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    <p>Just iterate over pairs, using <code>zip()</code>, which is much more readable.</p> <p><strong>UPDATE:</strong> for python 2.x, use <code>itertools.izip</code> instead as it is more efficient!</p> <pre><code>from itertools import izip for prev, next in izip(tmp, tmp[1:]): print 'seq: ', next[1], 'prev seq:', prev[1], 'variance: ', next[1]-prev[1] </code></pre> <p>which can also use value unpacking to avoid the index:</p> <pre><code>for (_, prev), (_, next) in izip(tmp, tmp[1:]): print 'seq: ', next, 'prev seq:', prev, 'variance: ', next-prev </code></pre> <p>Or, if you really need the first iteration too</p> <pre><code>for prev, next in izip(tmp, tmp[:1] + tmp): print 'seq: ', next[1], 'prev seq:', prev[1], 'variance: ', next[1]-prev[1] </code></pre> <p><strong>EDIT</strong></p> <p>If you want to avoid the creation of a list in the second argument you can also use an explicit iterator:</p> <pre><code>itr = iter(tmp) itr.next() # here I assume tmp is not empty, otherwise an exception will be thrown for prev, next in izip(tmp, itr): print 'seq: ', next[1], 'prev seq:', prev[1], 'variance: ', next[1]-prev[1] </code></pre> <p><em>Note</em>: This zip pattern is useful in similar problems too. For example to extract successive triplets from a list:</p> <pre><code>xs = range(9) triplets = zip(xs[::3], xs[1::3], xs[2::3]) # python 2.x, zip returns a list print xs # [0, 1, 2, 3, 4, 5, 6, 7, 8] print triplets # [(0, 1, 2), (3, 4, 5), (6, 7, 8)] </code></pre> <p>Also note that in <a href="http://diveintopython3.org/porting-code-to-python-3-with-2to3.html#zip" rel="nofollow noreferrer">python 3 zip returns an iterator</a>, similar to <code>itertools.izip</code>.</p>
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    1. POOptimal way to access a value from the last iteration in a loop
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    1. COthanks for the response, just tested this (Your edited code) and it seems the slowest of all the options (even slower then my original code above)
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    2. COAs noted, it might be because zip builds a full list in memory in python 2.x. Anyway, in that case you should use an explicit loop. Too bad, IMHO this is the optimal solution (optimal in the "desired pythonic way" sense). Good luck!
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    3. COSCRAP THAT, python 2.x has `itertools.izip` :p, please time again
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