[jira] [Assigned] (SPARK-12703) Spark KMeans Documentation Python Api

["Apache Spark (JIRA)" <ji...@apache.org>]

     [ https://issues.apache.org/jira/browse/SPARK-12703?page=com.atlassian.jira.plugin.system.issuetabpanels:all-tabpanel ]

Apache Spark reassigned SPARK-12703:
------------------------------------

    Assignee:     (was: Apache Spark)

> 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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