发电机输出长度 [英] Length of generator output

问题描述

Python提供了一种很好的方法来获取渴望的可迭代的长度，即len(x).但是对于以生成器理解和函数表示的惰性可迭代对象，我找不到任何类似的东西.当然，写这样的东西并不难:

Python provides a nice method for getting length of an eager iterable, len(x) that is. But I couldn't find anything similar for lazy iterables represented by generator comprehensions and functions. Of course, it is not hard to write something like:

def iterlen(x):
  n = 0
  try:
    while True:
      next(x)
      n += 1
  except StopIteration: pass
  return n

但是我无法摆脱正在重新使用自行车的感觉.

But I can't get rid of a feeling that I'm reimplementing a bicycle.

(当我键入该函数时，我想到了一个想法:也许确实没有这样的函数，因为它破坏了"它的论点.不过，对于我的情况来说这不是问题).

(While I was typing the function, a thought struck my mind: maybe there really is no such function, because it "destroys" its argument. Not an issue for my case, though).

P.S .:关于第一个答案-是的，类似len(list(x))的方法也可以使用，但是会大大增加内存的使用量.

P.S.: concerning the first answers - yes, something like len(list(x)) would work too, but that drastically increases the usage of memory.

P.P.S .:重新检查...忽略P.S.，似乎我在尝试时出错，它可以正常工作.对不起，麻烦了.

P.P.S.: re-checked... Disregard the P.S., seems I made a mistake while trying that, it works fine. Sorry for the trouble.

推荐答案

没有一个是因为在一般情况下您不能这样做-如果您有一个惰性无限生成器怎么办?例如:

There isn't one because you can't do it in the general case - what if you have a lazy infinite generator? For example:

def fib():
    a, b = 0, 1
    while True:
        a, b = b, a + b
        yield a

这永远不会终止，但会生成斐波那契数.您可以通过调用next()获得任意数量的斐波那契数.

This never terminates but will generate the Fibonacci numbers. You can get as many Fibonacci numbers as you want by calling next().

如果您真的需要知道项的数量，那么无论如何都无法一次线性地遍历它们，因此只能使用常规列表之类的不同数据结构.

If you really need to know the number of items there are, then you can't iterate through them linearly one time anyway, so just use a different data structure such as a regular list.

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