Yielding until all needed values are yielded, is t

Question

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.

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

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/

Answer 1

In analogy with the take function in Haskell, you can build a 'limited' generator based upon another generator:

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

You can further this idea with an "until" generator, a "while", ...

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

And use it like

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

...

Answer 2

You can call close() on the generator object. This way, a GeneratorExit exception is raised within the generator and further calls to its next() method will raise 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

Answer 3

As you have already seen,

TypeError: 'generator' object is unsubscriptable

And the way you have written devtrue it shouldn't stop. If you need that capacity you could:

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

or far more simply:

y = [True] * 5

If you make an infinite generator, it will generate infinitely.

Answer 4

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

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