Integer To Bitfield As A List

I've created a method to convert an int to a bitfield (in a list) and it works, but I'm sure there is more elegant solution- I've just been staring at it for to long. I'm curious,

Solution 1:

How about this:

def bitfield(n):
    return [int(digit) for digit in bin(n)[2:]] # [2:] to chop off the "0b" part 

This gives you

>>> bitfield(123)
[1, 1, 1, 1, 0, 1, 1]
>>> bitfield(255)
[1, 1, 1, 1, 1, 1, 1, 1]
>>> bitfield(1234567)
[1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1]

This only works for positive integers, though.

EDIT:

Conversion to int using int() is a bit overkill here. This is a lot faster:

def bitfield(n):
    return [1 if digit=='1' else 0 for digit in bin(n)[2:]]

See the timings:

>>> import timeit
>>> timeit.timeit("[int(digit) for digit in bin(123)[2:]]")
7.895014818543946
>>> timeit.timeit("[123 >> i & 1 for i in range(7,-1,-1)]")
2.966295244250407
>>> timeit.timeit("[1 if digit=='1' else 0 for digit in bin(123)[2:]]")
1.7918431924733795

Solution 2:

This doesn't use bin:

 b = [n >> i & 1 for i in range(7,-1,-1)]

and this is how to handle any integer this way:

 b = [n >> i & 1 for i in range(n.bit_length() - 1,-1,-1)]

See bit_length.

If you want index 0 of the list to correspond to the lsb of the int, change the range order, i.e.

b = [n >> i & 1 for i in range(0, n.bit_length()-1)]

Note also that using n.bit_length() can be a point of failure if you're trying to represent fixed length binary values. It returns the minimum number of bits to represent n.

Solution 3:

Try

>>>n=1794
>>>bitfield=list(bin(n))[2:]
>>>bitfield
['1', '1', '1', '0', '0', '0', '0', '0', '0', '1', '0']

This does not work for negative n though and as you see gives you a list of strings

Solution 4:

Does not work for negative values

>>> import numpy as np
>>> [int(x) for x in np.binary_repr(123)]
[1, 1, 1, 1, 0, 1, 1]

Solution 5:

Array with fixed length

Array with fixed length:

>>> '{0:07b}'.format(12)
'0001100'

Do you believe string is also an array? No? See:

>>> [int(x) for x in '{0:07b}'.format(12)]
[0, 0, 0, 1, 1, 0, 0]