返回一组项目的 Python 生成器

Question

我正在尝试制作一个生成器，它可以返回列表中的多个连续项目，该列表仅“移动”一个索引。类似于 DSP 中的移动平均滤波器。例如，如果我有列表：

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

我希望这个输出：

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

我已经编写了代码，但它不适用于过滤器和生成器等。如果我需要提供大量单词，恐怕它也会因内存而中断。

函数gen：

def gen(enumobj, n):
    for idx,val in enumerate(enumobj):
        try:
            yield tuple(enumobj[i] for i in range(idx, idx + n))
        except:
            break

以及示例代码：

words = ['aaa','bb','c','dddddd','eeee','ff','g','h','iiiii','jjj','kk','lll','m','m','ooo']
w = filter(lambda x: len(x) > 1, words)

# It's working with list
print('\nList:')
g = gen(words, 4)
for i in g: print(i)

# It's not working with filetrs / generators etc.
print('\nFilter:')
g = gen(w, 4)
for i in g: print(i)

的列表不会产生任何东西。代码应该中断，因为无法索引过滤器对象。当然，其中一个答案是强制列出：list(w)。但是，我正在为该函数寻找更好的代码。如何更改它以便函数也可以接受过滤器等。我担心列表中大量数据的内存。

谢谢

Answer 1

使用迭代器，您需要跟踪已读取的值。 n 大小的列表可以解决问题。将下一个值附加到列表中，并在每次产量后丢弃最上面的项目。

import itertools

def gen(enumobj, n):
    # we need an iterator for the `next` call below. this creates
    # an iterator from an iterable such as a list, but leaves
    # iterators alone.
    enumobj = iter(enumobj)
    # cache the first n objects (fewer if iterator is exhausted)
    cache = list(itertools.islice(enumobj, n))
    # while we still have something in the cache...
    while cache:
        yield cache
        # drop stale item
        cache.pop(0)
        # try to get one new item, stopping when iterator is done
        try:
            cache.append(next(enumobj))
        except StopIteration:
            # pass to emit progressively smaller units
            #pass
            # break to stop when fewer than `n` items remain
            break

words = ['aaa','bb','c','dddddd','eeee','ff','g','h','iiiii','jjj','kk','lll','m','m','ooo']
w = filter(lambda x: len(x) > 1, words)

# It's working with list
print('\nList:')
g = gen(words, 4)
for i in g: print(i)

# now it works with iterators
print('\nFilter:')
g = gen(w, 4)
for i in g: print(i)