Metadata-Version: 2.1
Name: sparqldataframe
Version: 0.1.1
Summary: Get a Pandas dataframe from SPARQL queries
Home-page: https://github.com/njanakiev/sparqldataframe
Author: Nikolai Janakiev
Author-email: nikolai.janakiev@gmail.com
License: MIT
Platform: any
Classifier: Environment :: Console
Classifier: Intended Audience :: Developers
Classifier: Intended Audience :: Science/Research
Classifier: Intended Audience :: System Administrators
Classifier: License :: OSI Approved :: MIT License
Classifier: Operating System :: POSIX
Classifier: Operating System :: MacOS
Classifier: Operating System :: Unix
Classifier: Operating System :: Microsoft :: Windows
Classifier: Programming Language :: Python
Classifier: Programming Language :: Python :: 2
Classifier: Programming Language :: Python :: 3
Classifier: Topic :: Scientific/Engineering :: Information Analysis
Classifier: Topic :: Software Development :: Libraries :: Python Modules
Description-Content-Type: text/markdown
License-File: LICENSE

---
title: sparqldataframe
---

[![image](https://img.shields.io/pypi/v/sparqldataframe.svg)](https://pypi.python.org/pypi/sparqldataframe)

A Python library that can send
[SPARQL](https://en.wikipedia.org/wiki/SPARQL) queries to a SPARQL
endpoint and retrieve a [Pandas](http://pandas.pydata.org/) dataframe
from the result.

# Installation

``` bash
pip install sparqldataframe
```

# Usage

Here is an example how to run a SPARQL query on the
[Wikidata](http://wikidata.org/) endpoint:

``` python
import sparqldataframe

sparql_query = """
SELECT ?item ?itemLabel 
WHERE {
  ?item wdt:P31 wd:Q146.
  SERVICE wikibase:label { bd:serviceParam wikibase:language "[AUTO_LANGUAGE],en". }
}
"""
df = sparqldataframe.query("https://query.wikidata.org/sparql", sparql_query)
```

Wikidata and [DBPedia](http://dbpedia.org/) can be both used without
adding the SPARQL endpoint url by using the `wikidata_query()` and
`dbpedia_query()` functions respectively:

``` python
df = sparqldataframe.wikidata_query(sparql_query)
df = sparqldataframe.dbpedia_query(sparql_query)
```

# License

This project is licensed under the MIT license. See the
[LICENSE](LICENSE) for details.


