屈服直到所有需要的值都屈服，有没有办法让切片变得懒惰

问题描述

当生成器未完成值并且已读取所有需要的结果时，有没有办法停止屈服?我的意思是，生成器无需执行StopIteration就可以提供值.

Is there way to stop yielding when generator did not finish values and all needed results have been read? I mean that generator is giving values without ever doing StopIteration.

例如，这永远不会停止:(已修订)

For example, this never stops: (REVISED)

from random import randint
def devtrue():
    while True:
        yield True

answers=[False for _ in range(randint(100,100000))]
answers[::randint(3,19)]=devtrue()
print answers

我找到了此代码，但尚不了解如何在这种情况下应用它: http://code.activestate.com/recipes/576585-lazy -recursive-generator-function/

I found this code, but do not yet understand, how to apply it in this case:http://code.activestate.com/recipes/576585-lazy-recursive-generator-function/

推荐答案

这是我想出的最好方法，但是它仍然会进行两次切片以查找长度，并且需要将字符串号从split转换为int:

This is best I came up with, but it does still the slicing twice to find the length and need to convert string number from splitting to int:

from time import clock
from random import randint
a=[True for _ in range(randint(1000000,10000000))]
spacing=randint(3,101)
t=clock()
try:
    a[::spacing]=[False]
except ValueError as e:
    a[::spacing]=[False]*int(e.message.rsplit(' ',1)[-1])

print spacing,clock()-t

# baseline

t=clock()
a[::spacing]=[False]*len(a[::spacing])
print 'Baseline:',spacing,clock()-t

我将尽我所能尝试一下，但是它可能不会比根据递归公式进行长度算术更快.通过递归公式改进纯Python主筛子

I will try it to my prime sieve, but it is likely not to be faster than doing the length arithmetic from recurrence formula.Improving pure Python prime sieve by recurrence formula

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