我有一个像这样的值列表，
lst = [1, 2, 3, 4, 5, 6, 7, 8]
所需输出:
window size = 3
1 # first element in the list
forward = [2, 3, 4]
backward = []
2 # second element in the list
forward = [3, 4, 5]
backward = [1]
3 # third element in the list
forward = [4, 5, 6]
backward = [1, 2]
4 # fourth element in the list
forward = [5, 6, 7]
backward = [1, 2, 3]
5 # fifth element in the list
forward = [6, 7, 8]
backward = [2, 3, 4]
6 # sixth element in the list
forward = [7, 8]
backward = [3, 4, 5]
7 # seventh element in the list
forward = [8]
backward = [4, 5, 6]
8 # eight element in the list
forward = []
backward = [5, 6, 7]
让我们假设窗口大小为4，现在我想要的输出:
对于列表中的each_element，我希望前面有4个值，后面有4个值，忽略当前值。
我可以用它来获得值的滑动窗口，但这也不能给出正确的所需输出。
import more_itertools
list(more_itertools.windowed([1, 2, 3, 4, 5, 6, 7, 8], n=3))
方法1:枚举
较短的方法
代码:
window = 3
for backward, current in enumerate(range(len(arr)), start = 0-window):
if backward < 0:
backward = 0
print(arr[current+1:current+1+window], arr[backward:current])
输出:
[2, 3, 4], []
[3, 4, 5], [1]
[4, 5, 6], [1, 2]
[5, 6, 7], [1, 2, 3]
[6, 7, 8], [2, 3, 4]
[7, 8], [3, 4, 5]
[8], [4, 5, 6]
[], [5, 6, 7]
另一种方法
方法2:
代码:
arr = [1, 2, 3, 4, 5, 6, 7, 8]
window=3
for i in range(len(arr)):
forward = i + 1 + window
backward = i - window
if backward < 0:
backward =0
print(arr[i+1:forward], arr[backward:i])
输出:
[2, 3, 4] []
[3, 4, 5] [1]
[4, 5, 6] [1, 2]
[5, 6, 7] [1, 2, 3]
[6, 7, 8] [2, 3, 4]
[7, 8] [3, 4, 5]
[8] [4, 5, 6]
[] [5, 6, 7]
这应该会让你开始:
from dataclasses import dataclass
from typing import List
@dataclass
class Window:
index: int
backward: List[int]
forward: List[int]
def window(iterable, window_size, index):
backward = iterable[max(0, index - window_size):index]
forward = iterable[index + 1:min(len(iterable) - 1, index + 1 + window_size)]
return Window(index, backward, forward)
>>> window([1,2,3,4,5,6], 3, 0)
Window(index=0, backward=[], forward=[2, 3, 4])
>>> window([1,2,3,4,5,6], 3, 5)
Window(index=5, backward=[3, 4, 5], forward=[])
我还将添加一些检查索引和窗口大小是否有意义。
下面是一段简短，简洁，经过测试的代码
l = [1,2,3,4,5,6,7,8]
window = 3
forward = [l[i+1:i+1+window] for i in range(len(l)]
backward = [l[::-1][i+1:i+1+window] for i in range(len(l)] # first reverse the input list and do same as did in forward
out = zip(forward,backward[::-1]) # first reverse the backward list and zip two list into one
输出量
>>> forward
[[2, 3, 4], [3, 4, 5], [4, 5, 6], [5, 6, 7], [6, 7, 8], [7, 8], [8], []]
>>> backward
[[7, 6, 5], [6, 5, 4], [5, 4, 3], [4, 3, 2], [3, 2, 1], [2, 1], [1], []]
>>> out
[([2, 3, 4], []), ([3, 4, 5], [1]), ([4, 5, 6], [2, 1]), ([5, 6, 7], [3, 2, 1]), ([6, 7, 8], [4, 3, 2]), ([7, 8], [5, 4, 3]), ([8], [6, 5, 4]), ([], [7, 6, 5])]
