Python conversion from binary string to hexadecimal

Question

how can I perform a conversion of a binary string to the corresponding hex value in python

I have 0000 0100 1000 1101 and I want to get 048D
I'm using python2.6

Thanks for the help

Answer 1

Welcome to StackOverflow!

int given base 2 and then hex:

>>> int('010110', 2)
22
>>> hex(int('010110', 2))
'0x16'
>>> 

>>> hex(int('0000010010001101', 2))
'0x48d'

The doc of int:

int(x[, base]) -> integer

Convert a string or number to an integer, if possible.  A floating

point argument will be truncated towards zero (this does not include a string representation of a floating point number!) When converting a string, use the optional base. It is an error to supply a base when converting a non-string. If base is zero, the proper base is guessed based on the string content. If the argument is outside the integer range a long object will be returned instead.

The doc of hex:

hex(number) -> string

Return the hexadecimal representation of an integer or long

integer.

Answer 2

Assuming they are grouped by 4 and separated by whitespace. This preserves the leading 0.

b = '0000 0100 1000 1101'
h = ''.join(hex(int(a, 2))[2:] for a in b.split())

Answer 3

bstr = '0000 0100 1000 1101'.replace(' ', '')
hstr = '%0*X' % ((len(bstr) + 3) // 4, int(bstr, 2))

Answer 4

>>> import string
>>> s="0000 0100 1000 1101"
>>> ''.join([ "%x"%string.atoi(bin,2) for bin in s.split() ]  )
'048d'
>>>

or

>>> s="0000 0100 1000 1101"
>>> hex(string.atoi(s.replace(" ",""),2))
'0x48d'

Answer 5

You could use the format() built-in function like this:

"{0:0>4X}".format(int("0000010010001101", 2))