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    <p>Here's a lazy one-liner, also using itertools:</p> <pre><code>from itertools import compress, product def combinations(items): return ( set(compress(items,mask)) for mask in product(*[[0,1]]*len(items)) ) # alternative: ...in product([0,1], repeat=len(items)) ) </code></pre> <p>Main idea behind this answer: there are 2^N combinations -- same as the number of binary strings of length N. For each binary string, you pick all elements corresponding to a "1".</p> <pre><code>items=abc * mask=### | V 000 -&gt; 001 -&gt; c 010 -&gt; b 011 -&gt; bc 100 -&gt; a 101 -&gt; a c 110 -&gt; ab 111 -&gt; abc </code></pre> <p>Things to consider:</p> <ul> <li>This requires that you can call <code>len(...)</code> on <code>items</code> (workaround: if <code>items</code> is something like an iterable like a generator, turn it into a list first with <code>items=list(_itemsArg)</code>)</li> <li>This requires that the order of iteration on <code>items</code> is not random (workaround: don't be insane)</li> <li>This requires that the items are unique, or else <code>{2,2,1}</code> and <code>{2,1,1}</code> will both collapse to <code>{2,1}</code> (workaround: use <code>collections.Counter</code> as a drop-in replacement for <code>set</code>; it's basically a multiset... though you may need to later use <code>tuple(sorted(Counter(...).elements()))</code> if you need it to be hashable)</li> </ul> <hr> <p><strong>Demo</strong></p> <pre><code>&gt;&gt;&gt; list(combinations(range(4))) [set(), {3}, {2}, {2, 3}, {1}, {1, 3}, {1, 2}, {1, 2, 3}, {0}, {0, 3}, {0, 2}, {0, 2, 3}, {0, 1}, {0, 1, 3}, {0, 1, 2}, {0, 1, 2, 3}] &gt;&gt;&gt; list(combinations('abcd')) [set(), {'d'}, {'c'}, {'c', 'd'}, {'b'}, {'b', 'd'}, {'c', 'b'}, {'c', 'b', 'd'}, {'a'}, {'a', 'd'}, {'a', 'c'}, {'a', 'c', 'd'}, {'a', 'b'}, {'a', 'b', 'd'}, {'a', 'c', 'b'}, {'a', 'c', 'b', 'd'}] </code></pre>
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