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    <p>Your function <code>distance</code> is calculating the total distance, when you really only care about distance=1. The majority of cases you'll know it's >1 within a few characters, so you could return early and save a lot of time.</p> <p>Beyond that, there might be a better algorithm, but I can't think of it.</p> <p><b>Edit:</b> Another idea.</p> <p>You can make 2 cases, depending on whether the first character matches. If it doesn't match, the rest of the word has to match exactly, and you can test for that in one shot. Otherwise, do it similarly to what you were doing. You could even do it recursively, but I don't think that would be faster.</p> <pre><code>def DifferentByOne(word1, word2): if word1[0] != word2[0]: return word1[1:] == word2[1:] same = True for i in range(1, len(word1)): if word1[i] != word2[i]: if same: same = False else: return False return not same </code></pre> <p><b>Edit 2:</b> I've deleted the check to see if the strings are the same length, since you say it's redundant. Running <a href="https://stackoverflow.com/questions/788084/how-can-i-optimize-this-python-code/788168#788168">Ryan's tests</a> on my own code and on the is_neighbors function <a href="https://stackoverflow.com/questions/788084/how-can-i-optimize-this-python-code/788113#788113">provided by MizardX</a>, I get the following:</p> <ul> <li>Original distance(): 3.7 seconds</li> <li>My DifferentByOne(): 1.1 seconds</li> <li>MizardX's is_neighbors(): 3.7 seconds</li> </ul> <p><b>Edit 3:</b> (Probably getting into community wiki territory here, but...)</p> <p>Trying your final definition of is_neighbors() with izip instead of zip: 2.9 seconds.</p> <p>Here's my latest version, which still times at 1.1 seconds:</p> <pre><code>def DifferentByOne(word1, word2): if word1[0] != word2[0]: return word1[1:] == word2[1:] different = False for i in range(1, len(word1)): if word1[i] != word2[i]: if different: return False different = True return different </code></pre>
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    1. POHow can I optimize this Python code to generate all words with word-distance 1?
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    1. USMark Ransom
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    1. COif word1[0] != word2[0]: return word1[1:] == word2[1:] This is likely to make it slower because slicing an immutable string creates a new copy of the string in memory.
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    2. COThat seems slower than the original code, with a really basic benchmark.. Using the timeit module, distance("abc", "abd") took 4.03 seconds, DifferenceByOne("abc", "abd") took 5.02
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    3. COIs indexing each character also a slicing operation? That would probably be slow, too.
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