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    <p><strong>Starting with Python 2.6</strong> (and if you're on Python 3) you have a <strong>standard-library</strong> tool for this: <a href="https://docs.python.org/2/library/itertools.html#itertools.permutations" rel="noreferrer"><code>itertools.permutations</code></a>.</p> <pre><code>import itertools list(itertools.permutations([1, 2, 3])) </code></pre> <hr> <p>If you're using an <strong>older Python (&lt;2.6)</strong> for some reason or are just curious to know how it works, here's one nice approach, taken from <a href="http://code.activestate.com/recipes/252178/" rel="noreferrer">http://code.activestate.com/recipes/252178/</a>:</p> <pre><code>def all_perms(elements): if len(elements) &lt;=1: yield elements else: for perm in all_perms(elements[1:]): for i in range(len(elements)): # nb elements[0:1] works in both string and list contexts yield perm[:i] + elements[0:1] + perm[i:] </code></pre> <p>A couple of alternative approaches are listed in the documentation of <code>itertools.permutations</code>. Here's one:</p> <pre><code>def permutations(iterable, r=None): # permutations('ABCD', 2) --&gt; AB AC AD BA BC BD CA CB CD DA DB DC # permutations(range(3)) --&gt; 012 021 102 120 201 210 pool = tuple(iterable) n = len(pool) r = n if r is None else r if r &gt; n: return indices = range(n) cycles = range(n, n-r, -1) yield tuple(pool[i] for i in indices[:r]) while n: for i in reversed(range(r)): cycles[i] -= 1 if cycles[i] == 0: indices[i:] = indices[i+1:] + indices[i:i+1] cycles[i] = n - i else: j = cycles[i] indices[i], indices[-j] = indices[-j], indices[i] yield tuple(pool[i] for i in indices[:r]) break else: return </code></pre> <p>And another, based on <code>itertools.product</code>:</p> <pre><code>def permutations(iterable, r=None): pool = tuple(iterable) n = len(pool) r = n if r is None else r for indices in product(range(n), repeat=r): if len(set(indices)) == r: yield tuple(pool[i] for i in indices) </code></pre>
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    1. POHow to generate all permutations of a list in Python
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    1. USMateen Ulhaq
    1. USEli Bendersky
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    1. POHow to generate all permutations of a list in Python
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    1. COThis and other recursive solutions have a potential hazard of eating up all the RAM if the permutated list is big enough
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      1. USBoris Gorelik
    2. COThey also reach the recursion limit (and die) with large lists
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      1. USdbr
    3. CObgbg, dbr: Its using a generator, so the function itself won't eat up memory. Its left to you on how to consume the iterator returned by all_perms (say you could write each iteration to disk and not worry about memory). I know this post is old but I'm writing this for the benefit of everyone who reads it now. Also now, the best way would be to use itertools.permutations() as pointed out by many.
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      1. USJagtesh Chadha
 

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