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Jupyter metakernel for apache spark and scala

Project description

# spylon-kernel
[![Build Status](https://travis-ci.org/mariusvniekerk/spylon-kernel.svg?branch=master)](https://travis-ci.org/mariusvniekerk/spylon-kernel)
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This is an extremely early proof of concept for using the metakernel in combination with py4j to make a simpler
kernel for scala.

## Installation

On python 3.5+

```bash
pip install .
```

## Installing the jupyter kernel

To install the jupyter kernel install it using

```
python -m spylon_kernel install
```

## Using the kernel

The scala spark metakernl prodived a scala kernel by default.
At the first scala cell that is run a spark session will be constructed so that a user can interact with the
interpreter.

### Customizing the spark context

The launch arguments can be customized using the `%%init_spark` magic as follows

```python
%%init_spark
launcher.jars = ["file://some/jar.jar"]
launcher.master = "local[4]"
launcher.conf.spark.executor.cores = 8
```

### Other languages

Since this makes use of metakernel you can evaluate normal python code using the `%%python` magic. In addition once
the spark context has been created the `spark` variable will be added to your python ernvironment.

```python
%%python
df = spark.read.json("examples/src/main/resources/people.json")
```

To get completions for python, make sure that you have installed `jedi`

## Using as a magic

Spylon-kernel can be used as a magic in an existing ipykernel. This is the recommended solution when you want to write
relatively small blocks of scala.

```python
from spylon_kernel import register_ipython_magics
register_ipython_magics()
```

```scala
%%scala
val x = 8
x
```

## Using as a library

If you just want to send a string of scala code to the interpreter and evaluate it you can
do that too.

```python
from spylon_kernel import get_scala_interpreter

interp = get_scala_interpreter()

# Evaluate the result of a scala code block.
interp.interpret("""
val x = 8
x
""")

interp.last_result()
```

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