blocks|key|4818885|text|创建另一个迭代器的itertools解决方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4818886|from+itertools+import+takewhile
l+=+takewhile(lambda+x:+x+<+10000,+generate())|code-block|syntax|javascript|4818887|如果您确定需要一个列表，请将其包装在list()中：|4818888|l+=+list(takewhile(lambda+x:+x+<+10000,+generate()))|4818889|或者，如果你想要一个列表，比如发明轮子：|4818890|l+=+[]
for+x+in+generate():
++++if+x+<+10000:
++++++++l.append(x)
++++else:
++++++++break|4818891|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/2/library/itertools.html#itertools.takewhile^0|9|9|9|9|0|0|0|I|6|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@$9|13|A|14|1|15]]|E|$]]|$1|F|3|G|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1B|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|1C|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|1D|8|@]|D|@]|E|$]]]|T|$U|$5|V|W|X|E|$Y|Z]]]]

An <a href="http://docs.python.org/2/library/itertools.html#itertools.takewhile"><code>itertools</code></a> solution to create another iterator:

<pre><code>from itertools import takewhile
l = takewhile(lambda x: x &lt; 10000, generate())
</code></pre>

Wrap it in <code>list()</code> if you are sure you want a list:

<pre><code>l = list(takewhile(lambda x: x &lt; 10000, generate()))
</code></pre>

Or if you want a list and like inventing wheels:

<pre><code>l = []
for x in generate():
 if x &lt; 10000:
 l.append(x)
 else:
 break
</code></pre>

blocks|key|498378|text|只需对无限生成器使用计数器：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|498379|gen=Generate()++++++++#+your+generator+function+example
l=[gen.next()+for+i+in+range(100)]|code-block|syntax|javascript|498380|但是因为它是一个生成器，所以使用生成器表达式：|498381|seq=(gen.next()+for+i+in+xrange(100))++#need+x+in+xrange+in+Python+2.x;+3.x+use+range|498382|498383|编辑|offset|length|style|BOLD|498384|好的，那就使用一个控制器：|498385|def+controler(gen,limit):
++++n=gen.next()
++++while+n<limit:
++++++++yield+n
++++++++n=gen.next()

seq=[i+for+i+in+controler(Generate(),100)]|498386|entityMap^0|0|0|0|0|0|0|2|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Z|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|11|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|12|8|@$N|13|O|14|P|Q]]|9|@]|A|$]]|$1|R|3|S|5|6|7|15|8|@]|9|@]|A|$]]|$1|T|3|U|5|D|7|16|8|@]|9|@]|A|$E|F]]|$1|V|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|W|$]]

Just use a counter for an infinite generator:

<pre><code>gen=Generate() # your generator function example
l=[gen.next() for i in range(100)]
</code></pre>

But since it is a generator, use a generator expression:

<pre><code>seq=(gen.next() for i in xrange(100)) #need x in xrange in Python 2.x; 3.x use range
</code></pre>

<hr>

Edit

OK, then just use a controller:

<pre><code>def controler(gen,limit):
 n=gen.next()
 while n&lt;limit:
 yield n
 n=gen.next()

seq=[i for i in controler(Generate(),100)]
</code></pre>

blocks|key|498388|text|将您的生成器包装在另一个生成器中：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|498389|def+no_more_than(limit):
++++def+limiter(gen):
++++++++for+item+in+gen:
++++++++++++if+item+>+limit:
++++++++++++++++break
++++++++++++yield+item
++++return+limiter

def+fib():
++++a,b+=+1,1
++++while+1:
++++++++yield+a
++++++++a,b+=+b,a%2Bb


cutoff_at_100+=+no_more_than(100)
print+list(cutoff_at_100(fib()))|code-block|syntax|javascript|498390|打印：|498391|[1,+1,+2,+3,+5,+8,+13,+21,+34,+55,+89]|498392|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Wrap your generator within another generator:

<pre><code>def no_more_than(limit):
 def limiter(gen):
 for item in gen:
 if item &gt; limit:
 break
 yield item
 return limiter

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


cutoff_at_100 = no_more_than(100)
print list(cutoff_at_100(fib()))
</code></pre>

Prints:

