I can see how the linked documentation could be confusing:
"Aggregate function: returns the number of items in a group."

What it doesn't mention is that it returns the number of rows for which the
given column is non-null.

Xinh

On Wed, Jun 22, 2016 at 9:31 AM, Takeshi Yamamuro <linguin....@gmail.com>
wrote:

> Hi,
>
> An argument for `functions.count` is needed for per-column counting;
> df.groupBy($"a").agg(count($"b"))
>
> // maropu
>
> On Thu, Jun 23, 2016 at 1:27 AM, Ted Yu <yuzhih...@gmail.com> wrote:
>
>> See the first example in:
>>
>> http://www.w3schools.com/sql/sql_func_count.asp
>>
>> On Wed, Jun 22, 2016 at 9:21 AM, Jakub Dubovsky <
>> spark.dubovsky.ja...@gmail.com> wrote:
>>
>>> Hey Ted,
>>>
>>> thanks for reacting.
>>>
>>> I am refering to both of them. They both take column as parameter
>>> regardless of its type. Intuition here is that count should take no
>>> parameter. Or am I missing something?
>>>
>>> Jakub
>>>
>>> On Wed, Jun 22, 2016 at 6:19 PM, Ted Yu <yuzhih...@gmail.com> wrote:
>>>
>>>> Are you referring to the following method in
>>>> sql/core/src/main/scala/org/apache/spark/sql/functions.scala :
>>>>
>>>>   def count(e: Column): Column = withAggregateFunction {
>>>>
>>>> Did you notice this method ?
>>>>
>>>>   def count(columnName: String): TypedColumn[Any, Long] =
>>>>
>>>> On Wed, Jun 22, 2016 at 9:06 AM, Jakub Dubovsky <
>>>> spark.dubovsky.ja...@gmail.com> wrote:
>>>>
>>>>> Hey sparkers,
>>>>>
>>>>> an aggregate function *count* in *org.apache.spark.sql.functions*
>>>>> package takes a *column* as an argument. Is this needed for
>>>>> something? I find it confusing that I need to supply a column there. It
>>>>> feels like it might be distinct count or something. This can be seen in 
>>>>> latest
>>>>> documentation
>>>>> <http://spark.apache.org/docs/latest/api/scala/index.html#org.apache.spark.sql.functions$>
>>>>> .
>>>>>
>>>>> I am considering filling this in spark bug tracker. Any opinions on
>>>>> this?
>>>>>
>>>>> Thanks
>>>>>
>>>>> Jakub
>>>>>
>>>>>
>>>>
>>>
>>
>
>
> --
> ---
> Takeshi Yamamuro
>

Reply via email to