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    <p>This probably won't win any efficiency awards, but you get style points for the list comprehensions.</p> <p>This is how I would approach the problem. Make a list of a sliding window of size 3.</p> <pre><code>&gt;&gt;&gt; nums = [1, 3, 5, 6, 7, 8, 9, 10, 15, 19, 20, 22, 23, 24, 26, 27, 28, 32, 33, 35, 37, 38, 39, 40, 41, 42, 43, 44, 47, 48] &gt;&gt;&gt; [nums[i:i+3] for i in xrange(len(nums))] [[1, 3, 5], [3, 5, 6], [5, 6, 7], [6, 7, 8], [7, 8, 9], [8, 9, 10], [9, 10, 15], [10, 15, 19], [15, 19, 20], [19, 20, 22], [20, 22, 23], [22, 23, 24], [23, 24, 26], [24, 26, 27], [26, 27, 28], [27, 28, 32], [28, 32, 33], [32, 33, 35], [33, 35, 37], [35, 37, 38], [37, 38, 39], [38, 39, 40], [39, 40, 41], [40, 41, 42], [41, 42, 43], [42, 43, 44], [43, 44, 47], [44, 47, 48], [47, 48], [48]] </code></pre> <p>Next step, get rid of the consecutive items, which is trivially easy now. This predicate will get cleverly filter out the consecutive items.</p> <pre><code>&gt;&gt;&gt; [nums[i] for i in xrange(len(nums)) if nums[i:i+3] != range(nums[i],nums[i]+3)] [1, 3, 9, 10, 15, 19, 20, 23, 24, 27, 28, 32, 33, 35, 43, 44, 47, 48] </code></pre> <p>EDIT:</p> <p>Eric brought up a good point, the solution above does not work entirely. If you want this to work, then predicate is going to need some beefing up. First, I derived these equations. They perform the windowing operation. Convince yourself that they are true:</p> <pre><code>a = [1,2,3,4,5] i = 2 a[i-0:i+3] == range(a[i-0], a[i]+3) # left a[i-1:i+2] == range(a[i-1], a[i]+2) # center a[i-2:i+1] == range(a[i-2], a[i]+1) # right </code></pre> <p>Then you <em>could</em> jam it in there sideways...</p> <pre><code>[a for i,a in enumerate(nums) if all(nums[i-j:i+k] != range(nums[i-j], nums[i]+k) for j,k in zip(xrange(0,3,1), xrange(3,0,-1)))] </code></pre> <p>But if you don't want to get shot, pull out the predicate into a function:</p> <pre><code>consec_to_buddies = lambda i, xs: ( xs[i-0:i+3] == range(xs[i-0], xs[i]+3) or xs[i-1:i+2] == range(xs[i-1], xs[i]+2) or xs[i-2:i+1] == range(xs[i-2], xs[i]+1) ) [a for i,a in enumerate(nums) if not consec_to_buddies(i, nums)] </code></pre> <p>Again, this isn't the most efficient, as you will be calculating the predicate for every item, even if you already know you are taking it out. The price you pay for elegance :)</p>
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    1. POEliminating consecutive number combinations from a result set
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    1. COThis is an awesome list comprehension but only removes the *first* of the three consecutive. (Example: `22, 23, 24` -> `23, 24`)
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