问题描述
我想知道是否有比当前方法更好的方法.
I would like to know if there is a better way to do this than my current method.
我试图将整数表示为位列表，并且仅当整数为<时才将其填充为8位. 128 :
I'm trying to represent an integer as a list of bits and left pad it to 8 bits only if the integer is < 128:
Example input: 0x15
Desired output: [0, 0, 0, 1, 0, 1, 0, 1]
我通过以下方式进行操作:
I do it in the following way:
input = 0x15
output = deque([int(i) for i in list(bin(input))[2:]])
while len(output) != 8:
output.appendleft(0)
在python中有更好的方法吗?
Is there a better way to do this in python?
我想将任何整数转换为二进制列表.仅当数字需要少于8位表示时，才应填充到8.
I would like to convert any integer to a binary-list. Pad to 8 only if the number requires less than 8 bits to represent.
Another Example input: 0x715
Desired output: [1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1]
推荐答案
input = 0x15
output = [int(x) for x in '{:08b}'.format(input)]
{0:0=8b}'.format(0x15)以binary格式表示您的input，其中0 padding为8位数字，然后使用列表推导来创建位列表.
{0:0=8b}'.format(0x15) represents your input in binary format with 0 padding to 8 digits, then using list comprehension to create a list of bits.
或者，您可以使用map函数:
Alternatively, you can use map function:
output = map(int, [x for x in '{:08b}'.format(0x15)])
可变位宽，
如果要使位数可变，这是一种方法:
If you want to make number of bits variable, here is one way:
width = 8 #8bit width
output = [int(x) for x in '{:0{size}b}'.format(0x15,size=width)]
output = map(int, [x for x in '{:0{size}b}'.format(0x15,size=width)])
这已在Python 2.7中进行了测试
This was tested in Python 2.7
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