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POPython: Determine if an unsorted list is contained in a 'list of lists', regardless of the order to the elements
 

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  1. POPython: Determine if an unsorted list is contained in a 'list of lists', regardless of the order to the elements
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    <p>I have a similar question to this question: <a href="https://stackoverflow.com/questions/8866652/determine-if-2-lists-have-the-same-elements-regardless-of-order">Determine if 2 lists have the same elements, regardless of order?</a></p> <p>What is the best/quickest way to determine whether an unsorted list <code>list1</code> is contained in a 'list of lists' <code>myListOfLists</code>, regardless of the order to the elements in <code>list1</code>? My attempt is wrapped up in the function <code>doSomething(...)</code> which I call many times:</p> <pre class="lang-py prettyprint-override"><code>def doSomething(myListOfLists, otherInputs): list1 = [] ... # do something here with `otherInputs' ... # which gives `list1' some values # now only append `list1' to `myListOfLists' if it doesn't already exist # and if it does exist, remove it removeFromList = False for myList in myListOfLists: if sorted(list1) == sorted(myList): removeFromList = True break if removeFromList: myListOfLists.remove(list1) else: myListOfLists.append(list1) return myListOfLists </code></pre> <p>The problem with this is that I need to run the function <code>doSomething(...)</code> approximately 1.0e5 times. As <code>myListOfLists</code> gets bigger with every call of <code>doSomething(...)</code> this becomes massively time consuming.</p> <p>EDIT:</p> <p>Some clarification of the task. Let me give an example of the desired output here:</p> <pre><code>a = [] doSomething(a, [1,2,3]) &gt;&gt; a = [1,2,3] </code></pre> <p>Because <code>[1,2,3]</code> is not in <code>a</code>, it is appended to <code>a</code>.</p> <pre><code>doSomething(a, [3,4,5]) &gt;&gt; a = [[1,2,3], [3,4,5]] </code></pre> <p>Because <code>[3,4,5]</code> is not in <code>a</code>, it is appended to <code>a</code>.</p> <pre><code>doSomething(a, [1,2,3]) &gt;&gt;[3,4,5] </code></pre> <p>Because <code>[1,2,3]</code> <strong>is</strong> in <code>a</code>, it is removed from <code>a</code>.</p> <p>EDIT:</p> <p>All lists have the same length.</p>
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    1. COOne optimization at first glance: Don't sort `list1` again every iteration, but instead sort it once before the loop, and store it.
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    2. COsort list1 outside the loop !
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    3. COWhat do the contents of the inner lists look like? You may benefit from converting them to tuples / sets temporarily so that they can be more conveniently compared.
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      1. USg.d.d.c
 

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