将位写入二进制文件

Question

我有 23 位表示为一个字符串，我需要将此字符串作为 4 个字节写入二进制文件。最后一个字节始终为 0。以下代码有效（Python 3.3），但感觉不是很优雅（我对 Python 和编程还比较陌生）。你有什么让它变得更好的秘诀吗？似乎 for 循环可能很有用，但我如何在循环中进行切片而不得到 IndexError？请注意，当我将位提取到一个字节中时，我颠倒了位顺序。

from array import array

bin_array = array("B")
bits = "10111111111111111011110"    #Example string. It's always 23 bits
byte1 = bits[:8][::-1]
byte2 = bits[8:16][::-1]
byte3 = bits[16:][::-1]
bin_array.append(int(byte1, 2))
bin_array.append(int(byte2, 2))
bin_array.append(int(byte3, 2))
bin_array.append(0)

with open("test.bnr", "wb") as f:
    f.write(bytes(bin_array))

# Writes [253, 255, 61, 0] to the file

Answer 1

你可以把它当作一个int，然后创建4个字节如下：

>>> bits = "10111111111111111011110"
>>> int(bits[::-1], 2).to_bytes(4, 'little')
b'\xfd\xff=\x00'

Answer 2

struct 模块正是为这类事情而设计的——考虑以下内容，其中将字节转换分解为一些不必要的中间步骤，以便更清楚地理解它：

import struct

bits = "10111111111111111011110"  # example string. It's always 23 bits
int_value = int(bits[::-1], base=2)
bin_array = struct.pack('i', int_value)
with open("test.bnr", "wb") as f:
    f.write(bin_array)

一种更难阅读但更短的方法是：

bits = "10111111111111111011110"  # example string. It's always 23 bits
with open("test.bnr", "wb") as f:
    f.write(struct.pack('i', int(bits[::-1], 2)))

Answer 3

from array import array

bin_array = array("B")
bits = "10111111111111111011110"

bits = bits + "0" * (32 - len(bits))  # Align bits to 32, i.e. add "0" to tail
for index in range(0, 32, 8):
    byte = bits[index:index + 8][::-1]
    bin_array.append(int(byte, 2))

with open("test.bnr", "wb") as f:
    f.write(bytes(bin_array))

Answer 4

您可以使用re.findall方法在一行中执行拆分：

>>>bits = "10111111111111111011110"
>>>import re
>>>re.findall(r'\d{1,8}', bits)
['10111111', '11111111', '1011110']

作为一种算法，您可以将bits 填充到长度为32，然后使用re.findall 方法将其分组为八位字节：

>>> bits
'10111111111111111011110000000000'
>>> re.findall(r'\d{8}', bits)
['10111111', '11111111', '10111100', '00000000']

你的代码应该是这样的：

import re
from array import array

bin_array = array("B")
bits = "10111111111111111011110".ljust(32, '0')  # pad it to length 32

for octect in re.findall(r'\d{8}', bits): # split it in 4 octects
    bin_array.append(int(octect[::-1], 2)) # reverse them and append it

with open("test.bnr", "wb") as f:
    f.write(bytes(bin_array))