我在Python语言中有这样一个列表:[1,0,0,0,0,0,0,0]
。我是否可以像我输入0b10000000那样将其转换为整数(即转换为128)?我还需要将像[1,1,0,0,0,0,0,0,1,0,0,0,0,0,0,0]
这样的序列转换为整数(这里它将返回0b1100000010000000,即259)。如果需要,list的长度始终是8的倍数。
发布于 2012-09-17 22:25:17
您可以使用位移位:
out = 0
for bit in bitlist:
out = (out << 1) | bit
这很容易击败A.R.S.提出的"int cast“方法,或者Steven Rumbalski提出的带有查找的修改后的cast方法:
>>> 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
发布于 2012-09-17 22:33:45
使用bitstring模块的...or
>>> from bitstring import BitArray
>>> bitlist=[1,0,0,0,0,0,0,0]
>>> b = BitArray(bitlist)
>>> b.uint
128
发布于 2012-09-18 05:18:25
我发现了一种方法,虽然他的解决方案当然更漂亮,但它的性能略高于马提金·皮特斯的解决方案。实际上,我对结果有点惊讶,但不管怎样...
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
https://stackoverflow.com/questions/12461361
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