是否有无限发电机的表达式？ [英] Is there an expression for an infinite generator?

问题描述

是否存在可以产生无限元素的直接生成器表达式？

Is there a straight-forward generator expression that can yield infinite elements?

这是一个纯粹的理论问题。这里不需要实用答案：）

This is a purely theoretical question. No need for a "practical" answer here :)

例如，很容易制作一个有限的发电机： / p>

For example, it is easy to make a finite generator:

my_gen = (0 for i in xrange(42))

然而，要创建一个无限的，我需要用虚假函数污染我的命名空间：

However, to make an infinite one I need to "pollute" my namespace with a bogus function:

def _my_gen():
    while True:
        yield 0
my_gen = _my_gen()

在单独的文件中执行操作并且 import -ing稍后不计算。

Doing things in a separate file and import-ing later doesn't count.

我也知道 itertools.repeat 就是这样做的。我很好奇是否有没有这个的单线解决方案。

I also know that itertools.repeat does exactly this. I'm curious if there is a one-liner solution without that.

推荐答案

for x in iter(int, 1): pass

• 双参数 iter =零参数可调用+标记值


• int（）始终返回 0


• Two-argument iter = zero-argument callable + sentinel value
• int() always returns 0

因此， iter（int，1 ）是一个无限的迭代器。这个特定的主题显然有很多变化（特别是一旦你将 lambda 添加到混音中）。特别注意的一个变体是 iter（f，object（）），因为使用新创建的对象作为sentinel值几乎保证无限迭代器，无论使用哪个可调用对象第一个论点。

Therefore, iter(int, 1) is an infinite iterator. There are obviously a huge number of variations on this particular theme (especially once you add lambda into the mix). One variant of particular note is iter(f, object()), as using a freshly created object as the sentinel value almost guarantees an infinite iterator regardless of the callable used as the first argument.

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