如何在Python中将字符串转换为十六进制字符串？

Question

I have a bit-string of 32 characters that I need to represent as hexadecimal in Python. For example, the string "10000011101000011010100010010111" needs to also be output as "83A1A897" .

Any suggestions on how to best go about this in Python?

Answer 1

To format to hexadecimal you can use the hex function:

>>> hex(int('10000011101000011010100010010111', 2))
0x83a1a897

Or to get it in exactly the format you requested:

>>> '%08X' % int('10000011101000011010100010010111', 2)
83A1A897

Answer 2

>>> binary = '10010111'
>>> int(binary,2)
151
>>> hex(int(binary,2))
'0x97'

I hope this helps!

Answer 3

You can do this very easy with build in functions. The first thing you want to do is convert your binary to an integer:

>> int("1010",2)
10

The second step then would be to represent this as hex:

>> "%04X" % int("1010",2)
'000A'

in case you don't want any predefined length of the hex string then just use:

>> "%X" % int("1010",2)
'A'
>> "0x%X" % int("1010",2)
'0xA'

Answer 4

Well we could string format just like Mark Byers said.Or in other way we could string format in another method like given below:

>>> print('{0:x}'.format(0b10000011101000011010100010010111))
83a1a897

To make the alphabets between the hex in upper case try this:

>>> print('{0:X}'.format(0b10000011101000011010100010010111))
83A1A897

Hope this is helpful.

Answer 5

To read in a number in any base use the builtin int function with the optional second parameter specifying the base (in this case 2).

To convert a number to a string of its hexadecimal form just use the hex function.

>>> number=int("10000011101000011010100010010111",2)
>>> print hex(number)
0x83a1a897L