整数到位域作为列表 [英] Integer to bitfield as a list

问题描述

我已经创建了一种将 int 转换为位域(在列表中)的方法，并且它可以工作，但是我敢肯定还有更优雅的解决方案-我一直在盯着它会持续很长时间.

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.

我很好奇，如何将 int 转换为 list 中表示的位域?

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)

推荐答案

这是怎么回事:

def bitfield(n):
    return [int(digit) for digit in bin(n)[2:]] # [2:] to chop off the "0b" part 

这给你

>>> 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]

这仅适用于正整数.

使用 int()转换为 int 有点夸张.这要快得多:

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:]]

查看时间:

>>> 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

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