为什么读取一个字节比从文件读取2,3,4,...字节慢20倍?

问题描述 投票:14回答:1

我一直试图理解readseek之间的权衡。对于小的“跳跃”,读取不需要的数据比使用seek跳过它更快。

在定时不同的读取/搜索块大小以找到临界点时,我遇到了一个奇怪的现象:read(1)read(2)read(3)等慢约20倍。这种效果对于不同的读取方法是相同的,例如read()readinto()

为什么会这样?

搜索以下第2/3行的时间结果:

2 x buffered 1 byte readinto bytearray

环境:

Python 3.5.2 |Continuum Analytics, Inc.| (default, Jul  5 2016, 11:45:57) [MSC v.1900 32 bit (Intel)]

时间结果:

Non-cachable binary data ingestion (file object blk_size = 8192):
- 2 x buffered 0 byte readinto bytearray:
      robust mean: 6.01 µs +/- 377 ns
      min: 3.59 µs
- Buffered 0 byte seek followed by 0 byte readinto:
      robust mean: 9.31 µs +/- 506 ns
      min: 6.16 µs
- 2 x buffered 4 byte readinto bytearray:
      robust mean: 14.4 µs +/- 6.82 µs
      min: 2.57 µs
- 2 x buffered 7 byte readinto bytearray:
      robust mean: 14.5 µs +/- 6.76 µs
      min: 3.08 µs
- 2 x buffered 2 byte readinto bytearray:
      robust mean: 14.5 µs +/- 6.77 µs
      min: 3.08 µs
- 2 x buffered 5 byte readinto bytearray:
      robust mean: 14.5 µs +/- 6.76 µs
      min: 3.08 µs
- 2 x buffered 3 byte readinto bytearray:
      robust mean: 14.5 µs +/- 6.73 µs
      min: 2.57 µs
- 2 x buffered 49 byte readinto bytearray:
      robust mean: 14.5 µs +/- 6.72 µs
      min: 2.57 µs
- 2 x buffered 6 byte readinto bytearray:
      robust mean: 14.6 µs +/- 6.76 µs
      min: 3.08 µs
- 2 x buffered 343 byte readinto bytearray:
      robust mean: 15.3 µs +/- 6.43 µs
      min: 3.08 µs
- 2 x buffered 2401 byte readinto bytearray:
      robust mean: 138 µs +/- 247 µs
      min: 4.11 µs
- Buffered 7 byte seek followed by 7 byte readinto:
      robust mean: 278 µs +/- 333 µs
      min: 15.4 µs
- Buffered 3 byte seek followed by 3 byte readinto:
      robust mean: 279 µs +/- 333 µs
      min: 14.9 µs
- Buffered 1 byte seek followed by 1 byte readinto:
      robust mean: 279 µs +/- 334 µs
      min: 15.4 µs
- Buffered 2 byte seek followed by 2 byte readinto:
      robust mean: 279 µs +/- 334 µs
      min: 15.4 µs
- Buffered 4 byte seek followed by 4 byte readinto:
      robust mean: 279 µs +/- 334 µs
      min: 15.4 µs
- Buffered 49 byte seek followed by 49 byte readinto:
      robust mean: 281 µs +/- 336 µs
      min: 14.9 µs
- Buffered 6 byte seek followed by 6 byte readinto:
      robust mean: 281 µs +/- 337 µs
      min: 15.4 µs
- 2 x buffered 1 byte readinto bytearray:
      robust mean: 282 µs +/- 334 µs
      min: 17.5 µs
- Buffered 5 byte seek followed by 5 byte readinto:
      robust mean: 282 µs +/- 338 µs
      min: 15.4 µs
- Buffered 343 byte seek followed by 343 byte readinto:
      robust mean: 283 µs +/- 340 µs
      min: 15.4 µs
- Buffered 2401 byte seek followed by 2401 byte readinto:
      robust mean: 309 µs +/- 373 µs
      min: 15.4 µs
- Buffered 16807 byte seek followed by 16807 byte readinto:
      robust mean: 325 µs +/- 423 µs
      min: 15.4 µs
- 2 x buffered 16807 byte readinto bytearray:
      robust mean: 457 µs +/- 558 µs
      min: 16.9 µs
- Buffered 117649 byte seek followed by 117649 byte readinto:
      robust mean: 851 µs +/- 1.08 ms
      min: 15.9 µs
- 2 x buffered 117649 byte readinto bytearray:
      robust mean: 1.29 ms +/- 1.63 ms
      min: 18 µs

基准代码:

from _utils import BenchmarkResults

from timeit import timeit, repeat
import gc
import os
from contextlib import suppress
from math import floor
from random import randint

### Configuration

FILE_NAME = 'test.bin'
r = 5000
n = 100

reps = 1

chunk_sizes = list(range(7)) + [7**x for x in range(1,7)]

results = BenchmarkResults(description = 'Non-cachable binary data ingestion')


