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    <p>We can do away with questions 2 and 3 by answering question 1, but understanding question 3 is important so I'll do that first and then show how you can avoid it entirely:</p> <h2>Question 3</h2> <p>As to question 3, it's because when you loop over a list in python and make changes to it <em>within the loop</em>, the list that you loop over changes.</p> <p>From the <a href="http://docs.python.org/tutorial/controlflow.html" rel="nofollow">python docs on control flow (for statement section)</a>:</p> <blockquote> <p>It is <strong>not safe</strong> to modify the sequence being iterated over in the loop (this can only happen for mutable sequence types, such as lists).</p> </blockquote> <p>Say your list is <code>[a,b,c,d]</code> and you loop through it with <code>for el in List</code>. Say <code>el</code> is currently <code>a</code> and you do <code>List.remove(el)</code>.</p> <p>Now, your list is <code>[b,c,d]</code>. However, the iterator points to the second element in the list (since it's done the first), which is now <code>c</code>. In essence, you've skipped <code>b</code>. So the problem is that you are modifying the list you are iterating over.</p> <p>There are a few ways to fix this: if your <code>List</code> is not expensive to duplicate, you could make a copy. So iterate over <code>List[:]</code> but remove from <code>List</code>.</p> <p>But suppose it's expensive to make copies of <code>List</code> all the time. Then what you do is iterate over it <strong>backwards</strong>. Note the <code>reversed</code> below:</p> <pre><code>for el in reversed(List): if el.count('$') &lt; n: counter+=1 List.remove(el) return(List) </code></pre> <p>In the example above, suppose we iterate backwards over <code>List</code>. The iterator starts at <code>d</code>, and then goes to <code>c</code>. Suppose we remove <code>c</code>, so that <code>List=[a,b,d]</code>. Since the iterator is going <em>backwards</em>, it now points to element <code>b</code>, so we haven't skipped anything.</p> <p>Basically, this avoids modifying bits of the list you have yet to iterate over.</p> <h2>Questions 1 &amp; 2</h2> <p>If I understand your question correctly, you basically want to choose <code>n</code> out of <code>m</code> positions, where <code>m</code> is the length of the string (<code>abcde</code>), and place a '$' in each of these <code>n</code> positions.</p> <p>In that case, you can use the <code>itertools</code> module to do that.</p> <pre><code>import itertools def Mismatch(Search,n): SearchL = list(Search) List = [] # hold output # print list of indices to replace with '$' idxs = itertools.combinations(range(len(SearchL)),n) # for each combination `idx` in idxs, replace str[idx] with '$': for idx in idxs: str = SearchL[:] # make a copy for i in idx: str[i]='$' List.append( ''.join(str) ) # convert back to string return List </code></pre> <p>Let's look at how this works:</p> <ol> <li>turn the <code>Search</code> string into a list so it can be iterated over, create empty <code>List</code> to hold results.</li> <li><p><code>idxs = itertools.combinations(range(len(SearchL)),n)</code> says "find all subsets of length n in the set <code>[0,1,2,3,...,length-of-search-string -1]</code>. Try </p> <pre><code>idxs = itertools.combinations(range(5),4) for idx in idxs: print idx </code></pre> <p>to see what I mean.</p></li> <li>Each element of <code>idxs</code> is a tuple of <code>n</code> indices from 0 to <code>len(SearchL)-1</code> (e.g. <code>(0,1,2,4)</code>. Replace the i'th character of <code>SearchL</code> with a '$' for each <code>i</code> in the tuple.</li> <li>Convert the result back into a string and add it to <code>List</code>.</li> </ol> <p>As an example:</p> <pre><code>Mismatch('abcde',3) ['$$$de', '$$c$e', '$$cd$', '$b$$e', '$b$d$', '$bc$$', 'a$$$e', 'a$$d$', 'a$c$$', 'ab$$$'] Mismatch('abcde',4) # note, the code you had made lots of duplicates. ['$$$$e', '$$$d$', '$$c$$', '$b$$$', 'a$$$$'] </code></pre>
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    1. COThanks a lot for answering question 3! That was a very lucid explanation that helps me understand python list iterations a bit better. The reversed solution is very nice!
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    2. COAww....great. I figured there was something like itertools out there. Thanks so much for not only the solutions but the thoughtful explanations!
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