf = f.read(buf_size)
while buf:
lines += buf.count("\n")
buf = f.read(buf_size)
return lines
def buf_count_newlines_gen(fname):
def _make_gen(reader):
b = reader(2 ** 16)
while b:
yield b
b = reader(2 ** 16)
with open(fname, "rb") as f:
count = sum(buf.count(b"\n") for buf in _make_gen(f.raw.read))
return count
def wc_l(fname):
return int(subprocess.check_output(["wc", "-l", fname]).split()[0])
def sum_partial(fname):
with open(fname) as f:
count = sum(x.count("\n") for x in iter(partial(f.read, 2 ** 16), ""))
return count
def read_count(fname):
return open(fname).read().count("\n")
b = perfplot.bench(
setup=setup,
kernels=[
for_enumerate,
sum1,
mmap_count,
for_open,
wc_l,
buf_count_newlines,
buf_count_newlines_gen,
sum_partial,
read_count,
],
n_range=[2 ** k for k in range(27)],
xlabel="num lines",
)
b.save("out.png")
b.show()
其他回答
我得到了一个小(4-8%)的改进,这个版本重用了一个常量缓冲区,所以它应该避免任何内存或GC开销:
lines = 0
buffer = bytearray(2048)
with open(filename) as f:
while f.readinto(buffer) > 0:
lines += buffer.count('\n')
您可以调整缓冲区大小,可能会看到一些改进。
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
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。
我发现你可以。
f = open("data.txt")
linecout = len(f.readlines())
会给你答案吗
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。