Python:按顺序执行步骤

Question

我正在尝试制作一个完全可切片的斐波那契序列，但我陷入了切片的步骤功能。这是我到目前为止的代码:

class Fib:
    def __init__(self, start, end):
        self.start = start
        self.end = end
    def _fib(self, index):
        items = [0,1]
        n = 2
        while n <= index:
            items.append(items[n-1]+items[n-2])
            n += 1
        return items
    def __len__(self):
        return self.end - self.start
    def __reversed__(self):
        items = self._fib(self.end)
        return reversed(items)
    def __count__(self, item):
        items = self._fib(self.end)
        return items.count(item)
    def __getitem__(self, index):
        if isinstance(index, slice):
            return Fib(slice.start, slice.stop-1)
        if index > len(self):
            raise StopIteration
        if index == 0:
            return 0
        if index == 1:
            return 1
        if index < 1:
            return list(reversed(self._fib(self.end)))[abs(index)]
        items = self._fib(index)
        return items.pop()

它似乎适用于常规切片、反转和按负数索引。但是，当我尝试使用步骤时，例如 Fib(0, 100)[::2]，我收到一个错误，我很难解释。

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "fibonacci.py", line 22, in __getitem__
    return Fib(slice.start, slice.stop-1)
TypeError: unsupported operand type(s) for -: 'member_descriptor' and 'int'

编辑:我最终这样做了......

class Fib:
def __init__(self, start, end):
    self.start = start
    self.end = end
def _fib(self, index):
    items = [0,1]
    n = 2
    while n <= index:
        items.append(items[n-1]+items[n-2])
        n += 1
    return items
def __len__(self):
    return self.end - self.start
def __reversed__(self):
    items = self._fib(self.end)
    return reversed(items)
def __count__(self, item):
    items = self._fib(self.end)
    return items.count(item)
def __getitem__(self, index):
    if isinstance(index, slice):
        _start, _stop, _step = (0, self.end, 1)
        if index.start is not None:
            _start = index.start
        if index.stop is not None:
            _stop = index.stop
        if index.step is not None:
            _step = index.step
        abs_stop = _stop
        abs_start = _start
        if _stop < 0:
            abs_stop = self.end + _stop
        if _start < 0:
            abs_start = self.end + _start
        _n = max(abs_start, abs_stop) - 1
        return self._fib(_n)[_start:_stop:_step]
    if index > len(self):
        raise StopIteration
    if index == 0:
        return 0
    if index == 1:
        return 1
    if index < 1:
        return list(reversed(self._fib(self.end)))[abs(index)]
    items = self._fib(index)
    return items.pop()

Answer 1

应该是return Fib(index.start,index.stop-1)。然后，由于切片具有 None(如果未提供)，因此您应该通过将 index.start 替换为 0(如果它是 None)来解决此问题 和 index.stop 以及序列的长度。