写布尔字符串到二进制文件？ [英] Write boolean string to binary file?

问题描述

我有一串布尔值，我想用这些布尔值作为位来创建一个二进制文件。这是我正在做的：

 ＃首先追加字符串0以使其长度为8的倍数b $ b while len（boolString）％8！= 0：
 boolString + ='0'
 
＃将字符串逐字节写入文件
i = 0 
 while我< len（boolString）/ 8：
 byte = int（boolString [i * 8：（i + 1）* 8]，2）
 outputFile.write（'％c'％byte）
 
i + = 1 
  

但是这会产生输出1个字节， 。什么是更有效的方法呢？解决方案这是另一个答案，这次使用工业强度的效用函数 PyCrypto - Python加密工具包，在2.6版本（最新的稳定版本）中，它定义在 pycrypto-2.6 / lib / Crypto / Util / number.py 。

：

      经过仔细的基准测试后，Barry Warsaw提供了改进的转换函数


 导入结构
 
 def long_to_bytes（n，blocksize = 0）：
long_to_bytes（n：long，blocksize：int ）：string 
将一个长整数转换为一个字节字符串
 
如果给出了可选的blocksize并且大于0，则用$二进制零填充
字节的前面，长度是
 blocksize的倍数。
经过多次测试，这个算法被认为是最快的
s = b（''）
n = long（n）
 pack = struct.pack 
而n> 0：
 s = pack（'> I'，n& 0xffffffffL）+ s 
 n = n>> （b）如果s [i]！= b（'\ 000'）[0]：
 break 
 else：
＃只在n == 0时发生
s = b（'\000'）
i = 0 
s = s [i：] 
＃添加一些填充字节。这可以做得更有效率w.r.t. 
＃de-padding在上面完成，但感叹... 
如果blocksize> 0和len（s）％blocksize：
s =（blocksize-len（s）％blocksize）* b（'\ 000'）+ s 
 return s 
  




I have a string of booleans and I want to create a binary file using these booleans as bits. This is what I am doing:

# first append the string with 0s to make its length a multiple of 8
while len(boolString) % 8 != 0:
    boolString += '0'

# write the string to the file byte by byte
i = 0
while i < len(boolString) / 8:
    byte = int(boolString[i*8 : (i+1)*8], 2)
    outputFile.write('%c' % byte)

    i += 1

But this generates the output 1 byte at a time and is slow. What would be a more efficient way to do it?

解决方案

Here's another answer, this time using an industrial-strength utility function from the PyCrypto - The Python Cryptography Toolkit where, in version 2.6 (the current latest stable release), it's defined inpycrypto-2.6/lib/Crypto/Util/number.py.

The comments preceeding it say:
    Improved conversion functions contributed by Barry Warsaw, after careful benchmarking

import struct

def long_to_bytes(n, blocksize=0):
    """long_to_bytes(n:long, blocksize:int) : string
    Convert a long integer to a byte string.

    If optional blocksize is given and greater than zero, pad the front of the
    byte string with binary zeros so that the length is a multiple of
    blocksize.
    """
    # after much testing, this algorithm was deemed to be the fastest
    s = b('')
    n = long(n)
    pack = struct.pack
    while n > 0:
        s = pack('>I', n & 0xffffffffL) + s
        n = n >> 32
    # strip off leading zeros
    for i in range(len(s)):
        if s[i] != b('\000')[0]:
            break
    else:
        # only happens when n == 0
        s = b('\000')
        i = 0
    s = s[i:]
    # add back some pad bytes.  this could be done more efficiently w.r.t. the
    # de-padding being done above, but sigh...
    if blocksize > 0 and len(s) % blocksize:
        s = (blocksize - len(s) % blocksize) * b('\000') + s
    return s

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