使用以下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))
输出:
0b110100001100101011011000110110001101111
如果我将其放到该站点(位于右侧站点)中,则会得到hello回信。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,然后将其与相应的值进行匹配bin(ord(character))或以其他方式进行匹配。真正在寻找更简单的东西。
bin(ord(character))
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'
>>> n = int('0b110100001100101011011000110110001101111', 2) >>> n.to_bytes((n.bit_length() + 7) // 8, 'big').decode() 'hello'
要在Python中支持所有Unicode字符3:
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'
这是与Python 2/3兼容的单源版本:
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