### Setup

FILE_SIZE = int(100e6)

# remove left over test file
with suppress(FileNotFoundError):
    os.unlink(FILE_NAME)

# determine how large a file needs to be to not fit in memory
gc.collect()
try:
    while True:
        data = bytearray(FILE_SIZE)
        del data
        FILE_SIZE *= 2
        gc.collect()
except MemoryError:
    FILE_SIZE *= 2
    print('Using file with {} GB'.format(FILE_SIZE / 1024**3))

# check enough data in file
required_size = sum(chunk_sizes)*2*2*reps*r
print('File size used: {} GB'.format(required_size / 1024**3))
assert required_size <= FILE_SIZE


# create test file
with open(FILE_NAME, 'wb') as file:
    buffer_size = int(10e6)
    data = bytearray(buffer_size)
    for i in range(int(FILE_SIZE / buffer_size)):
        file.write(data)

# read file once to try to force it into system cache as much as possible
from io import DEFAULT_BUFFER_SIZE
buffer_size = 10*DEFAULT_BUFFER_SIZE
buffer = bytearray(buffer_size)
with open(FILE_NAME, 'rb') as file:
    bytes_read = True
    while bytes_read:
        bytes_read = file.readinto(buffer)
    blk_size = file.raw._blksize

results.description += ' (file object blk_size = {})'.format(blk_size)

file = open(FILE_NAME, 'rb')

### Benchmarks

setup = \
"""
# random seek to avoid advantageous starting position biasing results
file.seek(randint(0, file.raw._blksize), 1)
"""

read_read = \
"""
file.read(chunk_size)
file.read(chunk_size)
"""

seek_seek = \
"""
file.seek(buffer_size, 1)
file.seek(buffer_size, 1)
"""

seek_read = \
"""
file.seek(buffer_size, 1)
file.read(chunk_size)
"""

read_read_timings = {}
seek_seek_timings = {}
seek_read_timings = {}
for chunk_size in chunk_sizes:
    read_read_timings[chunk_size] = []
    seek_seek_timings[chunk_size] = []
    seek_read_timings[chunk_size] = []

for j in range(r):
    #file.seek(0)
    for chunk_size in chunk_sizes:
        buffer = bytearray(chunk_size)
        read_read_timings[chunk_size].append(timeit(read_read, setup, number=reps, globals=globals()))
        #seek_seek_timings[chunk_size].append(timeit(seek_seek, setup, number=reps, globals=globals()))
        seek_read_timings[chunk_size].append(timeit(seek_read, setup, number=reps, globals=globals()))

for chunk_size in chunk_sizes:
    results['2 x buffered {} byte readinto bytearray'.format(chunk_size)] = read_read_timings[chunk_size]
    #results['2 x buffered {} byte seek'.format(chunk_size)] = seek_seek_timings[chunk_size]
    results['Buffered {} byte seek followed by {} byte readinto'.format(chunk_size, chunk_size)] = seek_read_timings[chunk_size]


### Cleanup
file.close()
os.unlink(FILE_NAME)

results.show()
results.save()
python file io benchmarking
1个回答
1
投票

逐字节读取文件句柄通常比读取chunked慢。

通常,每个read()调用都对应于Python中的C read()调用。总结果涉及请求下一个char的系统调用。对于2 kb的文件,这意味着对内核的2000次调用;每个都需要一个函数调用,请求内核,然后等待响应,通过返回传递。

最值得注意的是awaiting response,系统调用将阻塞,直到您的呼叫在队列中得到确认,因此您必须等待。

呼叫越少越好,因此更多字节更快;这就是为什么缓冲io在相当常见的用途。

在python中,缓冲可以由io.BufferedReader提供,也可以通过buffering上的open关键字参数提供。


0
投票

我在处理与EEPROM接口的arduinos时遇到过类似的情况。基本上,为了从芯片或数据结构写入或读取,您必须发送写入/读取启用命令,发送起始位置,然后获取第一个字符。但是,如果你获取多个字节,大多数芯片将自动递增其目标地址寄存器。因此,开始读/写操作存在一些开销。这是区别:

  • 开始沟通
  • 发送读启用
  • 发送读命令
  • 发送地址1
  • 从目标1中获取数据
  • 结束通讯
  • 开始沟通
  • 发送读启用
  • 发送读命令
  • 发送地址2
  • 从目标2中获取数据
  • 结束通讯

  • 开始沟通
  • 发送读启用
  • 发送读命令
  • 发送地址1
  • 从目标1中获取数据
  • 从目标2中获取数据
  • 结束通讯

就机器指令而言,一次读取多个位/字节会清除大量开销。更糟糕的是,当发送读/写使能后,某些芯片要求您空闲几个时钟周期,让机械过程物理地将晶体管移动到位以启用读或写。

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