Python 3.x Shifting Ranges

Question

假设我有这样的范围:

x = range(10)

它将具有以下值作为列表:

list(x)      # Prints [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

我想改变这个范围(可能多次)并迭代结果,例如

             #        [7, 8, 9, 0, 1, 2, 3, 4, 5, 6]

创建等效列表不是问题.但我想知道是否有可能创建这样的东西作为一个范围来节省内存中的一些空间,当然如果解决方案可以达到如下性能,那将是很好的:

for i in range(1000000)

Answer 1

您可以将范围包装在生成器表达式中,即时应用shift和modulo:

def shifted_range(rangeob, shift):
    size, shift = rangeob.stop, shift * rangeob.step
    return ((i + shift) % size for i in rangeob)

演示:

>>> def shifted_range(rangeob, shift):
...     size, shift = rangeob.stop, shift * rangeobj.step
...     return ((i + shift) % size for i in rangeob)
... 
>>> range_10 = range(10)
>>> list(shifted_range(range_10, 3))
[3, 4, 5, 6, 7, 8, 9, 0, 1, 2]
>>> list(shifted_range(range_10, 7))
[7, 8, 9, 0, 1, 2, 3, 4, 5, 6]
>>> range_10_2 = range(0, 10, 2)
>>> list(shifted_range(range_10_2, 4))
[8, 0, 2, 4, 6]

你也可以把它变成包装对象:

class RangeShift:
    def __init__(self, rangeob, shift):
        self._range = rangeob
        self.shift = shift

    @property
    def start(self):
        r = self._range
        return (r.start + self.shift * r.step) % r.stop

    @property
    def stop(self):
        r = self._range
        return (r.stop + self.shift * r.step) % r.stop

    def index(self, value):
        idx = self._range.index(value)
        return (idx - self.shift) % len(self._range)

    def __getattr__(self, attr):
        return getattr(self._range, attr)

    def __getitem__(self, index):
        r = self._range
        return (r[index] + self.shift * r.step) % r.stop

    def __len__(self):
        return len(self._range)

    def __iter__(self):
        size, shift = self._range.stop, self.shift * self._range.step
        return ((i + shift) % size for i in self._range)

这将表现得与原始范围相似,但是对所有生成的值应用移位.它甚至可以让你改变转变!

演示:

>>> range_10 = range(10)
>>> shifted = RangeShift(range_10, 7)
>>> len(shifted)
10
>>> shifted.start
7
>>> shifted.stop
7
>>> shifted.step
1
>>> shifted[3]
0
>>> shifted[8]
5
>>> list(shifted)
[7, 8, 9, 0, 1, 2, 3, 4, 5, 6]
>>> shifted.shift = 3
>>> list(shifted)
[3, 4, 5, 6, 7, 8, 9, 0, 1, 2]
>>> range_10_2 = range(0, 10, 2)
>>> shifted_10_2 = RangeShift(range_10_2, 4)
>>> list(shifted_10_2)
[8, 0, 2, 4, 6]

此包装器现在支持的最佳技巧:反转移位范围:

>>> list(reversed(shifted))
[2, 1, 0, 9, 8, 7, 6, 5, 4, 3]
>>> list(reversed(shifted_10_2))
[6, 4, 2, 0, 8]