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https://issues.apache.org/jira/browse/SPARK-12703?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=15089844#comment-15089844
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Joseph K. Bradley commented on SPARK-12703:
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You're right that it shouldn't be computing sqrt. Would you mind sending a
little PR to fix it? Thanks!
> Spark KMeans Documentation Python Api
> -------------------------------------
>
> Key: SPARK-12703
> URL: https://issues.apache.org/jira/browse/SPARK-12703
> Project: Spark
> Issue Type: Documentation
> Components: MLlib
> Reporter: Anton
> Priority: Minor
> Original Estimate: 5m
> Remaining Estimate: 5m
>
> In the documentation of Spark's Kmeans - python api:
> http://spark.apache.org/docs/latest/mllib-clustering.html#k-means
> the cost of the final result is calculated using the 'error()' function where
> its returning:
> {quote}
> return sqrt(sum([x**2 for x in (point - center)]))
> {quote}
> As I understand, it's wrong to use sqrt() and it should be omitted:
> {quote} return sum([x**2 for x in (point - center)]).{quote}
> Please refer to :
> https://en.wikipedia.org/wiki/K-means_clustering#Description
> Where you can see that the power is canceling the square.
> What do you think? It's minor but wasted me a few min to understand why the
> result isn't what I'm expecting.
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