Converting from hex to binary without losing leading 0's python

Question

I have a hex value in a string like

h = '00112233aabbccddee'

I know I can convert this to binary with:

h = bin(int(h, 16))[2:]

However, this loses the leading 0's. Is there anyway to do this conversion without losing the 0's? Or is the best way to do this just to count the number of leading 0's before the conversion then add it in afterwards.

Solution 1

I don't think there is a way to keep those leading zeros by default.

Each hex digit translates to 4 binary digits, so the length of the new string should be exactly 4 times the size of the original.

h_size = len(h) * 4

Then, you can use .zfill to fill in zeros to the size you want:

h = ( bin(int(h, 16))[2:] ).zfill(h_size)

Solution 2

This is actually quite easy in Python, since it doesn't have any limit on the size of integers. Simply prepend a '1' to the hex string, and strip the corresponding '1' from the output.

>>> h = '00112233aabbccddee'
>>> bin(int(h, 16))[2:]  # old way
'1000100100010001100111010101010111011110011001101110111101110'
>>> bin(int('1'+h, 16))[3:]  # new way
'000000000001000100100010001100111010101010111011110011001101110111101110'

Solution 3

Basically the same but padding to 4 bindigits each hexdigit

''.join(bin(int(c, 16))[2:].zfill(4) for c in h)

Comments

[root]

I think this gave me 3 extra 0's for some reason

[root]

Ah nevermind it works perfectly. I wasn't counting the 0's before the first non zero hex number. Thanks.

[jfs]

@root: unrelated: don't reuse h name for different things e.g., use bits name for the result: bits = bin(int(h, 16))[2:].zfill(len(h) * 4)

[Alex Stewart]

Been looking for exactly this for an hour!

[foobar]

Why is this not an accepted answer.

[Mark Ransom]

@foobar you'll notice it was posted 3 years after the accepted answer - don't remember what prompted me to make an answer.