[
https://issues.apache.org/jira/browse/SPARK-12703?page=com.atlassian.jira.plugin.system.issuetabpanels:all-tabpanel
]
Anton updated SPARK-12703:
--------------------------
Description:
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.
was:
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} 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.
> 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.
--
This message was sent by Atlassian JIRA
(v6.3.4#6332)
---------------------------------------------------------------------
To unsubscribe, e-mail: [email protected]
For additional commands, e-mail: [email protected]
