I agree with Klein. If you do not need to exploit the entire Wikipedia database, requests through a python scraping library (like Beautiful Soup) are certainly sufficient and easy to set up. With an aleatory algorithm to select the "ids" you can create a fine sample.
PCL

Le 07/10/13 19:31, Klein,Max a écrit :
Hi Susan,

Do you need the entire database diff'd? I.e. all edits ever. Or are you interested in a particular subset of the diffs? It would help to know your purpose.

For instance I am interested in diffs around specific articles for specific dates to study news events. So I calculate the diffs myself using python on page histories rather than the entire database.

Maximilian Klein
Wikipedian in Residence, OCLC
+17074787023


From: wiki-research-l-bounces@lists.wikimedia.org <wiki-research-l-bounces@lists.wikimedia.org> on behalf of Susan Biancani <inacnaib@gmail.com>
Sent: Thursday, October 03, 2013 10:06 PM
To: wiki-research-l@lists.wikimedia.org
Subject: [Wiki-research-l] diffdb formatted Wikipedia dump
 
I'm looking for a dump from English Wikipedia in diff format (i.e. each entry is the text that was added/deleted since the last edit, rather than each entry is the current state of the page).

The Summer of Research folks provided a handy guide to how to create such a dataset from the standard complete dumps here: http://meta.wikimedia.org/wiki/WSoR_datasets/revision_diff
But the time estimate they give is prohibitive for me (20-24 hours for each dump file--there are currently 158--running on 24 cores). I'm a grad student in a social science department, and don't have access to extensive computing power. I've been paying out of pocket for AWS, but this would get expensive.

There is a diff-format dataset available, but only through April, 2011 (here: http://dumps.wikimedia.org/other/diffdb/). I'd like to get a diff-format dataset for January, 2010- March, 2013 (or, for everything up to March, 2013).

Does anyone know if such a dataset exists somewhere? Any leads or suggestions would be much appreciated!

Susan


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