跳过滑动窗口

Question

给定的 [1,2,3,4,5,6,7,8,9,10]，一次获得3个项目的滑动窗口以获取：

[(1, 2, 3), (2, 3, 4), (3, 4, 5), (4, 5, 6), (5, 6, 7), (6, 7, 8), (7, 8, 9), (8, 9, 10)]

来自https://stackoverflow.com/q/42220614/610569，可以使用以下方式实现序列的滑动窗口：

def per_window(sequence, n=1):
    """
    Returns a sliding window.
    From https://stackoverflow.com/q/42220614/610569
        >>> list(per_window([1,2,3,4], n=2))
        [(1, 2), (2, 3), (3, 4)]
        >>> list(per_window([1,2,3,4], n=3))
        [(1, 2, 3), (2, 3, 4)]
    """
    start, stop = 0, n
    seq = list(sequence)
    while stop <= len(seq):
        yield tuple(seq[start:stop])
        start += 1
        stop += 1

但是，如果我想放入滑动窗口中有一些约束，我只想获取存在某个元素的窗口。

假设我只需要包含4个窗口，我可以喜欢：

>>> [window for window in per_window(x, 3) if 4 in window]
[((2, 3, 4), (3, 4, 5), (4,5,6)]

但是，以某种方式，循环仍然必须通过整个Windows列表进行处理并检查IF条件。

我可以通过寻找4的位置并将输入限制为per_window，例如

来进行一些跳过。

# Input sequence.
x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
# Window size.
n = 3
# Constraint.
c = 4 
# Set the index to 0
i = 0
while i < len(x)-n:
    i = x.index(4, i)
    # First window where the constraint is met.
    left = i - (n-1)
    if left > 0:
        print (list(per_window(x[left:i], 3)))
    right = i + n
    if right < len(x):
        print (list(per_window(x[i:right], 3)))
    i = right

(请注意上面的IFS不起作用的代码=((

而不是在per_window函数之外找到索引，而是另一种方法可以在per_window函数中添加此类约束？

---

编辑

阅读 @raymondhettinger的答案：

def skipping_window(sequence, target, n=3):
    """
    Return a sliding window with a constraint to check that
    target is inside the window.
    From https://stackoverflow.com/q/43626525/610569
    """
    start, stop = 0, n
    seq = list(sequence)
    while stop <= len(seq):
        subseq = seq[start:stop]
        if target in subseq:
            yield tuple(seq[start:stop])
        start += 1
        stop += 1
        # Fast forwarding the start.
        # Find the next window which contains the target.
        try:
            # `seq.index(target, start) - (n-1)` would be the next
            # window where the constraint is met.
            start = max(seq.index(target, start) - (n-1), start)
            stop = start + n
        except ValueError:
            break

[out]：

>>> x = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> list(skipping_window(x, 4, 3))
[(2, 3, 4), (3, 4, 5), (4, 5, 6)]

Answer 1

而不是在per_window函数之外找到索引，而是另一种方法可以在per_window函数中添加这样的约束？

是的，您可以在收益率之前添加条件：

def per_window(sequence, target, n=1):
    start, stop = 0, n
    seq = list(sequence)
    while stop <= len(seq):
        subseq = seq[start:stop]
        if target in subseq:
            yield tuple(subseq)
        start += 1
        stop += 1