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Posted to users@zeppelin.apache.org by Ruslan Dautkhanov <da...@gmail.com> on 2017/12/11 05:19:57 UTC
IPython is available, use IPython for PySparkInterpreter
Getting "IPython is available, use IPython for PySparkInterpreter" warning
after starting pyspark interpreter.
How do I default %pyspark to ipython?
Tried to change to
"class": "org.apache.zeppelin.spark.PySparkInterpreter",
to
"class": "org.apache.zeppelin.spark.IPySparkInterpreter",
in interpreter.json but this gets overwritten back to PySparkInterpreter.
Also tried to change to zeppelin.pyspark.python to ipython with no luck too.
Is there is a documented way to default pyspark interpreter to ipython?
Glanced over PR-2474 but can't quickly get what I am missing.
Thanks.
Re: IPython is available, use IPython for PySparkInterpreter
Posted by Ruslan Dautkhanov <da...@gmail.com>.
Makes sense
Thank you Jeff !
On Sun, Dec 10, 2017 at 11:24 PM Jeff Zhang <zj...@gmail.com> wrote:
>
> I am afraid currently there's no way to make ipython as default of
> %pyspark, but you can use %ipyspark to use ipython without this warning
> message.
>
> But making ipython as default is on my plan, for now I try to keep
> backward compatibility as much as possible, so only use ipython when it is
> available, otherwise still use the old python interpreter implementation.
> I will change ipython as default and the original python implementation as
> fallback when ipython interpreter become much more mature.
>
>
>
>
> Ruslan Dautkhanov <da...@gmail.com>于2017年12月11日周一 下午1:20写道：
>
>> Getting "IPython is available, use IPython for PySparkInterpreter"
>> warning after starting pyspark interpreter.
>>
>> How do I default %pyspark to ipython?
>>
>> Tried to change to
>> "class": "org.apache.zeppelin.spark.PySparkInterpreter",
>> to
>> "class": "org.apache.zeppelin.spark.IPySparkInterpreter",
>> in interpreter.json but this gets overwritten back to PySparkInterpreter.
>>
>> Also tried to change to zeppelin.pyspark.python to ipython with no luck
>> too.
>>
>> Is there is a documented way to default pyspark interpreter to ipython?
>> Glanced over PR-2474 but can't quickly get what I am missing.
>>
>>
>> Thanks.
>>
>>
>>
Re: IPython is available, use IPython for PySparkInterpreter
Posted by Jeff Zhang <zj...@gmail.com>.
I am afraid currently there's no way to make ipython as default of
%pyspark, but you can use %ipyspark to use ipython without this warning
message.
But making ipython as default is on my plan, for now I try to keep backward
compatibility as much as possible, so only use ipython when it is
available, otherwise still use the old python interpreter implementation.
I will change ipython as default and the original python implementation as
fallback when ipython interpreter become much more mature.
Ruslan Dautkhanov <da...@gmail.com>于2017年12月11日周一 下午1:20写道：
> Getting "IPython is available, use IPython for PySparkInterpreter"
> warning after starting pyspark interpreter.
>
> How do I default %pyspark to ipython?
>
> Tried to change to
> "class": "org.apache.zeppelin.spark.PySparkInterpreter",
> to
> "class": "org.apache.zeppelin.spark.IPySparkInterpreter",
> in interpreter.json but this gets overwritten back to PySparkInterpreter.
>
> Also tried to change to zeppelin.pyspark.python to ipython with no luck
> too.
>
> Is there is a documented way to default pyspark interpreter to ipython?
> Glanced over PR-2474 but can't quickly get what I am missing.
>
>
> Thanks.
>
>
>