[jira] [Commented] (SPARK-17130) SparseVectors.apply and SparseVectors.toArray have different returns when creating with a illegal indices

["Sean Owen (JIRA)" <ji...@apache.org>]

    [ https://issues.apache.org/jira/browse/SPARK-17130?page=com.atlassian.jira.plugin.system.issuetabpanels:comment-tabpanel&focusedCommentId=15426490#comment-15426490 ] 

Sean Owen commented on SPARK-17130:
-----------------------------------

Yeah, didn't you just comment on https://github.com/apache/spark/pull/14555 ? that's already being fixed there. This is a duplicate of SPARK-16965

> SparseVectors.apply and SparseVectors.toArray have different returns when creating with a illegal indices
> ---------------------------------------------------------------------------------------------------------
>
>                 Key: SPARK-17130
>                 URL: https://issues.apache.org/jira/browse/SPARK-17130
>             Project: Spark
>          Issue Type: Bug
>          Components: ML, MLlib
>    Affects Versions: 1.6.2, 2.0.0
>         Environment: spark 1.6.1 + scala
>            Reporter: Jon Zhong
>            Priority: Minor
>
> One of my colleagues ran into a bug of SparseVectors. He called the Vectors.sparse(size: Int, indices: Array[Int], values: Array[Double]) without noticing that the indices are assumed to be ordered.
> The vector he created has all value of 0.0 (without any warning), if we try to get value via apply method. However, SparseVector.toArray will generates a array using a method that is order insensitive. Hence, you will get a 0.0 when you call apply method, while you can get correct result using toArray or toDense method. The result of SparseVector.toArray is actually misleading.
> It could be safer if there is a validation of indices in the constructor or at least make the returns of apply method and toArray method the same.



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