3

我知道 Python 有一些惰性实现,因此,我想知道是否可以在 Python 中使用循环编程。

如果不是,为什么?

4

2 回答 2

5

我认为您的意思是协同例程,而不是协同递归。是的,这在 Python 中是完全可能的,因为PEP 342: Coroutines via Enhanced Generators已经实现。

典型的例子是消费者装饰器:

def consumer(func):
    def wrapper(*args,**kw):
        gen = func(*args, **kw)
        next(gen)
        return gen
    wrapper.__name__ = func.__name__
    wrapper.__dict__ = func.__dict__
    wrapper.__doc__  = func.__doc__
    return wrapper

使用这样consumer的,让我们链接过滤器并通过它们推送信息,充当管道:

from itertools import product

@consumer
def thumbnail_pager(pagesize, thumbsize, destination):
    while True:
        page = new_image(pagesize)
        rows, columns = pagesize / thumbsize
        pending = False
        try:
            for row, column in product(range(rows), range(columns)):
                thumb = create_thumbnail((yield), thumbsize)
                page.write(
                    thumb, col * thumbsize.x, row * thumbsize.y
                )
                pending = True
        except GeneratorExit:
            # close() was called, so flush any pending output
            if pending:
                destination.send(page)

            # then close the downstream consumer, and exit
            destination.close()
            return
        else:
            # we finished a page full of thumbnails, so send it
            # downstream and keep on looping
            destination.send(page)

@consumer
def jpeg_writer(dirname):
    fileno = 1
    while True:
        filename = os.path.join(dirname,"page%04d.jpg" % fileno)
        write_jpeg((yield), filename)
        fileno += 1


# Put them together to make a function that makes thumbnail
# pages from a list of images and other parameters.      
#
def write_thumbnails(pagesize, thumbsize, images, output_dir):
    pipeline = thumbnail_pager(
        pagesize, thumbsize, jpeg_writer(output_dir)
    )

    for image in images:
        pipeline.send(image)

    pipeline.close()

核心原则是python生成器yield表达式;后者让生成器从调用者那里接收信息。

编辑:啊,共同递归确实是一个不同的概念。请注意,Wikipedia 文章使用 python作为示例,此外,还使用了 python generators

于 2012-11-06T11:49:44.290 回答
1

你试过了吗?

def a(x):
    if x == 1: return
    print "a", x
    b(x - 1)

def b(x):
    if x == 1: return
    print "b", x
    a(x - 1)

a(10)

作为旁注,python没有尾递归,这将失败x > 1000(尽管这个限制是可配置的)

于 2012-11-06T11:47:24.923 回答