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    <h1>(New Answer) Using Numpy</h1> <pre><code>ranges = np.vstack((a,b)) ranges.sort(0) # List of non-overlapping ranges nonoverlapping = (ranges[1:,0] - ranges[:-1,1] &gt; 1).nonzero()[0] # Starts are 0, and all the starts not overlapped by their predecessor starts = np.hstack(([0], nonoverlapping + 1)) # Ends are -1 and all the ends who aren't overlapped by their successor ends = np.hstack(( nonoverlapping, [-1])) # Result result = np.vstack((ranges[starts, 0], ranges[ends, 1])).T </code></pre> <p><strong>(Old answer) Using lists and sets</strong></p> <pre><code>import numpy as np import itertools def ranges(s): """ Converts a list of integers into start, end pairs """ for a, b in itertools.groupby(enumerate(s), lambda(x, y): y - x): b = list(b) yield b[0][1], b[-1][1] def intersect(*args): """ Converts any number of numpy arrays containing start, end pairs into a set of indexes """ s = set() for start, end in np.vstack(args): s = s | set(range(start,end+1)) return s a = np.array([[5,7],[12, 18],[20,29]]) b = np.array([[2,4],[8,11],[33,35]]) result = np.array(list(ranges(intersect(a,b)))) </code></pre> <h1>References</h1> <ul> <li><a href="https://stackoverflow.com/questions/6821156/how-to-find-range-overlap-in-python">How to find range overlap in python?</a></li> <li><a href="http://docs.python.org/2/library/sets.html" rel="nofollow noreferrer">http://docs.python.org/2/library/sets.html</a></li> <li><a href="https://stackoverflow.com/questions/4628333/converting-a-list-of-integers-into-range-in-python">converting a list of integers into range in python</a></li> <li><a href="http://docs.python.org/2/library/itertools.html#itertools.groupby" rel="nofollow noreferrer">http://docs.python.org/2/library/itertools.html#itertools.groupby</a></li> </ul>
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    1. COthanks for the proposed solution
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    2. COWhat did you find to not work with your new solution? I was poking at it right now...
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    3. COWith the "non overlapping" caveat, I'd say it works fine, doesn't it?
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