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    <p><strong>Warning</strong> This is a bit convoluted but does the job. I will use an example to explain it.</p> <p>Let say <code>expensive_function = math.sin</code></p> <p><code>infinite generator = collections.count(0.1,0.1)</code></p> <p>then </p> <pre><code>[z for z in (y if y &lt; 5 else next(iter([])) for y in (math.sin(x) for x in itertools.count(0.1,0.1)))] </code></pre> <p>is</p> <pre><code>[0.09983341664682815, 0.19866933079506122, 0.2955202066613396, 0.3894183423086505, 0.479425538604203] </code></pre> <p>So your problem boils down to</p> <pre><code>[z for z in (y if y &lt; 0.5 else next(iter([])) \ for y in (expensive_function(x) for x in generator))] </code></pre> <p>The trick is to force a <code>StopIteration</code> from a generator and nothing elegant than <code>next(iter([]))</code></p> <p>Here <code>expensive_function</code> is only called once per iteration.</p> <p>Extend the Infinite Generator with a Finite Generator, with the Stop Condition. As the generator won't allow <code>raise StopIteration</code>, we opt for a convoluted way i.e. <code>next(iter([]))</code> And now you have a Finite Generator, which can be used in a List Comprehension</p> <p>As OP was concerned with the application of the above method for a non-<a href="http://en.wikipedia.org/wiki/Monotonic_function" rel="nofollow">monotonic</a> function here is a fictitious non-monotonic function</p> <p>Expensive Non-Monotonic Function <code>f(x) = random.randint(1,100)*x</code></p> <p>Stop Condition = <code>&lt; 7</code></p> <pre><code>[z for z in (y if y &lt; 7 else next(iter([])) for y in (random.randint(1,10)*x for x in itertools.count(0.1,0.1)))] [0.9, 0.6000000000000001, 1.8000000000000003, 4.0, 0.5, 6.0, 4.8999999999999995, 3.1999999999999997, 3.5999999999999996, 5.999999999999999] </code></pre> <p>Btw: <code>sin</code> in true sense is non-monotonic over the entire range <code>(0,2pi)</code></p>
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    1. POPossible to capture the returned value from a Python list comprehension for use a condition?
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    1. USEthan Furman
    1. USAbhijit
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    1. COCrazy! This is subtly different, because `sin` is a non-monotonic function (I didn't say that `expensive_function` is monotonic, but it is!) and this halts **the first time** the condition is not true, not continuing **for all conditions** where the condition is true. That said, if it were non-monotonic, this would result in an infinite evaluation...
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      1. USJoe
    2. CO@Joe, This will work even for non-monotonic function. See my update
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