Integer to bitfield as a list

问题

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)

回答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

回答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.

回答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

回答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]

回答5:

I'm doing this for my program where you specify a template to get your values from an int:

def field(template, value):
    sums = [int(v) if v.__class__==str else len(bin(v))-2 for v in template]
    return [(value>> (sum(sums[:i]) if i else 0) )&(~(~0<<int(t)) if t.__class__==str else t) for i,t in enumerate(template)]

how to use
in the template, specify ints relating to your bit-sizes:

field([0b1,0b111,0b1111], 204) #>>> [0, 6, 12]

or you can specify the bit-size of each value needed using strings: (noob friendly)

field(['1','3','4'], 204) #>>> [0, 6, 12]

EDIT: and vice versa: (separate code)

field(['1','3','4'], [0, 6, 12]) #>>> 204
field([0b1,0b111,0b1111], [0,3,9]) #>>> 150

the code:

def field(template, value):
    res = 0
    for t, v in zip(template, value)[::-1]: res = (res << (t.bit_length() if t.__class__ is int else int(t)) )|v
    return res

EDIT2: faster code^

回答6:

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]

来源：https://stackoverflow.com/questions/10321978/integer-to-bitfield-as-a-list