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    <p>When the list is so small there is no significant difference between the two. If the input list can grow large then there is a worse problem: you're iterating over the whole list, while you could stop at the first element. You could accomplish this with a for loop, but if you want to use a comprehension-like statement, here come <em>generator expressions</em>:</p> <pre><code># like list comprehensions but with () instead of [] gen = (b for a, b in foo if a == 'b') my_element = next(gen) </code></pre> <p>or simply:</p> <pre><code>my_element = next(b for a, b in foo if a == 'b') </code></pre> <p>If you want to learn more about generator expressions give a look at <a href="http://www.python.org/dev/peps/pep-0289/" rel="nofollow noreferrer">PEP 289</a>.</p> <hr> <p>Note that even with generators and iterators you have more than one choice.</p> <pre><code># Python 3: my_element = next(filter(lambda x: x[0] == 'b', foo)) # Python 2: from itertools import ifilter my_element = next(ifilter(lambda (x, y): x == 'b', foo)) </code></pre> <p>I personally don't like and don't recommend this because it is much less readable. It turns out that this is actually slower than my first snippet, but more in general using <code>filter()</code> instead of a generator expression might be faster in some special cases.</p> <p>In any case if you need benchmarking your code, I recommend using <a href="http://docs.python.org/2/library/timeit.html" rel="nofollow noreferrer">the <code>timeit</code> module</a>.</p>
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    1. CO`next(v for k,v in foo if k == "b")` -- using `next()` rather than `.next()` -- is more forward-compatible.
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    2. COThank you, updated the answer.
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    3. COIn my timings, the `ifilter` version loses if it has to check more than two items.
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      1. USJanne Karila
 

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