将二进制转换为ASCII,反之亦然

Question

使用此代码获取字符串并将其转换为二进制:

bin(reduce(lambda x, y: 256*x+y, (ord(c) for c in 'hello'), 0))

这个输出:

0b110100001100101011011000110110001101111

哪个,如果我把它放到这个网站(在右边的网站上),我得到了我的hello回信息.我想知道它使用什么方法.我知道我可以将二进制字符串拼接成8位,然后将其与相应的值bin(ord(character))或其他方式匹配.真的在寻找更简单的东西.

Answer 1

对于[ -~]Python 2 范围内的ASCII字符:

>>> import binascii
>>> bin(int(binascii.hexlify('hello'), 16))
'0b110100001100101011011000110110001101111'

相反:

>>> n = int('0b110100001100101011011000110110001101111', 2)
>>> binascii.unhexlify('%x' % n)
'hello'

在Python 3.2+中:

>>> bin(int.from_bytes('hello'.encode(), 'big'))
'0b110100001100101011011000110110001101111'

相反:

>>> n = int('0b110100001100101011011000110110001101111', 2)
>>> n.to_bytes((n.bit_length() + 7) // 8, 'big').decode()
'hello'

要支持Python 3中的所有Unicode字符:

def text_to_bits(text, encoding='utf-8', errors='surrogatepass'):
    bits = bin(int.from_bytes(text.encode(encoding, errors), 'big'))[2:]
    return bits.zfill(8 * ((len(bits) + 7) // 8))

def text_from_bits(bits, encoding='utf-8', errors='surrogatepass'):
    n = int(bits, 2)
    return n.to_bytes((n.bit_length() + 7) // 8, 'big').decode(encoding, errors) or '\0'

这是单源Python 2/3兼容版本:

import binascii

def text_to_bits(text, encoding='utf-8', errors='surrogatepass'):
    bits = bin(int(binascii.hexlify(text.encode(encoding, errors)), 16))[2:]
    return bits.zfill(8 * ((len(bits) + 7) // 8))

def text_from_bits(bits, encoding='utf-8', errors='surrogatepass'):
    n = int(bits, 2)
    return int2bytes(n).decode(encoding, errors)

def int2bytes(i):
    hex_string = '%x' % i
    n = len(hex_string)
    return binascii.unhexlify(hex_string.zfill(n + (n & 1)))

例

>>> text_to_bits('hello')
'0110100001100101011011000110110001101111'
>>> text_from_bits('110100001100101011011000110110001101111') == u'hello'
True

Answer 2

仅内置python

这是一个简单字符串的纯python方法,留给后人留下来.

def string2bits(s=''):
    return [bin(ord(x))[2:].zfill(8) for x in s]

def bits2string(b=None):
    return ''.join([chr(int(x, 2)) for x in b])

s = 'Hello, World!'
b = string2bits(s)
s2 = bits2string(b)

print 'String:'
print s

print '\nList of Bits:'
for x in b:
    print x

print '\nString:'
print s2

String:
Hello, World!

List of Bits:
01001000
01100101
01101100
01101100
01101111
00101100
00100000
01010111
01101111
01110010
01101100
01100100
00100001

String:
Hello, World!

Answer 3

我不确定你认为除了逐个字符之外你怎么能做到这一点 - 它本身就是一个逐个字符的操作.肯定有代码可以为你做这个,但没有比逐个字符做更简单的方法.

首先,您需要去除0b前缀,并对字符串进行左零填充,使其长度可以被8整除,以便将位串分成容易的字符:

bitstring = bitstring[2:]
bitstring = -len(bitstring) % 8 * '0' + bitstring

然后将字符串分成八个二进制数字的块,将它们转换为ASCII字符,然后将它们连接回一个字符串:

string_blocks = (bitstring[i:i+8] for i in range(0, len(bitstring), 8))
string = ''.join(chr(int(char, 2)) for char in string_blocks)

如果你真的想把它当成一个数字,你仍然需要考虑这样一个事实:如果你想从左到右而不是从右到左,最左边的字符长度最多为七位数.