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屈服直到所有需要的值都屈服,有没有办法让slice变得懒惰

lazy-sequences slice variable-assignment generator python

Tony Veijalainen · 技术社区 · 15 年前

当生成器未完成值并且所有需要的结果都已读取时,是否有方法停止生成?我的意思是生成器在不做StopIteration的情况下给出值。

例如,这从未停止:(修订)

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

http://code.activestate.com/recipes/576585-lazy-recursive-generator-function/

4 回复 | 直到 15 年前

Ferdinand Beyer 15 年前

你可以打电话 close() GeneratorExit 在生成器中引发异常,并进一步调用其 next() StopIteration :

>>> def test():
...     while True:
...         yield True
... 
>>> gen = test()
>>> gen
<generator object test at ...>
>>> gen.next()
True
>>> gen.close()
>>> gen.next()
Traceback (most recent call last):
  ...
StopIteration

msw 15 年前

如你所见,

TypeError: 'generator' object is unsubscriptable

还有你的写作方式 devtrue

def bounded_true(count)
   while count > 0:
       yield True
       count -= 1

或者更简单地说:

y = [True] * 5

xtofl Adam Rosenfield 15 年前

类似于 take 在Haskell函数中,您可以基于另一个生成器构建“有限”生成器:

def take(n,gen):
    '''borrowed concept from functional languages'''
togo=n
while togo > 0:
    yield gen.next()
    togo = togo - 1

def naturalnumbers():
    ''' an unlimited series of numbers '''
    i=0
    while True:
        yield i
        i=i+1

for n in take(10, naturalnumbers() ):
   print n

def gen_until( condition, gen ):
   g=gen.next()
   while( not condition(g) ):
      yield g
      g=gen.next()

像这样使用它

for i in gen_until( lambda x: x*x>100, naturalnumbers() ):
  print i

Community Mohan Dere 9 年前

这是我想到的最好的方法,但它仍然会进行两次切片以找到长度,并且需要将字符串编号从拆分转换为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

Improving pure Python prime sieve by recurrence formula