跳过滑动窗口

Question

给定[1,2,3,4,5,6,7,8,9,10]，一次获得一个由3项组成的滑动窗口：

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

在https://stackoverflow.com/q/42220614/610569中，可以通过以下方法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

但是如果我想在滑动窗口中设置一些约束，而我只想得到存在某个元素的窗口。

假设我只想要包含4的窗口，我可以这样做：

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

但不知怎么的，循环仍然必须处理整个窗口列表，并检查if条件。

我可以通过寻找4的位置来跳过，并将输入限制在per_window上。

# 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

(请注意上面使用ifs不工作的代码=( )

除了在per_window函数之外查找索引之外，还有其他方法在per_window函数中添加这样的约束吗？

编辑

在阅读了@RaymondHettinger的答案之后：

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

输出

>>> 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

除了在per_window函数之外查找索引之外，还有其他方法在per_window函数中添加这样的约束吗？

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