Python 3二进制到十六进制带填充

Question

In Python 3 how would I turn binary into a hexadecimal with padding for example "00001111" would equal "0F" i would like it to include a 0 with no 0x attached aswell. I use this to convert binary to hexadecimal:

def bin2hex(binary):
return ''.join((hex(int(binary[i:i+8], 2))[2:] for i in range(0, len(binary), 8)))

print(bin2hex("00001111"))

Result: "F"

But it does not include a 0.

Answer 1

You made a mistake: a hexadecimal character is equivalent to four bits, not eight:

def bin2hex(binary):
    return ''.join(hex(int(binary[i:i+], 2))[2:] for i in range(0, len(binary), ))

That being said, I think you make things too complicated, you can simply use Python's string formatting:

def bin2hex(binary):
    return '{:02x}'.format(int(binary, 2))

Or for an arbitrary number of bits (dividable by 4):

def bin2hex(binary):
    return '{:0{}x}'.format(int(binary, 2), len(binary)//4)

of for a number of bits that is not per se dividable by 4:

def bin2hex(binary):
    return '{:0{}x}'.format(int(binary, 2), (len(binary)+3)//4)

or with string interpolation , like @HåkenLid said:

def bin2hex(binary):
    return f'{int(binary, 2):0{(len(binary)+3)//4}x}'

For example:

>>> bin2hex('0001')
'1'
>>> bin2hex('00011100')
'1c'
>>> bin2hex('000111001011')
'1cb'
>>> bin2hex('1101000111001011')
'd1cb'

Answer 2

One simple way to do that would be to use the zfill string method.

def bin2hex(binary):
    return hex(int(binary, 2))[2:].zfill(len(binary)//4)

bin2hex('00001111')
# >>> 0f

bin2hex('000000001111')
# >>> 00f