Integer to bitfield as a list

bit-fields, list, python

Solution

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

Problem

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, how would you convert a `int` to a bitfield represented in a `list`? ``` def get(self): results = [] results.append(1 if (self.bits & 1) else 0) results.append(1 if (self.bits & 2) else 0) results.append(1 if (self.bits & 4) else 0) results.append(1 if (self.bits & 8) else 0) results.append(1 if (self.bits & 16) else 0) results.append(1 if (self.bits & 32) else 0) results.append(1 if (self.bits & 64) else 0) results.append(1 if (self.bits & 128) else 0) return results def set(self, pin, direction): pin -= 1 if pin not in range(0, 8): raise ValueError if direction: self.bits |= (2 ** pin) else: self.bits &=~(2 ** pin) ```

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