在python中处理二进制文字

Question

我正在尝试将代码从php转换为python。我有一个二进制文字列表 -

['0b0', '0b0', '0b0', '0b0', '0b0', '0b0', '0b100', '0b1001100']

连接时等于0000000000000000000000000000000000000000000000000000010001001100。

int("0000000000000000000000000000000000000000000000000000010001001100",2)

给出1100

如何从列表中进行此操作。无法连接二进制文字。

Answer 1

>>> l = ['0b0', '0b0', '0b0', '0b0', '0b0', '0b0', '0b100', '0b1001100']
>>> int("".join("%02x" % int(x,0) for x in l), 16)
1100

Python将0b0101理解为二进制文字，因此我使用int('0b0101', 0)将每个部分转换为int。然后我以一种方便的格式（十六进制的两位数）格式化它，连接它们，并将它们解释为十六进制整数。

Answer 2

您需要使用zfill方法用正确数量的零填充元素

li = ['0b0', '0b0', '0b0', '0b0', '0b0', '0b0', '0b100', '0b1001100']
zero_padded = [x[2:].zfill(8) for x in li]
print ''.join(zero_padded)

输出

0000000000000000000000000000000000000000000000000000010001001100

Answer 3

只需剥离0b：

binary = ''.join(x[2:] for x in yourlist)
print binary
print int(binary, 2)

Answer 4

这对你来说是否丑陋：

int(''.join([i for sl in [s[2:].zfill(8) for s in l] for i in sl]),2)

（似乎有效。）

也许这更具可读性：

l = ['0b0', '0b0', '0b0', '0b0', '0b0', '0b0', '0b100', '0b1001100']
n = 0
for s in l:
    n = n*256 + int(s,2)
print n