在Python中将位列表转换为整数

Question

我在Python中有这样的列表:[1,0,0,0,0,0,0,0].我可以将它转换为整数,就像我输入0b10000000(即转换为128)一样吗？我还需要将序列转换[1,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0]为整数(这里它将返回0b1100000010000000,即259).如果有必要,列表长度始终是8的倍数.

Answer 1

你可以使用bitshifting:

out = 0
for bit in bitlist:
    out = (out << 1) | bit

这很容易击败ARS提出的"int cast"方法,或Steven Rumbalski提出的修改后的查找:

>>> def intcaststr(bitlist):
...     return int("".join(str(i) for i in bitlist), 2)
... 
>>> def intcastlookup(bitlist):
...     return int(''.join('01'[i] for i in bitlist), 2)
... 
>>> def shifting(bitlist):
...     out = 0
...     for bit in bitlist:
...         out = (out << 1) | bit
...     return out
... 
>>> timeit.timeit('convert([1,0,0,0,0,0,0,0])', 'from __main__ import intcaststr as convert', number=100000)
0.5659139156341553
>>> timeit.timeit('convert([1,0,0,0,0,0,0,0])', 'from __main__ import intcastlookup as convert', number=100000)
0.4642159938812256
>>> timeit.timeit('convert([1,0,0,0,0,0,0,0])', 'from __main__ import shifting as convert', number=100000)
0.1406559944152832

Answer 2

...或使用bitstring模块

>>> from bitstring import BitArray
>>> bitlist=[1,0,0,0,0,0,0,0]
>>> b = BitArray(bitlist)
>>> b.uint
128

Answer 3

试试这个单线：

int("".join(str(i) for i in my_list), 2)

如果您关心速度/效率，请查看 Martijn Pieters 的解决方案。

Answer 4

我遇到了一种方法，该方法比Martijn Pieters解决方案稍胜一筹，尽管他的解决方案当然更漂亮。我实际上对结果感到有些惊讶，但是无论如何...

import timeit

bit_list = [1,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0]

def mult_and_add(bit_list):
    output = 0
    for bit in bit_list:
        output = output * 2 + bit
    return output

def shifting(bitlist):
     out = 0
     for bit in bitlist:
         out = (out << 1) | bit
     return out

n = 1000000

t1 = timeit.timeit('convert(bit_list)', 'from __main__ import mult_and_add as convert, bit_list', number=n)
print "mult and add method time is : {} ".format(t1)
t2 = timeit.timeit('convert(bit_list)', 'from __main__ import shifting as convert, bit_list', number=n)
print "shifting method time is : {} ".format(t2)

结果：

mult and add method time is : 1.69138722958 
shifting method time is : 1.94066818592 

Answer 5

这个怎么样：

out = sum([b<<i for i, b in enumerate(my_list)])

或按相反顺序：

out = sum([b<<i for i, b in enumerate(my_list[::-1])])