3

我有一个序列,窗口大小和步骤:

seq = [0,1,2,3,4]
n=4
step=2

from more_itertools import windowed
list(windowed([0,1,2,3,4], n, fillvalue=0, step=step))

结果:

[(0, 1, 2, 3), (2, 3, 4, 0)]

但是我需要:

[(0, 1, 2, 3), (2, 3, 4, 0), (4, 0, 0, 0)]

请帮我找到解决方案

4

4 回答 4

3

这也应该适用于可迭代对象,而不仅仅是序列:

from itertools import islice

def sliding_window(seq, n, step, fillvalue=None):
    it = iter(seq)
    values = tuple(islice(it, n))
    while values:
        yield values + (n-len(values)) * (fillvalue, )
        values = values[step:] + tuple(islice(it, step))

函数输出:

print(list(sliding_window(seq, n, step, fillvalue=0)))
# [(0, 1, 2, 3), (2, 3, 4, 0), (4, 0, 0, 0)]

其中大部分是从原始 itertools 的滑动窗口配方中借来的。

于 2019-04-09T10:55:29.283 回答
3

只需编写自己的windowed函数:

def windowed(iterable, size, fillvalue=None, step=1):
    for i in range(0, len(iterable), step):
        window = iterable[i:i+size]
        window += [fillvalue] * (size - len(window))
        yield window
>>> list(windowed([0,1,2,3,4], 4, fillvalue=0, step=2))
[[0, 1, 2, 3], [2, 3, 4, 0], [4, 0, 0, 0]]
于 2019-04-09T10:37:19.350 回答
2

使用padded怎么样?

seq = [0,1,2,3,4]
n=4
step=2

from more_itertools import windowed, padded
list(windowed(padded(seq, 0, n=n, next_multiple=True), n, step=step))
于 2019-04-09T10:45:37.363 回答
0

考虑more_itertools.stagger

给定

import itertools as it

import more_itertools as mit


iterable = [0, 1, 2, 3, 5]

代码

从滑动窗口获取所有结果:

windows = list(mit.stagger(iterable, offsets=(0, 1, 2, 3), longest=True, fillvalue=0))
windows
# [(0, 1, 2, 3), (1, 2, 3, 5), (2, 3, 5, 0), (3, 5, 0, 0), (5, 0, 0, 0)]

接下来,过滤掉想要的结果:

[w for i, w in enumerate(windows) if not (i % 2)]
# [(0, 1, 2, 3), (2, 3, 5, 0), (5, 0, 0, 0)]

切片可迭代:

list(it.islice(windows, 0, None, 2))
# [(0, 1, 2, 3), (2, 3, 5, 0), (5, 0, 0, 0)]
于 2019-04-12T23:41:07.387 回答