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POk-means clustering implementation in python, running out of memory
 

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  1. POk-means clustering implementation in python, running out of memory
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    <h3> Note: updates/solutions at the bottom of this question</h3> <p>As part of a product recommendation engine, I'm trying to segment my users based on their product preferences starting with using the k-means clustering algorithm.</p> <p>My data is a dictionary of the form:</p> <pre><code>prefs = { 'user_id_1': { 1L: 3.0f, 2L: 1.0f, }, 'user_id_2': { 4L: 1.0f, 8L: 1.5f, }, } </code></pre> <p>where the product ids are the longs, and ratings are floats. the data is sparse. I currently have about 60,000 users, most of whom have only rated a handful of products. The dictionary of values for each user is implemented using a defaultdict(float) to simplify the code.</p> <p>My implementation of k-means clustering is as follows:</p> <pre><code>def kcluster(prefs,sim_func=pearson,k=100,max_iterations=100): from collections import defaultdict users = prefs.keys() centroids = [prefs[random.choice(users)] for i in range(k)] lastmatches = None for t in range(max_iterations): print 'Iteration %d' % t bestmatches = [[] for i in range(k)] # Find which centroid is closest for each row for j in users: row = prefs[j] bestmatch=(0,0) for i in range(k): d = simple_pearson(row,centroids[i]) if d &lt; bestmatch[1]: bestmatch = (i,d) bestmatches[bestmatch[0]].append(j) # If the results are the same as last time, this is complete if bestmatches == lastmatches: break lastmatches=bestmatches centroids = [defaultdict(float) for i in range(k)] # Move the centroids to the average of their members for i in range(k): len_best = len(bestmatches[i]) if len_best &gt; 0: items = set.union(*[set(prefs[u].keys()) for u in bestmatches[i]]) for user_id in bestmatches[i]: row = prefs[user_id] for m in items: if row[m] &gt; 0.0: centroids[i][m]+=(row[m]/len_best) return bestmatches </code></pre> <p>As far as I can tell, the algorithm is handling the first part (assigning each user to its nearest centroid) fine.</p> <p>The problem is when doing the next part, taking the average rating for each product in each cluster and using these average ratings as the centroids for the next pass.</p> <p>Basically, before it's even managed to do the calculations for the first cluster (i=0), the algorithm bombs out with a MemoryError at this line:</p> <pre><code>if row[m] &gt; 0.0: centroids[i][m]+=(row[m]/len_best) </code></pre> <p>Originally the division step was in a seperate loop, but fared no better.</p> <p>This is the exception I get:</p> <pre><code>malloc: *** mmap(size=16777216) failed (error code=12) *** error: can't allocate region *** set a breakpoint in malloc_error_break to debug </code></pre> <p>Any help would be greatly appreciated.</p> <hr> <h3>Update: Final algorithms</h3> <p>Thanks to the help recieved here, this is my fixed algorithm. If you spot anything blatantly wrong please add a comment.</p> <p>First, the simple_pearson implementation</p> <pre><code>def simple_pearson(v1,v2): si = [val for val in v1 if val in v2] n = len(si) if n==0: return 0.0 sum1 = 0.0 sum2 = 0.0 sum1_sq = 0.0 sum2_sq = 0.0 p_sum = 0.0 for v in si: sum1+=v1[v] sum2+=v2[v] sum1_sq+=pow(v1[v],2) sum2_sq+=pow(v2[v],2) p_sum+=(v1[v]*v2[v]) # Calculate Pearson score num = p_sum-(sum1*sum2/n) temp = (sum1_sq-pow(sum1,2)/n) * (sum2_sq-pow(sum2,2)/n) if temp &lt; 0.0: temp = -temp den = sqrt(temp) if den==0: return 1.0 r = num/den return r </code></pre> <p>A simple method to turn simple_pearson into a distance value:</p> <pre><code>def distance(v1,v2): return 1.0-simple_pearson(v1,v2) </code></pre> <p>And finally, the k-means clustering implementation:</p> <pre><code>def kcluster(prefs,k=21,max_iterations=50): from collections import defaultdict users = prefs.keys() centroids = [prefs[u] for u in random.sample(users, k)] lastmatches = None for t in range(max_iterations): print 'Iteration %d' % t bestmatches = [[] for i in range(k)] # Find which centroid is closest for each row for j in users: row = prefs[j] bestmatch=(0,2.0) for i in range(k): d = distance(row,centroids[i]) if d &lt;= bestmatch[1]: bestmatch = (i,d) bestmatches[bestmatch[0]].append(j) # If the results are the same as last time, this is complete if bestmatches == lastmatches: break lastmatches=bestmatches centroids = [defaultdict(float) for i in range(k)] # Move the centroids to the average of their members for i in range(k): len_best = len(bestmatches[i]) if len_best &gt; 0: for user_id in bestmatches[i]: row = prefs[user_id] for m in row: centroids[i][m]+=row[m] for key in centroids[i].keys(): centroids[i][key]/=len_best # We may have made the centroids quite dense which significantly # slows down subsequent iterations, so we delete values below a # threshold to speed things up if centroids[i][key] &lt; 0.001: del centroids[i][key] return centroids, bestmatches </code></pre>
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    1. COI am doing this on a Windows Vista laptop with 4GB and the memory use appears to be about 100MB using 100k users. so I am not getting the problem you describe. However, I do get this: >>> print kcluster(prefs,k=100,max_iterations=100) Iteration 0 Traceback (most recent call last): File "<stdin>", line 1, in <module> File "<stdin>", line 38, in kcluster KeyError: 3 So maybe there is something wrong with your algorithm. Or the indentation: I could not cut and paste your code from SO without reformatting. Might have got that wrong.
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