<pre><code>[1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
</code></pre>

blocks|key|1468571|text|用xrange(10000)替换Generate()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1468572|entityMap^0|1|D|G|A|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

replace <code>Generate()</code> with <code>xrange(10000)</code>

blocks|key|5301931|text|itertools.takewhile只有在遇到不满足谓词的项时才起作用。如果需要从可能的无序迭代器返回所有值，我建议使用Python2.x的itertools.ifilter，如下所示|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5301932|from+itertools+import+ifilter
f+=+ifilter(lambda+x:+x+<+400,+gen())
f.next()|code-block|syntax|javascript|5301933|这过滤了一个生成器，如所希望的那样产生0到400之间的随机整数。|5301934|Python3.x中弃用了FWIW+itertools.ifilter，取而代之的是内置的filter()，后者的迭代语法略有不同|5301935|f+=+filter(lambda+x:+x+<+400,+gen())
next(f)|5301936|entityMap^0|0|J|1Z|H|0|0|0|I|H|19|8|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|Q|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|R|$]]

<code>itertools.takewhile</code> will only work until it comes across an item that does not fulfill the predicate. If you need to return all values from a possibly unordered iterable, I'd recommend using <code>itertools.ifilter</code> for Python 2.x as in

<pre><code>from itertools import ifilter
f = ifilter(lambda x: x &lt; 400, gen())
f.next()
</code></pre>

This filtered a generator yielding random integers between 0 and 400 as hoped.

FWIW <code>itertools.ifilter</code> was deprecated in Python 3.x in favour of the built-in <code>filter()</code> which has slightly different syntax for iterating

<pre><code>f = filter(lambda x: x &lt; 400, gen())
next(f)
</code></pre>

blocks|key|5301951|text|把它包在一个范围内的zip生成器上：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5301952|gen+=+range(100_000_000_000)
limit+=+10
(z[1]+for+z+in+zip(range(limit),+gen))|code-block|syntax|javascript|5301953|zip会创建一个元组，这就是z[1]的原因|offset|length|style|CODE|5301954|这可以在for循环上使用：|5301955|for+g+in+(z[1]+for+z+in+zip(range(limit),+gen)):
++++print(g)|5301956|或者，您可以使用lambda|5301957|wrap+=+lambda+gen,+limit:+(z[1]+for+z+in+zip(range(limit),+gen))
for+g+in+wrap(gen,+10):
++++print(g)|5301958|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.7/library/functions.html#zip|1|https://docs.python.org/3.7/reference/expressions.html#lambda^0|0|0|E|4|0|3|0|0|4|3|0|0|8|6|1|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|16|8|@$I|17|J|18|K|L]]|9|@$I|19|J|1A|1|1B]]|A|$]]|$1|M|3|N|5|6|7|1C|8|@$I|1D|J|1E|K|L]]|9|@]|A|$]]|$1|O|3|P|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|Q|3|R|5|6|7|1G|8|@]|9|@$I|1H|J|1I|1|1J]]|A|$]]|$1|S|3|T|5|D|7|1K|8|@]|9|@]|A|$E|F]]|$1|U|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|V|$W|$5|X|Y|Z|A|$10|11]]|12|$5|X|Y|Z|A|$10|13]]]]

Wrap it on a zip generator of a range of the limit:
<pre><code>gen = range(100_000_000_000)
limit = 10
(z[1] for z in zip(range(limit), gen))
</code></pre>
<a href="https://docs.python.org/3.7/library/functions.html#zip" rel="nofollow noreferrer">zip</a> creates a tuple, that is the reason for <code>z[1]</code>
This may be used on <code>for</code> loops:
<pre><code>for g in (z[1] for z in zip(range(limit), gen)):
 print(g)
</code></pre>
Or you could use <a href="https://docs.python.org/3.7/reference/expressions.html#lambda" rel="nofollow noreferrer">lambda</a>:
<pre><code>wrap = lambda gen, limit: (z[1] for z in zip(range(limit), gen))
for g in wrap(gen, 10):
 print(g)
</code></pre>

I have a generator that will keep giving numbers that follow a specific formula. For sake of argument let's say this is the function:

<pre><code># this is not the actual generator, just an example
def Generate():
 i = 0
 while 1:
 yield i
 i+=1 
</code></pre>

I then want to get a list of numbers from that generator that are below a certain threshold. I'm trying to figure out a pythonic way of doing this. I don't want to edit the function definition. I realize you could just use a while loop with your cutoff as the condition, but I'm wondering if there is a better way. I gave this a try, but soon realized why it wouldn't work.

<pre><code>l = [x for x in Generate() x&lt;10000] # will go on infinitely
</code></pre>

So is there a correct way of doing this.

Thanks

Python Generator Cutoff

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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我有一个生成器，它会不断给出遵循特定公式的数字。为了便于讨论，我们假设这是一个函数：# this is not the actual generator, just an exampledef Generate():    i = 0    while 1:        yield i        i+=1    ...

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