blocks|key|782194|text|您可以在生成器对象上调用close()。这样，就会在生成器中引发GeneratorExit异常，并且对其next()方法的进一步调用将引发StopIteration|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|782195|>>>+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|code-block|syntax|javascript|782196|entityMap^0|C|7|W|D|1G|6|1X|D|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

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

<pre><code>&gt;&gt;&gt; def test():
... while True:
... yield True
... 
&gt;&gt;&gt; gen = test()
&gt;&gt;&gt; gen
&lt;generator object test at ...&gt;
&gt;&gt;&gt; gen.next()
True
&gt;&gt;&gt; gen.close()
&gt;&gt;&gt; gen.next()
Traceback (most recent call last):
 ...
StopIteration
</code></pre>

blocks|key|782234|text|正如你已经看到的，|type|unstyled|depth|inlineStyleRanges|entityRanges|data|782235|TypeError:+'generator'+object+is+unsubscriptable|code-block|syntax|javascript|782236|而且您编写devtrue的方式不应该停止。如果您需要该容量，您可以：|offset|length|style|CODE|782237|def+bounded_true(count)
+++while+count+>+0:
+++++++yield+True
+++++++count+-=+1|782238|或者更简单地说：|782239|y+=+[True]+*+5|782240|如果你做一个无限生成器，它就会无限生成。|782241|entityMap^0|0|0|5|7|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Y|8|@$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|14|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|V|$]]

As you have already seen, 

<pre><code>TypeError: 'generator' object is unsubscriptable
</code></pre>

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

<pre><code>def bounded_true(count)
 while count &gt; 0:
 yield True
 count -= 1
</code></pre>

or far more simply:

<pre><code>y = [True] * 5
</code></pre>

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

blocks|key|1527673|text|这是我想出的最好的方法，但它仍然会进行两次切片来找到长度，并且需要将字符串数字从still转换为int：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1527674|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|code-block|syntax|javascript|1527675|我会尝试我的素数筛子，但它可能不会比从递归公式中进行长度算术更快。Improving+pure+Python+prime+sieve+by+recurrence+formula|offset|length|1527676|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/3285443/improving-pure-python-prime-sieve-by-recurrence-formula^0|0|0|X|1J|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

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:

<pre><code>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
</code></pre>

I will try it to my prime sieve, but it is likely not to be faster than doing the length arithmetic from recurrence formula.
<a href="https://stackoverflow.com/questions/3285443/improving-pure-python-prime-sieve-by-recurrence-formula">Improving pure Python prime sieve by recurrence formula</a>

blocks|key|1531152|text|与Haskell中的take函数类似，您可以基于另一个生成器构建一个“有限的”生成器：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1531153|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%2B1

for+n+in+take(10,+naturalnumbers()+):
+++print+n|code-block|syntax|javascript|1531154|你可以用一个“直到”生成器，一个"while"，...|1531155|def+gen_until(+condition,+gen+):
+++g=gen.next()
+++while(+not+condition(g)+):
++++++yield+g
++++++g=gen.next()|1531156|并像这样使用它|1531157|for+i+in+gen_until(+lambda+x:+x*x>100,+naturalnumbers()+):
++print+i|1531158|..。|1531159|entityMap^0|A|4|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|S|3|T|5|6|7|14|8|@]|D|@]|E|$]]|$1|U|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|V|$]]

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

<pre><code>def take(n,gen):
 '''borrowed concept from functional languages'''
togo=n
while togo &gt; 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
</code></pre>

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

<pre><code>def gen_until( condition, gen ):
 g=gen.next()
 while( not condition(g) ):
 yield g
 g=gen.next()
</code></pre>

And use it like

<pre><code>for i in gen_until( lambda x: x*x&gt;100, naturalnumbers() ):
 print i
</code></pre>

...

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)

<pre><code>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
</code></pre>

I found this code, but do not yet understand, how to apply it in this case:
<a href="http://code.activestate.com/recipes/576585-lazy-recursive-generator-function/" rel="nofollow noreferrer">http://code.activestate.com/recipes/576585-lazy-recursive-generator-function/</a>

Yielding until all needed values are yielded, is there way to make slice to become lazy

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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当生成器没有完成值并且所有需要的结果都被读取时，有没有办法停止让步？我的意思是生成器在不做StopIteration的情况下给出值。例如，这永远不会停止：(修订)from random import randintdef devtrue():    while True:        yield Trueanswer...

问直到得到所有需要的值，有没有办法让切片变得懒惰？
EN

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问直到得到所有需要的值，有没有办法让切片变得懒惰？EN