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    <pre><code>import itertools as it import heapq import pprint def imerge(*iterables): ''' http://code.activestate.com/recipes/491285-iterator-merge/ Author: Raymond Hettinger Merge multiple sorted inputs into a single sorted output. Equivalent to: sorted(itertools.chain(*iterables)) &gt;&gt;&gt; list(imerge([1,3,5,7], [0,2,4,8], [5,10,15,20], [], [25])) [0, 1, 2, 3, 4, 5, 5, 7, 8, 10, 15, 20, 25] ''' heappop, siftup, _StopIteration = heapq.heappop, heapq._siftup, StopIteration h = [] h_append = h.append for it in map(iter, iterables): try: next = it.next h_append([next(), next]) except _StopIteration: pass heapq.heapify(h) while 1: try: while 1: v, next = s = h[0] # raises IndexError when h is empty yield v s[0] = next() # raises StopIteration when exhausted siftup(h, 0) # restore heap condition except _StopIteration: heappop(h) # remove empty iterator except IndexError: return a = ( 2, 3, 4, ) b = (1, ) c = ( 5,) d = (1, 3, 5,) def tag(iterator,val): for elt in iterator: yield elt,val def expand(group): dct=dict((tag,val)for val,tag in group) result=[dct.get(tag,None) for tag in range(4)] return result pprint.pprint( [ expand(group) for key,group in it.groupby( imerge(*it.imap(tag,(a,b,c,d),it.count())), key=lambda x:x[0] )]) </code></pre> <p>Explanation:</p> <ul> <li>Life would be easier if we mergesort the iterators. This can be done with <code>imerge</code></li> <li><p><code>itertools.groupby</code> gives us the desired grouping if we feed it the<br> result from <code>imerge</code>. The rest is just niggling details. </p> <pre><code>pprint.pprint( [ list(group) for key,group in it.groupby( imerge(a,b,c,d)) ] ) # [[1, 1], [2], [3, 3], [4], [5, 5]] </code></pre></li> <li>From the output above, it is clear we need to keep track of the source of each value -- (did the value come from <code>a</code>, or <code>b</code>, etc.). That way we can pad the output with <code>None</code>s in the right places.</li> <li><p>To do that I used <code>it.imap(tag,(a,b,c,d),it.count())</code>. <code>tag(a)</code> returns an iterator which yields values from <code>a</code> along with a counter value.</p> <pre><code>&gt;&gt;&gt; list(tag(a,0)) # [(2, 0), (3, 0), (4, 0)] </code></pre></li> <li><p>The output now looks like this:</p> <pre><code>pprint.pprint( [ list(group) for key,group in it.groupby( imerge(*it.imap(tag,(a,b,c,d),it.count())), key=lambda x:x[0] )]) # [[(1, 1), (1, 3)], [(2, 0)], [(3, 0), (3, 3)], [(4, 0)], [(5, 2), (5, 3)]] </code></pre></li> <li>Finally, we use <code>expand(group)</code> to change <code>[(1, 1), (1, 3)]</code> into <code>[None, 1, None, 1]</code>.</li> </ul>
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    1. POJoining multiple iteratorars by a key
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    1. USunutbu
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      1. VTUpMod
    1. COInteresting solution. Made my think about the problem in a very different way.
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      1. USGary van der Merwe
 

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