可变增量滑动窗口 - Python

发布于 2024-11-27 13:12:53 字数 659 浏览 3 评论 0原文

我正在尝试使用Python中的滑动窗口函数来比较很长的值列表。我找到的滑动窗口函数的代码如下：

from itertools import islice

idlist = [1, 2, 3, 4, 5, 6, 7, 8, 9]
list = []

def window(seq, n=2):
    "Returns a sliding window (of width n) over data from the iterable"
    "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result    
    for elem in it:
        result = result[1:] + (elem,)
        yield result

for i in window(idlist, n=2):
    list.append(i)

print list

我的问题是，如何修改此代码，以便将窗口的增量（生成每个元组后移动的量）从 1 更改为更大的整数，说5还是50？我知道如何更改窗口的大小，但不知道增量。谢谢！

原文

I am trying to use the sliding window function in Python to compare a very long list of values. The code I have found for the sliding window function is below:

from itertools import islice

idlist = [1, 2, 3, 4, 5, 6, 7, 8, 9]
list = []

def window(seq, n=2):
    "Returns a sliding window (of width n) over data from the iterable"
    "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result    
    for elem in it:
        result = result[1:] + (elem,)
        yield result

for i in window(idlist, n=2):
    list.append(i)

print list

My question is, how would I modify this code so I could change the increment of the window (the amount it moves after each tuple is generated) from 1 to a much greater integer, say 5 or 50? I know how to change the size of the window, but not the increment.
Thanks!

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逆流 2024-12-04 13:12:53

您不必更改增量，您可以获取每个第 n 个元素：

# taking every 3rd element moves the start by 3
print list(islice(window(idlist, n=2),None,None,3))

未完全优化，但很简单。

You don't have to change the increment, you can take every n'th element:

# taking every 3rd element moves the start by 3
print list(islice(window(idlist, n=2),None,None,3))

Not fully optimized, but simple.

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女皇必胜 2024-12-04 13:12:53

提示：next 函数可用于从迭代器获取下一个元素。您需要在每次迭代中获取并附加多个元素（我认为这就是困难；您肯定知道如何将窗口的另一端向前移动不同的量:)）。

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公布 2024-12-04 13:12:53

也许这可以解决问题

L=[1,2,3,4,5]

def window(L, n=2, jump=1):
    lenght = len(L)
    assert n <= lenght
    for i in range(0,lenght-n+1,jump):
        yield tuple(L[i:i+n])

A=[]
for i in window(L, n=3, jump=1):
    A.append(i)

print A

maybe this solve the problem

L=[1,2,3,4,5]

def window(L, n=2, jump=1):
    lenght = len(L)
    assert n <= lenght
    for i in range(0,lenght-n+1,jump):
        yield tuple(L[i:i+n])

A=[]
for i in window(L, n=3, jump=1):
    A.append(i)

print A

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