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copied!<p>I would do something like this:</p> <pre><code>from collections import defaultdict a = [1, 2, 3] b = [1, 2, 3, 1, 2, 3] # Build up the count of occurrences in b d = defaultdict(int) for bb in b: d[bb] += 1 # Remove one for each occurrence in a for aa in a: d[aa] -= 1 # Create a list for all elements that still have a count of one or more result = [] for k, v in d.iteritems(): if v > 0: result += [k] * v </code></pre> <p>Or, if you are willing to be slightly more obscure:</p> <pre><code>from operator import iadd result = reduce(iadd, [[k] * v for k, v in d.iteritems() if v > 0], []) </code></pre> <p>defaultdict generates a count of the occurrences of each key. Once it has been built up from <code>b</code>, it is decremented for each occurrence of a key in <code>a</code>. Then we print out the elements that are still left over, allowing them to occur multiple times.</p> <p>defaultdict works with python 2.6 and up. If you are using a later python (2.7 and up, I believe), you can look into <a href="http://docs.python.org/library/collections.html#collections.Counter" rel="nofollow"><code>collections.Counter</code></a>.</p> <hr> <p>Later: you can also generalize this and create subtractions of counter-style defaultdicts:</p> <pre><code>from collections import defaultdict from operator import iadd a = [1, 2, 3, 4, 5, 6] b = [1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3] def build_dd(lst): d = defaultdict(int) for item in lst: d[item] += 1 return d def subtract_dd(left, right): return {k: left[k] - v for k, v in right.iteritems()} db = build_dd(b) da = build_dd(a) result = reduce(iadd, [[k] * v for k, v in subtract_dd(db, da).iteritems() if v > 0], []) print result </code></pre> <p>But the <code>reduce</code> expression is pretty obscure now.</p> <hr> <p>Later still: in python 2.7 and later, using <code>collections.Counter</code>, it looks like this:</p> <pre><code>from collections import Counter base = [1, 2, 3] missing = [4, 5, 6] extra = [7, 8, 9] a = base + missing b = base * 4 + extra result = Counter(b) - Counter(a) print result assert result == dict([(k, 3) for k in base] + [(k, 1) for k in extra]) </code></pre>

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