By the way, is there somewhere where I could find total
active editors of
all wikisources?
All projects agreggated? Not as far as I know. More than a matter of
calculation I think is a matter of definition. I might be wrong though and
this might already have been discussed.
Active editor metric is defined "per wiki" to my knowledge:
https://meta.wikimedia.org/wiki/Research:Wikistats_metrics/Editors
For all projects you could add "all active editors for all projects" and
that is doable. Now, in that case you will not count someone with 2 edits
in eswiki and (using same login) another 3 edits on dewiki if in order to
become an "active editor" you need 5 edits in one wiki.
Phab ticket filed:
https://phabricator.wikimedia.org/T188194
On Thu, Feb 22, 2018 at 1:37 AM, mathieu stumpf guntz <
psychoslave(a)culture-libre.org> wrote:
Hi Nuria,
Thank you for the report, and congratulation to all people involved in
releasing this tool. It would be fine to also have map with editiors and
active editors.
By the way, is there somewhere where I could find total active editors of
all wikisources? Surely I could sum that through API (providing that it
exposes such a data for each language), but it would be fine that everybody
could have a straight forward access to this kind of cross language data. I
think there real are usecases for that, actually I'm looking for number of
active wikisourcerer to evaluate number of attendees we might set for the
next Wikisource conference.
Cheers.
Le 14/02/2018 à 23:15, Nuria Ruiz a écrit :
Hello from Analytics team:
Just a brief note to announce that Wikistats 2.0 includes data about
pageviews per project per country for the current month.
Take a look, pageviews for Spanish Wikipedia this current
month:https://stats.wikimedia.org/v2/#/es.wikipedia.org/reading/pageviews-b…
Data is also available programatically vi APIs:
https://wikitech.wikimedia.org/wiki/Analytics/AQS/Pageviews#Pageviews_split…
We will be deploying small UI tweaks during this week but please explore
and let us know what you think.
Thanks,
Nuria
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